A mixture of ideal gases has the following molar analysis CO

A mixture of ideal gases has the following molar analysis:

CO2 0.12

O2 0.04

N2 0.82

CO 0.02

Determine the analysis of this mixture on a mass basis, determine the molecular mass of the mixture and the gas constant for the mixture (on a mass basis). Ans. 30.08; 0.276 kJ/(kg-K)

Solution

Moles of CO2 = 0.12

Mass of CO2 = 0.12*44 = 5.28 g

Moles of O2 = 0.04

Mass of O2 = 0.04*32 = 1.28 g

Moles of N2 = 0.82

Mass of N2 = 0.82*28 = 22.96 g

Moles of CO = 0.02

Mass of CO = 0.02*28 = 0.56 g

Total mass of the mixture = 5.28 + 1.28 + 22.96 +0.56 = 30.08 g

Mass fraction of CO2 = 5.28/30.08 =0.176 (R= 0.189)

Mass fraction of O2 = 1.28/30.08 = 0.043 (R = 0.260)

Mass fraction of N2 = 22.96/30.08 = 0.763 (R = 0.297)

Mass fraction of CO = 0.56/30.08 =0.019 (R= 0.297)

Therefore R of mixture

R = 0.176*0.189 + 0.043*0.260 + 0.763*0.297 + 0.019*0.297

R = 0.276 kJ/kg-K