Question

the vapour pressure of two pure liquids A and B are 200 and 400 Torr respectively at 300K. A liquid solution (ideal) of A and B for which the mole fraction of A is 0.40 is contained in a cylinder. The composition of components A and B in vapour phase after equilibrium is reached respectively is ​

Answer 1

Answer:

Based on Raoult's law:

$p=x_{a} p_{a} ^{0} +x_{b} p_{b} ^{0}$

p=(0.40)200+(1-0.40)400

$\frac{1}{p} =\frac{x_{a}^{'}  }{p_{a}^{0}  } +\frac{x_{b}^{'}  }{p_{b}^{0}  }$

$\frac{1}{320} =\frac{x_{a} ^{'} }{200} +\frac{1-x_{a}^{'}  }{400}$

$\frac{1}{320} =\frac{400x_{a}^{'}+200(1-x_{a} ^{'})   }{80000}$

$250=400x_{a} ^{'} +200-200x_{a} ^{'}$

$250=200+200x_{a} ^{'}$

$50=200x_{a} ^{'}$

$x_{a} ^{'} =0.25$

$x_{b} ^{'} =1-0.25=0.75$

Answer 2

Dear Students,

◆ Answer -

In vapour phase after equilibrium -

X'A = 0.25

X'B = 0.75

● Explanation -

# Given -

p°A = 200 torr

p°B = 400 torr

XA = 0.4

# Solution -

Mole fraction of B in solution -

XB = 1 - XA

XB = 1 - 0.4

XB = 0.6

Partial pressure of A in solution -

pA = XA.p°A

pA = 0.4 × 200

pA = 80 torr

Partial pressure of B in solution -

pB = XB.p°B

pB = 0.6 × 400

pB = 240 torr

Mole fraction of A in vapor -

X'A = pA / (pA+pB)

X'A = 80 / (80+240)

X'A = 0.25

Mole fraction of B in vapor -

X'B = 1 - X'A

X'B = 1 - 0.25

X'B = 0.75

Hope this helped you.