Final temperature (T 2 ) of the methane is to be calculated when some amount of heat is given to the to the methane. Concept Introduction : The final temperature is calculated from below equation where amount of heat is given: Q = n Δ H = R ∫ T 0 T Δ C P R d T . ...(1) Where Δ H is heat of reaction at any temperature T . and ∫ T 0 T Δ C ∘ P R d T = A T 0 ( τ − 1 ) + B 2 T 0 2 ( τ 2 − 1 ) + C 3 T 0 3 ( τ 3 − 1 ) + D T 0 ( τ − 1 τ ) .....(2) Where τ = T T 0

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Answer 1

Question

Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(a)

Expert Solution

Answer to Problem 4.3P

$T=626.817 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dmethane 1.702 9.081 −2.164 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.702×373.15×(τ−1)+9.081×10−32373.152(τ2−1) +−2.164×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=635.1013(τ−1)+632.223(τ2−1)−37.479(τ3−1)∫373.15TΔC∘PRdT=635.1013τ−635.1013+632.223τ2−632.223−37.479τ3+37.479∫373.15TΔC∘PRdT=−37.479τ3+632.223τ2+635.1013τ−1229.8453$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K =−37.479τ3+632.223τ2+635.1013τ−1229.845337.479τ3−632.223τ2−635.1013τ+2673.194=0$

On solving,

$τ=1.6798,−2.412, 17.6$

Considering positive value of temperature $τ=1.6798$

Hence,

$τ=TT0=T100+273.15=1.6798T=626.817 K$

(b)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(b)

Expert Solution

Answer to Problem 4.3P

$T=538.455 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethane 1.131 19.225 −5.561 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.131×373.15×(τ−1)+19.225×10−32373.152(τ2−1) +−5.561×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=422.033(τ−1)+1338.453(τ2−1)−96.31(τ3−1)∫373.15TΔC∘PRdT=422.033τ−422.033+1338.453τ2−1338.453−96.31τ3+96.31∫373.15TΔC∘PRdT=−96.31τ3+1338.453τ2+422.033τ−1664.176$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−96.31τ3+1338.453τ2+422.033τ−1664.17696.31τ3−1338.453τ2−422.033τ+3107.525=0$

On solving,

$τ=1.443,−1.592, 14.0458$

Considering positive value of temperature $τ=1.443$

Hence,

$τ=TT0=T100+273.15=1.443T=538.455 K$

(c)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(c)

Expert Solution

Answer to Problem 4.3P

$T=493.3043 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.213×373.15×(τ−1)+28.785×10−32373.152(τ2−1) +−8.824×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=452.631(τ−1)+2004.025(τ2−1)−152.825(τ3−1)∫373.15TΔC∘PRdT=452.631τ−452.631+2004.025τ2−2004.025−152.825τ3+152.825∫373.15TΔC∘PRdT=−152.825τ3+2004.025τ2+452.631τ−2303.831$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−152.825τ3+2004.025τ2+452.631τ−2303.831152.825τ3−2004.025τ2−452.631τ+3747.18=0$

On solving,

$τ=1.322,−1.405, 13.197$

Considering positive value of temperature $τ=1.322$

Hence,

$τ=TT0=T100+273.15=1.322T=493.3043 K$

(d)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(d)

Expert Solution

Answer to Problem 4.3P

$T=466.81 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -butane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.935×373.15×(τ−1)+36.915×10−32373.152(τ2−1) +−11.402×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=722.045(τ−1)+2570.039(τ2−1)−197.474(τ3−1)∫373.15TΔC∘PRdT=722.045τ−722.045+2570.039τ2−2570.039−197.474τ3+197.474∫373.15TΔC∘PRdT=−197.474τ3+2570.039τ2+722.045τ−3094.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−197.474τ3+2570.039τ2+722.045τ−3094.61197.474τ3−2570.039τ2−722.045τ+4537.96=0$

On solving,

$τ=1.251,−1.396, 13.16$

Considering positive value of temperature $τ=1.251$

Hence,

$τ=TT0=T100+273.15=1.251T=466.81 K$

(e)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(e)

Expert Solution

Answer to Problem 4.3P

$T=438.45 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -hexanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 3.025 53.722 −16.791 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.025×373.15×(τ−1)+53.722×10−32373.152(τ2−1) +−16.791×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1128.78(τ−1)+3740.15(τ2−1)−290.807(τ3−1)∫373.15TΔC∘PRdT=1128.78τ−1128.78+3740.15τ2−3740.15−290.807τ3+290.807∫373.15TΔC∘PRdT=−290.807τ3+3740.15τ2+1128.78τ−4578.12$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−290.807τ3+3740.15τ2+1128.78τ−4578.12290.807τ3−3740.15τ2−1128.78τ+6021.471=0$

On solving,

$τ=1.175,−1.351, 13.04$

Considering positive value of temperature $τ=1.175$

Hence,

$τ=TT0=T100+273.15=1.175T=438.45 K$

(f)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(f)

Expert Solution

Answer to Problem 4.3P

$T=371.695 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -octanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 4.108 70.567 −22.208 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.108×373.15×(τ−1)+70.567×10−32373.152(τ2−1) +−22.208×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1532.9002(τ−1)+4912.91(τ2−1)−384.63(τ3−1)∫373.15TΔC∘PRdT=1532.9002τ−1532.9002+4912.91τ2−4912.91−384.63τ3+384.63∫373.15TΔC∘PRdT=−384.63τ3+4912.91τ2+1532.9002τ−6061.184$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−384.63τ3+4912.91τ2+1532.9002τ−6061.184384.63τ3−4912.91τ2−1532.9002τ+6021.471=0$

On solving,

$τ=0.9961,−1.21, 12.987$

Considering positive value of temperature $τ=0.9961$

Hence,

$τ=TT0=T100+273.15=0.9961T=371.695 K$

(g)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(g)

Expert Solution

Answer to Problem 4.3P

$T=513.08 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropylene 1.637 22.706 −6.915 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.637×373.15×(τ−1)+22.706×10−32373.152(τ2−1) +−6.915×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=610.847(τ−1)+1580.8(τ2−1)−119.763(τ3−1)∫373.15TΔC∘PRdT=610.847τ−610.847+1580.8τ2−1580.8−119.763τ3+119.763∫373.15TΔC∘PRdT=−119.763τ3+1580.8τ2+610.847τ−2071.88$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−119.763τ3+1580.8τ2+610.847τ−2071.88119.763τ3−1580.8τ2−610.847τ+3515.233=0$

On solving,

$τ=1.375,−1.592, 13.416$

Considering positive value of temperature $τ=1.375$

Hence,

$τ=TT0=T100+273.15=1.375T=513.08 K$

(h)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(h)

Expert Solution

Answer to Problem 4.3P

$T=457.855 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-pentene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−pentene 2.691 39.753 −12.447 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.691×373.15×(τ−1)+39.753×10−32373.152(τ2−1) +−12.447×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1004.147(τ−1)+2767.62(τ2−1)−215.57(τ3−1)∫373.15TΔC∘PRdT=1004.147τ−1004.147+2767.62τ2−2767.62−215.57τ3+215.57∫373.15TΔC∘PRdT=−215.57τ3+2767.62τ2+1004.147τ−3556.197$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−215.57τ3+2767.62τ2+1004.147τ−3556.197215.57τ3−2767.62τ2−1004.147τ+4999.546=0$

On solving,

$τ=1.227,−1.447, 13.059$

Considering positive value of temperature $τ=1.227$

Hence,

$τ=TT0=T100+273.15=1.227T=457.855 K$

(i)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(i)

Expert Solution

Answer to Problem 4.3P

$T=433.97 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-heptenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−heptene 3.768 56.588 −17.847 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.768×373.15×(τ−1)+56.588×10−32373.152(τ2−1) +−17.847×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1406.03(τ−1)+3939.68(τ2−1)−309.097(τ3−1)∫373.15TΔC∘PRdT=1406.03τ−1406.03+3939.68τ2−3939.68−309.097τ3+309.097∫373.15TΔC∘PRdT=−309.097τ3+3939.68τ2+1406.03τ−5036.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−309.097τ3+3939.68τ2+1406.03τ−5036.61309.097τ3−3939.68τ2−1406.03τ+6479.96=0$

On solving,

$τ=1.163,−1.389, 12.97$

Considering positive value of temperature $τ=1.163$

Hence,

$τ=TT0=T100+273.15=1.163T=433.97 K$

(j)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(j)

Expert Solution

Answer to Problem 4.3P

$T=426.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-Octenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−Octene 4.324 64.96 −20.521 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.324×373.15×(τ−1)+64.96×10−32373.152(τ2−1) +−20.521×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1613.5(τ−1)+4522.55(τ2−1)−355.41(τ3−1)∫373.15TΔC∘PRdT=1613.5τ−1613.5+4522.55τ2−4522.55−355.41τ3+355.41∫373.15TΔC∘PRdT=−355.41τ3+4522.55τ2+1613.5τ−5780.64$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−355.41τ3+4522.55τ2+1613.5τ−5780.64355.41τ3−4522.55τ2−1613.5τ+7223.99=0$

On solving,

$τ=1.143,−1.37, 12.95$

Considering positive value of temperature $τ=1.143$

Hence,

$τ=TT0=T100+273.15=1.143T=426.51 K$

(k)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(k)

Expert Solution

Answer to Problem 4.3P

$T=595.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for acetylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dacetylene 6.132 1.952 0 −1.299$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=6.132×373.15×(τ−1)+1.952×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.299×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2288.15(τ−1)+135.9(τ2−1)−348.12(τ−1τ)∫373.15TΔC∘PRdT=2288.15τ−2288.15+135.9τ2−135.9−348.12+348.12τ∫373.15TΔC∘PRdT=135.9τ2+2288.15τ−2772.17+348.12τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=135.9τ2+2288.15τ−2772.17+348.12τ135.9τ2+2288.15τ−4215.519+348.12τ=0135.9τ3+2288.15τ2−4215.519τ+348.12τ=0$

On solving,

$τ=1.596,0.099, −18.52$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.596T=595.55 K$

(l)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(l)

Expert Solution

Answer to Problem 4.3P

$T=476.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for benzenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dbenzene −0.206 39.064 −13.301 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=−0.206×373.15×(τ−1)+39.064×10−32373.152(τ2−1) +−13.301×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=−76.87(τ−1)+2719.65(τ2−1)−230.36(τ3−1)∫373.15TΔC∘PRdT=−76.87τ+76.87+2719.65τ2−2719.65−230.36τ3+230.36∫373.15TΔC∘PRdT=−230.36τ3+2719.65τ2−76.87τ−2412.42$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−230.36τ3+2719.65τ2−76.87τ−2412.42230.36τ3−2719.65τ2+76.87τ+3855.769=0$

On solving,

$τ=1.277,−1.12, 11.65$

Considering positive value of temperature $τ=1.277$

Hence,

$τ=TT0=T100+273.15=1.277T=476.51 K$

(m)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(m)

Expert Solution

Answer to Problem 4.3P

$T=380.613 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethanol in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethanol 3.518 20.001 −6.002 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.518×373.15×(τ−1)+20.001×10−32373.152(τ2−1) +−6.002×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1312.74(τ−1)+1392.48(τ2−1)−103.95(τ3−1)∫373.15TΔC∘PRdT=1312.74τ−1312.74+1392.48τ2−1392.48−103.95τ3+103.95∫373.15TΔC∘PRdT=−103.95τ3+2719.65τ2+1312.74τ−2601.27$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−103.95τ3+2719.65τ2+1312.74τ−2601.27103.95τ3−2719.65τ2−1312.74τ+4044.62=0$

On solving,

$τ=1.02,−1.44, 26.58$

Considering positive value of temperature $τ=1.02$

Hence,

$τ=TT0=T100+273.15=1.02T=380.613 K$

(n)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(n)

Expert Solution

Answer to Problem 4.3P

$T=445.91 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for styrene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dstyrene 2.050 50.192 −16.662 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.050×373.15×(τ−1)+50.192×10−32373.152(τ2−1) +−16.662×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=764.96(τ−1)+3494.39(τ2−1)−288.57(τ3−1)∫373.15TΔC∘PRdT=764.96τ−764.96+3494.39τ2−3494.39−288.57τ3+288.57∫373.15TΔC∘PRdT=−288.57τ3+3494.39τ2+764.96τ−3970.78$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−288.57τ3+3494.39τ2+764.96τ−3970.78288.57τ3−3494.39τ2−764.96τ+5414.126=0$

On solving,

$τ=1.195,−1.287, 12.2$

Considering positive value of temperature $τ=1.195$

Hence,

$τ=TT0=T100+273.15=1.195T=445.91 K$

(o)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(o)

Expert Solution

Answer to Problem 4.3P

$T=643.68 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for formaldehyde in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dformaldehyde 2.264 7.022 −1.877 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.264×373.15×(τ−1)+7.022×10−32373.152(τ2−1) +−1.877×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=844.81(τ−1)+488.88(τ2−1)−32.51(τ3−1)∫373.15TΔC∘PRdT=844.81τ−844.81+488.88τ2−488.88−32.51τ3+32.51∫373.15TΔC∘PRdT=-32.51τ3+488.88τ2+844.81τ−1301.18$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=-32.51τ3+488.88τ2+844.81τ−1301.1832.51τ3−488.88τ2−844.81τ+2744.53=0$

On solving,

$τ=1.725,−3, 16.31$

Considering positive value of temperature $τ=1.725$

Hence,

$τ=TT0=T100+273.15=1.725T=643.68 K$

(p)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(p)

Expert Solution

Answer to Problem 4.3P

$T=416.06 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dammonia 3.578 3.020 0 −0.186$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.578×373.15×(τ−1)+3.020×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.186×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1335.13(τ−1)+210.25(τ2−1)−49.85(τ−1τ)∫373.15TΔC∘PRdT=1335.13τ−1335.13+210.25τ2−210.25−49.85+49.85τ∫373.15TΔC∘PRdT=210.25τ2+1335.13τ−1595.23+49.85τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=210.25τ2+1335.13τ−1595.23+49.85τ1210.25τ2+1335.13τ−3038.579+49.85τ=01210.25τ3+1335.13τ2−3038.579τ+49.85τ=0$

On solving,

$τ=1.115,0.0165, −2.23$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.115T=416.06 K$

(q)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(q)

Expert Solution

Answer to Problem 4.3P

$T=764.96 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon monoxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO 3.376 0.557 0 −0.031$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.376×373.15×(τ−1)+0.557×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.031×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1259.75(τ−1)+38.78(τ2−1)−8.31(τ−1τ)∫373.15TΔC∘PRdT=1259.75τ−1259.75+38.78τ2−38.78−8.31+8.31τ∫373.15TΔC∘PRdT=38.78τ2+1259.75τ−1306.84+8.31τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=38.78τ2+1259.75τ−1306.84+8.31τ38.78τ2+1259.75τ−2750.19+8.31τ=038.78τ3+1259.75τ2−2750.19τ+8.31τ=0$

On solving,

$τ=2.05,0.003, −34.54$

Considering positive value of temperature $τ=2.05$

Hence,

$τ=TT0=T100+273.15=2.05T=764.96 K$

(r)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(r)

Expert Solution

Answer to Problem 4.3P

$T=634.355 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO2 5.457 1.045 0 −1.157$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.457×373.15×(τ−1)+1.045×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.157×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2036.28(τ−1)+72.75(τ2−1)−310.06(τ−1τ)∫373.15TΔC∘PRdT=2036.28τ−2036.28+72.75τ2−72.75−310.06+310.06τ∫373.15TΔC∘PRdT=72.75τ2+2036.28τ−2419.09+310.06τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=72.75τ2+2036.28τ−2419.09+310.06τ72.75τ2+2036.28τ−3862.44+310.06τ=072.75τ3+2036.28τ2−3862.44τ+310.06τ=0$

On solving,

$τ=1.7,0.084, −29.78$

Considering positive value of temperature $τ=1.7$

Hence,

$τ=TT0=T100+273.15=1.7T=634.355 K$

(s)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(s)

Expert Solution

Answer to Problem 4.3P

$T=626.1457 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for sulphur dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DSO2 5.699 0.801 0 −1.015$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.699×373.15×(τ−1)+0.801×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.015×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2126.58(τ−1)+55.77(τ2−1)−272(τ−1τ)∫373.15TΔC∘PRdT=2126.58τ−2126.58+55.77τ2−55.77−272+272τ∫373.15TΔC∘PRdT=55.77τ2+2126.58τ−2454.36+272τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=55.77τ2+2126.58τ−2454.36+272τ55.77τ2+2126.58τ−3887.71+272τ=055.77τ3+2126.58τ2−3887.71τ+272τ=0$

On solving,

$τ=1.678,0.073, −39.88$

Considering positive value of temperature $τ=1.678$

Hence,

$τ=TT0=T100+273.15=1.678T=626.1457 K$

(t)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(t)

Expert Solution

Answer to Problem 4.3P

$T=706.37 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DH2O 3.47 1.450 0 0.121$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.470×373.15×(τ−1)+1.450×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.121×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1294.83(τ−1)+100.95(τ2−1)+32.43(τ−1τ)∫373.15TΔC∘PRdT=1294.83τ−1294.83+100.95τ2−100.95+32.43−32.43τ∫373.15TΔC∘PRdT=100.95τ2+1294.83τ−1363.35−32.43τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=100.95τ2+1294.83τ−1363.35−32.43τ100.95τ2+1294.83τ−2796.7−32.43τ=0100.95τ3+1294.83τ2−2796.7τ−32.43τ=0$

