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:

See explanation

Explanation:

Number of moles of each gas is

Nitrogen = 15/28 = 0.536 kmoles

Helium = 5/4  = 1.25 kmoles

Methane = 60/16 = 3.75 kmoles

Ethane =    20/30 = 0.67 kmoles

Total number of moles =  0.536 kmoles + 1.25 kmoles + 3.75 kmoles +  0.67 kmoles = 6.206 kmoles

Mole fraction of each gas;

Nitrogen = 0.536 kmoles/6.206 kmoles = 0.086

Helium = 1.25 kmoles/6.206 kmoles = 0.201

Methane = 3.75 kmoles/6.206 kmoles =0.604

Ethane = 0.67 kmoles/6.206 kmoles =0.108

Partial pressure of each gas;

Nitrogen = 0.086 * 1200 kPa = 103.2 kPa

Helium = 0.201 * 1200 kPa = 241.2 kPa

Methane = 0.604 * 1200 kPa = 724.8 kPa

Ethane = 0.108 * 1200 kPa = 129.6 kPa

Apparent specific heat at constant pressure;

Cp = (0.15 * 1.039) + (0.05 * 5.1926) + ( 0.6 * 2.2537) + (0.2 * 1.7662)

Cp = 2.12 KJ Kg-1 K-1

Cv = Cp- Ru/M

Cv= 2.12 - 8.314/16.12 = 1.604 KJ Kg-1 K-1