Question

The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane with a total mixture molecular weight of 16.12 kg/kmole. Determine the mole fraction of each constituent, the partial pressure of each constituent when the mixture pressure is 1200 kPa and the apparent specific heats of the mixture when the mixture is at room temperature.

Answer 1

Answer:

Explanation:

mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20

mole fraction  N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30

mole fraction  N₂ : He : CH₄ : C₂H₆ : : .5357  : 1.25 : 3.75 : .67

Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057

mole fraction of N₂ =  .5357 / 6.2057 = .0877

mole fraction of He = 1.25 / 6.2057 = .20

mole fraction of CH₄ = 3.75 / 6.2057 = .6043

mole fraction of C₂H₆ = .67 / 6.2057 = .108

Partial pressure = total pressure x mole fraction

Partial pressure of N₂ = 1200 kPa  x .0877 = 105.24 kPa

Partial pressure of He = 1200 kPa  x .20  = 240 kPa

Partial pressure of CH₄ = 1200 kPa  x  .6043  = 725.16 kPa

Partial pressure of C₂H₆ = 1200 kPa  x .108    = 129.6 kPa