Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
Question
Book Icon
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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(a)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=626.817K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dmethane1.7029.0812.1640

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=1.702×373.15×(τ1)+9.081×1032373.152(τ21)+2.164×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=635.1013(τ1)+632.223(τ21)37.479(τ31)373.15TΔCPRdT=635.1013τ635.1013+632.223τ2632.22337.479τ3+37.479373.15TΔCPRdT=37.479τ3+632.223τ2+635.1013τ1229.8453

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=37.479τ3+632.223τ2+635.1013τ1229.845337.479τ3632.223τ2635.1013τ+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=626.817K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(b)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=538.455K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dethane1.13119.2255.5610

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=1.131×373.15×(τ1)+19.225×1032373.152(τ21)+5.561×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=422.033(τ1)+1338.453(τ21)96.31(τ31)373.15TΔCPRdT=422.033τ422.033+1338.453τ21338.45396.31τ3+96.31373.15TΔCPRdT=96.31τ3+1338.453τ2+422.033τ1664.176

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=96.31τ3+1338.453τ2+422.033τ1664.17696.31τ31338.453τ2422.033τ+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=538.455K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(c)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=493.3043K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dpropane1.21328.7858.8240

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=1.213×373.15×(τ1)+28.785×1032373.152(τ21)+8.824×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=452.631(τ1)+2004.025(τ21)152.825(τ31)373.15TΔCPRdT=452.631τ452.631+2004.025τ22004.025152.825τ3+152.825373.15TΔCPRdT=152.825τ3+2004.025τ2+452.631τ2303.831

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=152.825τ3+2004.025τ2+452.631τ2303.831152.825τ32004.025τ2452.631τ+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=493.3043K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(d)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=466.81K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dnbutane1.93536.91511.4020

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=1.935×373.15×(τ1)+36.915×1032373.152(τ21)+11.402×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=722.045(τ1)+2570.039(τ21)197.474(τ31)373.15TΔCPRdT=722.045τ722.045+2570.039τ22570.039197.474τ3+197.474373.15TΔCPRdT=197.474τ3+2570.039τ2+722.045τ3094.61

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=197.474τ3+2570.039τ2+722.045τ3094.61197.474τ32570.039τ2722.045τ+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=466.81K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(e)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=438.45K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dnhexane3.02553.72216.7910

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=3.025×373.15×(τ1)+53.722×1032373.152(τ21)+16.791×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=1128.78(τ1)+3740.15(τ21)290.807(τ31)373.15TΔCPRdT=1128.78τ1128.78+3740.15τ23740.15290.807τ3+290.807373.15TΔCPRdT=290.807τ3+3740.15τ2+1128.78τ4578.12

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=290.807τ3+3740.15τ2+1128.78τ4578.12290.807τ33740.15τ21128.78τ+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=438.45K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(f)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=371.695K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dnhexane4.10870.56722.2080

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=4.108×373.15×(τ1)+70.567×1032373.152(τ21)+22.208×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=1532.9002(τ1)+4912.91(τ21)384.63(τ31)373.15TΔCPRdT=1532.9002τ1532.9002+4912.91τ24912.91384.63τ3+384.63373.15TΔCPRdT=384.63τ3+4912.91τ2+1532.9002τ6061.184

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=384.63τ3+4912.91τ2+1532.9002τ6061.184384.63τ34912.91τ21532.9002τ+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=371.695K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(g)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=513.08K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dpropylene1.63722.7066.9150

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=1.637×373.15×(τ1)+22.706×1032373.152(τ21)+6.915×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=610.847(τ1)+1580.8(τ21)119.763(τ31)373.15TΔCPRdT=610.847τ610.847+1580.8τ21580.8119.763τ3+119.763373.15TΔCPRdT=119.763τ3+1580.8τ2+610.847τ2071.88

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=119.763τ3+1580.8τ2+610.847τ2071.88119.763τ31580.8τ2610.847τ+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=513.08K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(h)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=457.855K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105D1pentene2.69139.75312.4470

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=2.691×373.15×(τ1)+39.753×1032373.152(τ21)+12.447×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=1004.147(τ1)+2767.62(τ21)215.57(τ31)373.15TΔCPRdT=1004.147τ1004.147+2767.62τ22767.62215.57τ3+215.57373.15TΔCPRdT=215.57τ3+2767.62τ2+1004.147τ3556.197

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=215.57τ3+2767.62τ2+1004.147τ3556.197215.57τ32767.62τ21004.147τ+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=457.855K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(i)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=433.97K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105D1heptene3.76856.58817.8470

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=3.768×373.15×(τ1)+56.588×1032373.152(τ21)+17.847×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=1406.03(τ1)+3939.68(τ21)309.097(τ31)373.15TΔCPRdT=1406.03τ1406.03+3939.68τ23939.68309.097τ3+309.097373.15TΔCPRdT=309.097τ3+3939.68τ2+1406.03τ5036.61

