Question

Using Henry's Law, solve the following problem:

The vapour pressure of a pure liquid A is 23 kPa at 20 ºC and the Henry’s law constant of B in liquid A is 73 kPa. Calculate the vapour pressure of each component, the total pressure, and the composition of the vapour phase when the mole fraction of B is 0.066 on the assumption that the conditions of the ideal - dilute solution are satisfied at this concentration

Answer 1

Vopour pressure of a liquid in pure form at given temprature is its henry constant

So whrn B mole fraction is 0.066

Then vapour pressure of A

Pa= 23k ( 1- 0.066)

Pa = 21.48 k pa

Vapour pressure of B

Pb = 73k x 0.066 = 4.82k pa

Total pressure = pa + Pb

Pt = 21.48 + 4.82 = 26.3 k pa

In vapour phase

Mole frection of a = Pa/ pt =21.48/26.3 = 0.81

mole frection of B = 1- 0.81 = 0.19

May it helps you., thank you