Question

The dew-point temperature of a gas mixture with a molar composition of 40.0% n-hexane, 25.0% n-heptane, and 35.0% air at 1.00 atm.

Answer 1

Given: Mole fraction of n-hexane in vapor phase, y₁=40%=0.40

Mole fraction of n-heptane in vapor phase, y₂=25%=0.25

Pressure of the gas mixture, P=1 atm=760 mmHg

Method 1:

Assuming ideal gas mixture, the Raoult's law is

where x_i is the Mole fraction of component in liquid phase

y_i is the Mole fraction of component in vapor phase

P is the Pressure

p_i* is the Vapor pressure

T_dp is the Dew point temperature

At dew point temperature, the sum of liquid phase mole fraction is equal to 1.

Vapor pressure is otained using Antoine equation

where A, B and C are Antoine constants

T is the Temperature

Antoine constants for n-hexane and n-heptane is obtained from Antoine Equation Constants table.

For n-hexane, A=6.88555, B=1175.817 and C=224.867

For n-heptane, A=6.90253, B=1267.828 and C=216.823

Solving the above expression using a solver

Method 2:Dew Point Temperature by trial and error method

Consider the expression

A trial and error method is used to obtain the dew point temperature until f(T_dp) is approximately equal to zero.

T (0C)	f(Tdp)
50	1.092
65	0.200
67	0.119
69	0.045
70	0.010
71	-0.023

Interpolate the temperature 70⁰C and 71⁰C to obtain f(T_dp)=0.

f(T_dp)=0 at T=70.30⁰C

Thus,Dew point temperature is T_dp=70.3⁰C