The vapor pressure of methanol at 25 ∘ C is to be determined. Concept introduction: The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed 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 vapor pressures, and T 1 & T 2 are the temperatures. Conversion factor forkilo joule per mole to joule per mole: 1 kJ / mol = 10 3 J / mol

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Answer 1

Question

Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at $25∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

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

Conversion factor forkilo joule per mole to joule per mole:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 141AP

Solution: $127 mm Hg$

Explanation of Solution

To convert 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.

Convert the temperature intokelvin:

$T1 =(65oC+273.15)=338.15 KT2 =(25oC+273.15)=298.15 K$

The reaction is as follows:

$CH3OH(l)→CH3OH(g)$

The molar enthalpy of vaporization is as follows:

$ΔHvap=ΔHf∘[CH3OH(g)]−ΔHf∘[CH3OH(l)]=−201.2 kJ/mol−(−238.7 kJ/mol)=37.5 kJ/mol$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol37.5 kJ/mol=37.5×103 J/mol$

The relation between pressure and temperature is as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Substitute the values of pressure, temperatures, and enthalpy of vaporization,

$ln(P1760 mm Hg)=37.5×103 J/mol8.314 J/K.mol(1338.15 K −1298.15 K)=4510.5(298.15 K−338.15 K338.15 K ×298.15 K)=−1.79$

Usingantilogarithm on both sides as follows:

$ln(P1760 mm Hg)=−1.79P1760 mm Hg=exp(−1.79)=0.1669$

Rearrange the above expression to discern the value of pressure.

$P1=0.1669×760 mm Hg=127 mm Hg$

Conclusion

The vapor pressure of methanol at $25∘C$ is $127 mm Hg$ .

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Answer 2

Question

Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at $25∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

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

Conversion factor forkilo joule per mole to joule per mole:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 141AP

Solution: $127 mm Hg$

Explanation of Solution

To convert 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.

Convert the temperature intokelvin:

$T1 =(65oC+273.15)=338.15 KT2 =(25oC+273.15)=298.15 K$

The reaction is as follows:

$CH3OH(l)→CH3OH(g)$

The molar enthalpy of vaporization is as follows:

$ΔHvap=ΔHf∘[CH3OH(g)]−ΔHf∘[CH3OH(l)]=−201.2 kJ/mol−(−238.7 kJ/mol)=37.5 kJ/mol$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol37.5 kJ/mol=37.5×103 J/mol$

The relation between pressure and temperature is as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Substitute the values of pressure, temperatures, and enthalpy of vaporization,

$ln(P1760 mm Hg)=37.5×103 J/mol8.314 J/K.mol(1338.15 K −1298.15 K)=4510.5(298.15 K−338.15 K338.15 K ×298.15 K)=−1.79$

Usingantilogarithm on both sides as follows:

$ln(P1760 mm Hg)=−1.79P1760 mm Hg=exp(−1.79)=0.1669$

Rearrange the above expression to discern the value of pressure.

$P1=0.1669×760 mm Hg=127 mm Hg$

Conclusion

The vapor pressure of methanol at $25∘C$ is $127 mm Hg$ .

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Answer 3

Question

Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at $25∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

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

Conversion factor forkilo joule per mole to joule per mole:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 141AP

Solution: $127 mm Hg$

Explanation of Solution

To convert 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.

Convert the temperature intokelvin:

$T1 =(65oC+273.15)=338.15 KT2 =(25oC+273.15)=298.15 K$

The reaction is as follows:

$CH3OH(l)→CH3OH(g)$

The molar enthalpy of vaporization is as follows:

$ΔHvap=ΔHf∘[CH3OH(g)]−ΔHf∘[CH3OH(l)]=−201.2 kJ/mol−(−238.7 kJ/mol)=37.5 kJ/mol$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol37.5 kJ/mol=37.5×103 J/mol$

The relation between pressure and temperature is as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Substitute the values of pressure, temperatures, and enthalpy of vaporization,

$ln(P1760 mm Hg)=37.5×103 J/mol8.314 J/K.mol(1338.15 K −1298.15 K)=4510.5(298.15 K−338.15 K338.15 K ×298.15 K)=−1.79$

Usingantilogarithm on both sides as follows:

$ln(P1760 mm Hg)=−1.79P1760 mm Hg=exp(−1.79)=0.1669$

Rearrange the above expression to discern the value of pressure.

$P1=0.1669×760 mm Hg=127 mm Hg$

Conclusion

The vapor pressure of methanol at $25∘C$ is $127 mm Hg$ .

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Answer 4

Question

Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at $25∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

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

Conversion factor forkilo joule per mole to joule per mole:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 141AP

Solution: $127 mm Hg$

Explanation of Solution

To convert 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.

Convert the temperature intokelvin:

$T1 =(65oC+273.15)=338.15 KT2 =(25oC+273.15)=298.15 K$

The reaction is as follows:

$CH3OH(l)→CH3OH(g)$

The molar enthalpy of vaporization is as follows:

$ΔHvap=ΔHf∘[CH3OH(g)]−ΔHf∘[CH3OH(l)]=−201.2 kJ/mol−(−238.7 kJ/mol)=37.5 kJ/mol$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol37.5 kJ/mol=37.5×103 J/mol$

The relation between pressure and temperature is as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Substitute the values of pressure, temperatures, and enthalpy of vaporization,

$ln(P1760 mm Hg)=37.5×103 J/mol8.314 J/K.mol(1338.15 K −1298.15 K)=4510.5(298.15 K−338.15 K338.15 K ×298.15 K)=−1.79$

Usingantilogarithm on both sides as follows:

$ln(P1760 mm Hg)=−1.79P1760 mm Hg=exp(−1.79)=0.1669$

Rearrange the above expression to discern the value of pressure.

$P1=0.1669×760 mm Hg=127 mm Hg$

Conclusion

The vapor pressure of methanol at $25∘C$ is $127 mm Hg$ .

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Answer 5

Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at $25∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

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

Conversion factor forkilo joule per mole to joule per mole:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 141AP

Solution: $127 mm Hg$

Explanation of Solution

To convert 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.

Convert the temperature intokelvin:

$T1 =(65oC+273.15)=338.15 KT2 =(25oC+273.15)=298.15 K$

The reaction is as follows:

$CH3OH(l)→CH3OH(g)$

The molar enthalpy of vaporization is as follows:

$ΔHvap=ΔHf∘[CH3OH(g)]−ΔHf∘[CH3OH(l)]=−201.2 kJ/mol−(−238.7 kJ/mol)=37.5 kJ/mol$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol37.5 kJ/mol=37.5×103 J/mol$

The relation between pressure and temperature is as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Substitute the values of pressure, temperatures, and enthalpy of vaporization,

$ln(P1760 mm Hg)=37.5×103 J/mol8.314 J/K.mol(1338.15 K −1298.15 K)=4510.5(298.15 K−338.15 K338.15 K ×298.15 K)=−1.79$

Usingantilogarithm on both sides as follows:

$ln(P1760 mm Hg)=−1.79P1760 mm Hg=exp(−1.79)=0.1669$

Rearrange the above expression to discern the value of pressure.

$P1=0.1669×760 mm Hg=127 mm Hg$

Conclusion

The vapor pressure of methanol at $25∘C$ is $127 mm Hg$ .

Answer 6

Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at $25∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

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

Conversion factor forkilo joule per mole to joule per mole:

$1 kJ/mol = 103 J/mol$

Expert Solution & Answer

Answer to Problem 141AP

Solution: $127 mm Hg$

Explanation of Solution

To convert 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.

Convert the temperature intokelvin:

$T1 =(65oC+273.15)=338.15 KT2 =(25oC+273.15)=298.15 K$

The reaction is as follows:

$CH3OH(l)→CH3OH(g)$

The molar enthalpy of vaporization is as follows:

$ΔHvap=ΔHf∘[CH3OH(g)]−ΔHf∘[CH3OH(l)]=−201.2 kJ/mol−(−238.7 kJ/mol)=37.5 kJ/mol$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol37.5 kJ/mol=37.5×103 J/mol$

The relation between pressure and temperature is as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Substitute the values of pressure, temperatures, and enthalpy of vaporization,

$ln(P1760 mm Hg)=37.5×103 J/mol8.314 J/K.mol(1338.15 K −1298.15 K)=4510.5(298.15 K−338.15 K338.15 K ×298.15 K)=−1.79$

Usingantilogarithm on both sides as follows:

$ln(P1760 mm Hg)=−1.79P1760 mm Hg=exp(−1.79)=0.1669$

Rearrange the above expression to discern the value of pressure.

$P1=0.1669×760 mm Hg=127 mm Hg$

Conclusion

The vapor pressure of methanol at $25∘C$ is $127 mm Hg$ .

Answer 7

Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at $25∘C$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation, expressed as:

$log(P1P2)=ΔHvapR(1T2−1T1)$

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

Conversion factor forkilo joule per mole to joule per mole:

$1 kJ/mol = 103 J/mol$

Answer 8

Expert Solution & Answer

Answer to Problem 141AP

Solution: $127 mm Hg$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

To convert 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.

Convert the temperature intokelvin:

$T1 =(65oC+273.15)=338.15 KT2 =(25oC+273.15)=298.15 K$

The reaction is as follows:

$CH3OH(l)→CH3OH(g)$

The molar enthalpy of vaporization is as follows:

$ΔHvap=ΔHf∘[CH3OH(g)]−ΔHf∘[CH3OH(l)]=−201.2 kJ/mol−(−238.7 kJ/mol)=37.5 kJ/mol$

Convert the heat in joule per mole as follows:

$1 kJ/mol = 103 J/mol37.5 kJ/mol=37.5×103 J/mol$

The relation between pressure and temperature is as follows:

$log(P1P2)=ΔHvapR(1T2−1T1)$

Substitute the values of pressure, temperatures, and enthalpy of vaporization,

$ln(P1760 mm Hg)=37.5×103 J/mol8.314 J/K.mol(1338.15 K −1298.15 K)=4510.5(298.15 K−338.15 K338.15 K ×298.15 K)=−1.79$

Usingantilogarithm on both sides as follows:

$ln(P1760 mm Hg)=−1.79P1760 mm Hg=exp(−1.79)=0.1669$

Rearrange the above expression to discern the value of pressure.

$P1=0.1669×760 mm Hg=127 mm Hg$

Answer 12

Conclusion

The vapor pressure of methanol at $25∘C$ is $127 mm Hg$ .

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 ...

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Answer to Problem 141AP

Explanation of Solution

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Chapter 11 Solutions