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The work done by an adiabatic process?

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    On my formula sheet I have this:
    W=(C/1-y)(Vf^(1-y)-Vi^(1-y))
    y=Cp/Cv

    2. Relevant equations



    3. The attempt at a solution

    I am confused about what the C in C/1-y stands for. My textbook does not even have this formula and I realized that C is not Cp or Cv...
     
  2. jcsd
  3. Oct 11, 2011 #2

    Andrew Mason

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    The C is usually expressed as K as in: [itex]PV^\gamma = K[/itex]. So it can be written:

    [tex]W = \frac{P_iV_i^\gamma}{1-\gamma}(V_f^{1-\gamma} - V_i^{1-\gamma})[/tex]

    AM
     
  4. Oct 11, 2011 #3
    Ahh ok thanks! But why is it Vi and Pi, why not Vf and Pf? Is there a reason?
     
  5. Oct 11, 2011 #4

    Andrew Mason

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    You can make it Vf and Pf or any V, P during the adiabatic process if you like:

    [tex]P_iV_i^\gamma = P_fV_f^\gamma = PV^\gamma = K = \text{ constant}[/tex]

    AM
     
  6. Oct 11, 2011 #5
    Okay, so for an adiabatic process, we cannot use the typical ideal gas law P1V1=P2V2 right?

    What about PV/T=PV/T?
     
  7. Oct 12, 2011 #6

    Andrew Mason

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    If PV=nRT and PV = constant then T would be constant (assuming n is constant). But in an adiabatic process that does work this cannot be the case: dQ = 0 -> dU = -dW where dW is the work done by the gas. So P1V1 ≠ P2V2.

    In a reversible adiabatic process involving an ideal gas, the condition [itex]PV^\gamma = K[/itex] holds. But so does PV=nRT. Substituting P = nRT/V the adiabatic condition becomes:

    [tex]PV^\gamma = nRTV^{\gamma-1} = K[/tex]

    and since n and R are constant (assuming no change in the amount of gas) we have:

    [tex]TV^{\gamma-1} = K/nR = K' = \text{constant}[/tex] and

    [tex]PV/T = nR = \text{constant}[/tex]

    AM
     
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