devarshi-dt-logo

Question:

A mixture of two miscible liquids A and B is distilled under equilibrium conditions at 1 atm pressure. The mole fraction of A in solution and vapour phase are 0.30 and 0.60 respectively. Assuming ideal behaviour of the solution and the vapour, calculate the ratio of the vapour pressure of pure A to that of pure B.

1.85

4.0

3.5

2.5

Solution:

In solution xA = 0.30; xB = 0.70
In vapour phase, yA = 0.60; yB = 0.40
Using Dalton's law and Raoult's law
yA = 0.60 = PA/PT = PA/(PA + PB) = 0.30poA/(0.30poA + 0.70poB)
yB = 0.70poB/(0.30poA + 0.70poB)
yA/yB = 0.60/0.40 = 0.30poA/0.70poB
poA/poB = 0.60 × 0.70 / 0.40 × 0.30 = 3.5
Option B is correct.