On solving,

$τ=1.893,−0.011, −14.708$

Considering positive value of temperature $τ=1.893$

Hence,

$τ=TT0=T100+273.15=1.893T=706.37 K$

(u)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(u)

Expert Solution

Answer to Problem 4.3P

$T=770.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for nitrogen in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DN2 3.280 0.593 0 0.040$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.280×373.15×(τ−1)+0.593×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.040×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1223.93(τ−1)+41.29(τ2−1)+10.72(τ−1τ)∫373.15TΔC∘PRdT=1223.93τ−1223.93+41.29τ2−41.29+10.72−10.72τ∫373.15TΔC∘PRdT=41.29τ2+1223.93τ−1254.5−10.72τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=41.29τ2+1223.93τ−1254.5−10.72τ41.29τ2+1223.93τ−2697.85−10.72τ=041.29τ3+1223.93τ2−2697.85τ−10.72τ=0$

On solving,

$τ=2.065,−0.003, −31.703$

Considering positive value of temperature $τ=2.065$

Hence,

$τ=TT0=T100+273.15=2.065T=770.55 K$

(v)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(v)

Expert Solution

Answer to Problem 4.3P

$T=654.13 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for hydrogen cyanide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DHCN 4.736 1.359 0 −0.725$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.736×373.15×(τ−1)+1.359×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.725×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1767.24(τ−1)+94.61(τ2−1)−194.29(τ−1τ)∫373.15TΔC∘PRdT=1767.24τ−1767.24+94.61τ2−94.61−194.29+194.29τ∫373.15TΔC∘PRdT=94.61τ2+1767.24τ−2056.14+194.29τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=94.61τ2+1767.24τ−2056.14+194.29τ94.61τ2+1767.24τ−3499.49+194.29τ=094.61τ2+1767.24τ−3499.49τ+194.29τ=0$

On solving,

$τ=1.753,0.057, −20.49$

Considering positive value of temperature $τ=1.753$

Hence,

$τ=TT0=T100+273.15=1.753T=654.13 K$

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Answer 2

Question

Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(a)

Expert Solution

Answer to Problem 4.3P

$T=626.817 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dmethane 1.702 9.081 −2.164 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.702×373.15×(τ−1)+9.081×10−32373.152(τ2−1) +−2.164×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=635.1013(τ−1)+632.223(τ2−1)−37.479(τ3−1)∫373.15TΔC∘PRdT=635.1013τ−635.1013+632.223τ2−632.223−37.479τ3+37.479∫373.15TΔC∘PRdT=−37.479τ3+632.223τ2+635.1013τ−1229.8453$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K =−37.479τ3+632.223τ2+635.1013τ−1229.845337.479τ3−632.223τ2−635.1013τ+2673.194=0$

On solving,

$τ=1.6798,−2.412, 17.6$

Considering positive value of temperature $τ=1.6798$

Hence,

$τ=TT0=T100+273.15=1.6798T=626.817 K$

(b)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(b)

Expert Solution

Answer to Problem 4.3P

$T=538.455 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethane 1.131 19.225 −5.561 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.131×373.15×(τ−1)+19.225×10−32373.152(τ2−1) +−5.561×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=422.033(τ−1)+1338.453(τ2−1)−96.31(τ3−1)∫373.15TΔC∘PRdT=422.033τ−422.033+1338.453τ2−1338.453−96.31τ3+96.31∫373.15TΔC∘PRdT=−96.31τ3+1338.453τ2+422.033τ−1664.176$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−96.31τ3+1338.453τ2+422.033τ−1664.17696.31τ3−1338.453τ2−422.033τ+3107.525=0$

On solving,

$τ=1.443,−1.592, 14.0458$

Considering positive value of temperature $τ=1.443$

Hence,

$τ=TT0=T100+273.15=1.443T=538.455 K$

(c)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(c)

Expert Solution

Answer to Problem 4.3P

$T=493.3043 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.213×373.15×(τ−1)+28.785×10−32373.152(τ2−1) +−8.824×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=452.631(τ−1)+2004.025(τ2−1)−152.825(τ3−1)∫373.15TΔC∘PRdT=452.631τ−452.631+2004.025τ2−2004.025−152.825τ3+152.825∫373.15TΔC∘PRdT=−152.825τ3+2004.025τ2+452.631τ−2303.831$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−152.825τ3+2004.025τ2+452.631τ−2303.831152.825τ3−2004.025τ2−452.631τ+3747.18=0$

On solving,

$τ=1.322,−1.405, 13.197$

Considering positive value of temperature $τ=1.322$

Hence,

$τ=TT0=T100+273.15=1.322T=493.3043 K$

(d)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(d)

Expert Solution

Answer to Problem 4.3P

$T=466.81 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -butane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.935×373.15×(τ−1)+36.915×10−32373.152(τ2−1) +−11.402×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=722.045(τ−1)+2570.039(τ2−1)−197.474(τ3−1)∫373.15TΔC∘PRdT=722.045τ−722.045+2570.039τ2−2570.039−197.474τ3+197.474∫373.15TΔC∘PRdT=−197.474τ3+2570.039τ2+722.045τ−3094.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−197.474τ3+2570.039τ2+722.045τ−3094.61197.474τ3−2570.039τ2−722.045τ+4537.96=0$

On solving,

$τ=1.251,−1.396, 13.16$

Considering positive value of temperature $τ=1.251$

Hence,

$τ=TT0=T100+273.15=1.251T=466.81 K$

(e)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(e)

Expert Solution

Answer to Problem 4.3P

$T=438.45 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -hexanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 3.025 53.722 −16.791 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.025×373.15×(τ−1)+53.722×10−32373.152(τ2−1) +−16.791×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1128.78(τ−1)+3740.15(τ2−1)−290.807(τ3−1)∫373.15TΔC∘PRdT=1128.78τ−1128.78+3740.15τ2−3740.15−290.807τ3+290.807∫373.15TΔC∘PRdT=−290.807τ3+3740.15τ2+1128.78τ−4578.12$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−290.807τ3+3740.15τ2+1128.78τ−4578.12290.807τ3−3740.15τ2−1128.78τ+6021.471=0$

On solving,

$τ=1.175,−1.351, 13.04$

Considering positive value of temperature $τ=1.175$

Hence,

$τ=TT0=T100+273.15=1.175T=438.45 K$

(f)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(f)

Expert Solution

Answer to Problem 4.3P

$T=371.695 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -octanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 4.108 70.567 −22.208 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.108×373.15×(τ−1)+70.567×10−32373.152(τ2−1) +−22.208×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1532.9002(τ−1)+4912.91(τ2−1)−384.63(τ3−1)∫373.15TΔC∘PRdT=1532.9002τ−1532.9002+4912.91τ2−4912.91−384.63τ3+384.63∫373.15TΔC∘PRdT=−384.63τ3+4912.91τ2+1532.9002τ−6061.184$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−384.63τ3+4912.91τ2+1532.9002τ−6061.184384.63τ3−4912.91τ2−1532.9002τ+6021.471=0$

On solving,

$τ=0.9961,−1.21, 12.987$

Considering positive value of temperature $τ=0.9961$

Hence,

$τ=TT0=T100+273.15=0.9961T=371.695 K$

(g)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(g)

Expert Solution

Answer to Problem 4.3P

$T=513.08 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropylene 1.637 22.706 −6.915 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.637×373.15×(τ−1)+22.706×10−32373.152(τ2−1) +−6.915×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=610.847(τ−1)+1580.8(τ2−1)−119.763(τ3−1)∫373.15TΔC∘PRdT=610.847τ−610.847+1580.8τ2−1580.8−119.763τ3+119.763∫373.15TΔC∘PRdT=−119.763τ3+1580.8τ2+610.847τ−2071.88$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−119.763τ3+1580.8τ2+610.847τ−2071.88119.763τ3−1580.8τ2−610.847τ+3515.233=0$

On solving,

$τ=1.375,−1.592, 13.416$

Considering positive value of temperature $τ=1.375$

Hence,

$τ=TT0=T100+273.15=1.375T=513.08 K$

(h)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(h)

Expert Solution

Answer to Problem 4.3P

$T=457.855 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-pentene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−pentene 2.691 39.753 −12.447 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.691×373.15×(τ−1)+39.753×10−32373.152(τ2−1) +−12.447×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1004.147(τ−1)+2767.62(τ2−1)−215.57(τ3−1)∫373.15TΔC∘PRdT=1004.147τ−1004.147+2767.62τ2−2767.62−215.57τ3+215.57∫373.15TΔC∘PRdT=−215.57τ3+2767.62τ2+1004.147τ−3556.197$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−215.57τ3+2767.62τ2+1004.147τ−3556.197215.57τ3−2767.62τ2−1004.147τ+4999.546=0$

On solving,

$τ=1.227,−1.447, 13.059$

Considering positive value of temperature $τ=1.227$

Hence,

$τ=TT0=T100+273.15=1.227T=457.855 K$

(i)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(i)

Expert Solution

Answer to Problem 4.3P

$T=433.97 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-heptenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−heptene 3.768 56.588 −17.847 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.768×373.15×(τ−1)+56.588×10−32373.152(τ2−1) +−17.847×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1406.03(τ−1)+3939.68(τ2−1)−309.097(τ3−1)∫373.15TΔC∘PRdT=1406.03τ−1406.03+3939.68τ2−3939.68−309.097τ3+309.097∫373.15TΔC∘PRdT=−309.097τ3+3939.68τ2+1406.03τ−5036.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−309.097τ3+3939.68τ2+1406.03τ−5036.61309.097τ3−3939.68τ2−1406.03τ+6479.96=0$

On solving,

$τ=1.163,−1.389, 12.97$

Considering positive value of temperature $τ=1.163$

Hence,

$τ=TT0=T100+273.15=1.163T=433.97 K$

(j)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(j)

Expert Solution

Answer to Problem 4.3P

$T=426.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-Octenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−Octene 4.324 64.96 −20.521 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.324×373.15×(τ−1)+64.96×10−32373.152(τ2−1) +−20.521×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1613.5(τ−1)+4522.55(τ2−1)−355.41(τ3−1)∫373.15TΔC∘PRdT=1613.5τ−1613.5+4522.55τ2−4522.55−355.41τ3+355.41∫373.15TΔC∘PRdT=−355.41τ3+4522.55τ2+1613.5τ−5780.64$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−355.41τ3+4522.55τ2+1613.5τ−5780.64355.41τ3−4522.55τ2−1613.5τ+7223.99=0$

On solving,

$τ=1.143,−1.37, 12.95$

Considering positive value of temperature $τ=1.143$

Hence,

$τ=TT0=T100+273.15=1.143T=426.51 K$

(k)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(k)

Expert Solution

Answer to Problem 4.3P

$T=595.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for acetylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dacetylene 6.132 1.952 0 −1.299$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=6.132×373.15×(τ−1)+1.952×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.299×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2288.15(τ−1)+135.9(τ2−1)−348.12(τ−1τ)∫373.15TΔC∘PRdT=2288.15τ−2288.15+135.9τ2−135.9−348.12+348.12τ∫373.15TΔC∘PRdT=135.9τ2+2288.15τ−2772.17+348.12τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=135.9τ2+2288.15τ−2772.17+348.12τ135.9τ2+2288.15τ−4215.519+348.12τ=0135.9τ3+2288.15τ2−4215.519τ+348.12τ=0$

On solving,

$τ=1.596,0.099, −18.52$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.596T=595.55 K$

(l)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(l)

Expert Solution

Answer to Problem 4.3P

$T=476.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for benzenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dbenzene −0.206 39.064 −13.301 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=−0.206×373.15×(τ−1)+39.064×10−32373.152(τ2−1) +−13.301×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=−76.87(τ−1)+2719.65(τ2−1)−230.36(τ3−1)∫373.15TΔC∘PRdT=−76.87τ+76.87+2719.65τ2−2719.65−230.36τ3+230.36∫373.15TΔC∘PRdT=−230.36τ3+2719.65τ2−76.87τ−2412.42$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−230.36τ3+2719.65τ2−76.87τ−2412.42230.36τ3−2719.65τ2+76.87τ+3855.769=0$

On solving,

$τ=1.277,−1.12, 11.65$

Considering positive value of temperature $τ=1.277$

Hence,

$τ=TT0=T100+273.15=1.277T=476.51 K$

(m)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(m)

Expert Solution

Answer to Problem 4.3P

$T=380.613 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethanol in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethanol 3.518 20.001 −6.002 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.518×373.15×(τ−1)+20.001×10−32373.152(τ2−1) +−6.002×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1312.74(τ−1)+1392.48(τ2−1)−103.95(τ3−1)∫373.15TΔC∘PRdT=1312.74τ−1312.74+1392.48τ2−1392.48−103.95τ3+103.95∫373.15TΔC∘PRdT=−103.95τ3+2719.65τ2+1312.74τ−2601.27$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−103.95τ3+2719.65τ2+1312.74τ−2601.27103.95τ3−2719.65τ2−1312.74τ+4044.62=0$

On solving,

$τ=1.02,−1.44, 26.58$

Considering positive value of temperature $τ=1.02$

Hence,

$τ=TT0=T100+273.15=1.02T=380.613 K$

(n)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(n)

Expert Solution

Answer to Problem 4.3P

$T=445.91 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for styrene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dstyrene 2.050 50.192 −16.662 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.050×373.15×(τ−1)+50.192×10−32373.152(τ2−1) +−16.662×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=764.96(τ−1)+3494.39(τ2−1)−288.57(τ3−1)∫373.15TΔC∘PRdT=764.96τ−764.96+3494.39τ2−3494.39−288.57τ3+288.57∫373.15TΔC∘PRdT=−288.57τ3+3494.39τ2+764.96τ−3970.78$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−288.57τ3+3494.39τ2+764.96τ−3970.78288.57τ3−3494.39τ2−764.96τ+5414.126=0$

On solving,

$τ=1.195,−1.287, 12.2$

Considering positive value of temperature $τ=1.195$

Hence,

$τ=TT0=T100+273.15=1.195T=445.91 K$

(o)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(o)

Expert Solution

Answer to Problem 4.3P

$T=643.68 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for formaldehyde in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dformaldehyde 2.264 7.022 −1.877 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.264×373.15×(τ−1)+7.022×10−32373.152(τ2−1) +−1.877×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=844.81(τ−1)+488.88(τ2−1)−32.51(τ3−1)∫373.15TΔC∘PRdT=844.81τ−844.81+488.88τ2−488.88−32.51τ3+32.51∫373.15TΔC∘PRdT=-32.51τ3+488.88τ2+844.81τ−1301.18$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=-32.51τ3+488.88τ2+844.81τ−1301.1832.51τ3−488.88τ2−844.81τ+2744.53=0$

On solving,

$τ=1.725,−3, 16.31$

Considering positive value of temperature $τ=1.725$

Hence,

$τ=TT0=T100+273.15=1.725T=643.68 K$

(p)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(p)

Expert Solution

Answer to Problem 4.3P

$T=416.06 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dammonia 3.578 3.020 0 −0.186$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.578×373.15×(τ−1)+3.020×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.186×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1335.13(τ−1)+210.25(τ2−1)−49.85(τ−1τ)∫373.15TΔC∘PRdT=1335.13τ−1335.13+210.25τ2−210.25−49.85+49.85τ∫373.15TΔC∘PRdT=210.25τ2+1335.13τ−1595.23+49.85τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=210.25τ2+1335.13τ−1595.23+49.85τ1210.25τ2+1335.13τ−3038.579+49.85τ=01210.25τ3+1335.13τ2−3038.579τ+49.85τ=0$

On solving,

$τ=1.115,0.0165, −2.23$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.115T=416.06 K$

(q)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(q)

Expert Solution

Answer to Problem 4.3P

$T=764.96 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon monoxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO 3.376 0.557 0 −0.031$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.376×373.15×(τ−1)+0.557×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.031×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1259.75(τ−1)+38.78(τ2−1)−8.31(τ−1τ)∫373.15TΔC∘PRdT=1259.75τ−1259.75+38.78τ2−38.78−8.31+8.31τ∫373.15TΔC∘PRdT=38.78τ2+1259.75τ−1306.84+8.31τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=38.78τ2+1259.75τ−1306.84+8.31τ38.78τ2+1259.75τ−2750.19+8.31τ=038.78τ3+1259.75τ2−2750.19τ+8.31τ=0$

On solving,

$τ=2.05,0.003, −34.54$

Considering positive value of temperature $τ=2.05$

Hence,

$τ=TT0=T100+273.15=2.05T=764.96 K$

(r)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(r)

Expert Solution

Answer to Problem 4.3P

$T=634.355 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO2 5.457 1.045 0 −1.157$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.457×373.15×(τ−1)+1.045×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.157×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2036.28(τ−1)+72.75(τ2−1)−310.06(τ−1τ)∫373.15TΔC∘PRdT=2036.28τ−2036.28+72.75τ2−72.75−310.06+310.06τ∫373.15TΔC∘PRdT=72.75τ2+2036.28τ−2419.09+310.06τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=72.75τ2+2036.28τ−2419.09+310.06τ72.75τ2+2036.28τ−3862.44+310.06τ=072.75τ3+2036.28τ2−3862.44τ+310.06τ=0$

