A gas mixture at 350 K and 300 kPa has the following mole fr

A gas mixture at 350 K and 300 kPa has the following mole fractions: 65% N_2, 20% O_2, and 15% CO_2. Determine the mass fraction and partial pressure of each gas.

Solution

Let total no. of moles = 100

Moles of N2 = 65, Moles of O2 = 20, Moles of CO2 = 15

Molecular mass of N2 = 14*2 = 28

Mol. mass of O2 = 16*2 = 32


mol. mass of CO2 = 12+16*2 = 44

Mass of N2 = 65/28 = 2.32
Mass of O2 = 20/32 = 0.63
Mass of CO2 = 15/44 = 0.34

Total mass = 2.32 + 0.63 + 0.34 = 3.29

Mass fraction of N2 = 2.32/3.29 = 0.71
Mass frac of O2 = 0.63/3.29 = 0.19
Mass frac of CO2 = 0.34/3.29 = 0.1

Ideal gas equation, P = nRT/V

P_N2 = 0.65*8314*350/V
P_O2 = 0.20*8314*350/V
P_CO2 = 0.15*8314*350/V

Sum of partial pressures = total pressure

So, 8314*350/V*(.65+.20+.15) = 300*10^3
V = 9.7 m3

So, P_N2 = .65*8314*350/9.7 = 195 kPa
P_O2 = 0.2*8314*350/9.7 = 60 kPa
P_CO2 = 0.15*8314*350/9.7 = 45 kPa