An ideal mixture of two liquids A and B is put in a cylinder containing piston. Piston is pulled out…

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 evaporated?
  1. 0.22 atm
  2. 0.431 atm
  3. 0.667 atm
  4. 1 atm

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

Asked in: JEE-TOPICTESTS-CHEMISTRY

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