1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Nov 1, 2011 #1
    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

    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.
  2. jcsd
  3. Nov 2, 2011 #2


    User Avatar
    Homework Helper

    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 WG=PdV. That of an external force is -PdV. So you determine the work of the external agent.

    Nice derivation!:smile:

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

    Last edited: Nov 2, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook