Proof of work done by an ideal gas in a quasi-static adiabatic expansion

jrklx250s

1. The problem statement, all variables and given/known data
Prove that the work done by an ideal gas with constant heat capacities during a quasi-static adiabatic expansion is equal to

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

where Y = gamma, which is heat capacity at constant pressure over heat capacity at constant volume


2. Relevant equations



3. The attempt at a solution
Alright so this is my attempt and im not sure where to go from here...

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)]

Any suggestions and help would be greatly appreciative.

ehild

As this is an expansion, and Q=0, the work of the gas is positive. The formula results in negative work, as Pi/Pf>1 and (γ-1)/γ >0, so it should be the external work.

The elementary work of gas is W_G=PdV. That of an external force is -PdV. So you determine the work of the external agent.

Nice derivation!

Factor out PfVf and use that PiVi^γ=PfVf^γ to replace Vi/Vf
by (Pf/Pi)^1/γ.

ehild