Question

A solution of liquids A and B having vapour pressure in pure state $P_A^0$ and $P_B^0$. The solution contains 30% mole of A which is in equilibrium with 60% mole of A in vapour phase. If $P_B^0$ is 2 cm, the $P_A^0$ in cm is :

Answer 1

The mole fraction of A in liquid phase is $X_A=\frac{30}{30+70}=0.3$.

The mole fraction of B in the liquid phase is $X_B=1-0.3=0.7$.

In the vapour phase, the mole fraction of A is $Y_A=0.6$.

In the vapor phase, the mole fraction of B is $Y_B=1-0.6=0.4$.

Total pressure P is $P=P_A^0{X}_A+P_B^0{X}_B=P_A^0\times 0.3+2cm\times 0.7=0.3P_A^0+1.4cm$

The partial pressure of B in the vapor is:

$1.4cm=0.4(0.3P_A^0+1.4cm)$

$1.4cm=0.12P_A^o+0.56cm$

$P_A^0=7cm$