Question

By measuring the equilibrium between liquid and vapour phases of a solution at 30.0 degree C at 1.00 atm, it was found that xA = 0.220 when yA = 0.314. Calculate the activities and activity coefficients of both components in this solution on the Raoult's law basis. The vapour pressures of the pure components at this temperature are: pA = 73.0 kPa and PB 92.1 kPa. (xA is the mole fraction in the liquid and yA the mole fraction in the vapour.)

Answer 1

Answer:The activities of A and B are and 0.44 and 0.76 the activity coefficients are 2 and 0.97 respectively.

Step-by-step explanation:

Given that at equilibrium

Xa = 0.220

Ya = 0.314

Vapour pressure of a =Pa* = 73kPa

Vapour pressure of b = Pb*= 92.1kPa

Total pressure Pt = 1atm= 101.325kPa

From Partial Pressure Law

Pt = Pa+Pb. ..... Eqn(1)

Where Pa and Pb are partial pressures of a and b.

Pa= Ya * Pt

=0.314 * 101.328

Pa = 31.82kPa

From Eqn(1)

Pb = Pt -Pa

= 101.325- 31.82

=69.51kPa

From Raoult's Law

Aa = Pa/Pa*

Where Aa is the activity of component a

= 31.82/73

Aa=0.44

Also Ab is the activity of component b

Ab=69.51/92.1

=0.76

The activity coefficients of each component is given by :

Yi = Ai/Xi

For component a

Ya = Aa/Xa

= 0.44/0.220

=2

For component b

Yb = Ab/Xb

Xb is unknown.

But at equilibrium Xa+Xb =1

Therefore, Xb= 1-0.22 =0.78

Yb= 0.755/0.78

= 0.97