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=309.097τ3+3939.68τ2+1406.03τ5036.61309.097τ33939.68τ21406.03τ+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=433.97K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(j)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=426.51K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105D1Octene4.32464.9620.5210

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=4.324×373.15×(τ1)+64.96×1032373.152(τ21)+20.521×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=1613.5(τ1)+4522.55(τ21)355.41(τ31)373.15TΔCPRdT=1613.5τ1613.5+4522.55τ24522.55355.41τ3+355.41373.15TΔCPRdT=355.41τ3+4522.55τ2+1613.5τ5780.64

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=355.41τ3+4522.55τ2+1613.5τ5780.64355.41τ34522.55τ21613.5τ+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=426.51K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(k)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=595.55K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dacetylene6.1321.95201.299

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=6.132×373.15×(τ1)+1.952×1032373.152(τ21)+03×373.153(τ31)+1.299×105373.15×(τ1τ)373.15TΔCPRdT=2288.15(τ1)+135.9(τ21)348.12(τ1τ)373.15TΔCPRdT=2288.15τ2288.15+135.9τ2135.9348.12+348.12τ373.15TΔCPRdT=135.9τ2+2288.15τ2772.17+348.12τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=135.9τ2+2288.15τ2772.17+348.12τ135.9τ2+2288.15τ4215.519+348.12τ=0135.9τ3+2288.15τ24215.519τ+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=595.55K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(l)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=476.51K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dbenzene0.20639.06413.3010

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=0.206×373.15×(τ1)+39.064×1032373.152(τ21)+13.301×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=76.87(τ1)+2719.65(τ21)230.36(τ31)373.15TΔCPRdT=76.87τ+76.87+2719.65τ22719.65230.36τ3+230.36373.15TΔCPRdT=230.36τ3+2719.65τ276.87τ2412.42

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=230.36τ3+2719.65τ276.87τ2412.42230.36τ32719.65τ2+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=476.51K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(m)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=380.613K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dethanol3.51820.0016.0020

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=3.518×373.15×(τ1)+20.001×1032373.152(τ21)+6.002×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=1312.74(τ1)+1392.48(τ21)103.95(τ31)373.15TΔCPRdT=1312.74τ1312.74+1392.48τ21392.48103.95τ3+103.95373.15TΔCPRdT=103.95τ3+2719.65τ2+1312.74τ2601.27

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=103.95τ3+2719.65τ2+1312.74τ2601.27103.95τ32719.65τ21312.74τ+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=380.613K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(n)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=445.91K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dstyrene2.05050.19216.6620

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=2.050×373.15×(τ1)+50.192×1032373.152(τ21)+16.662×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=764.96(τ1)+3494.39(τ21)288.57(τ31)373.15TΔCPRdT=764.96τ764.96+3494.39τ23494.39288.57τ3+288.57373.15TΔCPRdT=288.57τ3+3494.39τ2+764.96τ3970.78

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=288.57τ3+3494.39τ2+764.96τ3970.78288.57τ33494.39τ2764.96τ+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=445.91K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(o)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=643.68K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dformaldehyde2.2647.0221.8770

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=2.264×373.15×(τ1)+7.022×1032373.152(τ21)+1.877×1063×373.153(τ31)+0373.15×(τ1τ)373.15TΔCPRdT=844.81(τ1)+488.88(τ21)32.51(τ31)373.15TΔCPRdT=844.81τ844.81+488.88τ2488.8832.51τ3+32.51373.15TΔCPRdT=-32.51τ3+488.88τ2+844.81τ1301.18

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=-32.51τ3+488.88τ2+844.81τ1301.1832.51τ3488.88τ2844.81τ+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=643.68K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(p)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=416.06K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105Dammonia3.5783.02000.186

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=3.578×373.15×(τ1)+3.020×1032373.152(τ21)+03×373.153(τ31)+0.186×105373.15×(τ1τ)373.15TΔCPRdT=1335.13(τ1)+210.25(τ21)49.85(τ1τ)373.15TΔCPRdT=1335.13τ1335.13+210.25τ2210.2549.85+49.85τ373.15TΔCPRdT=210.25τ2+1335.13τ1595.23+49.85τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=210.25τ2+1335.13τ1595.23+49.85τ1210.25τ2+1335.13τ3038.579+49.85τ=01210.25τ3+1335.13τ23038.579τ+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=416.06K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(q)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=764.96K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105DCO3.3760.55700.031

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=3.376×373.15×(τ1)+0.557×1032373.152(τ21)+03×373.153(τ31)+0.031×105373.15×(τ1τ)373.15TΔCPRdT=1259.75(τ1)+38.78(τ21)8.31(τ1τ)373.15TΔCPRdT=1259.75τ1259.75+38.78τ238.788.31+8.31τ373.15TΔCPRdT=38.78τ2+1259.75τ1306.84+8.31τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=38.78τ2+1259.75τ1306.84+8.31τ38.78τ2+1259.75τ2750.19+8.31τ=038.78τ3+1259.75τ22750.19τ+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=764.96K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(r)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=634.355K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105DCO25.4571.04501.157

