Chapter 11, Problem 141AP

Interpretation Introduction

Interpretation:

The vapor pressure of methanol at ${25}^{\circ }\text{C }$ is to be determined.

Concept introduction:

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

$\log \left(\frac{P_1}{P_2}\right)=\frac{\Delta {\text{H}}_{vap}}{\text{R}}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Here, $\Delta {\text{H}}_{vap}$ is the molar heat of vaporization, $\text{R}$ 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:

$1\text{ kJ}/\text{mol = }{10}^3\text{ J}/\text{mol}$

Answer to Problem 141AP

Solution: $127\text{ mm Hg}$

Explanation of Solution

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

${\text{K = }}^{\text{o}}\text{C}+273$

Here, $\text{K}$ is the temperature in kelvin, ${}^{\text{o}}\text{C}$ is the temperature in Celsius.

Convert the temperature intokelvin:

$\begin{array}{c}{T}_1=\left({\text{65}}^{\text{o}}\text{C}+273.15\right)\\ =\text{338}\text{.15 K}\\ T_2=\left({\text{25}}^{\text{o}}\text{C}+273.15\right)\\ =\text{298}\text{.15 K}\end{array}$

The reaction is as follows:

${\text{CH}}_3\text{OH}\left(l\right)\to {\text{CH}}_3\text{OH}\left(g\right)$

The molar enthalpy of vaporization is as follows:

$\begin{array}{c}\Delta H_{vap}=\Delta H_{\text{f}}^{\circ }\left[{\text{CH}}_3\text{OH}\left(g\right)\right]-\Delta H_{\text{f}}^{\circ }\left[{\text{CH}}_3\text{OH}\left(l\right)\right]\\ =-201.2\text{ kJ/mol}-\left(-238.7\text{ kJ/mol}\right)\\ =37.5\text{ kJ/mol}\end{array}$

Convert the heat in joule per mole as follows:

$\begin{array}{c}1\text{ kJ}/\text{mol = }{10}^3\text{ J}/\text{mol}\\ \text{37}\text{.5 kJ}/\text{mol}=37.5\times {10}^3\text{ J}/\text{mol}\end{array}$

The relation between pressure and temperature is as follows:

$\log \left(\frac{P_1}{P_2}\right)=\frac{\Delta {\text{H}}_{vap}}{\text{R}}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

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

$\begin{array}{c}\ln \left(\frac{P_1}{\text{760 mm Hg}}\right)=\frac{37.5\times {10}^3\text{ J}/\text{mol}}{8.314\text{ J}/\text{K}.\text{mol}}\left(\frac{1}{\text{338}\text{.15 K }}-\frac{1}{\text{298}\text{.15 K}}\right)\\ =4510.5\left(\frac{298.15\text{ K}-338.15\text{ K}}{\text{338}\text{.15 K }\times 298.15\text{ K}}\right)\\ =-1.79\end{array}$

Usingantilogarithm on both sides as follows:

$\begin{array}{c}\ln \left(\frac{P_1}{\text{760 mm Hg}}\right)=-1.79\\ \frac{P_1}{\text{760 mm Hg}}=\exp \left(-1.79\right)\\ =0.1669\end{array}$

Rearrange the above expression to discern the value of pressure.

$\begin{array}{c}{P}_1=0.1669\times \text{760 mm Hg}\\ =127\text{ mm Hg}\end{array}$

Conclusion

The vapor pressure of methanol at ${25}^{\circ }\text{C }$ is $127\text{ mm Hg}$.