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Proving PV^gamma = Constant

  1. Oct 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Proving PV^gamma = constant In adiabatic expansion. Q = 0

    2. Relevant equations
    N/A

    3. The attempt at a solution

    ΔEint = W

    nCvdT = PdV = nRT / V dV

    ∫Cv/T dT = ∫R/V dV

    Cv ln(T2/T1) = R ln(V2/V1)

    ln(T2/T1) = (R/Cv) ln(V2/V1)

    T2/T1 = (V2/V1)R⋅gamma / Cp

    P2V2 / P1V1 = (V2/V1)R⋅gamma / Cp

    I did some extra rearranging after this point, but it appears that the equation in this form cannot get PV^gamma = constant or for P1V1^gamma = P2V2^gamma. Is there a mistake somewhere?

    Thanks.
     
    Last edited: Oct 29, 2016
  2. jcsd
  3. Oct 29, 2016 #2

    TSny

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    Watch the signs. Does W stand for the work done by the gas or the work done on the gas?
     
  4. Oct 29, 2016 #3
    Well it is the differentiated form of work, but I guess that doesn't matter since the rate of change of work done on the gas is negative right?

    I'll try adding a negative sign to that and see if it works. Thanks
     
  5. Oct 29, 2016 #4

    TSny

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    To see if dT and dV have the same sign, think about what happens to the temperature during an adiabatic expansion.
     
  6. Oct 29, 2016 #5
    Okay, so temperature will decrease as energy is lost in the form of work on the environment due to expansion of volume right?
     
  7. Oct 29, 2016 #6

    TSny

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    Yes.

    Some people write the first law for an infinitesimal reversible step as dE = dQ - dWby where dWby is the work done by the gas: dWby = +PdV.

    Others write it as dE = dQ + dWon where dWon is the work done on the gas: dWon = - PdV

    Either way, dE = dQ - PdV.
     
  8. Oct 29, 2016 #7
    Alright, I got this to work correctly, I had to use the Cv + R identity which cancelled out the Cp in the denominator. However this gave me P1V1^gamma = P2V2^gamma. Would it be possible to rearrange this somehow to make just one side equal to a constant?
     
  9. Oct 29, 2016 #8

    TSny

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    State 2 can be thought of as any state during the adiabatic process. So for any state (P, V) during the process, you have P V γ = P1V1γ.
     
  10. Oct 29, 2016 #9
    Yeah, it's just that I saw a proof somewhere where they got PVgamma = econstant which ended up being PVgamma = constant. Anyways, it's pretty much the same result.

    Thanks for the help
     
  11. Oct 29, 2016 #10

    TSny

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    Instead of doing definite integrals from state 1 to state 2, you could do indefinite integrals. This will lead to P V γ = const where the const is related to the constants of integration of the indefinite integrals.
     
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