Natural gas is a mixture of methane, ethane, propane, and butane...as well as ot

Question

Natural gas is a mixture of methane, ethane, propane, and butane...as well as other components. Consider natural gas, modeled as an ideal gas, with an equivalent molecular weight of 23.6 kg/kmol and an equivalent constant specific heat of c_p=2.01 kJ.kg*K. The gas is slowly compressed in a frictionless, adiabatic process from an initial volume of 212cm^3 to a final volume of 98cm^3. The initial pressure is 39 kPa and the initial temperature is 15 degrees Celsius.

a) Find the final temperature and pressure of the gas.

b) Find the work done to compress the gas.

Explanation / Answer

Let 1 be the initial state and 2 be the final state.

P1=39 KPa V1=212 c.c T1=15 degree C = 288 K V2=98 c.c

In an adiabatic process, P.V^r = Constant (where r(gamma)=Specific heat ratio)

Cp = 2.01 KJ/kg.K

R= 8.314 KJ/Kmole.K = (8.314/23.6) KJ/kg.K = 0.3523 KJ/kg.K

Cv = Cp - R = 1.6577 KJ/kg.K

r = Cp/Cv = 1.2125

a)

So, according to adiabatic process relation, P1.V1^r = P2.V2^r

P2 = P1.(V1/V2)^r = (39KPa)*(212/98)^1.2125 = 39*2.5487 = 99.3993 KPa

By ideal gas equation, P1.V1/T1 = P2.V2/T2, hence

T2 = (P2.V2.T1)/(P1.V1) = 339.3137 K

b)

Work done in adiabatic process = W= m.Cp.(T2 - T1)

Here m = n.M

n can be got from ideal gas equation as, n= P1.V1/(R.T1) = (39x 10^3 x 212 x 10^-6)/(8.314 x 288)

= 3.453 x 10^-3 moles = 3.453 x 23.6 x 10^-6 kg = 81.4908 x 10^-6 kg

Work done = 81.4908 x 10^-6 x 2.01 x (339.3137 - 288) = 8.4050 x 10^-3 KJ = 8.4050 J

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