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=5.457×373.15×(τ1)+1.045×1032373.152(τ21)+03×373.153(τ31)+1.157×105373.15×(τ1τ)373.15TΔCPRdT=2036.28(τ1)+72.75(τ21)310.06(τ1τ)373.15TΔCPRdT=2036.28τ2036.28+72.75τ272.75310.06+310.06τ373.15TΔCPRdT=72.75τ2+2036.28τ2419.09+310.06τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=72.75τ2+2036.28τ2419.09+310.06τ72.75τ2+2036.28τ3862.44+310.06τ=072.75τ3+2036.28τ23862.44τ+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=634.355K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(s)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=626.1457K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105DSO25.6990.80101.015

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=5.699×373.15×(τ1)+0.801×1032373.152(τ21)+03×373.153(τ31)+1.015×105373.15×(τ1τ)373.15TΔCPRdT=2126.58(τ1)+55.77(τ21)272(τ1τ)373.15TΔCPRdT=2126.58τ2126.58+55.77τ255.77272+272τ373.15TΔCPRdT=55.77τ2+2126.58τ2454.36+272τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=55.77τ2+2126.58τ2454.36+272τ55.77τ2+2126.58τ3887.71+272τ=055.77τ3+2126.58τ23887.71τ+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=626.1457K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(t)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=706.37K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105DH2O3.471.45000.121

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=3.470×373.15×(τ1)+1.450×1032373.152(τ21)+03×373.153(τ31)+0.121×105373.15×(τ1τ)373.15TΔCPRdT=1294.83(τ1)+100.95(τ21)+32.43(τ1τ)373.15TΔCPRdT=1294.83τ1294.83+100.95τ2100.95+32.4332.43τ373.15TΔCPRdT=100.95τ2+1294.83τ1363.3532.43τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=100.95τ2+1294.83τ1363.3532.43τ100.95τ2+1294.83τ2796.732.43τ=0100.95τ3+1294.83τ22796.7τ32.43τ=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=706.37K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(u)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=770.55K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105DN23.2800.59300.040

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=3.280×373.15×(τ1)+0.593×1032373.152(τ21)+03×373.153(τ31)+0.040×105373.15×(τ1τ)373.15TΔCPRdT=1223.93(τ1)+41.29(τ21)+10.72(τ1τ)373.15TΔCPRdT=1223.93τ1223.93+41.29τ241.29+10.7210.72τ373.15TΔCPRdT=41.29τ2+1223.93τ1254.510.72τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=41.29τ2+1223.93τ1254.510.72τ41.29τ2+1223.93τ2697.8510.72τ=041.29τ3+1223.93τ22697.85τ10.72τ=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=770.55K

(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=RT0TΔCPRdT.     ...(1)

Where ΔH is heat of reaction at any temperature T .

and

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)   .....(2)

Where τ=TT0

(v)

Expert Solution
Check Mark

Answer to Problem 4.3P

  T=654.13K

Explanation of Solution

Given information:

  Q= 12 kJ/mol

  initial temperature(T1) =1000C. So,

  T0=1000CT0=1000C+273.15=373.15K

  T=?oC=T+273.15K

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

  iA103B106C105DHCN4.7361.35900.725

  gas constant R in SI unit is 8.314JmolK

Put values in equation (2)

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)373.15TΔCPRdT=4.736×373.15×(τ1)+1.359×1032373.152(τ21)+03×373.153(τ31)+0.725×105373.15×(τ1τ)373.15TΔCPRdT=1767.24(τ1)+94.61(τ21)194.29(τ1τ)373.15TΔCPRdT=1767.24τ1767.24+94.61τ294.61194.29+194.29τ373.15TΔCPRdT=94.61τ2+1767.24τ2056.14+194.29τ

Given that

  Q=12kJmolQ=12kJmol×1000 J1kJ=12000Jmol

Now from equation (1),

  12000Jmol=RT0TΔ C PRdT12000Jmol=8.314JmolK×T0TΔCPRdT1443.349K=T0TΔCPRdT

Put in abovE

  1443.349K=94.61τ2+1767.24τ2056.14+194.29τ94.61τ2+1767.24τ3499.49+194.29τ=094.61τ2+1767.24τ3499.49τ+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=654.13K

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Introduction to Chemical Engineering Thermodynami...
Chemical Engineering
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:McGraw-Hill Education
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemical Engineering
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Text book image
Elements of Chemical Reaction Engineering (5th Ed...
Chemical Engineering
ISBN:9780133887518
Author:H. Scott Fogler
Publisher:Prentice Hall
Text book image
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:9781119285915
Author:Seborg
Publisher:WILEY
Text book image
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:9781285061238
Author:Lokensgard, Erik
Publisher:Delmar Cengage Learning
Text book image
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:9780072848236
Author:Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:McGraw-Hill Companies, The