Week 12 - Ideal Gas Mixtures

Example 1 - Mixing

5 kg of N2 is mixed with 2 kg of an unknown gas. The resulting mixture occupies a volume of 4.0 m3 at 0.2 MPa and 80oC. Both gases and the mixture are ideal gases. Determine (a) the molecular mass of the unknown gas, and (b) the mole fraction of each gas present in the mixture.

Solution:

Given: pM = 0.2 MPa; TM = 80oC; VM = 4.0 m3; mN2 = 5 kg; mu = 2 kg.

The total mass of the mixture of gases is 7 kg. From the ideal gas law, we can find

R = pV/mT = (200 kPa) (4.0 m3)/(7 kg)(353 K) = 0.324 kJ/kg-K

The Molecular mass of the mixture can then be found by dividing the universal ideal gas constant by this number : M = Ru/R = 25.68 kg/kmole. Then, using the statement that the molecular mass of the mixture is equal to the inverse of the sum of each component's mass fraction divided by the molecular mass of that component, we obtain:

M = [(mf/M)N2 + (mf/M)u]-1

The mass fractions are just the ratio of the mass of the component to the total mass...so mfN2 = 0.714, mfu = 0.286. Substituting and solving (MN2 = 28 kg/kmole), we obtain Mu = 21.3 kg/kmole.

(a) Mu = 21.3 kg/kmole

The mole fractions for each component are found from yi = mfi M/Mi. So, the mole fractions are

(b) yN2 = 0.655
yu = 0.345

Example 2 - Work

Methane, hydrogen, and propane, all ideal gases, are mixed together in equal parts by mass to create a new fuel gas. The gas is then adiabatically compressed from 10oC to 0.1 m3 at 2 MPa, 25oC. Determine the work required for the compression process.

Solution:

Given: mfCH4 = mfH2 = mfC3H8 = 0.333. (Equal parts by mass.)
T1 = 10oC, V2 = 0.1 m3, p2 = 2 MPa, T2 = 25oC.

The process is adaiabatic (Q = 0), and we will assume that there is no change in kinetic or potential energy. Therefore, for this closed system, the first law gives

W = -m (u2 - u1) = m cv (T2 - T1)

The constant volume specific heat of the mixture, is the sum of the mass fractions of each components times their specific heat of each component. If we assume that the gases have constant specific heats,

cv = [(mf)(cv)]CH4 + [(mf)(cv)]H2 + [(mf)(cv)]C3H8

cv = (0.333)(1.709) + (0.333)(10.19) + (0.333)(1.502) = 4.46 kJ/kg-K

The molecular mass of the mixture can be found by taking the inverse of the sum of the mass fractions divided by the molecular mass of each component. This yields M = 5.13 kg/kmole. The gas specific ideal gas constant is then R = Ru/M = 1.62 kJ/kg-K. The mass of the mixture can then be found from the ideal gas law at the second state:

m = pV/RT = (2000 kPa)(0.1 m3)/((1.62 kJ/kg-K)(298 K) = 0.414 kg.

The work used to compress the gas is then

W = - (0.414 kg)(4.46 kJ/kg-K)(25 - 10)K = -27.7 kg

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