Question

The plots of $\frac{1}{X_A}vs.\frac{1}{Y_A}$ (where, $X_A$ and $Y_A$ are the mole fraction of liquid A in liquid and vapour phase respectively) is linear with slope and intercepts respectively

• A. $P_A^0/P_B^0$ and $\frac{(P_A^0-P_B^0)}{P_B^0}$
• B. $P_A^0/P_B^0$ and $\frac{(P_B^0-P_A^0)}{P_B^0}$
• C. $P_B^0/P_A^0$ and $\frac{(P_A^0-P_B^0)}{P_B^0}$
• D. $P_B^0/P_A^0$ and $\frac{(P_B^0-P_A^0)}{P_B^0}$

Answer 1

The correct option is B $P_A^0/P_B^0$ and $\frac{(P_B^0-P_A^0)}{P_B^0}$

The plots of $\frac{1}{X_A}vs.\frac{1}{Y_A}$ (where, $X_A$ and $Y_A$ are the mole fraction of liquid A in liquid and vapour phase respectively) is linear with slope and intercepts equal to $P_A^0/P_B^0$ and $\frac{(P_B^0-P_A^0)}{P_B^0}$ respectively.

$P_A^{\prime }=P_A^0{X}_A$ and $P_B^{\prime }=P_B^0{X}_B$
$P_A^{\prime }=P_M{Y}_A$ and $P_B^{\prime }=P_M{Y}_B$
$\therefore \frac{P_A}{Y_A}=\frac{P_B}{Y_B}$
or
$\frac{P_A^0{X}_A}{Y_A}=\frac{P_B^0{X}_B}{Y_B}=\frac{P_B^0(1-X_A)}{(1-Y_A)}$
or
$\frac{P_B^0}{X_A}=\frac{P_A^0}{Y_A}+(P_B^0-P_A^0)$
or
$\frac{1}{X_A}=\frac{1}{Y_A}\cdot \frac{P_A^0}{P_B^0}+\frac{(P_B^0-P_A^0)}{P_B^0}$
or
$y=mx+C$
$\therefore Slope=m=\frac{P_A^0}{P_B^0}$ and intercept $C=\frac{(P_B^0-P_A^0)}{P_B^0}$.