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IRREVERSIBLE adiabatic work

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    So I came down with something and missed some class and am making it up from others notes which arent entirely clear plus the long weekend for turkey day and Im really hoping someone can help me make some sense of this...

    I am trying to come up with the derivations for irreversible adiabatic expansions/work, ill try to explain as much as possible and hopefully it will be sufficient for someone to see what im getting at give me a hand






    2. Relevant equations

    The notes begin with adiabatic reversible work
    Du=Cv,mdT
    dU=dq+dW -->q=0 for adiabatic so dU=dW
    pex=Pf=nRT/V
    ∴nCv,mdT=-nRTdV/V

    the notes then integrate each side wrt T & V end then exponentiate each side to arrive at v1T1c=v2T2c where c =Cv,m/R

    they then get sloppy to where i cant make anything of it and arrive at Pivi[itex]\gamma[/itex]Pfvf[itex]\gamma[/itex] where[itex]\gamma[/itex]=Cpm/Cvm


    Then there is a huge note that says DO NOT USE THIS IF WORK IS IRREV and that the instructor wants us to derive similar for irreversible work Hint: (it changes where dU=dw)

    3. The attempt at a solution

    I tried working on this all morning and didnt make any progress...integrating dV/V to get the ln(vf/vi) is how we were taught to do reversible work and for non-reversible it was just w=-PexΔV

    I dont see any way this could turn into similar expressions relate T,P&V

    Someone please help! If i see a derivation a couple times they make sense but I can never figure this out solo.

    Thank you so much
     
  2. jcsd
  3. Oct 14, 2013 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    The derivation can be found here.

    You cannot derive a general expression for irreversible processes since they are all different. For reversible processes, the internal pressure is equal to the external pressure so the work done against the external pressure is ∫PextdV = ∫PintdV = ∫nRTdV/V. But if the external pressure is lower than the internal pressure of the gas (irreversible) you can't use this.

    For example, if the external pressure is constant, the work done by the gas is W = PextΔV. If you apply the first law, Q = ΔU + W where W is the work done by the gas, then (since Q=0) ΔU = -W so ΔU = nCvΔT = -PextΔV . From that you could work out the relationship between P and V in such a process. But, as I said, you need to know some details about the process before you can relate P and V.

    AM
     
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