On solving,

$τ=1.7,0.084, −29.78$

Considering positive value of temperature $τ=1.7$

Hence,

$τ=TT0=T100+273.15=1.7T=634.355 K$

(s)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(s)

Expert Solution

Answer to Problem 4.3P

$T=626.1457 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for sulphur dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DSO2 5.699 0.801 0 −1.015$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.699×373.15×(τ−1)+0.801×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.015×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2126.58(τ−1)+55.77(τ2−1)−272(τ−1τ)∫373.15TΔC∘PRdT=2126.58τ−2126.58+55.77τ2−55.77−272+272τ∫373.15TΔC∘PRdT=55.77τ2+2126.58τ−2454.36+272τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=55.77τ2+2126.58τ−2454.36+272τ55.77τ2+2126.58τ−3887.71+272τ=055.77τ3+2126.58τ2−3887.71τ+272τ=0$

On solving,

$τ=1.678,0.073, −39.88$

Considering positive value of temperature $τ=1.678$

Hence,

$τ=TT0=T100+273.15=1.678T=626.1457 K$

(t)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(t)

Expert Solution

Answer to Problem 4.3P

$T=706.37 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DH2O 3.47 1.450 0 0.121$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.470×373.15×(τ−1)+1.450×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.121×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1294.83(τ−1)+100.95(τ2−1)+32.43(τ−1τ)∫373.15TΔC∘PRdT=1294.83τ−1294.83+100.95τ2−100.95+32.43−32.43τ∫373.15TΔC∘PRdT=100.95τ2+1294.83τ−1363.35−32.43τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=100.95τ2+1294.83τ−1363.35−32.43τ100.95τ2+1294.83τ−2796.7−32.43τ=0100.95τ3+1294.83τ2−2796.7τ−32.43τ=0$

On solving,

$τ=1.893,−0.011, −14.708$

Considering positive value of temperature $τ=1.893$

Hence,

$τ=TT0=T100+273.15=1.893T=706.37 K$

(u)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(u)

Expert Solution

Answer to Problem 4.3P

$T=770.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for nitrogen in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DN2 3.280 0.593 0 0.040$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.280×373.15×(τ−1)+0.593×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.040×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1223.93(τ−1)+41.29(τ2−1)+10.72(τ−1τ)∫373.15TΔC∘PRdT=1223.93τ−1223.93+41.29τ2−41.29+10.72−10.72τ∫373.15TΔC∘PRdT=41.29τ2+1223.93τ−1254.5−10.72τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=41.29τ2+1223.93τ−1254.5−10.72τ41.29τ2+1223.93τ−2697.85−10.72τ=041.29τ3+1223.93τ2−2697.85τ−10.72τ=0$

On solving,

$τ=2.065,−0.003, −31.703$

Considering positive value of temperature $τ=2.065$

Hence,

$τ=TT0=T100+273.15=2.065T=770.55 K$

(v)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(v)

Expert Solution

Answer to Problem 4.3P

$T=654.13 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for hydrogen cyanide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DHCN 4.736 1.359 0 −0.725$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.736×373.15×(τ−1)+1.359×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.725×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1767.24(τ−1)+94.61(τ2−1)−194.29(τ−1τ)∫373.15TΔC∘PRdT=1767.24τ−1767.24+94.61τ2−94.61−194.29+194.29τ∫373.15TΔC∘PRdT=94.61τ2+1767.24τ−2056.14+194.29τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=94.61τ2+1767.24τ−2056.14+194.29τ94.61τ2+1767.24τ−3499.49+194.29τ=094.61τ2+1767.24τ−3499.49τ+194.29τ=0$

On solving,

$τ=1.753,0.057, −20.49$

Considering positive value of temperature $τ=1.753$

Hence,

$τ=TT0=T100+273.15=1.753T=654.13 K$

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Answer 3

Question

Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(a)

Expert Solution

Answer to Problem 4.3P

$T=626.817 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dmethane 1.702 9.081 −2.164 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.702×373.15×(τ−1)+9.081×10−32373.152(τ2−1) +−2.164×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=635.1013(τ−1)+632.223(τ2−1)−37.479(τ3−1)∫373.15TΔC∘PRdT=635.1013τ−635.1013+632.223τ2−632.223−37.479τ3+37.479∫373.15TΔC∘PRdT=−37.479τ3+632.223τ2+635.1013τ−1229.8453$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K =−37.479τ3+632.223τ2+635.1013τ−1229.845337.479τ3−632.223τ2−635.1013τ+2673.194=0$

On solving,

$τ=1.6798,−2.412, 17.6$

Considering positive value of temperature $τ=1.6798$

Hence,

$τ=TT0=T100+273.15=1.6798T=626.817 K$

(b)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(b)

Expert Solution

Answer to Problem 4.3P

$T=538.455 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethane 1.131 19.225 −5.561 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.131×373.15×(τ−1)+19.225×10−32373.152(τ2−1) +−5.561×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=422.033(τ−1)+1338.453(τ2−1)−96.31(τ3−1)∫373.15TΔC∘PRdT=422.033τ−422.033+1338.453τ2−1338.453−96.31τ3+96.31∫373.15TΔC∘PRdT=−96.31τ3+1338.453τ2+422.033τ−1664.176$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−96.31τ3+1338.453τ2+422.033τ−1664.17696.31τ3−1338.453τ2−422.033τ+3107.525=0$

On solving,

$τ=1.443,−1.592, 14.0458$

Considering positive value of temperature $τ=1.443$

Hence,

$τ=TT0=T100+273.15=1.443T=538.455 K$

(c)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(c)

Expert Solution

Answer to Problem 4.3P

$T=493.3043 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.213×373.15×(τ−1)+28.785×10−32373.152(τ2−1) +−8.824×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=452.631(τ−1)+2004.025(τ2−1)−152.825(τ3−1)∫373.15TΔC∘PRdT=452.631τ−452.631+2004.025τ2−2004.025−152.825τ3+152.825∫373.15TΔC∘PRdT=−152.825τ3+2004.025τ2+452.631τ−2303.831$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−152.825τ3+2004.025τ2+452.631τ−2303.831152.825τ3−2004.025τ2−452.631τ+3747.18=0$

On solving,

$τ=1.322,−1.405, 13.197$

Considering positive value of temperature $τ=1.322$

Hence,

$τ=TT0=T100+273.15=1.322T=493.3043 K$

(d)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(d)

Expert Solution

Answer to Problem 4.3P

$T=466.81 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -butane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.935×373.15×(τ−1)+36.915×10−32373.152(τ2−1) +−11.402×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=722.045(τ−1)+2570.039(τ2−1)−197.474(τ3−1)∫373.15TΔC∘PRdT=722.045τ−722.045+2570.039τ2−2570.039−197.474τ3+197.474∫373.15TΔC∘PRdT=−197.474τ3+2570.039τ2+722.045τ−3094.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−197.474τ3+2570.039τ2+722.045τ−3094.61197.474τ3−2570.039τ2−722.045τ+4537.96=0$

On solving,

$τ=1.251,−1.396, 13.16$

Considering positive value of temperature $τ=1.251$

Hence,

$τ=TT0=T100+273.15=1.251T=466.81 K$

(e)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(e)

Expert Solution

Answer to Problem 4.3P

$T=438.45 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -hexanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 3.025 53.722 −16.791 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.025×373.15×(τ−1)+53.722×10−32373.152(τ2−1) +−16.791×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1128.78(τ−1)+3740.15(τ2−1)−290.807(τ3−1)∫373.15TΔC∘PRdT=1128.78τ−1128.78+3740.15τ2−3740.15−290.807τ3+290.807∫373.15TΔC∘PRdT=−290.807τ3+3740.15τ2+1128.78τ−4578.12$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−290.807τ3+3740.15τ2+1128.78τ−4578.12290.807τ3−3740.15τ2−1128.78τ+6021.471=0$

On solving,

$τ=1.175,−1.351, 13.04$

Considering positive value of temperature $τ=1.175$

Hence,

$τ=TT0=T100+273.15=1.175T=438.45 K$

(f)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(f)

Expert Solution

Answer to Problem 4.3P

$T=371.695 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -octanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 4.108 70.567 −22.208 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.108×373.15×(τ−1)+70.567×10−32373.152(τ2−1) +−22.208×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1532.9002(τ−1)+4912.91(τ2−1)−384.63(τ3−1)∫373.15TΔC∘PRdT=1532.9002τ−1532.9002+4912.91τ2−4912.91−384.63τ3+384.63∫373.15TΔC∘PRdT=−384.63τ3+4912.91τ2+1532.9002τ−6061.184$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−384.63τ3+4912.91τ2+1532.9002τ−6061.184384.63τ3−4912.91τ2−1532.9002τ+6021.471=0$

On solving,

$τ=0.9961,−1.21, 12.987$

Considering positive value of temperature $τ=0.9961$

Hence,

$τ=TT0=T100+273.15=0.9961T=371.695 K$

(g)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(g)

Expert Solution

Answer to Problem 4.3P

$T=513.08 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropylene 1.637 22.706 −6.915 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.637×373.15×(τ−1)+22.706×10−32373.152(τ2−1) +−6.915×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=610.847(τ−1)+1580.8(τ2−1)−119.763(τ3−1)∫373.15TΔC∘PRdT=610.847τ−610.847+1580.8τ2−1580.8−119.763τ3+119.763∫373.15TΔC∘PRdT=−119.763τ3+1580.8τ2+610.847τ−2071.88$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−119.763τ3+1580.8τ2+610.847τ−2071.88119.763τ3−1580.8τ2−610.847τ+3515.233=0$

On solving,

$τ=1.375,−1.592, 13.416$

Considering positive value of temperature $τ=1.375$

Hence,

$τ=TT0=T100+273.15=1.375T=513.08 K$

(h)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(h)

Expert Solution

Answer to Problem 4.3P

$T=457.855 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-pentene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−pentene 2.691 39.753 −12.447 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.691×373.15×(τ−1)+39.753×10−32373.152(τ2−1) +−12.447×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1004.147(τ−1)+2767.62(τ2−1)−215.57(τ3−1)∫373.15TΔC∘PRdT=1004.147τ−1004.147+2767.62τ2−2767.62−215.57τ3+215.57∫373.15TΔC∘PRdT=−215.57τ3+2767.62τ2+1004.147τ−3556.197$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−215.57τ3+2767.62τ2+1004.147τ−3556.197215.57τ3−2767.62τ2−1004.147τ+4999.546=0$

On solving,

$τ=1.227,−1.447, 13.059$

Considering positive value of temperature $τ=1.227$

Hence,

$τ=TT0=T100+273.15=1.227T=457.855 K$

(i)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(i)

Expert Solution

Answer to Problem 4.3P

$T=433.97 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-heptenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−heptene 3.768 56.588 −17.847 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.768×373.15×(τ−1)+56.588×10−32373.152(τ2−1) +−17.847×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1406.03(τ−1)+3939.68(τ2−1)−309.097(τ3−1)∫373.15TΔC∘PRdT=1406.03τ−1406.03+3939.68τ2−3939.68−309.097τ3+309.097∫373.15TΔC∘PRdT=−309.097τ3+3939.68τ2+1406.03τ−5036.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−309.097τ3+3939.68τ2+1406.03τ−5036.61309.097τ3−3939.68τ2−1406.03τ+6479.96=0$

On solving,

$τ=1.163,−1.389, 12.97$

Considering positive value of temperature $τ=1.163$

Hence,

$τ=TT0=T100+273.15=1.163T=433.97 K$

(j)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(j)

Expert Solution

Answer to Problem 4.3P

$T=426.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-Octenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−Octene 4.324 64.96 −20.521 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.324×373.15×(τ−1)+64.96×10−32373.152(τ2−1) +−20.521×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1613.5(τ−1)+4522.55(τ2−1)−355.41(τ3−1)∫373.15TΔC∘PRdT=1613.5τ−1613.5+4522.55τ2−4522.55−355.41τ3+355.41∫373.15TΔC∘PRdT=−355.41τ3+4522.55τ2+1613.5τ−5780.64$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−355.41τ3+4522.55τ2+1613.5τ−5780.64355.41τ3−4522.55τ2−1613.5τ+7223.99=0$

On solving,

$τ=1.143,−1.37, 12.95$

Considering positive value of temperature $τ=1.143$

Hence,

$τ=TT0=T100+273.15=1.143T=426.51 K$

(k)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(k)

Expert Solution

Answer to Problem 4.3P

$T=595.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for acetylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dacetylene 6.132 1.952 0 −1.299$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=6.132×373.15×(τ−1)+1.952×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.299×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2288.15(τ−1)+135.9(τ2−1)−348.12(τ−1τ)∫373.15TΔC∘PRdT=2288.15τ−2288.15+135.9τ2−135.9−348.12+348.12τ∫373.15TΔC∘PRdT=135.9τ2+2288.15τ−2772.17+348.12τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=135.9τ2+2288.15τ−2772.17+348.12τ135.9τ2+2288.15τ−4215.519+348.12τ=0135.9τ3+2288.15τ2−4215.519τ+348.12τ=0$

On solving,

$τ=1.596,0.099, −18.52$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.596T=595.55 K$

(l)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(l)

Expert Solution

Answer to Problem 4.3P

$T=476.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for benzenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dbenzene −0.206 39.064 −13.301 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=−0.206×373.15×(τ−1)+39.064×10−32373.152(τ2−1) +−13.301×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=−76.87(τ−1)+2719.65(τ2−1)−230.36(τ3−1)∫373.15TΔC∘PRdT=−76.87τ+76.87+2719.65τ2−2719.65−230.36τ3+230.36∫373.15TΔC∘PRdT=−230.36τ3+2719.65τ2−76.87τ−2412.42$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−230.36τ3+2719.65τ2−76.87τ−2412.42230.36τ3−2719.65τ2+76.87τ+3855.769=0$

On solving,

$τ=1.277,−1.12, 11.65$

Considering positive value of temperature $τ=1.277$

Hence,

$τ=TT0=T100+273.15=1.277T=476.51 K$

(m)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(m)

Expert Solution

Answer to Problem 4.3P

$T=380.613 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethanol in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethanol 3.518 20.001 −6.002 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.518×373.15×(τ−1)+20.001×10−32373.152(τ2−1) +−6.002×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1312.74(τ−1)+1392.48(τ2−1)−103.95(τ3−1)∫373.15TΔC∘PRdT=1312.74τ−1312.74+1392.48τ2−1392.48−103.95τ3+103.95∫373.15TΔC∘PRdT=−103.95τ3+2719.65τ2+1312.74τ−2601.27$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−103.95τ3+2719.65τ2+1312.74τ−2601.27103.95τ3−2719.65τ2−1312.74τ+4044.62=0$

On solving,

$τ=1.02,−1.44, 26.58$

Considering positive value of temperature $τ=1.02$

Hence,

$τ=TT0=T100+273.15=1.02T=380.613 K$

(n)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(n)

Expert Solution

Answer to Problem 4.3P

$T=445.91 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for styrene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dstyrene 2.050 50.192 −16.662 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.050×373.15×(τ−1)+50.192×10−32373.152(τ2−1) +−16.662×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=764.96(τ−1)+3494.39(τ2−1)−288.57(τ3−1)∫373.15TΔC∘PRdT=764.96τ−764.96+3494.39τ2−3494.39−288.57τ3+288.57∫373.15TΔC∘PRdT=−288.57τ3+3494.39τ2+764.96τ−3970.78$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−288.57τ3+3494.39τ2+764.96τ−3970.78288.57τ3−3494.39τ2−764.96τ+5414.126=0$

On solving,

$τ=1.195,−1.287, 12.2$

Considering positive value of temperature $τ=1.195$

Hence,

$τ=TT0=T100+273.15=1.195T=445.91 K$

(o)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(o)

Expert Solution

Answer to Problem 4.3P

$T=643.68 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for formaldehyde in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dformaldehyde 2.264 7.022 −1.877 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.264×373.15×(τ−1)+7.022×10−32373.152(τ2−1) +−1.877×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=844.81(τ−1)+488.88(τ2−1)−32.51(τ3−1)∫373.15TΔC∘PRdT=844.81τ−844.81+488.88τ2−488.88−32.51τ3+32.51∫373.15TΔC∘PRdT=-32.51τ3+488.88τ2+844.81τ−1301.18$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=-32.51τ3+488.88τ2+844.81τ−1301.1832.51τ3−488.88τ2−844.81τ+2744.53=0$

On solving,

$τ=1.725,−3, 16.31$

Considering positive value of temperature $τ=1.725$

Hence,

$τ=TT0=T100+273.15=1.725T=643.68 K$

(p)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(p)

Expert Solution

Answer to Problem 4.3P

$T=416.06 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dammonia 3.578 3.020 0 −0.186$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.578×373.15×(τ−1)+3.020×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.186×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1335.13(τ−1)+210.25(τ2−1)−49.85(τ−1τ)∫373.15TΔC∘PRdT=1335.13τ−1335.13+210.25τ2−210.25−49.85+49.85τ∫373.15TΔC∘PRdT=210.25τ2+1335.13τ−1595.23+49.85τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=210.25τ2+1335.13τ−1595.23+49.85τ1210.25τ2+1335.13τ−3038.579+49.85τ=01210.25τ3+1335.13τ2−3038.579τ+49.85τ=0$

