Question

The plot of $P_T$ vs $y_A$ at constant temperature in the pressure composition diagram is
$P_A\to \text{Total pressure}$
$y_A\to \text{Mole fraction of A in vapour phase}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Pressure composition phase diagram is the superimpose plot of $P_T$ vs $Y_A$ (curve) and $P_T$ vs $X_A$ (straight line).

From raoult's Law we have:
$p_A=p_A^{\circ }{\chi }_A\Rightarrow {\chi }_A=\frac{p_A}{p_A^{\circ }}....eqn\left(i\right)p_B=p_B^{\circ }{\chi }_B\Rightarrow {\chi }_B=\frac{p_B}{p_B^{\circ }}....eqn\left(ii\right)$
where $p_A,p_B$ are the partial pressures of A and B repsectively
Also by dalton's law , we have :
$p_A=y_A\times P_T{p}_B=y_B\times P_T$

Put the values of $p_A,p_B$ in eqn (i) and eqn (ii)

${\chi }_A=\frac{y_A\times P_T}{p_A^{\circ }}$

${\chi }_B=\frac{y_B\times P_T}{p_B^{\circ }}$

Also ${\chi }_A+{\chi }_B=1$

So, $\frac{1}{P_T}=\frac{y_A}{p_A^{\circ }}+\frac{y_B}{p_B^{\circ }}\frac{1}{P_T}=\frac{y_A}{p_A^{\circ }}+\frac{1-y_A}{p_B^{\circ }}$
Solving,

$P_T=\frac{p_A^{\circ }\times p_B^{\circ }}{p_A^{\circ }+(p_B^{\circ }-p_A^{\circ })\times y_A}$

Thus, Plot of $P_T$ vs $y_A$ is