Question

a liquid isture of A and B is placed in a cyliner and piston arrangement. the piston is slowly pulled out isothermally so vol od liquid decreases and vapour increases. when liquid is negligible, mole fraction of A is 0.4. if P0A is 0.4 ans P0B is 1.2, the totalpressure at which this piwuid id almsor evaporated is

• A. 0.334 atm
• B. 0.667 atm
• C. 1 atm
• D. 2 atm

Answer 1

Answer: (B)

0.667 atm

Answer 2

Let the liquid mole fractions be
x_A for A and x_B = 1 – x_A for B.

According to Raoult’s law, the total pressure is
P_T = x_A·P0_A + (1–x_A)·P0_B
= x_A·0.4 + (1–x_A)·1.2

The vapor phase composition is given by modified Raoult’s law (using Dalton’s law):
y_A = (x_A·P0_A)⁄P_T = 0.4

Substitute P_T:
x_A·0.4 ⁄ [x_A·0.4 + (1–x_A)·1.2] = 0.4

Solve for x_A:
0.4·x_A = 0.4·[0.4·x_A + 1.2(1–x_A)]
0.4·x_A = 0.16·x_A + 0.48 – 0.48·x_A
0.4·x_A = 0.48 – 0.32·x_A
0.4·x_A + 0.32·x_A = 0.48
0.72·x_A = 0.48
x_A = 0.48⁄0.72 = 2⁄3 ≈ 0.667

Now, calculate the total pressure:
P_T = (2⁄3)·0.4 + (1⁄3)·1.2
= 0.2667 + 0.4 = 0.667 atm

Thus, the total pressure when the liquid is almost evaporated is 0.667 atm.