On solving,

$τ=1.115,0.0165, −2.23$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.115T=416.06 K$

(q)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(q)

Expert Solution

Answer to Problem 4.3P

$T=764.96 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon monoxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO 3.376 0.557 0 −0.031$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.376×373.15×(τ−1)+0.557×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.031×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1259.75(τ−1)+38.78(τ2−1)−8.31(τ−1τ)∫373.15TΔC∘PRdT=1259.75τ−1259.75+38.78τ2−38.78−8.31+8.31τ∫373.15TΔC∘PRdT=38.78τ2+1259.75τ−1306.84+8.31τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=38.78τ2+1259.75τ−1306.84+8.31τ38.78τ2+1259.75τ−2750.19+8.31τ=038.78τ3+1259.75τ2−2750.19τ+8.31τ=0$

On solving,

$τ=2.05,0.003, −34.54$

Considering positive value of temperature $τ=2.05$

Hence,

$τ=TT0=T100+273.15=2.05T=764.96 K$

(r)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(r)

Expert Solution

Answer to Problem 4.3P

$T=634.355 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO2 5.457 1.045 0 −1.157$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.457×373.15×(τ−1)+1.045×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.157×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2036.28(τ−1)+72.75(τ2−1)−310.06(τ−1τ)∫373.15TΔC∘PRdT=2036.28τ−2036.28+72.75τ2−72.75−310.06+310.06τ∫373.15TΔC∘PRdT=72.75τ2+2036.28τ−2419.09+310.06τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=72.75τ2+2036.28τ−2419.09+310.06τ72.75τ2+2036.28τ−3862.44+310.06τ=072.75τ3+2036.28τ2−3862.44τ+310.06τ=0$

On solving,

$τ=1.7,0.084, −29.78$

Considering positive value of temperature $τ=1.7$

Hence,

$τ=TT0=T100+273.15=1.7T=634.355 K$

(s)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(s)

Expert Solution

Answer to Problem 4.3P

$T=626.1457 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for sulphur dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DSO2 5.699 0.801 0 −1.015$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.699×373.15×(τ−1)+0.801×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.015×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2126.58(τ−1)+55.77(τ2−1)−272(τ−1τ)∫373.15TΔC∘PRdT=2126.58τ−2126.58+55.77τ2−55.77−272+272τ∫373.15TΔC∘PRdT=55.77τ2+2126.58τ−2454.36+272τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=55.77τ2+2126.58τ−2454.36+272τ55.77τ2+2126.58τ−3887.71+272τ=055.77τ3+2126.58τ2−3887.71τ+272τ=0$

On solving,

$τ=1.678,0.073, −39.88$

Considering positive value of temperature $τ=1.678$

Hence,

$τ=TT0=T100+273.15=1.678T=626.1457 K$

(t)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(t)

Expert Solution

Answer to Problem 4.3P

$T=706.37 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DH2O 3.47 1.450 0 0.121$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.470×373.15×(τ−1)+1.450×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.121×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1294.83(τ−1)+100.95(τ2−1)+32.43(τ−1τ)∫373.15TΔC∘PRdT=1294.83τ−1294.83+100.95τ2−100.95+32.43−32.43τ∫373.15TΔC∘PRdT=100.95τ2+1294.83τ−1363.35−32.43τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=100.95τ2+1294.83τ−1363.35−32.43τ100.95τ2+1294.83τ−2796.7−32.43τ=0100.95τ3+1294.83τ2−2796.7τ−32.43τ=0$

On solving,

$τ=1.893,−0.011, −14.708$

Considering positive value of temperature $τ=1.893$

Hence,

$τ=TT0=T100+273.15=1.893T=706.37 K$

(u)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(u)

Expert Solution

Answer to Problem 4.3P

$T=770.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for nitrogen in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DN2 3.280 0.593 0 0.040$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.280×373.15×(τ−1)+0.593×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.040×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1223.93(τ−1)+41.29(τ2−1)+10.72(τ−1τ)∫373.15TΔC∘PRdT=1223.93τ−1223.93+41.29τ2−41.29+10.72−10.72τ∫373.15TΔC∘PRdT=41.29τ2+1223.93τ−1254.5−10.72τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=41.29τ2+1223.93τ−1254.5−10.72τ41.29τ2+1223.93τ−2697.85−10.72τ=041.29τ3+1223.93τ2−2697.85τ−10.72τ=0$

On solving,

$τ=2.065,−0.003, −31.703$

Considering positive value of temperature $τ=2.065$

Hence,

$τ=TT0=T100+273.15=2.065T=770.55 K$

(v)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(v)

Expert Solution

Answer to Problem 4.3P

$T=654.13 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for hydrogen cyanide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DHCN 4.736 1.359 0 −0.725$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.736×373.15×(τ−1)+1.359×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.725×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1767.24(τ−1)+94.61(τ2−1)−194.29(τ−1τ)∫373.15TΔC∘PRdT=1767.24τ−1767.24+94.61τ2−94.61−194.29+194.29τ∫373.15TΔC∘PRdT=94.61τ2+1767.24τ−2056.14+194.29τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=94.61τ2+1767.24τ−2056.14+194.29τ94.61τ2+1767.24τ−3499.49+194.29τ=094.61τ2+1767.24τ−3499.49τ+194.29τ=0$

On solving,

$τ=1.753,0.057, −20.49$

Considering positive value of temperature $τ=1.753$

Hence,

$τ=TT0=T100+273.15=1.753T=654.13 K$

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Answer 4

Question

Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(a)

Expert Solution

Answer to Problem 4.3P

$T=626.817 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dmethane 1.702 9.081 −2.164 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.702×373.15×(τ−1)+9.081×10−32373.152(τ2−1) +−2.164×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=635.1013(τ−1)+632.223(τ2−1)−37.479(τ3−1)∫373.15TΔC∘PRdT=635.1013τ−635.1013+632.223τ2−632.223−37.479τ3+37.479∫373.15TΔC∘PRdT=−37.479τ3+632.223τ2+635.1013τ−1229.8453$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K =−37.479τ3+632.223τ2+635.1013τ−1229.845337.479τ3−632.223τ2−635.1013τ+2673.194=0$

On solving,

$τ=1.6798,−2.412, 17.6$

Considering positive value of temperature $τ=1.6798$

Hence,

$τ=TT0=T100+273.15=1.6798T=626.817 K$

(b)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(b)

Expert Solution

Answer to Problem 4.3P

$T=538.455 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethane 1.131 19.225 −5.561 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.131×373.15×(τ−1)+19.225×10−32373.152(τ2−1) +−5.561×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=422.033(τ−1)+1338.453(τ2−1)−96.31(τ3−1)∫373.15TΔC∘PRdT=422.033τ−422.033+1338.453τ2−1338.453−96.31τ3+96.31∫373.15TΔC∘PRdT=−96.31τ3+1338.453τ2+422.033τ−1664.176$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−96.31τ3+1338.453τ2+422.033τ−1664.17696.31τ3−1338.453τ2−422.033τ+3107.525=0$

On solving,

$τ=1.443,−1.592, 14.0458$

Considering positive value of temperature $τ=1.443$

Hence,

$τ=TT0=T100+273.15=1.443T=538.455 K$

(c)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(c)

Expert Solution

Answer to Problem 4.3P

$T=493.3043 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.213×373.15×(τ−1)+28.785×10−32373.152(τ2−1) +−8.824×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=452.631(τ−1)+2004.025(τ2−1)−152.825(τ3−1)∫373.15TΔC∘PRdT=452.631τ−452.631+2004.025τ2−2004.025−152.825τ3+152.825∫373.15TΔC∘PRdT=−152.825τ3+2004.025τ2+452.631τ−2303.831$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−152.825τ3+2004.025τ2+452.631τ−2303.831152.825τ3−2004.025τ2−452.631τ+3747.18=0$

On solving,

$τ=1.322,−1.405, 13.197$

Considering positive value of temperature $τ=1.322$

Hence,

$τ=TT0=T100+273.15=1.322T=493.3043 K$

(d)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(d)

Expert Solution

Answer to Problem 4.3P

$T=466.81 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -butane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.935×373.15×(τ−1)+36.915×10−32373.152(τ2−1) +−11.402×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=722.045(τ−1)+2570.039(τ2−1)−197.474(τ3−1)∫373.15TΔC∘PRdT=722.045τ−722.045+2570.039τ2−2570.039−197.474τ3+197.474∫373.15TΔC∘PRdT=−197.474τ3+2570.039τ2+722.045τ−3094.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−197.474τ3+2570.039τ2+722.045τ−3094.61197.474τ3−2570.039τ2−722.045τ+4537.96=0$

On solving,

$τ=1.251,−1.396, 13.16$

Considering positive value of temperature $τ=1.251$

Hence,

$τ=TT0=T100+273.15=1.251T=466.81 K$

(e)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(e)

Expert Solution

Answer to Problem 4.3P

$T=438.45 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -hexanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 3.025 53.722 −16.791 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.025×373.15×(τ−1)+53.722×10−32373.152(τ2−1) +−16.791×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1128.78(τ−1)+3740.15(τ2−1)−290.807(τ3−1)∫373.15TΔC∘PRdT=1128.78τ−1128.78+3740.15τ2−3740.15−290.807τ3+290.807∫373.15TΔC∘PRdT=−290.807τ3+3740.15τ2+1128.78τ−4578.12$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−290.807τ3+3740.15τ2+1128.78τ−4578.12290.807τ3−3740.15τ2−1128.78τ+6021.471=0$

On solving,

$τ=1.175,−1.351, 13.04$

Considering positive value of temperature $τ=1.175$

Hence,

$τ=TT0=T100+273.15=1.175T=438.45 K$

(f)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(f)

Expert Solution

Answer to Problem 4.3P

$T=371.695 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -octanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 4.108 70.567 −22.208 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.108×373.15×(τ−1)+70.567×10−32373.152(τ2−1) +−22.208×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1532.9002(τ−1)+4912.91(τ2−1)−384.63(τ3−1)∫373.15TΔC∘PRdT=1532.9002τ−1532.9002+4912.91τ2−4912.91−384.63τ3+384.63∫373.15TΔC∘PRdT=−384.63τ3+4912.91τ2+1532.9002τ−6061.184$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−384.63τ3+4912.91τ2+1532.9002τ−6061.184384.63τ3−4912.91τ2−1532.9002τ+6021.471=0$

On solving,

$τ=0.9961,−1.21, 12.987$

Considering positive value of temperature $τ=0.9961$

Hence,

$τ=TT0=T100+273.15=0.9961T=371.695 K$

(g)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(g)

Expert Solution

Answer to Problem 4.3P

$T=513.08 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropylene 1.637 22.706 −6.915 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.637×373.15×(τ−1)+22.706×10−32373.152(τ2−1) +−6.915×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=610.847(τ−1)+1580.8(τ2−1)−119.763(τ3−1)∫373.15TΔC∘PRdT=610.847τ−610.847+1580.8τ2−1580.8−119.763τ3+119.763∫373.15TΔC∘PRdT=−119.763τ3+1580.8τ2+610.847τ−2071.88$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−119.763τ3+1580.8τ2+610.847τ−2071.88119.763τ3−1580.8τ2−610.847τ+3515.233=0$

On solving,

$τ=1.375,−1.592, 13.416$

Considering positive value of temperature $τ=1.375$

Hence,

$τ=TT0=T100+273.15=1.375T=513.08 K$

(h)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(h)

Expert Solution

Answer to Problem 4.3P

$T=457.855 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-pentene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−pentene 2.691 39.753 −12.447 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.691×373.15×(τ−1)+39.753×10−32373.152(τ2−1) +−12.447×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1004.147(τ−1)+2767.62(τ2−1)−215.57(τ3−1)∫373.15TΔC∘PRdT=1004.147τ−1004.147+2767.62τ2−2767.62−215.57τ3+215.57∫373.15TΔC∘PRdT=−215.57τ3+2767.62τ2+1004.147τ−3556.197$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−215.57τ3+2767.62τ2+1004.147τ−3556.197215.57τ3−2767.62τ2−1004.147τ+4999.546=0$

On solving,

$τ=1.227,−1.447, 13.059$

Considering positive value of temperature $τ=1.227$

Hence,

$τ=TT0=T100+273.15=1.227T=457.855 K$

(i)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(i)

Expert Solution

Answer to Problem 4.3P

$T=433.97 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-heptenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−heptene 3.768 56.588 −17.847 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.768×373.15×(τ−1)+56.588×10−32373.152(τ2−1) +−17.847×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1406.03(τ−1)+3939.68(τ2−1)−309.097(τ3−1)∫373.15TΔC∘PRdT=1406.03τ−1406.03+3939.68τ2−3939.68−309.097τ3+309.097∫373.15TΔC∘PRdT=−309.097τ3+3939.68τ2+1406.03τ−5036.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−309.097τ3+3939.68τ2+1406.03τ−5036.61309.097τ3−3939.68τ2−1406.03τ+6479.96=0$

On solving,

$τ=1.163,−1.389, 12.97$

Considering positive value of temperature $τ=1.163$

Hence,

$τ=TT0=T100+273.15=1.163T=433.97 K$

(j)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(j)

Expert Solution

Answer to Problem 4.3P

$T=426.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-Octenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−Octene 4.324 64.96 −20.521 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.324×373.15×(τ−1)+64.96×10−32373.152(τ2−1) +−20.521×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1613.5(τ−1)+4522.55(τ2−1)−355.41(τ3−1)∫373.15TΔC∘PRdT=1613.5τ−1613.5+4522.55τ2−4522.55−355.41τ3+355.41∫373.15TΔC∘PRdT=−355.41τ3+4522.55τ2+1613.5τ−5780.64$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−355.41τ3+4522.55τ2+1613.5τ−5780.64355.41τ3−4522.55τ2−1613.5τ+7223.99=0$

On solving,

$τ=1.143,−1.37, 12.95$

Considering positive value of temperature $τ=1.143$

Hence,

$τ=TT0=T100+273.15=1.143T=426.51 K$

(k)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(k)

Expert Solution

Answer to Problem 4.3P

$T=595.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for acetylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dacetylene 6.132 1.952 0 −1.299$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=6.132×373.15×(τ−1)+1.952×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.299×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2288.15(τ−1)+135.9(τ2−1)−348.12(τ−1τ)∫373.15TΔC∘PRdT=2288.15τ−2288.15+135.9τ2−135.9−348.12+348.12τ∫373.15TΔC∘PRdT=135.9τ2+2288.15τ−2772.17+348.12τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=135.9τ2+2288.15τ−2772.17+348.12τ135.9τ2+2288.15τ−4215.519+348.12τ=0135.9τ3+2288.15τ2−4215.519τ+348.12τ=0$

On solving,

$τ=1.596,0.099, −18.52$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.596T=595.55 K$

(l)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(l)

Expert Solution

Answer to Problem 4.3P

$T=476.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for benzenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dbenzene −0.206 39.064 −13.301 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=−0.206×373.15×(τ−1)+39.064×10−32373.152(τ2−1) +−13.301×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=−76.87(τ−1)+2719.65(τ2−1)−230.36(τ3−1)∫373.15TΔC∘PRdT=−76.87τ+76.87+2719.65τ2−2719.65−230.36τ3+230.36∫373.15TΔC∘PRdT=−230.36τ3+2719.65τ2−76.87τ−2412.42$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−230.36τ3+2719.65τ2−76.87τ−2412.42230.36τ3−2719.65τ2+76.87τ+3855.769=0$

On solving,

$τ=1.277,−1.12, 11.65$

Considering positive value of temperature $τ=1.277$

Hence,

$τ=TT0=T100+273.15=1.277T=476.51 K$

(m)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(m)

Expert Solution

Answer to Problem 4.3P

$T=380.613 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethanol in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethanol 3.518 20.001 −6.002 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.518×373.15×(τ−1)+20.001×10−32373.152(τ2−1) +−6.002×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1312.74(τ−1)+1392.48(τ2−1)−103.95(τ3−1)∫373.15TΔC∘PRdT=1312.74τ−1312.74+1392.48τ2−1392.48−103.95τ3+103.95∫373.15TΔC∘PRdT=−103.95τ3+2719.65τ2+1312.74τ−2601.27$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−103.95τ3+2719.65τ2+1312.74τ−2601.27103.95τ3−2719.65τ2−1312.74τ+4044.62=0$

On solving,

$τ=1.02,−1.44, 26.58$

Considering positive value of temperature $τ=1.02$

Hence,

$τ=TT0=T100+273.15=1.02T=380.613 K$

(n)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(n)

Expert Solution

Answer to Problem 4.3P

$T=445.91 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for styrene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dstyrene 2.050 50.192 −16.662 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.050×373.15×(τ−1)+50.192×10−32373.152(τ2−1) +−16.662×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=764.96(τ−1)+3494.39(τ2−1)−288.57(τ3−1)∫373.15TΔC∘PRdT=764.96τ−764.96+3494.39τ2−3494.39−288.57τ3+288.57∫373.15TΔC∘PRdT=−288.57τ3+3494.39τ2+764.96τ−3970.78$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−288.57τ3+3494.39τ2+764.96τ−3970.78288.57τ3−3494.39τ2−764.96τ+5414.126=0$

