Question

The vapour pressures of pure liquids A and B are 400 mm Hg and 650 mm Hg respectively at 330 K. Find the composition of liquid and vapour if total vapour pressure of solution is 600 mm Hg.

Answer 1

Given : 

\(P_A^0\)= 400 mm Hg; 

\(P_B^0\) = 650 mm Hg, 

P_T = 600 mm Hg, 

T = 330 K; 

x_A = ? 

x_B = ?, 

y₁ = ? 

y₂ = ?

(x is mole fraction in liquid phase while y is mole fraction in vapour phase.)

P_T = (\(P_A^0\) - \(P_B^0\))x_B + \(P_A^0\)

600 = (650 – 400)x_B + 400

= 250x_B + 400

∴ x_B = \(\frac{600-400}{250}\) = 0.8

∵ x_A + x_B = 1

∴ x_A = 1 – x_B = 1 – 0.8 = 0.2

The composition of A and B in liquid mixture is, x_A = 0.2 and x_B = 0.8.

If P₁ and P₂ are vapour pressures (or partial pressures) of A and B in vapour phase then by Raoult’s law,

P₁ = x_A × \(P_A^0\) = 0.2 × 400 = 80 mm Hg

P₂ = x_A ×\(P_B^0\) = 0.8 × 650 = 520 mm Hg

If y₁ and y₂ are mole fractions of A and B respectively in vapour phase then, 

By Dalton’s law,

P₁ = y₁P_T

∴ y₁ = \(\frac{P_1}{P_T}\) = \(\frac{80}{600}\) = 0.1333

P₂ = y₂P_T

∴ y₂ = \(\frac{P_2}{P_T}\) = \(\frac{520}{600}\) = 0.8667

(or y₂ = 1 – y₁ = 1 – 0.1333 = 0.8667)

∴ Composition of liquid : 

x_A = 0.2 and x_B = 0.8

∴ Composition of vapour :

y_A = 0.1333 and y_B = 0.8667