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 idea

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

TestHub · Accepted Answer

In solution, x A ​ = 0.30 ; x B ​ = 0.70 In vapour phase, x A ′ ​ = 0.60 ; x B ′ ​ = 0.40 Using Dalton's law and Raoult's law ​ x A ′ ​ = 0.60 = P p A ​ ​ = p A ​ + p B ​ p A ​ ​ = 0.30 p A ∘ ​ + 0.70 p B ∘ ​ 0.30 p A ∘ ​ ​ x B ′ ​ = 0.40 = P p B ​ ​ = p A ​ + p B ​ p B ​ ​ = 0.30 p A ∘ ​ + 0.70 p B ∘ ​ 0.70 p B ∘ ​ ​ x B ′ ​ x A ′ ​ ​ = 0.40 0.60 ​ = 0.70 p B ∘ ​ 0.30 p A ∘ ​ ​ p B ∘ ​ p A ∘ ​ ​ = 0.40 × 0.30 0.60 × 0.70 ​ = 2 7 ​ = 3.5 ​

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