On solving,

$τ=1.195,−1.287, 12.2$

Considering positive value of temperature $τ=1.195$

Hence,

$τ=TT0=T100+273.15=1.195T=445.91 K$

(o)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(o)

Expert Solution

Answer to Problem 4.3P

$T=643.68 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for formaldehyde in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dformaldehyde 2.264 7.022 −1.877 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.264×373.15×(τ−1)+7.022×10−32373.152(τ2−1) +−1.877×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=844.81(τ−1)+488.88(τ2−1)−32.51(τ3−1)∫373.15TΔC∘PRdT=844.81τ−844.81+488.88τ2−488.88−32.51τ3+32.51∫373.15TΔC∘PRdT=-32.51τ3+488.88τ2+844.81τ−1301.18$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=-32.51τ3+488.88τ2+844.81τ−1301.1832.51τ3−488.88τ2−844.81τ+2744.53=0$

On solving,

$τ=1.725,−3, 16.31$

Considering positive value of temperature $τ=1.725$

Hence,

$τ=TT0=T100+273.15=1.725T=643.68 K$

(p)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(p)

Expert Solution

Answer to Problem 4.3P

$T=416.06 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dammonia 3.578 3.020 0 −0.186$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.578×373.15×(τ−1)+3.020×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.186×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1335.13(τ−1)+210.25(τ2−1)−49.85(τ−1τ)∫373.15TΔC∘PRdT=1335.13τ−1335.13+210.25τ2−210.25−49.85+49.85τ∫373.15TΔC∘PRdT=210.25τ2+1335.13τ−1595.23+49.85τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=210.25τ2+1335.13τ−1595.23+49.85τ1210.25τ2+1335.13τ−3038.579+49.85τ=01210.25τ3+1335.13τ2−3038.579τ+49.85τ=0$

On solving,

$τ=1.115,0.0165, −2.23$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.115T=416.06 K$

(q)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(q)

Expert Solution

Answer to Problem 4.3P

$T=764.96 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon monoxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO 3.376 0.557 0 −0.031$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.376×373.15×(τ−1)+0.557×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.031×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1259.75(τ−1)+38.78(τ2−1)−8.31(τ−1τ)∫373.15TΔC∘PRdT=1259.75τ−1259.75+38.78τ2−38.78−8.31+8.31τ∫373.15TΔC∘PRdT=38.78τ2+1259.75τ−1306.84+8.31τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=38.78τ2+1259.75τ−1306.84+8.31τ38.78τ2+1259.75τ−2750.19+8.31τ=038.78τ3+1259.75τ2−2750.19τ+8.31τ=0$

On solving,

$τ=2.05,0.003, −34.54$

Considering positive value of temperature $τ=2.05$

Hence,

$τ=TT0=T100+273.15=2.05T=764.96 K$

(r)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(r)

Expert Solution

Answer to Problem 4.3P

$T=634.355 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO2 5.457 1.045 0 −1.157$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.457×373.15×(τ−1)+1.045×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.157×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2036.28(τ−1)+72.75(τ2−1)−310.06(τ−1τ)∫373.15TΔC∘PRdT=2036.28τ−2036.28+72.75τ2−72.75−310.06+310.06τ∫373.15TΔC∘PRdT=72.75τ2+2036.28τ−2419.09+310.06τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=72.75τ2+2036.28τ−2419.09+310.06τ72.75τ2+2036.28τ−3862.44+310.06τ=072.75τ3+2036.28τ2−3862.44τ+310.06τ=0$

On solving,

$τ=1.7,0.084, −29.78$

Considering positive value of temperature $τ=1.7$

Hence,

$τ=TT0=T100+273.15=1.7T=634.355 K$

(s)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(s)

Expert Solution

Answer to Problem 4.3P

$T=626.1457 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for sulphur dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DSO2 5.699 0.801 0 −1.015$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.699×373.15×(τ−1)+0.801×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.015×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2126.58(τ−1)+55.77(τ2−1)−272(τ−1τ)∫373.15TΔC∘PRdT=2126.58τ−2126.58+55.77τ2−55.77−272+272τ∫373.15TΔC∘PRdT=55.77τ2+2126.58τ−2454.36+272τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=55.77τ2+2126.58τ−2454.36+272τ55.77τ2+2126.58τ−3887.71+272τ=055.77τ3+2126.58τ2−3887.71τ+272τ=0$

On solving,

$τ=1.678,0.073, −39.88$

Considering positive value of temperature $τ=1.678$

Hence,

$τ=TT0=T100+273.15=1.678T=626.1457 K$

(t)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(t)

Expert Solution

Answer to Problem 4.3P

$T=706.37 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DH2O 3.47 1.450 0 0.121$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.470×373.15×(τ−1)+1.450×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.121×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1294.83(τ−1)+100.95(τ2−1)+32.43(τ−1τ)∫373.15TΔC∘PRdT=1294.83τ−1294.83+100.95τ2−100.95+32.43−32.43τ∫373.15TΔC∘PRdT=100.95τ2+1294.83τ−1363.35−32.43τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=100.95τ2+1294.83τ−1363.35−32.43τ100.95τ2+1294.83τ−2796.7−32.43τ=0100.95τ3+1294.83τ2−2796.7τ−32.43τ=0$

On solving,

$τ=1.893,−0.011, −14.708$

Considering positive value of temperature $τ=1.893$

Hence,

$τ=TT0=T100+273.15=1.893T=706.37 K$

(u)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(u)

Expert Solution

Answer to Problem 4.3P

$T=770.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for nitrogen in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DN2 3.280 0.593 0 0.040$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.280×373.15×(τ−1)+0.593×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.040×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1223.93(τ−1)+41.29(τ2−1)+10.72(τ−1τ)∫373.15TΔC∘PRdT=1223.93τ−1223.93+41.29τ2−41.29+10.72−10.72τ∫373.15TΔC∘PRdT=41.29τ2+1223.93τ−1254.5−10.72τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=41.29τ2+1223.93τ−1254.5−10.72τ41.29τ2+1223.93τ−2697.85−10.72τ=041.29τ3+1223.93τ2−2697.85τ−10.72τ=0$

On solving,

$τ=2.065,−0.003, −31.703$

Considering positive value of temperature $τ=2.065$

Hence,

$τ=TT0=T100+273.15=2.065T=770.55 K$

(v)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(v)

Expert Solution

Answer to Problem 4.3P

$T=654.13 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for hydrogen cyanide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DHCN 4.736 1.359 0 −0.725$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.736×373.15×(τ−1)+1.359×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.725×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1767.24(τ−1)+94.61(τ2−1)−194.29(τ−1τ)∫373.15TΔC∘PRdT=1767.24τ−1767.24+94.61τ2−94.61−194.29+194.29τ∫373.15TΔC∘PRdT=94.61τ2+1767.24τ−2056.14+194.29τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=94.61τ2+1767.24τ−2056.14+194.29τ94.61τ2+1767.24τ−3499.49+194.29τ=094.61τ2+1767.24τ−3499.49τ+194.29τ=0$

On solving,

$τ=1.753,0.057, −20.49$

Considering positive value of temperature $τ=1.753$

Hence,

$τ=TT0=T100+273.15=1.753T=654.13 K$

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Answer 5

Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(a)

Expert Solution

Answer to Problem 4.3P

$T=626.817 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dmethane 1.702 9.081 −2.164 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.702×373.15×(τ−1)+9.081×10−32373.152(τ2−1) +−2.164×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=635.1013(τ−1)+632.223(τ2−1)−37.479(τ3−1)∫373.15TΔC∘PRdT=635.1013τ−635.1013+632.223τ2−632.223−37.479τ3+37.479∫373.15TΔC∘PRdT=−37.479τ3+632.223τ2+635.1013τ−1229.8453$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K =−37.479τ3+632.223τ2+635.1013τ−1229.845337.479τ3−632.223τ2−635.1013τ+2673.194=0$

On solving,

$τ=1.6798,−2.412, 17.6$

Considering positive value of temperature $τ=1.6798$

Hence,

$τ=TT0=T100+273.15=1.6798T=626.817 K$

(b)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(b)

Expert Solution

Answer to Problem 4.3P

$T=538.455 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethane 1.131 19.225 −5.561 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.131×373.15×(τ−1)+19.225×10−32373.152(τ2−1) +−5.561×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=422.033(τ−1)+1338.453(τ2−1)−96.31(τ3−1)∫373.15TΔC∘PRdT=422.033τ−422.033+1338.453τ2−1338.453−96.31τ3+96.31∫373.15TΔC∘PRdT=−96.31τ3+1338.453τ2+422.033τ−1664.176$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−96.31τ3+1338.453τ2+422.033τ−1664.17696.31τ3−1338.453τ2−422.033τ+3107.525=0$

On solving,

$τ=1.443,−1.592, 14.0458$

Considering positive value of temperature $τ=1.443$

Hence,

$τ=TT0=T100+273.15=1.443T=538.455 K$

(c)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(c)

Expert Solution

Answer to Problem 4.3P

$T=493.3043 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.213×373.15×(τ−1)+28.785×10−32373.152(τ2−1) +−8.824×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=452.631(τ−1)+2004.025(τ2−1)−152.825(τ3−1)∫373.15TΔC∘PRdT=452.631τ−452.631+2004.025τ2−2004.025−152.825τ3+152.825∫373.15TΔC∘PRdT=−152.825τ3+2004.025τ2+452.631τ−2303.831$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−152.825τ3+2004.025τ2+452.631τ−2303.831152.825τ3−2004.025τ2−452.631τ+3747.18=0$

On solving,

$τ=1.322,−1.405, 13.197$

Considering positive value of temperature $τ=1.322$

Hence,

$τ=TT0=T100+273.15=1.322T=493.3043 K$

(d)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(d)

Expert Solution

Answer to Problem 4.3P

$T=466.81 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -butane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.935×373.15×(τ−1)+36.915×10−32373.152(τ2−1) +−11.402×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=722.045(τ−1)+2570.039(τ2−1)−197.474(τ3−1)∫373.15TΔC∘PRdT=722.045τ−722.045+2570.039τ2−2570.039−197.474τ3+197.474∫373.15TΔC∘PRdT=−197.474τ3+2570.039τ2+722.045τ−3094.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−197.474τ3+2570.039τ2+722.045τ−3094.61197.474τ3−2570.039τ2−722.045τ+4537.96=0$

On solving,

$τ=1.251,−1.396, 13.16$

Considering positive value of temperature $τ=1.251$

Hence,

$τ=TT0=T100+273.15=1.251T=466.81 K$

(e)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(e)

Expert Solution

Answer to Problem 4.3P

$T=438.45 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -hexanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 3.025 53.722 −16.791 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.025×373.15×(τ−1)+53.722×10−32373.152(τ2−1) +−16.791×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1128.78(τ−1)+3740.15(τ2−1)−290.807(τ3−1)∫373.15TΔC∘PRdT=1128.78τ−1128.78+3740.15τ2−3740.15−290.807τ3+290.807∫373.15TΔC∘PRdT=−290.807τ3+3740.15τ2+1128.78τ−4578.12$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−290.807τ3+3740.15τ2+1128.78τ−4578.12290.807τ3−3740.15τ2−1128.78τ+6021.471=0$

On solving,

$τ=1.175,−1.351, 13.04$

Considering positive value of temperature $τ=1.175$

Hence,

$τ=TT0=T100+273.15=1.175T=438.45 K$

(f)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(f)

Expert Solution

Answer to Problem 4.3P

$T=371.695 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -octanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 4.108 70.567 −22.208 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.108×373.15×(τ−1)+70.567×10−32373.152(τ2−1) +−22.208×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1532.9002(τ−1)+4912.91(τ2−1)−384.63(τ3−1)∫373.15TΔC∘PRdT=1532.9002τ−1532.9002+4912.91τ2−4912.91−384.63τ3+384.63∫373.15TΔC∘PRdT=−384.63τ3+4912.91τ2+1532.9002τ−6061.184$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−384.63τ3+4912.91τ2+1532.9002τ−6061.184384.63τ3−4912.91τ2−1532.9002τ+6021.471=0$

On solving,

$τ=0.9961,−1.21, 12.987$

Considering positive value of temperature $τ=0.9961$

Hence,

$τ=TT0=T100+273.15=0.9961T=371.695 K$

(g)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(g)

Expert Solution

Answer to Problem 4.3P

$T=513.08 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropylene 1.637 22.706 −6.915 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.637×373.15×(τ−1)+22.706×10−32373.152(τ2−1) +−6.915×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=610.847(τ−1)+1580.8(τ2−1)−119.763(τ3−1)∫373.15TΔC∘PRdT=610.847τ−610.847+1580.8τ2−1580.8−119.763τ3+119.763∫373.15TΔC∘PRdT=−119.763τ3+1580.8τ2+610.847τ−2071.88$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−119.763τ3+1580.8τ2+610.847τ−2071.88119.763τ3−1580.8τ2−610.847τ+3515.233=0$

On solving,

$τ=1.375,−1.592, 13.416$

Considering positive value of temperature $τ=1.375$

Hence,

$τ=TT0=T100+273.15=1.375T=513.08 K$

(h)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(h)

Expert Solution

Answer to Problem 4.3P

$T=457.855 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-pentene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−pentene 2.691 39.753 −12.447 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.691×373.15×(τ−1)+39.753×10−32373.152(τ2−1) +−12.447×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1004.147(τ−1)+2767.62(τ2−1)−215.57(τ3−1)∫373.15TΔC∘PRdT=1004.147τ−1004.147+2767.62τ2−2767.62−215.57τ3+215.57∫373.15TΔC∘PRdT=−215.57τ3+2767.62τ2+1004.147τ−3556.197$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−215.57τ3+2767.62τ2+1004.147τ−3556.197215.57τ3−2767.62τ2−1004.147τ+4999.546=0$

On solving,

$τ=1.227,−1.447, 13.059$

Considering positive value of temperature $τ=1.227$

Hence,

$τ=TT0=T100+273.15=1.227T=457.855 K$

(i)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(i)

Expert Solution

Answer to Problem 4.3P

$T=433.97 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-heptenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−heptene 3.768 56.588 −17.847 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.768×373.15×(τ−1)+56.588×10−32373.152(τ2−1) +−17.847×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1406.03(τ−1)+3939.68(τ2−1)−309.097(τ3−1)∫373.15TΔC∘PRdT=1406.03τ−1406.03+3939.68τ2−3939.68−309.097τ3+309.097∫373.15TΔC∘PRdT=−309.097τ3+3939.68τ2+1406.03τ−5036.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−309.097τ3+3939.68τ2+1406.03τ−5036.61309.097τ3−3939.68τ2−1406.03τ+6479.96=0$

On solving,

$τ=1.163,−1.389, 12.97$

Considering positive value of temperature $τ=1.163$

Hence,

$τ=TT0=T100+273.15=1.163T=433.97 K$

(j)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(j)

Expert Solution

Answer to Problem 4.3P

$T=426.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-Octenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−Octene 4.324 64.96 −20.521 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.324×373.15×(τ−1)+64.96×10−32373.152(τ2−1) +−20.521×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1613.5(τ−1)+4522.55(τ2−1)−355.41(τ3−1)∫373.15TΔC∘PRdT=1613.5τ−1613.5+4522.55τ2−4522.55−355.41τ3+355.41∫373.15TΔC∘PRdT=−355.41τ3+4522.55τ2+1613.5τ−5780.64$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−355.41τ3+4522.55τ2+1613.5τ−5780.64355.41τ3−4522.55τ2−1613.5τ+7223.99=0$

On solving,

$τ=1.143,−1.37, 12.95$

Considering positive value of temperature $τ=1.143$

Hence,

$τ=TT0=T100+273.15=1.143T=426.51 K$

(k)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(k)

Expert Solution

Answer to Problem 4.3P

$T=595.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for acetylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dacetylene 6.132 1.952 0 −1.299$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=6.132×373.15×(τ−1)+1.952×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.299×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2288.15(τ−1)+135.9(τ2−1)−348.12(τ−1τ)∫373.15TΔC∘PRdT=2288.15τ−2288.15+135.9τ2−135.9−348.12+348.12τ∫373.15TΔC∘PRdT=135.9τ2+2288.15τ−2772.17+348.12τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=135.9τ2+2288.15τ−2772.17+348.12τ135.9τ2+2288.15τ−4215.519+348.12τ=0135.9τ3+2288.15τ2−4215.519τ+348.12τ=0$

On solving,

$τ=1.596,0.099, −18.52$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.596T=595.55 K$

(l)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(l)

Expert Solution

Answer to Problem 4.3P

$T=476.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for benzenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dbenzene −0.206 39.064 −13.301 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=−0.206×373.15×(τ−1)+39.064×10−32373.152(τ2−1) +−13.301×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=−76.87(τ−1)+2719.65(τ2−1)−230.36(τ3−1)∫373.15TΔC∘PRdT=−76.87τ+76.87+2719.65τ2−2719.65−230.36τ3+230.36∫373.15TΔC∘PRdT=−230.36τ3+2719.65τ2−76.87τ−2412.42$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−230.36τ3+2719.65τ2−76.87τ−2412.42230.36τ3−2719.65τ2+76.87τ+3855.769=0$

On solving,

$τ=1.277,−1.12, 11.65$

Considering positive value of temperature $τ=1.277$

Hence,

$τ=TT0=T100+273.15=1.277T=476.51 K$

(m)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(m)

