The vapor pressure of benzene at 60.6 ∘ C is to be determined. Concept introduction: The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as: log ( P 1 P 2 ) = Δ H v a p R ( 1 T 2 − 1 T 1 ) Here, Δ H v a p is the molar heat of vaporization, R is the gas constant, P 1 & P 2 are the vapour pressures, and T 1 & T 2 are the temperatures. To convert a temperature from degree Celsius to Kelvin, the expression is as follows: K = o C + 273 Here, K is the temperature in Kelvin, o C is the temperature in Celsius. To convert heat from kilo joule per mole to joule per mole, the expression is as follows: 1 kJ / mol = 10 3 J / mol

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Answer 1

Question

Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Here, $ΔHvap$ is the molar heat of vaporization, $R$ is the gas constant, $P1 & P2$ are the vapour pressures, and $T1&T2$ are the temperatures.

To convert a temperature from degree Celsius to Kelvin, the expression is as follows:

$K = oC+273$

Here, $K$ is the temperature in Kelvin, $oC$ is the temperature in Celsius.

To convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 32QP

Solution: $331 mm Hg$

Explanation of Solution

Given information: The vapour pressure of benzene is $P1, 40.1 mm Hg$ .

The temperatures of benzene are $7.6oC and 60.6oC$ .

The molar heat of vaporization, $ΔHvap=31.0 kJ/mol$

The gas constant, $R=8.314 J/K.mol$

Convert the temperature in Kelvin:

$T1 = 7.6oC+273.15= 280.75 KT2 = 60.6oC+273.15= 333.75 K$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol31.0 kJ/mol=31.0×103 J/mol=3.1×104 J/mol$

Now, calculate $P2$ as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)=3728.65(280.75 K−333.75 K333.75 K×280.75 K)=−5393700.31×3728.65=−2.109$

Take antilogarithm on both the sides as follows:

$ln(40.1 mm HgP2)=−2.10940.1 mm HgP2=exp(−2.109)=0.1213$

Rearrange the above expression to obtain the vapor pressure as follows:

$P2=40.1 mm Hg0.1213=330.58 mm Hg=331 mm Hg$

Conclusion

The vapour pressure of benzene at $60.6∘C$ is $331 mm Hg$ .

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Answer 2

Question

Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Here, $ΔHvap$ is the molar heat of vaporization, $R$ is the gas constant, $P1 & P2$ are the vapour pressures, and $T1&T2$ are the temperatures.

To convert a temperature from degree Celsius to Kelvin, the expression is as follows:

$K = oC+273$

Here, $K$ is the temperature in Kelvin, $oC$ is the temperature in Celsius.

To convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 32QP

Solution: $331 mm Hg$

Explanation of Solution

Given information: The vapour pressure of benzene is $P1, 40.1 mm Hg$ .

The temperatures of benzene are $7.6oC and 60.6oC$ .

The molar heat of vaporization, $ΔHvap=31.0 kJ/mol$

The gas constant, $R=8.314 J/K.mol$

Convert the temperature in Kelvin:

$T1 = 7.6oC+273.15= 280.75 KT2 = 60.6oC+273.15= 333.75 K$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol31.0 kJ/mol=31.0×103 J/mol=3.1×104 J/mol$

Now, calculate $P2$ as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)=3728.65(280.75 K−333.75 K333.75 K×280.75 K)=−5393700.31×3728.65=−2.109$

Take antilogarithm on both the sides as follows:

$ln(40.1 mm HgP2)=−2.10940.1 mm HgP2=exp(−2.109)=0.1213$

Rearrange the above expression to obtain the vapor pressure as follows:

$P2=40.1 mm Hg0.1213=330.58 mm Hg=331 mm Hg$

Conclusion

The vapour pressure of benzene at $60.6∘C$ is $331 mm Hg$ .

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Answer 3

Question

Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Here, $ΔHvap$ is the molar heat of vaporization, $R$ is the gas constant, $P1 & P2$ are the vapour pressures, and $T1&T2$ are the temperatures.

To convert a temperature from degree Celsius to Kelvin, the expression is as follows:

$K = oC+273$

Here, $K$ is the temperature in Kelvin, $oC$ is the temperature in Celsius.

To convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 32QP

Solution: $331 mm Hg$

Explanation of Solution

Given information: The vapour pressure of benzene is $P1, 40.1 mm Hg$ .

The temperatures of benzene are $7.6oC and 60.6oC$ .

The molar heat of vaporization, $ΔHvap=31.0 kJ/mol$

The gas constant, $R=8.314 J/K.mol$

Convert the temperature in Kelvin:

$T1 = 7.6oC+273.15= 280.75 KT2 = 60.6oC+273.15= 333.75 K$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol31.0 kJ/mol=31.0×103 J/mol=3.1×104 J/mol$

Now, calculate $P2$ as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)=3728.65(280.75 K−333.75 K333.75 K×280.75 K)=−5393700.31×3728.65=−2.109$

Take antilogarithm on both the sides as follows:

$ln(40.1 mm HgP2)=−2.10940.1 mm HgP2=exp(−2.109)=0.1213$

Rearrange the above expression to obtain the vapor pressure as follows:

$P2=40.1 mm Hg0.1213=330.58 mm Hg=331 mm Hg$

Conclusion

The vapour pressure of benzene at $60.6∘C$ is $331 mm Hg$ .

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Answer 4

Question

Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Here, $ΔHvap$ is the molar heat of vaporization, $R$ is the gas constant, $P1 & P2$ are the vapour pressures, and $T1&T2$ are the temperatures.

To convert a temperature from degree Celsius to Kelvin, the expression is as follows:

$K = oC+273$

Here, $K$ is the temperature in Kelvin, $oC$ is the temperature in Celsius.

To convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 32QP

Solution: $331 mm Hg$

Explanation of Solution

Given information: The vapour pressure of benzene is $P1, 40.1 mm Hg$ .

The temperatures of benzene are $7.6oC and 60.6oC$ .

The molar heat of vaporization, $ΔHvap=31.0 kJ/mol$

The gas constant, $R=8.314 J/K.mol$

Convert the temperature in Kelvin:

$T1 = 7.6oC+273.15= 280.75 KT2 = 60.6oC+273.15= 333.75 K$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol31.0 kJ/mol=31.0×103 J/mol=3.1×104 J/mol$

Now, calculate $P2$ as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)=3728.65(280.75 K−333.75 K333.75 K×280.75 K)=−5393700.31×3728.65=−2.109$

Take antilogarithm on both the sides as follows:

$ln(40.1 mm HgP2)=−2.10940.1 mm HgP2=exp(−2.109)=0.1213$

Rearrange the above expression to obtain the vapor pressure as follows:

$P2=40.1 mm Hg0.1213=330.58 mm Hg=331 mm Hg$

Conclusion

The vapour pressure of benzene at $60.6∘C$ is $331 mm Hg$ .

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Answer 5

Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Here, $ΔHvap$ is the molar heat of vaporization, $R$ is the gas constant, $P1 & P2$ are the vapour pressures, and $T1&T2$ are the temperatures.

To convert a temperature from degree Celsius to Kelvin, the expression is as follows:

$K = oC+273$

Here, $K$ is the temperature in Kelvin, $oC$ is the temperature in Celsius.

To convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 32QP

Solution: $331 mm Hg$

Explanation of Solution

Given information: The vapour pressure of benzene is $P1, 40.1 mm Hg$ .

The temperatures of benzene are $7.6oC and 60.6oC$ .

The molar heat of vaporization, $ΔHvap=31.0 kJ/mol$

The gas constant, $R=8.314 J/K.mol$

Convert the temperature in Kelvin:

$T1 = 7.6oC+273.15= 280.75 KT2 = 60.6oC+273.15= 333.75 K$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol31.0 kJ/mol=31.0×103 J/mol=3.1×104 J/mol$

Now, calculate $P2$ as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)=3728.65(280.75 K−333.75 K333.75 K×280.75 K)=−5393700.31×3728.65=−2.109$

Take antilogarithm on both the sides as follows:

$ln(40.1 mm HgP2)=−2.10940.1 mm HgP2=exp(−2.109)=0.1213$

Rearrange the above expression to obtain the vapor pressure as follows:

$P2=40.1 mm Hg0.1213=330.58 mm Hg=331 mm Hg$

Conclusion

The vapour pressure of benzene at $60.6∘C$ is $331 mm Hg$ .

