Question

A 1.000-L vessel contains a mixture of hydrogen, carbon monoxide and methane gases in unknown proportions at 23.3-bar and 25 °C. The gaseous mixture is found to have a mass of 13.70-g and a higher heating value of 494.90 kJ/mol. What is the composition of the gas?

Answer 1

Volume V = 1 L = 0.001 m³

Pressure P = 23.3 bar = 2330 kpa

Temperature = 25 °C = 25+273.15 = 298.15 K

Mass M = 13.7 g = 0.0137 kg

Assume this as ideal mixture, for ideal gas mole%=vol%

PV=nRT

where n = M/M_w

n - no of moles

M_w - Moleculat weight of mixture

R - gas constant 8.314 kJ/kmol.K

M_w = MRT/PV

=0.0137*8.314*298.15/2330*0.001

M_w=14.575 g/gmol

M_w= x_h2*Mw_h2 + x_co*Mw_co + x_ch4*Mw_ch4

14.575 = x_h2*2 + x_co*28 + x_ch4*16   -------- (1)

x_h2 - mole fraction of hydrogen

Mw_h2 - Molecular weight of hydrogen, 2

x_co - mole fraction of carbonmonoxide

Mw_co - molecular weight of carbonmonoxide, 28

x_ch4 - mole faction of methane

Mw_ch4 - molecular weight of methane, 16

Higher heating value gas mixture HHV_mix = x_h2*HHV_h2 + x_co*HHV_co + x_ch4*HHV_ch4

HHV_h2 higher heating value of H₂ = 284.162 kJ/mol

HHV_co higher heating value of CO = 284.479 kJ/mol

HHV_ch4 higher heating value of CH₄ = 886.15 kJ/mol

494.9 = x_h2*284.162 + x_co*284.479 + x_ch4*886.15 ---------- (2)

we know that sum of the fraction is equal to 1, so

1 = x_h2 + x_co+ x_ch4 --------(3)

Solving (1) , (2) & (3)

we get

x_h2 = 0.3548

x_co = 0.2952

x_ch4 = 0.35