Question

Substances a and b are both volatile liquids with p*a = 300 torr, p*b = 250 torr, and kb = 200 torr (concentration expressed in mole fraction). when xa = 0.9, bb = 2.22 mol kg−1, pa = 250 torr, and pb = 25 torr. calculate the activities and activity coefficients of a and
b. use the mole fraction, raoult's law basis system for a and the henry's law basis system (both mole fractions and molalities) for


b.

Answer 1

Calculating activity and activity coefficient:

I. using mole fraction and Raoult's law for a:

Activity α $\alpha a = \frac{Pa}{P*a}$

$\alpha a = \frac{250 torr}{300 torr} = 0.83$

and activity of a = activity coefficient of a x mole fraction of a

αa = γa . xa

γa = αa/xa = 0.83/0.90 = 0.926

II. using the henry's law basis system and:

1. Mole fraction:

αb = Pb/Kb = 25 torr/ 200 torr = 0.13

αb = γb . xb

γb = αb/xb = 0.13/ 0.1 = 1.3

2. Molality:

K*b = Xa.Ma.b0.Kb

(Xa, b0 and Kb are known)

Ma = Xb/Xa.bb

K*b = Xa.Ma.b0.Kb = Xa.Xb.b0.Kb/Xa.bb = Xb.b0.Kb/bb

K*b = (0.1)(1.0 mol/Kg)(200 torr)/ (2.22 mol/Kg) = 9.0 torr

αb = Pb/K*b = 25 torr / 9 torr = 2.8

αb = γb.bb/bo

γb = αb.bo/bb = (1.0 mol/Kg)(2.8) / (2.22 mol/Kg) = 1.3