Chapter 4, Problem 4D.12E

Interpretation Introduction

Interpretation:

The composition of the liquid mixture that boils at ${25}^{o}C$ when the pressure is 0.50 atm has to be determined and the composition of the vapour produced should be identified.

Concept introduction:

According to the Dalton’s law of partial pressure, the total pressure of a mixture of non-reacting gases in a system is the sum if their partial pressure.

$\begin{array}{c}{P}_{}=P_A+P_B+P_C+\mathrm{.......}+P_n\\ =x_A{P}_A^o+\left(1-x_A\right)P_B^o\\ \\ x_A=\frac{P-P_B^o}{P_A^o-P_B^o}\boxed{\begin{array}{l}{P}_A^{o}and{P}_B^o:vapourpressureofaAandBiftheywere\\ inpurestate.\\ x_A:MolefractionofA\end{array}}\end{array}$

The composition of vapour pressure can be determined using the equation,

$\begin{array}{l}{X}_A=\frac{x_A{P}_A^o}{P_B^o+\left(P_A^o-P_B^o\right)x_A}\boxed{\begin{array}{l}{P}_A^{o}and{P}_B^o:vapourpressureofaAandBiftheywere\\ inpurestate.\\ x_A:MolefractionofA\\ X_A:VapourCompositionofA\end{array}}\\ \end{array}$

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

Explanation of Solution

Given,

At ${90}^{o}C$ the vapour pressure of toluene is 53 kPa and that o-xylene is 20 kPa. The composition of the liquid mixture that boils at ${25}^{o}C$ when the pressure is 0.50 atm can be determined as follows,

$x_A=\frac{P-P_B^o}{P_A^o-P_B^o}\boxed{\begin{array}{l}{P}_A^{o}and{P}_B^o:vapourpressureofaAandBiftheywere\\ inpurestate.\\ x_A:MolefractionofA\end{array}}$

Substituting the known values,

$\begin{array}{c}{x}_A=\frac{50.66-20}{53-20}\\ =0.929\end{array}$

The composition of vapour pressure can be determined using the equation,

$\begin{array}{l}{X}_A=\frac{x_A{P}_A^o}{P_B^o+\left(P_A^o-P_B^o\right)x_A}\boxed{\begin{array}{l}{P}_A^{o}and{P}_B^o:vapourpressureofaAandBiftheywere\\ inpurestate.\\ x_A:MolefractionofA\\ X_A:VapourCompositionofA\end{array}}\\ \end{array}$

Substituting the known values,

$\begin{array}{c}{X}_A=\frac{0.929\times 53}{20+\left(53-20\right)\times 0.929}\\ =0.9719\\ X_B=1-0.9719\\ =0.0281\end{array}$

The compositions of vapour pressures of toluene and o-xylene at the given temperatures are, 0.9719 and 0.0281 respectively.