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Liquids A and B form an ideal solution at 300 K; the pure vapour pressures are 700 torr (A) and 420 torr (B). A solution of 1 mol A and 3 mol B is in…

Numerical answer

Liquids A and B form an ideal solution at

300\ \mathrm{K}

; the pure vapour pressures are

700\ \mathrm{torr}

(A) and

420\ \mathrm{torr}

(B). A solution of

1

mol A and

3

mol B is in equilibrium with its vapour. This vapour is completely condensed to a liquid, which is then allowed to equilibrate with its own vapour at

300\ \mathrm{K}

. The total vapour pressure above the condensed liquid, in torr, is

Show answer and solution

Answer: 520

Original liquid:

x_A=0.25

, so

P_A=0.25\times700=175

torr and

P_B=0.75\times420=315

torr, total

490

torr. Vapour mole fraction

y_A=175/490=5/14

. The condensate has this composition (

x_A=5/14

x_B=9/14

). Its vapour pressure:

P=\dfrac{5}{14}(700)+\dfrac{9}{14}(420)=250+270=520

torr.

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