Question

The plots of $\frac{1}{X_{A}}$ (on $y$-axis) $v s \frac{1}{Y_{A}}$ (on $x$-axis) (where $X_{A}$ and $Y_{A}$ are the mole fractions of liquid $A$ in liquid and vapour phase respectively) is linear with slope and y-intercept respectively.

• A. $\frac{P_{A}^{0}}{P_{B}^{0}}$ and $\frac{\left(P_{A}^{0}-P_{B}^{0}\right)}{P_{B}^{0}}$
• B. $\frac{P_{B}^{0}}{P_{A}^{0}}$ and $\frac{\left(P_{A}^{0}-P_{B}^{0}\right)}{P_{B}^{0}}$
• C. $\frac{P_{A}^{0}}{P_{B}^{0}}$ and $\frac{\left(P_{B}^{0}-P_{A}^{0}\right)}{P_{B}^{0}}$
• D. $\frac{P_{B}^{0}}{P_{A}^{0}}$ and $\frac{\left(P_{B}^{0}-P_{A}^{0}\right)}{P_{B}^{0}}$

Answer 1

Answer: (C)

$P = P _{A}{ }^{\circ} X _{A}+ P _{ B }{ }^{\circ}\left(1- X _{\Lambda}\right)$ and
$P _{ A }{ }^{\prime} X _{ A }= Y _{ A } P = Y _{ A }\left[ P _{ A }{ }^{\circ} X _{ A }+ P _{ B }{ }^{\circ}\left(1- X _{ A }\right)\right]$
so, $\frac{1}{ Y _{ A }}=1+\frac{ P _{ B }^{\circ}}{ P _{ A }^{\circ}}\left(\frac{1}{ X _{ A }}-1\right) $
so, $x =1+\frac{ P _{ B }^{\circ}}{ P _{ A }^{\circ}}( y -1)$
Hence $\Rightarrow (x-1) \frac{P_{B}^{\circ}}{P_{A}^{\circ}}+1=y $
so, $y=m x+C$ gives the result