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Reversible adiabatic expansion proof

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove the relationship between the pressure, P, and the temperature, T, for an ideal gas with a reversible adiabatic expansion. Base the proof on the first law of thermodynamics and the ideal gas law.

    The relationship is: T^(Cp,m/R)/P = constant

    Where R is the gas constant and Cp,m is the molar heat capacity.

    2. Relevant equations
    ΔU=Q-W
    pV=nRT

    Possibly relevant:
    Wrev=-pdV
    γ=Cp,m/Cv,m where γ is the heat capacity ratio.
    PVγ= constant

    3. The attempt at a solution
    I dont even know where to start explaining what I have tried... I have been kicking around a lot of things. I cant seem to manage to manipulate things so that temperature is to the power of anything, although I have burnt a lot of time trying to turn PVγ into some relation for T to the power of some derivative of gamma. I assume since gamma has Cp,m in it, that it is going to play in, and especially since it is already a power for volume, this flagged me in that direction, but I am just going around in circles.

    Any help in the right direction would be GREATLY appreciated.
     
  2. jcsd
  3. Jan 28, 2015 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Start with PVγ= constant and substitute nRT/P for V.

    AM
     
  4. Jan 28, 2015 #3
    Ok, so I end up with:
    P*(nRT/P)γ=c

    Which I then turn into:
    P*(nRT)γ/Pγ=c

    And then:
    (nRT)γ/Pγ-1=c

    Am I on the right track? Because I have been playing around with it, and I cant seem to extract nR out, nor am I having luck manipulating the powers to what I need.
     
  5. Jan 28, 2015 #4
    If you are supposed to use the first law, then you are expected to start with

    ##dU=nCvdT=-PdV##
     
  6. Jan 28, 2015 #5
    I finally got it nailed down. Chester was correct, the missing piece here was that dU=Cv,mdT=-PdV. An integration of that, and some minor modifying of terms produced what I was after. Thank you!
     
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