Chapter 11, Problem 32QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at ${60.6}^{\circ }\text{C }$ is to be determined.

Concept introduction:

The relationship between vapor pressure and temperature is given by Clausius- Clapeyron equation 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 vapour pressures, and $T_1\&T_2$ are the temperatures.

To convert a 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.

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

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

Answer to Problem 32QP

Solution: $331\text{ mm Hg}$

Explanation of Solution

Given information: The vapour pressure of benzene is $P_1,40.1\text{ mm Hg}$.

The temperatures of benzene are ${7.6}^{\text{o}}\text{C and }{60.6}^{\text{o}}\text{C }$.

The molar heat of vaporization, $\Delta {\text{H}}_{\text{vap}}=31.0\text{ kJ}/\text{mol}$

The gas constant, $\text{R}=8.314\text{ J}/\text{K}.\text{mol}$

Convert the temperature in Kelvin:

$\begin{array}{c}{T}_1=\text{ 7}{\text{.6}}^{\text{o}}\text{C}+273.15\\ =\text{ 280}\text{.75 K}\\ T_2=\text{ 60}{\text{.6}}^{\text{o}}\text{C}+273.15\\ =\text{ 333}\text{.75 K}\end{array}$

Convert the heat in joule per mole as follows:

$\begin{array}{c}1\text{ kJ}/\text{mol = }{10}^3\text{ J}/\text{mol}\\ 31.0\text{ kJ}/\text{mol}=31.0\times {10}^3\text{ J}/\text{mol}\\ =3.1\times {10}^4\text{ J}/\text{mol}\end{array}$

Now, calculate $P_2$ as follows:

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

$\ln \left(\frac{\text{40}\text{.1 mm Hg}}{P_2}\right)=\frac{3.1\times {10}^4\text{ J}/\text{mol}}{8.314\text{ J}/\text{K}.\text{mol}}\left(\frac{1}{\text{333}\text{.75 K}}-\frac{1}{\text{280}\text{.75 K}}\right)$

$\begin{array}{c}\ln \left(\frac{\text{40}\text{.1 mm Hg}}{P_2}\right)=\frac{3.1\times {10}^4\text{ J}/\text{mol}}{8.314\text{ J}/\text{K}.\text{mol}}\left(\frac{1}{\text{333}\text{.75 K}}-\frac{1}{\text{280}\text{.75 K}}\right)\\ =3728.65\left(\frac{280.75\text{ K}-333.75\text{ K}}{333.75\text{ K}\times 280.75\text{ K}}\right)\\ =\frac{-53}{93700.31}\times 3728.65\\ =-2.109\end{array}$

Take antilogarithm on both the sides as follows:

$\begin{array}{c}\ln \left(\frac{\text{40}\text{.1 mm Hg}}{P_2}\right)=-2.109\\ \frac{\text{40}\text{.1 mm Hg}}{P_2}=\exp \left(-2.109\right)\\ =0.1213\end{array}$

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

$\begin{array}{c}{P}_2=\frac{\text{40}\text{.1 mm Hg}}{0.1213}\\ =330.58\text{ mm Hg}\\ =\text{331 mm Hg}\end{array}$

Conclusion

The vapour pressure of benzene at ${60.6}^{\circ }\text{C }$ is $\text{331 mm Hg}$.