Answer 6

Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Here, $ΔHvap$ is the molar heat of vaporization, $R$ is the gas constant, $P1 & P2$ are the vapour pressures, and $T1&T2$ are the temperatures.

To convert a temperature from degree Celsius to Kelvin, the expression is as follows:

$K = oC+273$

Here, $K$ is the temperature in Kelvin, $oC$ is the temperature in Celsius.

To convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 32QP

Solution: $331 mm Hg$

Explanation of Solution

Given information: The vapour pressure of benzene is $P1, 40.1 mm Hg$ .

The temperatures of benzene are $7.6oC and 60.6oC$ .

The molar heat of vaporization, $ΔHvap=31.0 kJ/mol$

The gas constant, $R=8.314 J/K.mol$

Convert the temperature in Kelvin:

$T1 = 7.6oC+273.15= 280.75 KT2 = 60.6oC+273.15= 333.75 K$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol31.0 kJ/mol=31.0×103 J/mol=3.1×104 J/mol$

Now, calculate $P2$ as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)=3728.65(280.75 K−333.75 K333.75 K×280.75 K)=−5393700.31×3728.65=−2.109$

Take antilogarithm on both the sides as follows:

$ln(40.1 mm HgP2)=−2.10940.1 mm HgP2=exp(−2.109)=0.1213$

Rearrange the above expression to obtain the vapor pressure as follows:

$P2=40.1 mm Hg0.1213=330.58 mm Hg=331 mm Hg$

Conclusion

The vapour pressure of benzene at $60.6∘C$ is $331 mm Hg$ .

Answer 7

Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $60.6∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Here, $ΔHvap$ is the molar heat of vaporization, $R$ is the gas constant, $P1 & P2$ are the vapour pressures, and $T1&T2$ are the temperatures.

To convert a temperature from degree Celsius to Kelvin, the expression is as follows:

$K = oC+273$

Here, $K$ is the temperature in Kelvin, $oC$ is the temperature in Celsius.

To convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1 kJ/mol = 103 J/mol$

Answer 8

Expert Solution & Answer

Answer to Problem 32QP

Solution: $331 mm Hg$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information: The vapour pressure of benzene is $P1, 40.1 mm Hg$ .

The temperatures of benzene are $7.6oC and 60.6oC$ .

The molar heat of vaporization, $ΔHvap=31.0 kJ/mol$

The gas constant, $R=8.314 J/K.mol$

Convert the temperature in Kelvin:

$T1 = 7.6oC+273.15= 280.75 KT2 = 60.6oC+273.15= 333.75 K$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol31.0 kJ/mol=31.0×103 J/mol=3.1×104 J/mol$

Now, calculate $P2$ as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)$

$ln(40.1 mm HgP2)=3.1×104 J/mol8.314 J/K.mol(1333.75 K−1280.75 K)=3728.65(280.75 K−333.75 K333.75 K×280.75 K)=−5393700.31×3728.65=−2.109$

Take antilogarithm on both the sides as follows:

$ln(40.1 mm HgP2)=−2.10940.1 mm HgP2=exp(−2.109)=0.1213$

Rearrange the above expression to obtain the vapor pressure as follows:

$P2=40.1 mm Hg0.1213=330.58 mm Hg=331 mm Hg$

Answer 12

Conclusion

The vapour pressure of benzene at $60.6∘C$ is $331 mm Hg$ .

Answer 13

Ch. 11.1 - Prob. 1PPA Ch. 11.1 - Prob. 1PPB Ch. 11.1 - Prob. 1PPC Ch. 11.1 - Prob. 1CP Ch. 11.1 - Prob. 2CP Ch. 11.2 - Prob. 1PPA Ch. 11.2 - Prob. 1PPB Ch. 11.2 - Prob. 1PPC Ch. 11.2 - Prob. 1CP Ch. 11.2 - 11.2.2 Given the following information for ...

Concept explainers

Answer to Problem 32QP

Explanation of Solution

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Chapter 11 Solutions