Expert Solution

Answer to Problem 4.3P

$T=380.613 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethanol in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethanol 3.518 20.001 −6.002 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.518×373.15×(τ−1)+20.001×10−32373.152(τ2−1) +−6.002×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1312.74(τ−1)+1392.48(τ2−1)−103.95(τ3−1)∫373.15TΔC∘PRdT=1312.74τ−1312.74+1392.48τ2−1392.48−103.95τ3+103.95∫373.15TΔC∘PRdT=−103.95τ3+2719.65τ2+1312.74τ−2601.27$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−103.95τ3+2719.65τ2+1312.74τ−2601.27103.95τ3−2719.65τ2−1312.74τ+4044.62=0$

On solving,

$τ=1.02,−1.44, 26.58$

Considering positive value of temperature $τ=1.02$

Hence,

$τ=TT0=T100+273.15=1.02T=380.613 K$

(n)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(n)

Expert Solution

Answer to Problem 4.3P

$T=445.91 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for styrene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dstyrene 2.050 50.192 −16.662 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.050×373.15×(τ−1)+50.192×10−32373.152(τ2−1) +−16.662×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=764.96(τ−1)+3494.39(τ2−1)−288.57(τ3−1)∫373.15TΔC∘PRdT=764.96τ−764.96+3494.39τ2−3494.39−288.57τ3+288.57∫373.15TΔC∘PRdT=−288.57τ3+3494.39τ2+764.96τ−3970.78$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−288.57τ3+3494.39τ2+764.96τ−3970.78288.57τ3−3494.39τ2−764.96τ+5414.126=0$

On solving,

$τ=1.195,−1.287, 12.2$

Considering positive value of temperature $τ=1.195$

Hence,

$τ=TT0=T100+273.15=1.195T=445.91 K$

(o)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(o)

Expert Solution

Answer to Problem 4.3P

$T=643.68 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for formaldehyde in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dformaldehyde 2.264 7.022 −1.877 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.264×373.15×(τ−1)+7.022×10−32373.152(τ2−1) +−1.877×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=844.81(τ−1)+488.88(τ2−1)−32.51(τ3−1)∫373.15TΔC∘PRdT=844.81τ−844.81+488.88τ2−488.88−32.51τ3+32.51∫373.15TΔC∘PRdT=-32.51τ3+488.88τ2+844.81τ−1301.18$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=-32.51τ3+488.88τ2+844.81τ−1301.1832.51τ3−488.88τ2−844.81τ+2744.53=0$

On solving,

$τ=1.725,−3, 16.31$

Considering positive value of temperature $τ=1.725$

Hence,

$τ=TT0=T100+273.15=1.725T=643.68 K$

(p)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(p)

Expert Solution

Answer to Problem 4.3P

$T=416.06 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dammonia 3.578 3.020 0 −0.186$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.578×373.15×(τ−1)+3.020×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.186×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1335.13(τ−1)+210.25(τ2−1)−49.85(τ−1τ)∫373.15TΔC∘PRdT=1335.13τ−1335.13+210.25τ2−210.25−49.85+49.85τ∫373.15TΔC∘PRdT=210.25τ2+1335.13τ−1595.23+49.85τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=210.25τ2+1335.13τ−1595.23+49.85τ1210.25τ2+1335.13τ−3038.579+49.85τ=01210.25τ3+1335.13τ2−3038.579τ+49.85τ=0$

On solving,

$τ=1.115,0.0165, −2.23$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.115T=416.06 K$

(q)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(q)

Expert Solution

Answer to Problem 4.3P

$T=764.96 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon monoxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO 3.376 0.557 0 −0.031$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.376×373.15×(τ−1)+0.557×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.031×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1259.75(τ−1)+38.78(τ2−1)−8.31(τ−1τ)∫373.15TΔC∘PRdT=1259.75τ−1259.75+38.78τ2−38.78−8.31+8.31τ∫373.15TΔC∘PRdT=38.78τ2+1259.75τ−1306.84+8.31τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=38.78τ2+1259.75τ−1306.84+8.31τ38.78τ2+1259.75τ−2750.19+8.31τ=038.78τ3+1259.75τ2−2750.19τ+8.31τ=0$

On solving,

$τ=2.05,0.003, −34.54$

Considering positive value of temperature $τ=2.05$

Hence,

$τ=TT0=T100+273.15=2.05T=764.96 K$

(r)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(r)

Expert Solution

Answer to Problem 4.3P

$T=634.355 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO2 5.457 1.045 0 −1.157$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.457×373.15×(τ−1)+1.045×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.157×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2036.28(τ−1)+72.75(τ2−1)−310.06(τ−1τ)∫373.15TΔC∘PRdT=2036.28τ−2036.28+72.75τ2−72.75−310.06+310.06τ∫373.15TΔC∘PRdT=72.75τ2+2036.28τ−2419.09+310.06τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=72.75τ2+2036.28τ−2419.09+310.06τ72.75τ2+2036.28τ−3862.44+310.06τ=072.75τ3+2036.28τ2−3862.44τ+310.06τ=0$

On solving,

$τ=1.7,0.084, −29.78$

Considering positive value of temperature $τ=1.7$

Hence,

$τ=TT0=T100+273.15=1.7T=634.355 K$

(s)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(s)

Expert Solution

Answer to Problem 4.3P

$T=626.1457 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for sulphur dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DSO2 5.699 0.801 0 −1.015$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.699×373.15×(τ−1)+0.801×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.015×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2126.58(τ−1)+55.77(τ2−1)−272(τ−1τ)∫373.15TΔC∘PRdT=2126.58τ−2126.58+55.77τ2−55.77−272+272τ∫373.15TΔC∘PRdT=55.77τ2+2126.58τ−2454.36+272τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=55.77τ2+2126.58τ−2454.36+272τ55.77τ2+2126.58τ−3887.71+272τ=055.77τ3+2126.58τ2−3887.71τ+272τ=0$

On solving,

$τ=1.678,0.073, −39.88$

Considering positive value of temperature $τ=1.678$

Hence,

$τ=TT0=T100+273.15=1.678T=626.1457 K$

(t)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(t)

Expert Solution

Answer to Problem 4.3P

$T=706.37 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DH2O 3.47 1.450 0 0.121$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.470×373.15×(τ−1)+1.450×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.121×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1294.83(τ−1)+100.95(τ2−1)+32.43(τ−1τ)∫373.15TΔC∘PRdT=1294.83τ−1294.83+100.95τ2−100.95+32.43−32.43τ∫373.15TΔC∘PRdT=100.95τ2+1294.83τ−1363.35−32.43τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=100.95τ2+1294.83τ−1363.35−32.43τ100.95τ2+1294.83τ−2796.7−32.43τ=0100.95τ3+1294.83τ2−2796.7τ−32.43τ=0$

On solving,

$τ=1.893,−0.011, −14.708$

Considering positive value of temperature $τ=1.893$

Hence,

$τ=TT0=T100+273.15=1.893T=706.37 K$

(u)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(u)

Expert Solution

Answer to Problem 4.3P

$T=770.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for nitrogen in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DN2 3.280 0.593 0 0.040$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.280×373.15×(τ−1)+0.593×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.040×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1223.93(τ−1)+41.29(τ2−1)+10.72(τ−1τ)∫373.15TΔC∘PRdT=1223.93τ−1223.93+41.29τ2−41.29+10.72−10.72τ∫373.15TΔC∘PRdT=41.29τ2+1223.93τ−1254.5−10.72τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=41.29τ2+1223.93τ−1254.5−10.72τ41.29τ2+1223.93τ−2697.85−10.72τ=041.29τ3+1223.93τ2−2697.85τ−10.72τ=0$

On solving,

$τ=2.065,−0.003, −31.703$

Considering positive value of temperature $τ=2.065$

Hence,

$τ=TT0=T100+273.15=2.065T=770.55 K$

(v)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(v)

Expert Solution

Answer to Problem 4.3P

$T=654.13 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for hydrogen cyanide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DHCN 4.736 1.359 0 −0.725$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.736×373.15×(τ−1)+1.359×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.725×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1767.24(τ−1)+94.61(τ2−1)−194.29(τ−1τ)∫373.15TΔC∘PRdT=1767.24τ−1767.24+94.61τ2−94.61−194.29+194.29τ∫373.15TΔC∘PRdT=94.61τ2+1767.24τ−2056.14+194.29τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=94.61τ2+1767.24τ−2056.14+194.29τ94.61τ2+1767.24τ−3499.49+194.29τ=094.61τ2+1767.24τ−3499.49τ+194.29τ=0$

On solving,

$τ=1.753,0.057, −20.49$

Considering positive value of temperature $τ=1.753$

Hence,

$τ=TT0=T100+273.15=1.753T=654.13 K$

Answer 6

Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(a)

Expert Solution

Answer to Problem 4.3P

$T=626.817 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dmethane 1.702 9.081 −2.164 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.702×373.15×(τ−1)+9.081×10−32373.152(τ2−1) +−2.164×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=635.1013(τ−1)+632.223(τ2−1)−37.479(τ3−1)∫373.15TΔC∘PRdT=635.1013τ−635.1013+632.223τ2−632.223−37.479τ3+37.479∫373.15TΔC∘PRdT=−37.479τ3+632.223τ2+635.1013τ−1229.8453$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K =−37.479τ3+632.223τ2+635.1013τ−1229.845337.479τ3−632.223τ2−635.1013τ+2673.194=0$

On solving,

$τ=1.6798,−2.412, 17.6$

Considering positive value of temperature $τ=1.6798$

Hence,

$τ=TT0=T100+273.15=1.6798T=626.817 K$

Answer 7

Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 8

(a)

Expert Solution

Answer to Problem 4.3P

$T=626.817 K$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dmethane 1.702 9.081 −2.164 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.702×373.15×(τ−1)+9.081×10−32373.152(τ2−1) +−2.164×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=635.1013(τ−1)+632.223(τ2−1)−37.479(τ3−1)∫373.15TΔC∘PRdT=635.1013τ−635.1013+632.223τ2−632.223−37.479τ3+37.479∫373.15TΔC∘PRdT=−37.479τ3+632.223τ2+635.1013τ−1229.8453$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K =−37.479τ3+632.223τ2+635.1013τ−1229.845337.479τ3−632.223τ2−635.1013τ+2673.194=0$

On solving,

$τ=1.6798,−2.412, 17.6$

Considering positive value of temperature $τ=1.6798$

Hence,

$τ=TT0=T100+273.15=1.6798T=626.817 K$

Answer 13

(b)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(b)

Expert Solution

Answer to Problem 4.3P

$T=538.455 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethane 1.131 19.225 −5.561 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.131×373.15×(τ−1)+19.225×10−32373.152(τ2−1) +−5.561×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=422.033(τ−1)+1338.453(τ2−1)−96.31(τ3−1)∫373.15TΔC∘PRdT=422.033τ−422.033+1338.453τ2−1338.453−96.31τ3+96.31∫373.15TΔC∘PRdT=−96.31τ3+1338.453τ2+422.033τ−1664.176$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−96.31τ3+1338.453τ2+422.033τ−1664.17696.31τ3−1338.453τ2−422.033τ+3107.525=0$

On solving,

$τ=1.443,−1.592, 14.0458$

Considering positive value of temperature $τ=1.443$

Hence,

$τ=TT0=T100+273.15=1.443T=538.455 K$

Answer 14

(b)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 15

(b)

Expert Solution

Answer to Problem 4.3P

$T=538.455 K$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethane 1.131 19.225 −5.561 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.131×373.15×(τ−1)+19.225×10−32373.152(τ2−1) +−5.561×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=422.033(τ−1)+1338.453(τ2−1)−96.31(τ3−1)∫373.15TΔC∘PRdT=422.033τ−422.033+1338.453τ2−1338.453−96.31τ3+96.31∫373.15TΔC∘PRdT=−96.31τ3+1338.453τ2+422.033τ−1664.176$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−96.31τ3+1338.453τ2+422.033τ−1664.17696.31τ3−1338.453τ2−422.033τ+3107.525=0$

On solving,

$τ=1.443,−1.592, 14.0458$

Considering positive value of temperature $τ=1.443$

Hence,

$τ=TT0=T100+273.15=1.443T=538.455 K$

Answer 20

(c)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(c)

Expert Solution

Answer to Problem 4.3P

$T=493.3043 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.213×373.15×(τ−1)+28.785×10−32373.152(τ2−1) +−8.824×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=452.631(τ−1)+2004.025(τ2−1)−152.825(τ3−1)∫373.15TΔC∘PRdT=452.631τ−452.631+2004.025τ2−2004.025−152.825τ3+152.825∫373.15TΔC∘PRdT=−152.825τ3+2004.025τ2+452.631τ−2303.831$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−152.825τ3+2004.025τ2+452.631τ−2303.831152.825τ3−2004.025τ2−452.631τ+3747.18=0$

On solving,

$τ=1.322,−1.405, 13.197$

Considering positive value of temperature $τ=1.322$

Hence,

$τ=TT0=T100+273.15=1.322T=493.3043 K$

Answer 21

(c)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 22

(c)

Expert Solution

Answer to Problem 4.3P

$T=493.3043 K$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropane 1.213 28.785 −8.824 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.213×373.15×(τ−1)+28.785×10−32373.152(τ2−1) +−8.824×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=452.631(τ−1)+2004.025(τ2−1)−152.825(τ3−1)∫373.15TΔC∘PRdT=452.631τ−452.631+2004.025τ2−2004.025−152.825τ3+152.825∫373.15TΔC∘PRdT=−152.825τ3+2004.025τ2+452.631τ−2303.831$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−152.825τ3+2004.025τ2+452.631τ−2303.831152.825τ3−2004.025τ2−452.631τ+3747.18=0$

On solving,

$τ=1.322,−1.405, 13.197$

Considering positive value of temperature $τ=1.322$

Hence,

$τ=TT0=T100+273.15=1.322T=493.3043 K$

Answer 27

(d)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(d)

Expert Solution

Answer to Problem 4.3P

$T=466.81 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -butane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.935×373.15×(τ−1)+36.915×10−32373.152(τ2−1) +−11.402×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=722.045(τ−1)+2570.039(τ2−1)−197.474(τ3−1)∫373.15TΔC∘PRdT=722.045τ−722.045+2570.039τ2−2570.039−197.474τ3+197.474∫373.15TΔC∘PRdT=−197.474τ3+2570.039τ2+722.045τ−3094.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−197.474τ3+2570.039τ2+722.045τ−3094.61197.474τ3−2570.039τ2−722.045τ+4537.96=0$

On solving,

$τ=1.251,−1.396, 13.16$

Considering positive value of temperature $τ=1.251$

Hence,

$τ=TT0=T100+273.15=1.251T=466.81 K$

Answer 28

(d)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 29

(d)

Expert Solution

Answer to Problem 4.3P

$T=466.81 K$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -butane in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−butane 1.935 36.915 −11.402 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.935×373.15×(τ−1)+36.915×10−32373.152(τ2−1) +−11.402×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=722.045(τ−1)+2570.039(τ2−1)−197.474(τ3−1)∫373.15TΔC∘PRdT=722.045τ−722.045+2570.039τ2−2570.039−197.474τ3+197.474∫373.15TΔC∘PRdT=−197.474τ3+2570.039τ2+722.045τ−3094.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−197.474τ3+2570.039τ2+722.045τ−3094.61197.474τ3−2570.039τ2−722.045τ+4537.96=0$

On solving,

$τ=1.251,−1.396, 13.16$

Considering positive value of temperature $τ=1.251$

Hence,

$τ=TT0=T100+273.15=1.251T=466.81 K$

Answer 34

(e)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(e)

Expert Solution

Answer to Problem 4.3P

$T=438.45 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -hexanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 3.025 53.722 −16.791 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.025×373.15×(τ−1)+53.722×10−32373.152(τ2−1) +−16.791×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1128.78(τ−1)+3740.15(τ2−1)−290.807(τ3−1)∫373.15TΔC∘PRdT=1128.78τ−1128.78+3740.15τ2−3740.15−290.807τ3+290.807∫373.15TΔC∘PRdT=−290.807τ3+3740.15τ2+1128.78τ−4578.12$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−290.807τ3+3740.15τ2+1128.78τ−4578.12290.807τ3−3740.15τ2−1128.78τ+6021.471=0$

On solving,

$τ=1.175,−1.351, 13.04$

Considering positive value of temperature $τ=1.175$

Hence,

$τ=TT0=T100+273.15=1.175T=438.45 K$

Answer 35

(e)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 36

(e)

Expert Solution

Answer to Problem 4.3P

$T=438.45 K$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -hexanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 3.025 53.722 −16.791 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.025×373.15×(τ−1)+53.722×10−32373.152(τ2−1) +−16.791×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1128.78(τ−1)+3740.15(τ2−1)−290.807(τ3−1)∫373.15TΔC∘PRdT=1128.78τ−1128.78+3740.15τ2−3740.15−290.807τ3+290.807∫373.15TΔC∘PRdT=−290.807τ3+3740.15τ2+1128.78τ−4578.12$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−290.807τ3+3740.15τ2+1128.78τ−4578.12290.807τ3−3740.15τ2−1128.78τ+6021.471=0$

