Question

Calculate the mole fraction of ethane (C₂H₆) in water at 10ºC when the ethane concentration in air that is in equilibrium with the water is 10,000 ppm. The total pressure is 5.0 atm.

Answer 1

we will be using Antoine equation and Raoult's Law for solving this problem

T= 10 oC

Total P = 5 atm

conc. of ethane in air =10,000 ppm

1 ppm = 1 gas volume/106 air volume

mole % = Volume % = Pressure %

so,

mole fraction = volume of gas/( volume of gas+ volume of air)

(y) mole fraction of C2H6 in air = 10,000/(10,000+1,000,000) =0.09090909

Raoult's Law

yiP =xiPisat

For this first we need to know Pi sat of ethane at 10oC

for this use antoine equation for ethane

calculating vapour P

   where T in K & P in kPa

ln P=13.88 -1582/(283.15-13.76)

lnP=8.00747318

P=3003.318672 kPa

{as    1 atm= 101.325 kPa}

P=29.640450747594 atm

now solving Raoult's Law

0.09090909*5=xi*29.640450747594

xi =0.01533