Chemistry - Liquid Solution Question with Solution | TestHub

Given mole fraction of liquid 1 in liquid and vapour phase is $x_1$ and $y_1$ respectively.

$\begin{array}{rl}& {\mathrm{P}}_1={\mathrm{P}}_{\mathrm{T}}{y}_1={\mathrm{P}}_1^0{x}_1\\ & \frac{{\mathrm{P}}_{\mathrm{T}}}{x_1}=\frac{{\mathrm{P}}_1^0}{y_1}\\ & \frac{{\mathrm{P}}_2^0+x_1\left({\mathrm{P}}_1^0+{\mathrm{P}}_2^0\right)}{x_1}=\frac{{\mathrm{P}}_1^0}{y_1}\\ & \frac{{\mathrm{P}}_2^0}{x_1}+\left({\mathrm{P}}_1^0+{\mathrm{P}}_2^0\right)=\frac{{\mathrm{P}}_1^0}{y_1}\\ & \frac{{\mathrm{P}}_2^0}{x_1}=\frac{{\mathrm{P}}_1^0}{y_1}-\left({\mathrm{P}}_1^0+{\mathrm{P}}_2^0\right)\\ & \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{{\mathrm{P}}_1^0}{{\mathrm{P}}_2^0}\right)-\frac{\left({\mathrm{P}}_1^0+{\mathrm{P}}_2^0\right)}{{\mathrm{P}}_2^0}\end{array}$
$\begin{array}{rl}& \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{P_1^0}{P_2^0}\right)+\frac{\left(P_2^0-P_1^0\right)}{P_2^0}\\ & y=mx+c\\ & \text{ Slope }=\frac{P_1^0}{P_2^0};\text{ Intercept }=\frac{P_2^0-P_1^0}{P_2^0}\end{array}$