Question

In a mixture of A and B having vapour pressure of pure A and pure B are 400 mm Hg and 600 mm Hg respectively, the mole fraction of B in the liquid phase is 0.5. Calculate total vapour pressure and mole fraction of A and B in the vapour phase.

Answer 1

Given:

$$\begin{gathered} \text{The partial pressures of A}=\dot{P_A=400mgHg} \\ \text{The partial pressures of B}=P_B=600mgHg \end{gathered}$$

The mole fraction of B$=X_B=0.5$

The mole fraction of A$=X_A=1-X_B=1-0.5=0.5$

Since, $P_{total}=X_A\cdot {\dot{P}}_A+X_B\cdot {\dot{P}}_B$

$$\begin{gathered} \Rightarrow {\mathrm{P}}_{\mathrm{total}}=0.5\times 400+0.5\times 600 \\ \Rightarrow {\mathrm{P}}_{\mathrm{total}}=500\mathrm{mg}\mathrm{Hg} \end{gathered}$$

Now, the mole fraction of A in vapor $=Y_A=\frac{P_A}{P_{total}}=\frac{0.5\times 400}{500}=0.4$