twenty pounds of a gas mixture is contained in a tank at 30

twenty pounds of a gas mixture is contained in a tank at 30 psia and 70 F. The gas mixture has volumetric analysis of twenty percent carbon dioxide, thirty percent oxygen, and fifty percent nitrogen. (a) Using constant-specific heat theory, find the final temperature if 400 Btu of heat is added to the gas mixture

Solution

Volumetric analysis = molar analysis

CO2 = 20%

O2 = 30%

N2 = 50%

The mass of CO2 = 0.2*44 = 8.8

The mass of O2 = 0.30*32 =9.6

The mass of N2 = 0.5*28 = 14

The total mass of the mixure = 32.4

The mass fraction of CO2 = 8.8/32.4 = 27.16%

The mass fraction of O2 = 9.6/32.4 = 29.63%

The mass fraction of N2 = 14/32.4 = 43.21%

Mass of the mixture = 20 pounds = 9.072 kg

Cp of CO2 = 0.846kJ/kg/K

Cp of O2 =0.918 kJ/kg/K

Cp of N2 = 1.039 kJ/kg/K

The Cp of mixture is

Cp = 0.2716*0.846+0.2963*0.918+0.4321*1.039

Cp = 0.951 kJ/kg/K

Q = 400 Btu = 422.022 kJ

T1 = 70 F = 294.261 K

Therefore

Q = mCp*(T2-T1)

422.022 = 9.072*0.951*(T2-294.261)

T2 = 343.191

T2 = 158 F