An ideal mixture of two liquids A and B is put in a cylinder containing piston. Piston is pulled out isothermally so that the volume of liquid decreases but that of vapours increases. Negligibly small amount of liquid was left and mole-fraction of A in vapour is 0.4. If P A ∘ = 0.4 and P B ∘ = 1.2 atm at the experimental temperature, which of the following is the total pressure at which the liquid is almost evaporate

Step-by-step solution

Mole-fraction in vapour phase, Y for A Y A = 0.4 , Y B = 0.6 P A ∘ = 0.4 P B ∘ = 1.2 given P T = P A ∘ x A + P B ∘ x B ...(i) Y A = 0.4 = P A ∘ x A P T ⇒ 0.4 x A P T = 0.4 or P T = x A ...(ii) Y B = 0.6 = P B ∘ x B P T ⇒ 1.2 x B P T = 0.6 or P T = 2 x B ...(iii) Comparing (ii) and (iii), x A = 2 x B we know, x A + x B = 1 Hence, x B = 1 3 and x A = 2 3 Putting in Equation (i) P T = 0.4 × 2 3 + 1.2 × 1 3 = 0.667 atm