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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

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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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.

    ehild
     
    Last edited: Nov 2, 2011
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