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

Question

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