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Work done in a reversible adiabatic expansion

  1. Oct 18, 2009 #1
    Hi there!

    I have to determine the work done of a reversible adiabatic expansion. Becauce the system is adiabatic: Q = 0 so [tex]\Delta[/tex]U = Wrev

    Becauce both the pressure and the volume changes I can't use W = pex[tex]\Delta[/tex]V.

    1. The problem statement, all variables and given/known data
    Cv, Cp, Ti, Tf, Pi, Pf, Vi, Vf are known

    3. The attempt at a solution

    I thought at first that I could use [tex]\Delta[/tex]U = CV.[tex]\Delta[/tex]T but the volume is not constant so I don't know if I could use it.

    Who can help me find the right formula?

    Thanx in Advance!!
  2. jcsd
  3. Oct 18, 2009 #2


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    Hi Trowa, welcome to PF.

    (1) Because no heat goes in or out of the system, and because the process is reversible, one state variable remains constant. It's one that you haven't listed. What is it?

    (2) [itex]\Delta U =c_V\Delta T=(c_P-R)\Delta T[/itex] always holds for an ideal gas, and doesn't require constant volume, constant pressure, or any other condition.
  4. Oct 18, 2009 #3
    Thanx for the fast response.
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