On solving,

$τ=1.175,−1.351, 13.04$

Considering positive value of temperature $τ=1.175$

Hence,

$τ=TT0=T100+273.15=1.175T=438.45 K$

Answer 41

(f)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(f)

Expert Solution

Answer to Problem 4.3P

$T=371.695 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -octanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 4.108 70.567 −22.208 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.108×373.15×(τ−1)+70.567×10−32373.152(τ2−1) +−22.208×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1532.9002(τ−1)+4912.91(τ2−1)−384.63(τ3−1)∫373.15TΔC∘PRdT=1532.9002τ−1532.9002+4912.91τ2−4912.91−384.63τ3+384.63∫373.15TΔC∘PRdT=−384.63τ3+4912.91τ2+1532.9002τ−6061.184$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−384.63τ3+4912.91τ2+1532.9002τ−6061.184384.63τ3−4912.91τ2−1532.9002τ+6021.471=0$

On solving,

$τ=0.9961,−1.21, 12.987$

Considering positive value of temperature $τ=0.9961$

Hence,

$τ=TT0=T100+273.15=0.9961T=371.695 K$

Answer 42

(f)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 43

(f)

Expert Solution

Answer to Problem 4.3P

$T=371.695 K$

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for $n$ -octanein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dn−hexane 4.108 70.567 −22.208 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.108×373.15×(τ−1)+70.567×10−32373.152(τ2−1) +−22.208×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1532.9002(τ−1)+4912.91(τ2−1)−384.63(τ3−1)∫373.15TΔC∘PRdT=1532.9002τ−1532.9002+4912.91τ2−4912.91−384.63τ3+384.63∫373.15TΔC∘PRdT=−384.63τ3+4912.91τ2+1532.9002τ−6061.184$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−384.63τ3+4912.91τ2+1532.9002τ−6061.184384.63τ3−4912.91τ2−1532.9002τ+6021.471=0$

On solving,

$τ=0.9961,−1.21, 12.987$

Considering positive value of temperature $τ=0.9961$

Hence,

$τ=TT0=T100+273.15=0.9961T=371.695 K$

Answer 48

(g)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(g)

Expert Solution

Answer to Problem 4.3P

$T=513.08 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropylene 1.637 22.706 −6.915 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.637×373.15×(τ−1)+22.706×10−32373.152(τ2−1) +−6.915×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=610.847(τ−1)+1580.8(τ2−1)−119.763(τ3−1)∫373.15TΔC∘PRdT=610.847τ−610.847+1580.8τ2−1580.8−119.763τ3+119.763∫373.15TΔC∘PRdT=−119.763τ3+1580.8τ2+610.847τ−2071.88$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−119.763τ3+1580.8τ2+610.847τ−2071.88119.763τ3−1580.8τ2−610.847τ+3515.233=0$

On solving,

$τ=1.375,−1.592, 13.416$

Considering positive value of temperature $τ=1.375$

Hence,

$τ=TT0=T100+273.15=1.375T=513.08 K$

Answer 49

(g)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 50

(g)

Expert Solution

Answer to Problem 4.3P

$T=513.08 K$

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for propylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dpropylene 1.637 22.706 −6.915 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=1.637×373.15×(τ−1)+22.706×10−32373.152(τ2−1) +−6.915×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=610.847(τ−1)+1580.8(τ2−1)−119.763(τ3−1)∫373.15TΔC∘PRdT=610.847τ−610.847+1580.8τ2−1580.8−119.763τ3+119.763∫373.15TΔC∘PRdT=−119.763τ3+1580.8τ2+610.847τ−2071.88$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−119.763τ3+1580.8τ2+610.847τ−2071.88119.763τ3−1580.8τ2−610.847τ+3515.233=0$

On solving,

$τ=1.375,−1.592, 13.416$

Considering positive value of temperature $τ=1.375$

Hence,

$τ=TT0=T100+273.15=1.375T=513.08 K$

Answer 55

(h)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(h)

Expert Solution

Answer to Problem 4.3P

$T=457.855 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-pentene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−pentene 2.691 39.753 −12.447 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.691×373.15×(τ−1)+39.753×10−32373.152(τ2−1) +−12.447×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1004.147(τ−1)+2767.62(τ2−1)−215.57(τ3−1)∫373.15TΔC∘PRdT=1004.147τ−1004.147+2767.62τ2−2767.62−215.57τ3+215.57∫373.15TΔC∘PRdT=−215.57τ3+2767.62τ2+1004.147τ−3556.197$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−215.57τ3+2767.62τ2+1004.147τ−3556.197215.57τ3−2767.62τ2−1004.147τ+4999.546=0$

On solving,

$τ=1.227,−1.447, 13.059$

Considering positive value of temperature $τ=1.227$

Hence,

$τ=TT0=T100+273.15=1.227T=457.855 K$

Answer 56

(h)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 57

(h)

Expert Solution

Answer to Problem 4.3P

$T=457.855 K$

Answer 58

(h)

Expert Solution

Answer 59

(h)

Expert Solution

Answer 60

Expert Solution

Answer 61

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-pentene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−pentene 2.691 39.753 −12.447 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.691×373.15×(τ−1)+39.753×10−32373.152(τ2−1) +−12.447×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1004.147(τ−1)+2767.62(τ2−1)−215.57(τ3−1)∫373.15TΔC∘PRdT=1004.147τ−1004.147+2767.62τ2−2767.62−215.57τ3+215.57∫373.15TΔC∘PRdT=−215.57τ3+2767.62τ2+1004.147τ−3556.197$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−215.57τ3+2767.62τ2+1004.147τ−3556.197215.57τ3−2767.62τ2−1004.147τ+4999.546=0$

On solving,

$τ=1.227,−1.447, 13.059$

Considering positive value of temperature $τ=1.227$

Hence,

$τ=TT0=T100+273.15=1.227T=457.855 K$

Answer 62

(i)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(i)

Expert Solution

Answer to Problem 4.3P

$T=433.97 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-heptenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−heptene 3.768 56.588 −17.847 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.768×373.15×(τ−1)+56.588×10−32373.152(τ2−1) +−17.847×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1406.03(τ−1)+3939.68(τ2−1)−309.097(τ3−1)∫373.15TΔC∘PRdT=1406.03τ−1406.03+3939.68τ2−3939.68−309.097τ3+309.097∫373.15TΔC∘PRdT=−309.097τ3+3939.68τ2+1406.03τ−5036.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−309.097τ3+3939.68τ2+1406.03τ−5036.61309.097τ3−3939.68τ2−1406.03τ+6479.96=0$

On solving,

$τ=1.163,−1.389, 12.97$

Considering positive value of temperature $τ=1.163$

Hence,

$τ=TT0=T100+273.15=1.163T=433.97 K$

Answer 63

(i)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 64

(i)

Expert Solution

Answer to Problem 4.3P

$T=433.97 K$

Answer 65

(i)

Expert Solution

Answer 66

(i)

Expert Solution

Answer 67

Expert Solution

Answer 68

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-heptenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−heptene 3.768 56.588 −17.847 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.768×373.15×(τ−1)+56.588×10−32373.152(τ2−1) +−17.847×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1406.03(τ−1)+3939.68(τ2−1)−309.097(τ3−1)∫373.15TΔC∘PRdT=1406.03τ−1406.03+3939.68τ2−3939.68−309.097τ3+309.097∫373.15TΔC∘PRdT=−309.097τ3+3939.68τ2+1406.03τ−5036.61$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−309.097τ3+3939.68τ2+1406.03τ−5036.61309.097τ3−3939.68τ2−1406.03τ+6479.96=0$

On solving,

$τ=1.163,−1.389, 12.97$

Considering positive value of temperature $τ=1.163$

Hence,

$τ=TT0=T100+273.15=1.163T=433.97 K$

Answer 69

(j)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(j)

Expert Solution

Answer to Problem 4.3P

$T=426.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-Octenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−Octene 4.324 64.96 −20.521 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.324×373.15×(τ−1)+64.96×10−32373.152(τ2−1) +−20.521×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1613.5(τ−1)+4522.55(τ2−1)−355.41(τ3−1)∫373.15TΔC∘PRdT=1613.5τ−1613.5+4522.55τ2−4522.55−355.41τ3+355.41∫373.15TΔC∘PRdT=−355.41τ3+4522.55τ2+1613.5τ−5780.64$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−355.41τ3+4522.55τ2+1613.5τ−5780.64355.41τ3−4522.55τ2−1613.5τ+7223.99=0$

On solving,

$τ=1.143,−1.37, 12.95$

Considering positive value of temperature $τ=1.143$

Hence,

$τ=TT0=T100+273.15=1.143T=426.51 K$

Answer 70

(j)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 71

(j)

Expert Solution

Answer to Problem 4.3P

$T=426.51 K$

Answer 72

(j)

Expert Solution

Answer 73

(j)

Expert Solution

Answer 74

Expert Solution

Answer 75

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for 1-Octenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5D1−Octene 4.324 64.96 −20.521 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.324×373.15×(τ−1)+64.96×10−32373.152(τ2−1) +−20.521×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1613.5(τ−1)+4522.55(τ2−1)−355.41(τ3−1)∫373.15TΔC∘PRdT=1613.5τ−1613.5+4522.55τ2−4522.55−355.41τ3+355.41∫373.15TΔC∘PRdT=−355.41τ3+4522.55τ2+1613.5τ−5780.64$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−355.41τ3+4522.55τ2+1613.5τ−5780.64355.41τ3−4522.55τ2−1613.5τ+7223.99=0$

On solving,

$τ=1.143,−1.37, 12.95$

Considering positive value of temperature $τ=1.143$

Hence,

$τ=TT0=T100+273.15=1.143T=426.51 K$

Answer 76

(k)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(k)

Expert Solution

Answer to Problem 4.3P

$T=595.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for acetylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dacetylene 6.132 1.952 0 −1.299$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=6.132×373.15×(τ−1)+1.952×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.299×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2288.15(τ−1)+135.9(τ2−1)−348.12(τ−1τ)∫373.15TΔC∘PRdT=2288.15τ−2288.15+135.9τ2−135.9−348.12+348.12τ∫373.15TΔC∘PRdT=135.9τ2+2288.15τ−2772.17+348.12τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=135.9τ2+2288.15τ−2772.17+348.12τ135.9τ2+2288.15τ−4215.519+348.12τ=0135.9τ3+2288.15τ2−4215.519τ+348.12τ=0$

On solving,

$τ=1.596,0.099, −18.52$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.596T=595.55 K$

Answer 77

(k)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 78

(k)

Expert Solution

Answer to Problem 4.3P

$T=595.55 K$

Answer 79

(k)

Expert Solution

Answer 80

(k)

Expert Solution

Answer 81

Expert Solution

Answer 82

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for acetylenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dacetylene 6.132 1.952 0 −1.299$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=6.132×373.15×(τ−1)+1.952×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.299×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2288.15(τ−1)+135.9(τ2−1)−348.12(τ−1τ)∫373.15TΔC∘PRdT=2288.15τ−2288.15+135.9τ2−135.9−348.12+348.12τ∫373.15TΔC∘PRdT=135.9τ2+2288.15τ−2772.17+348.12τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=135.9τ2+2288.15τ−2772.17+348.12τ135.9τ2+2288.15τ−4215.519+348.12τ=0135.9τ3+2288.15τ2−4215.519τ+348.12τ=0$

On solving,

$τ=1.596,0.099, −18.52$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.596T=595.55 K$

Answer 83

(l)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(l)

Expert Solution

Answer to Problem 4.3P

$T=476.51 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for benzenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dbenzene −0.206 39.064 −13.301 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=−0.206×373.15×(τ−1)+39.064×10−32373.152(τ2−1) +−13.301×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=−76.87(τ−1)+2719.65(τ2−1)−230.36(τ3−1)∫373.15TΔC∘PRdT=−76.87τ+76.87+2719.65τ2−2719.65−230.36τ3+230.36∫373.15TΔC∘PRdT=−230.36τ3+2719.65τ2−76.87τ−2412.42$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−230.36τ3+2719.65τ2−76.87τ−2412.42230.36τ3−2719.65τ2+76.87τ+3855.769=0$

On solving,

$τ=1.277,−1.12, 11.65$

Considering positive value of temperature $τ=1.277$

Hence,

$τ=TT0=T100+273.15=1.277T=476.51 K$

Answer 84

(l)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 85

(l)

Expert Solution

Answer to Problem 4.3P

$T=476.51 K$

Answer 86

(l)

Expert Solution

Answer 87

(l)

Expert Solution

Answer 88

Expert Solution

Answer 89

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for benzenein equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dbenzene −0.206 39.064 −13.301 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=−0.206×373.15×(τ−1)+39.064×10−32373.152(τ2−1) +−13.301×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=−76.87(τ−1)+2719.65(τ2−1)−230.36(τ3−1)∫373.15TΔC∘PRdT=−76.87τ+76.87+2719.65τ2−2719.65−230.36τ3+230.36∫373.15TΔC∘PRdT=−230.36τ3+2719.65τ2−76.87τ−2412.42$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−230.36τ3+2719.65τ2−76.87τ−2412.42230.36τ3−2719.65τ2+76.87τ+3855.769=0$

On solving,

$τ=1.277,−1.12, 11.65$

Considering positive value of temperature $τ=1.277$

Hence,

$τ=TT0=T100+273.15=1.277T=476.51 K$

Answer 90

(m)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(m)

Expert Solution

Answer to Problem 4.3P

$T=380.613 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethanol in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethanol 3.518 20.001 −6.002 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.518×373.15×(τ−1)+20.001×10−32373.152(τ2−1) +−6.002×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1312.74(τ−1)+1392.48(τ2−1)−103.95(τ3−1)∫373.15TΔC∘PRdT=1312.74τ−1312.74+1392.48τ2−1392.48−103.95τ3+103.95∫373.15TΔC∘PRdT=−103.95τ3+2719.65τ2+1312.74τ−2601.27$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−103.95τ3+2719.65τ2+1312.74τ−2601.27103.95τ3−2719.65τ2−1312.74τ+4044.62=0$

On solving,

$τ=1.02,−1.44, 26.58$

Considering positive value of temperature $τ=1.02$

Hence,

$τ=TT0=T100+273.15=1.02T=380.613 K$

Answer 91

(m)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 92

(m)

Expert Solution

Answer to Problem 4.3P

$T=380.613 K$

Answer 93

(m)

Expert Solution

Answer 94

(m)

Expert Solution

Answer 95

Expert Solution

Answer 96

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ethanol in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dethanol 3.518 20.001 −6.002 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.518×373.15×(τ−1)+20.001×10−32373.152(τ2−1) +−6.002×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=1312.74(τ−1)+1392.48(τ2−1)−103.95(τ3−1)∫373.15TΔC∘PRdT=1312.74τ−1312.74+1392.48τ2−1392.48−103.95τ3+103.95∫373.15TΔC∘PRdT=−103.95τ3+2719.65τ2+1312.74τ−2601.27$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−103.95τ3+2719.65τ2+1312.74τ−2601.27103.95τ3−2719.65τ2−1312.74τ+4044.62=0$

On solving,

$τ=1.02,−1.44, 26.58$

Considering positive value of temperature $τ=1.02$

Hence,

$τ=TT0=T100+273.15=1.02T=380.613 K$

Answer 97

(n)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(n)

Expert Solution

Answer to Problem 4.3P

$T=445.91 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for styrene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dstyrene 2.050 50.192 −16.662 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.050×373.15×(τ−1)+50.192×10−32373.152(τ2−1) +−16.662×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=764.96(τ−1)+3494.39(τ2−1)−288.57(τ3−1)∫373.15TΔC∘PRdT=764.96τ−764.96+3494.39τ2−3494.39−288.57τ3+288.57∫373.15TΔC∘PRdT=−288.57τ3+3494.39τ2+764.96τ−3970.78$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−288.57τ3+3494.39τ2+764.96τ−3970.78288.57τ3−3494.39τ2−764.96τ+5414.126=0$

On solving,

$τ=1.195,−1.287, 12.2$

Considering positive value of temperature $τ=1.195$

Hence,

$τ=TT0=T100+273.15=1.195T=445.91 K$

Answer 98

(n)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 99

(n)

Expert Solution

Answer to Problem 4.3P

$T=445.91 K$

Answer 100

(n)

Expert Solution

Answer 101

(n)

Expert Solution

Answer 102

Expert Solution

Answer 103

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for styrene in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dstyrene 2.050 50.192 −16.662 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.050×373.15×(τ−1)+50.192×10−32373.152(τ2−1) +−16.662×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=764.96(τ−1)+3494.39(τ2−1)−288.57(τ3−1)∫373.15TΔC∘PRdT=764.96τ−764.96+3494.39τ2−3494.39−288.57τ3+288.57∫373.15TΔC∘PRdT=−288.57τ3+3494.39τ2+764.96τ−3970.78$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=−288.57τ3+3494.39τ2+764.96τ−3970.78288.57τ3−3494.39τ2−764.96τ+5414.126=0$

On solving,

$τ=1.195,−1.287, 12.2$

Considering positive value of temperature $τ=1.195$

Hence,

$τ=TT0=T100+273.15=1.195T=445.91 K$

Answer 104

(o)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(o)

