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Prove that work done by an ideal gas in an adiabatic process is W =[P(initial)*V(initial)- P(final)*V(final)]/(gamma-1), using integration [integral of PdV]

Prove that work done by an ideal gas in an adiabatic process is W =[P(initial)*V(initial)- P(final)*V(final)]/(gamma-1), using integration [integral of PdV]

Grade:11

1 Answers

Aman Bansal
592 Points
9 years ago

Dear Arnab,

In an adiabatic quasi-static process we can write the formula

PV^Y = constant

constant = K for simplification

Since its adiabatic no heat change so Q=0

Using the first law of thermo

Q= ΔU -W

We know that W = -PdV
and P= K/V^Y

so...
W = ΔU
W = -PdV
W = -(K/V^Y)*dV
W = -K∫(1/V^Y)*dV
W = -K[V^(1-Y)/(1-Y)]*∫dV
W = -(K/(1-Y))[Vf^(1-Y) - Vi^(1-Y)]
W = -(K/(1-Y))[Vf^(-Y)*Vf - Vi^(-Y)*Vi]
W = -(1/(1-Y))[((Vf*K)/(Vf^Y)) - ((Vi*K)/(Vi^Y))]

since Pi = K/Vi^Y and Pf = K/Vf^Y sub those in

W = -(1/(1-Y))(Vf*Pf - Vi*Pi)
Times this by (-1/-1)

and we get

W = (PfVf - PiVi)/(Y-1)

This is where I get to not sure where to go from here to make this into

W= (PfVf)/(Y-1)[1 - (Pi/Pf)^((Y-1)/Y)]

Best Of Luck...!!!!

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Thanks

Aman Bansal

Askiitian Expert

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