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` atm 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?

Options

A. 0.22 atm
B. 0.431 atm
C. 0.667 atm
D. 1 atm

Answer

C

Explanation

Mole fraction in vapour phase,
`Y_A = 0.4, Y_B = 0.6`
Given, `P_A^@ = 0.4 , P_B^@ = 1.2`
`P_T = P_A^@x_A+ P_B^@x_B` ...(i)
`Y_A = 0.4 = (P_A^@x_A)/(P_T) implies (0.4x_A)/(P_T) = 0.4`
or `P_T = x_A` ...(ii)
`Y_B = 0.4 = (P_B^@x_B)/(P_T) implies (0.2x_A)/(P_T) = 0.6`
or `P_T = 2x_B` ... (iii)
Comparing (ii) and (iii),
`x_A= 2x_B`
We know, `x_A+x_B= 1 implies x_B =1/3 and x_A = 2/3`
Putting in Eq. (i)
`P_T = 0.4xx2/3+1.2 xx 1/3 = 0.667` atm