Chapter 11, Problem 11.84QP

Interpretation Introduction

Interpretation:

The vapor pressure of benzene at $\text{60}\text{.6°C}$ has to be calculated.

Concept Introduction:

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state. Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid as follows –

$\begin{array}{l}\text{ln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]\\ \text{Where}{\text{P}}_{\text{1}}\text{=}\text{vapor}\text{pressure}\text{of}\text{the}\text{liquid}\text{}\text{at}\text{temperature}{\text{T}}_{\text{1}}\text{.}\\ {\text{P}}_{\text{2}}\text{=}\text{vapor}\text{pressure}\text{of}\text{the}\text{liquid}\text{}\text{at}\text{temperature}{\text{T}}_{\text{2}}\text{.}\\ {\text{ΔH}}_{\text{vap}}\text{=}\text{Molar}\text{heat}\text{of}\text{vaporization}\\ \text{R}\text{=}\text{Universal}\text{Gas constant}\text{.}\end{array}$

Answer to Problem 11.84QP

The vapor pressure of benzene at $\text{60}\text{.6°C}$ is calculated as $331mmHg.$

Explanation of Solution

Vapor pressure of benzene at $\text{7}{\text{.6}}^{\text{°}}\text{C}$ is given. Molar heat of vaporization of benzene is given. Vapor pressure of benzene at $\text{60}{\text{.6}}^{\text{°}}\text{C}$ is calculated using the formula $\text{ln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$ in which known values are substituted and the required parameter is obtained as follows –

$\begin{array}{l}\text{Given}\text{data:}{\text{ΔH}}_{\text{vap}}\text{=}\text{31}\text{.0}\text{kJ/mol}\\ {\text{P}}_{\text{1}}\text{=}\text{40}\text{.1}\text{mmHg}\\ {\text{T}}_{\text{1}}\text{=}\text{7}{\text{.6}}^{\text{°}}\text{C}\text{=}\text{280}\text{.8K}\\ {\text{T}}_{\text{2}}\text{=}\text{60}{\text{.6}}^{\text{°}}\text{C}\text{=}\text{333}\text{.8K}\\ {\text{P}}_2=?\end{array}$

$\begin{array}{l}\text{ln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]\\ \text{ln}\frac{\text{40}\text{.1}\text{mmHg}}{{\text{P}}_{\text{2}}}\text{=}\frac{\text{31}\text{.0}\text{×}{\text{10}}^{\text{3}}\text{J}}{\text{8}\text{.314}\text{J/K}\text{.mol}}\left(\frac{\text{1}}{\text{333}\text{.8}\text{K}}\text{-}\frac{\text{1}}{\text{280}\text{.8}\text{K}}\right)\\ \frac{\text{40}\text{.1}\text{mmHg}}{{\text{P}}_{\text{2}}}\text{=}{\text{e}}^{\frac{\text{31}\text{.0}\text{×}{\text{10}}^{\text{3}}\text{J}}{\text{8}\text{.314}\text{J/K}\text{.mol}}\left(\frac{\text{1}}{\text{333}\text{.8}\text{K}}\text{-}\frac{\text{1}}{\text{280}\text{.8}\text{K}}\right)}\\ \text{=}\text{0}\text{.121}\\ {\text{P}}_{\text{2}}\text{=}\frac{\text{40}\text{.1}\text{mmHg}}{\text{0}\text{.121}}\text{=}\text{331}\text{mmHg}\\ \end{array}$

Conclusion

The vapor pressure of benzene at $\text{60}{\text{.6}}^{°}\text{C}$ is calculated.