Question

Two liquids $A$ and $B$ form an ideal solution at temperature $T$. When the total vapour pressure above the solution if $600torr$, the amount fraction of $A$ in the vapour phase is $0.40$ and in the liquid phase is $0.6$. What are the vapour pressure of pure $A$ and $B$ at temperature $T$?

Answer 1

Let $P_A^{\circ }$ be vapour pressure of pure $A$ at temperature $T$.

Fraction of A in vapour phase$=\frac{P_A}{P_{total}}$

Let $x_A$ be mole fraction of A in liquid phase.

$\therefore 0.4=\frac{x_A\times P_A^{\circ }}{P_{total}}=\frac{0.6\times P_A^{\circ }}{600}$

$\Rightarrow P_A^{\circ }=400Torr$

Also, $P_{total}=P_A^{\circ }\cdot x_A+P_B^{\circ }\cdot x_B$

$\therefore 600=400\times 0.6+P_B^{\circ }\times (1-0.6)$

$\therefore P_B^{\circ }=\frac{600-240}{0.4}=900Torr$

Hence, $P_A^{\circ }=400Torr,P_B^{\circ }=900Torr$