Consider a binary solution of two volatile liquid components 1 and $2 . x_1$ and $y_1$ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as:

• A

  $\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$
• Correct answer
  B

  $\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$
• C

  $\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$
• D

  $\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$

Solution

Given mole fraction of liquid 1 in liquid and vapour phase is $x_1$ and $y_1$ respectively.
$\begin{aligned} & \mathrm{P}_1=\mathrm{P}_{\mathrm{T}} y_1=\mathrm{P}_1^0 x_1 \\ & \frac{\mathrm{P}_{\mathrm{T}}}{x_1}=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0+x_1\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)}{x_1}=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0}{x_1}+\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)=\frac{\mathrm{P}_1^0}{y_1} \\ & \frac{\mathrm{P}_2^0}{x_1}=\frac{\mathrm{P}_1^0}{y_1}-\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right) \\ & \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}\right)-\frac{\left(\mathrm{P}_1^0+\mathrm{P}_2^0\right)}{\mathrm{P}_2^0}\end{aligned}$ $\begin{aligned} & \frac{1}{x_1}=\frac{1}{y_2}\left(\frac{P_1^0}{P_2^0}\right)+\frac{\left(P_2^0-P_1^0\right)}{P_2^0} \\ & y=m x+c \\ & \text { Slope }=\frac{P_1^0}{P_2^0} ; \text { Intercept }=\frac{P_2^0-P_1^0}{P_2^0}\end{aligned}$

Hence, the correct answer is option: 'B'.