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Homework Help: Reversible adiabatic expansion

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm in a rutt for a tutorial question:

    The question is basically to show that the work done during a reversible adiabatic expansion of an ideal gas is

    W = (P1V1 – P2V2)/(1 ‐ γ) ........ Y is gamma

    2. Relevant equations

    3. The attempt at a solution

    I've got so far as to get W= (P1V1^γ – P2V2^γ)
    due to P1V1^γ being constant for a reversible adiabat
    also that W= -PdV

    But i havent a clue how to get the 1-Y at the bottom, me thinks its intergrating for V to get this but my maths is very bad so i dont know how to do this.
  2. jcsd
  3. Apr 15, 2010 #2
  4. Apr 16, 2010 #3

    Andrew Mason

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    Apply the first law: [itex]\Delta Q = \Delta U + W[/itex].

    Since [itex]\Delta Q = 0[/itex] and [itex]\Delta U = nC_v\Delta T[/itex] where [itex]C_v = R/(\gamma - 1)[/itex] you should be able to work it out quickly. (Hint: apply the ideal gas law: PV=nRT).

  5. Apr 16, 2010 #4
    still lost
  6. Apr 16, 2010 #5

    Andrew Mason

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    [tex]\Delta Q = \Delta U + W = 0[/tex]

    (1) [tex]\therefore W = - \Delta U[/tex]


    (2) [tex]\Delta U = nC_v\Delta T[/tex] and

    [tex]C_v = C_p/\gamma = (C_v+R)/\gamma[/tex] so:

    (3) [tex]C_v = R/(\gamma-1)[/tex]

    Substitute (3) into (2) and then just substitute PV for nRT

  7. Apr 17, 2010 #6
    Cheers, i got it
    The way you showed was much easier then the way i was going about it
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