Expert Solution

Answer to Problem 4.3P

$T=643.68 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for formaldehyde in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dformaldehyde 2.264 7.022 −1.877 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.264×373.15×(τ−1)+7.022×10−32373.152(τ2−1) +−1.877×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=844.81(τ−1)+488.88(τ2−1)−32.51(τ3−1)∫373.15TΔC∘PRdT=844.81τ−844.81+488.88τ2−488.88−32.51τ3+32.51∫373.15TΔC∘PRdT=-32.51τ3+488.88τ2+844.81τ−1301.18$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=-32.51τ3+488.88τ2+844.81τ−1301.1832.51τ3−488.88τ2−844.81τ+2744.53=0$

On solving,

$τ=1.725,−3, 16.31$

Considering positive value of temperature $τ=1.725$

Hence,

$τ=TT0=T100+273.15=1.725T=643.68 K$

Answer 105

(o)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 106

(o)

Expert Solution

Answer to Problem 4.3P

$T=643.68 K$

Answer 107

(o)

Expert Solution

Answer 108

(o)

Expert Solution

Answer 109

Expert Solution

Answer 110

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for formaldehyde in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dformaldehyde 2.264 7.022 −1.877 0$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=2.264×373.15×(τ−1)+7.022×10−32373.152(τ2−1) +−1.877×10−63×373.153(τ3−1)+0373.15×(τ−1τ)∫373.15TΔC∘PRdT=844.81(τ−1)+488.88(τ2−1)−32.51(τ3−1)∫373.15TΔC∘PRdT=844.81τ−844.81+488.88τ2−488.88−32.51τ3+32.51∫373.15TΔC∘PRdT=-32.51τ3+488.88τ2+844.81τ−1301.18$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=-32.51τ3+488.88τ2+844.81τ−1301.1832.51τ3−488.88τ2−844.81τ+2744.53=0$

On solving,

$τ=1.725,−3, 16.31$

Considering positive value of temperature $τ=1.725$

Hence,

$τ=TT0=T100+273.15=1.725T=643.68 K$

Answer 111

(p)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(p)

Expert Solution

Answer to Problem 4.3P

$T=416.06 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dammonia 3.578 3.020 0 −0.186$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.578×373.15×(τ−1)+3.020×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.186×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1335.13(τ−1)+210.25(τ2−1)−49.85(τ−1τ)∫373.15TΔC∘PRdT=1335.13τ−1335.13+210.25τ2−210.25−49.85+49.85τ∫373.15TΔC∘PRdT=210.25τ2+1335.13τ−1595.23+49.85τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=210.25τ2+1335.13τ−1595.23+49.85τ1210.25τ2+1335.13τ−3038.579+49.85τ=01210.25τ3+1335.13τ2−3038.579τ+49.85τ=0$

On solving,

$τ=1.115,0.0165, −2.23$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.115T=416.06 K$

Answer 112

(p)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 113

(p)

Expert Solution

Answer to Problem 4.3P

$T=416.06 K$

Answer 114

(p)

Expert Solution

Answer 115

(p)

Expert Solution

Answer 116

Expert Solution

Answer 117

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5Dammonia 3.578 3.020 0 −0.186$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.578×373.15×(τ−1)+3.020×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.186×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1335.13(τ−1)+210.25(τ2−1)−49.85(τ−1τ)∫373.15TΔC∘PRdT=1335.13τ−1335.13+210.25τ2−210.25−49.85+49.85τ∫373.15TΔC∘PRdT=210.25τ2+1335.13τ−1595.23+49.85τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=210.25τ2+1335.13τ−1595.23+49.85τ1210.25τ2+1335.13τ−3038.579+49.85τ=01210.25τ3+1335.13τ2−3038.579τ+49.85τ=0$

On solving,

$τ=1.115,0.0165, −2.23$

Considering positive value of temperature $τ=1.596$

Hence,

$τ=TT0=T100+273.15=1.115T=416.06 K$

Answer 118

(q)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(q)

Expert Solution

Answer to Problem 4.3P

$T=764.96 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon monoxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO 3.376 0.557 0 −0.031$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.376×373.15×(τ−1)+0.557×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.031×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1259.75(τ−1)+38.78(τ2−1)−8.31(τ−1τ)∫373.15TΔC∘PRdT=1259.75τ−1259.75+38.78τ2−38.78−8.31+8.31τ∫373.15TΔC∘PRdT=38.78τ2+1259.75τ−1306.84+8.31τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=38.78τ2+1259.75τ−1306.84+8.31τ38.78τ2+1259.75τ−2750.19+8.31τ=038.78τ3+1259.75τ2−2750.19τ+8.31τ=0$

On solving,

$τ=2.05,0.003, −34.54$

Considering positive value of temperature $τ=2.05$

Hence,

$τ=TT0=T100+273.15=2.05T=764.96 K$

Answer 119

(q)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 120

(q)

Expert Solution

Answer to Problem 4.3P

$T=764.96 K$

Answer 121

(q)

Expert Solution

Answer 122

(q)

Expert Solution

Answer 123

Expert Solution

Answer 124

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon monoxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO 3.376 0.557 0 −0.031$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.376×373.15×(τ−1)+0.557×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.031×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1259.75(τ−1)+38.78(τ2−1)−8.31(τ−1τ)∫373.15TΔC∘PRdT=1259.75τ−1259.75+38.78τ2−38.78−8.31+8.31τ∫373.15TΔC∘PRdT=38.78τ2+1259.75τ−1306.84+8.31τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=38.78τ2+1259.75τ−1306.84+8.31τ38.78τ2+1259.75τ−2750.19+8.31τ=038.78τ3+1259.75τ2−2750.19τ+8.31τ=0$

On solving,

$τ=2.05,0.003, −34.54$

Considering positive value of temperature $τ=2.05$

Hence,

$τ=TT0=T100+273.15=2.05T=764.96 K$

Answer 125

(r)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(r)

Expert Solution

Answer to Problem 4.3P

$T=634.355 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO2 5.457 1.045 0 −1.157$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.457×373.15×(τ−1)+1.045×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.157×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2036.28(τ−1)+72.75(τ2−1)−310.06(τ−1τ)∫373.15TΔC∘PRdT=2036.28τ−2036.28+72.75τ2−72.75−310.06+310.06τ∫373.15TΔC∘PRdT=72.75τ2+2036.28τ−2419.09+310.06τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=72.75τ2+2036.28τ−2419.09+310.06τ72.75τ2+2036.28τ−3862.44+310.06τ=072.75τ3+2036.28τ2−3862.44τ+310.06τ=0$

On solving,

$τ=1.7,0.084, −29.78$

Considering positive value of temperature $τ=1.7$

Hence,

$τ=TT0=T100+273.15=1.7T=634.355 K$

Answer 126

(r)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 127

(r)

Expert Solution

Answer to Problem 4.3P

$T=634.355 K$

Answer 128

(r)

Expert Solution

Answer 129

(r)

Expert Solution

Answer 130

Expert Solution

Answer 131

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for carbon dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DCO2 5.457 1.045 0 −1.157$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.457×373.15×(τ−1)+1.045×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.157×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2036.28(τ−1)+72.75(τ2−1)−310.06(τ−1τ)∫373.15TΔC∘PRdT=2036.28τ−2036.28+72.75τ2−72.75−310.06+310.06τ∫373.15TΔC∘PRdT=72.75τ2+2036.28τ−2419.09+310.06τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=72.75τ2+2036.28τ−2419.09+310.06τ72.75τ2+2036.28τ−3862.44+310.06τ=072.75τ3+2036.28τ2−3862.44τ+310.06τ=0$

On solving,

$τ=1.7,0.084, −29.78$

Considering positive value of temperature $τ=1.7$

Hence,

$τ=TT0=T100+273.15=1.7T=634.355 K$

Answer 132

(s)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(s)

Expert Solution

Answer to Problem 4.3P

$T=626.1457 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for sulphur dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DSO2 5.699 0.801 0 −1.015$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.699×373.15×(τ−1)+0.801×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.015×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2126.58(τ−1)+55.77(τ2−1)−272(τ−1τ)∫373.15TΔC∘PRdT=2126.58τ−2126.58+55.77τ2−55.77−272+272τ∫373.15TΔC∘PRdT=55.77τ2+2126.58τ−2454.36+272τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=55.77τ2+2126.58τ−2454.36+272τ55.77τ2+2126.58τ−3887.71+272τ=055.77τ3+2126.58τ2−3887.71τ+272τ=0$

On solving,

$τ=1.678,0.073, −39.88$

Considering positive value of temperature $τ=1.678$

Hence,

$τ=TT0=T100+273.15=1.678T=626.1457 K$

Answer 133

(s)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 134

(s)

Expert Solution

Answer to Problem 4.3P

$T=626.1457 K$

Answer 135

(s)

Expert Solution

Answer 136

(s)

Expert Solution

Answer 137

Expert Solution

Answer 138

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for sulphur dioxide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DSO2 5.699 0.801 0 −1.015$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=5.699×373.15×(τ−1)+0.801×10−32373.152(τ2−1) +03×373.153(τ3−1)+−1.015×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=2126.58(τ−1)+55.77(τ2−1)−272(τ−1τ)∫373.15TΔC∘PRdT=2126.58τ−2126.58+55.77τ2−55.77−272+272τ∫373.15TΔC∘PRdT=55.77τ2+2126.58τ−2454.36+272τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=55.77τ2+2126.58τ−2454.36+272τ55.77τ2+2126.58τ−3887.71+272τ=055.77τ3+2126.58τ2−3887.71τ+272τ=0$

On solving,

$τ=1.678,0.073, −39.88$

Considering positive value of temperature $τ=1.678$

Hence,

$τ=TT0=T100+273.15=1.678T=626.1457 K$

Answer 139

(t)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(t)

Expert Solution

Answer to Problem 4.3P

$T=706.37 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DH2O 3.47 1.450 0 0.121$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.470×373.15×(τ−1)+1.450×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.121×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1294.83(τ−1)+100.95(τ2−1)+32.43(τ−1τ)∫373.15TΔC∘PRdT=1294.83τ−1294.83+100.95τ2−100.95+32.43−32.43τ∫373.15TΔC∘PRdT=100.95τ2+1294.83τ−1363.35−32.43τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=100.95τ2+1294.83τ−1363.35−32.43τ100.95τ2+1294.83τ−2796.7−32.43τ=0100.95τ3+1294.83τ2−2796.7τ−32.43τ=0$

On solving,

$τ=1.893,−0.011, −14.708$

Considering positive value of temperature $τ=1.893$

Hence,

$τ=TT0=T100+273.15=1.893T=706.37 K$

Answer 140

(t)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 141

(t)

Expert Solution

Answer to Problem 4.3P

$T=706.37 K$

Answer 142

(t)

Expert Solution

Answer 143

(t)

Expert Solution

Answer 144

Expert Solution

Answer 145

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DH2O 3.47 1.450 0 0.121$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.470×373.15×(τ−1)+1.450×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.121×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1294.83(τ−1)+100.95(τ2−1)+32.43(τ−1τ)∫373.15TΔC∘PRdT=1294.83τ−1294.83+100.95τ2−100.95+32.43−32.43τ∫373.15TΔC∘PRdT=100.95τ2+1294.83τ−1363.35−32.43τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=100.95τ2+1294.83τ−1363.35−32.43τ100.95τ2+1294.83τ−2796.7−32.43τ=0100.95τ3+1294.83τ2−2796.7τ−32.43τ=0$

On solving,

$τ=1.893,−0.011, −14.708$

Considering positive value of temperature $τ=1.893$

Hence,

$τ=TT0=T100+273.15=1.893T=706.37 K$

Answer 146

(u)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(u)

Expert Solution

Answer to Problem 4.3P

$T=770.55 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for nitrogen in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DN2 3.280 0.593 0 0.040$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.280×373.15×(τ−1)+0.593×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.040×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1223.93(τ−1)+41.29(τ2−1)+10.72(τ−1τ)∫373.15TΔC∘PRdT=1223.93τ−1223.93+41.29τ2−41.29+10.72−10.72τ∫373.15TΔC∘PRdT=41.29τ2+1223.93τ−1254.5−10.72τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=41.29τ2+1223.93τ−1254.5−10.72τ41.29τ2+1223.93τ−2697.85−10.72τ=041.29τ3+1223.93τ2−2697.85τ−10.72τ=0$

On solving,

$τ=2.065,−0.003, −31.703$

Considering positive value of temperature $τ=2.065$

Hence,

$τ=TT0=T100+273.15=2.065T=770.55 K$

Answer 147

(u)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 148

(u)

Expert Solution

Answer to Problem 4.3P

$T=770.55 K$

Answer 149

(u)

Expert Solution

Answer 150

(u)

Expert Solution

Answer 151

Expert Solution

Answer 152

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for nitrogen in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DN2 3.280 0.593 0 0.040$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=3.280×373.15×(τ−1)+0.593×10−32373.152(τ2−1) +03×373.153(τ3−1)+0.040×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1223.93(τ−1)+41.29(τ2−1)+10.72(τ−1τ)∫373.15TΔC∘PRdT=1223.93τ−1223.93+41.29τ2−41.29+10.72−10.72τ∫373.15TΔC∘PRdT=41.29τ2+1223.93τ−1254.5−10.72τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=41.29τ2+1223.93τ−1254.5−10.72τ41.29τ2+1223.93τ−2697.85−10.72τ=041.29τ3+1223.93τ2−2697.85τ−10.72τ=0$

On solving,

$τ=2.065,−0.003, −31.703$

Considering positive value of temperature $τ=2.065$

Hence,

$τ=TT0=T100+273.15=2.065T=770.55 K$

Answer 153

(v)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

(v)

Expert Solution

Answer to Problem 4.3P

$T=654.13 K$

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for hydrogen cyanide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DHCN 4.736 1.359 0 −0.725$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.736×373.15×(τ−1)+1.359×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.725×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1767.24(τ−1)+94.61(τ2−1)−194.29(τ−1τ)∫373.15TΔC∘PRdT=1767.24τ−1767.24+94.61τ2−94.61−194.29+194.29τ∫373.15TΔC∘PRdT=94.61τ2+1767.24τ−2056.14+194.29τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=94.61τ2+1767.24τ−2056.14+194.29τ94.61τ2+1767.24τ−3499.49+194.29τ=094.61τ2+1767.24τ−3499.49τ+194.29τ=0$

On solving,

$τ=1.753,0.057, −20.49$

Considering positive value of temperature $τ=1.753$

Hence,

$τ=TT0=T100+273.15=1.753T=654.13 K$

Answer 154

(v)

Interpretation Introduction

Interpretation:

Final temperature $(T2)$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

$Q=nΔH=R∫T0TΔCPRdT$ . ...(1)

Where $ΔH$ is heat of reaction at any temperature $T$ .

and

$∫T0TΔC∘PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)$ .....(2)

Where $τ=TT0$

Answer 155

(v)

Expert Solution

Answer to Problem 4.3P

$T=654.13 K$

Answer 156

(v)

Expert Solution

Answer 157

(v)

Expert Solution

Answer 158

Expert Solution

Answer 159

Explanation of Solution

Given information:

$Q= 12 kJ/mol$

$initial temperature(T1) =1000C$ . So,

$T0=1000C⇒T0=1000C+273.15=373.15 K$

$T=?oC=T+273.15 K$

Values of above constants for hydrogen cyanide in equation (2) are given in appendix C table C.1 and noted down below:

$i A 103B 106 C 10−5DHCN 4.736 1.359 0 −0.725$

$gas constant R in SI unit is 8.314 Jmol K$

Put values in equation (2)

$∫T0TΔ C ∘ PRdT=AT0(τ−1)+B2T02(τ2−1)+C3T03(τ3−1)+DT0(τ−1τ)∫373.15TΔC∘PRdT=4.736×373.15×(τ−1)+1.359×10−32373.152(τ2−1) +03×373.153(τ3−1)+−0.725×105373.15×(τ−1τ)∫373.15TΔC∘PRdT=1767.24(τ−1)+94.61(τ2−1)−194.29(τ−1τ)∫373.15TΔC∘PRdT=1767.24τ−1767.24+94.61τ2−94.61−194.29+194.29τ∫373.15TΔC∘PRdT=94.61τ2+1767.24τ−2056.14+194.29τ$

Given that

$Q=12 kJmol⇒Q=12 kJmol×1000 J1 kJ=12000 Jmol$

Now from equation (1),

$12000 Jmol=R∫T0TΔ C PRdT12000 Jmol=8.314Jmol K×∫T0TΔCPRdT1443.349 K=∫T0TΔCPRdT$

Put in abovE

$1443.349 K=94.61τ2+1767.24τ−2056.14+194.29τ94.61τ2+1767.24τ−3499.49+194.29τ=094.61τ2+1767.24τ−3499.49τ+194.29τ=0$

On solving,

$τ=1.753,0.057, −20.49$

Considering positive value of temperature $τ=1.753$

Hence,

$τ=TT0=T100+273.15=1.753T=654.13 K$

Answer 160

Ch. 4 - Prob. 4.1P Ch. 4 - Prob. 4.2P Ch. 4 - Prob. 4.3P Ch. 4 - Prob. 4.4P Ch. 4 - Prob. 4.5P Ch. 4 - Prob. 4.6P Ch. 4 - Prob. 4.7P Ch. 4 - Prob. 4.8P Ch. 4 - Prob. 4.9P Ch. 4 - Prob. 4.10P

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

Answer to Problem 4.3P

Explanation of Solution

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