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Thermal - derive a work equation

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Show how W= (P2V2 - P1V1) / [itex]\gamma[/itex] -1

    can be derived using relations between PVgamma = constant, and W(1 to 2) = -[itex]\int[/itex] P(T,V) dV (from v1 to v2).


    2. Relevant equations
    I think we can use R = Cv ([itex]\gamma[/itex] - 1)


    3. The attempt at a solution
    Not sure how to start. The integral would be W = [itex]\int[/itex] P2V2 - P1V1 dV, but that means we'd have to do partial differential equations, and the problem is not meant to be so difficult.

    Any suggestions?
     
  2. jcsd
  3. Sep 11, 2011 #2

    Redbelly98

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    Actually, the formula for work is [itex]W = \int P \ dV[/itex]. You can use [itex]\ P \ V^{\gamma} = \mbox{constant} \ [/itex] in that integral.
     
  4. Sep 12, 2011 #3
    What exactly does it mean for PV[itex]\gamma[/itex] to equal a constant? It means that P and V are inversely proportional, right? I'm not sure what the [itex]\gamma[/itex] as an exponent of the V being constant means though.
     
  5. Sep 12, 2011 #4

    Redbelly98

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    Only if γ is equal to 1. Inversely proportional would mean that PV=PV1=constant.

    But since γ is not 1, P and V are not inversely proportional. So we have to leave it as
    [tex] P \ V^{\ \gamma} = \mbox{constant}[/tex]
    You can use that relation to substitude for P in the integral used to calculate W. If you do that substitution, then the integrand will be in terms of V and constants, and can be integrated.

    γ is just an exponent, the relation involving P and V γ is just how the relation between P and V is expressed for an ideal gas undergoing an adiabatic process.
     
  6. Sep 13, 2011 #5
    All right, by taking P to be constant and substituting PV = nRT, here's where I'm at:
    [itex] W = \int P dV = \int \frac{nRT}{V} dV = nR \int \frac{T}{V} dV [/itex]

    T is in the equation, so it's not just V that I can take an integral of.
    Did I have a wrong step?
     
  7. Sep 14, 2011 #6
    Actually, I'm sure I can substitute something into T in terms of V. I'll be back.
     
  8. Sep 14, 2011 #7

    Redbelly98

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    P is not constant. [itex] PV^{\ \gamma} [/itex] is a constant -- call it k, if you wish, and solve for P:
    [tex]PV^{\ \gamma} = k[/tex]
    Therefore,
    P = ____ ?​
     
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