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Work done in a reversible adiabatic change

  1. Nov 6, 2004 #1
    Can anyone help me with this?

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    Show that the work done in a reversible adiabatic change from an initial state (p1, V1) to a final state (p2, V2) is:

    W = ( p2V2 - p1V1) / (gamma - 1)

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    Here's what (little) I have done so far:

    Reversible => dW = -pdV
    Adiabatic => dq = 0
    gamma = Cp/Cv

    I know pV^gamma = constant, so p1V1^gamma = p2V2^gamma
     
    Last edited: Nov 6, 2004
  2. jcsd
  3. Nov 6, 2004 #2

    Clausius2

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    Science Advisor
    Gold Member

    As you stated, you have P(V) as an adiabatic process.

    The only thing you have to do is to substitute P(V) in the integral of work, and solve the integral.
     
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