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Arguing work is an inexact differential

  1. Feb 5, 2012 #1
    Hey,

    Given the fundamental thermodynamic relation (dN=0) ; TdS=pdV+dU, I have to argue that pdV is inexact. I know that dS and dU are exact differentials and that the integral of an exact differential around a closed loop will give a zero result, not too sure where to start, the question is only a couple of marks so the answer need only be brief but my understanding is limited as to why pdV, the work, has to be an inexact differential.

    Any help appreciated!

    Thanks.
     
  2. jcsd
  3. Feb 5, 2012 #2

    tiny-tim

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    Hey Sekonda! :smile:

    What would you add to pdV to make it an exact differential? :wink:
     
  4. Feb 5, 2012 #3
    Vdp?

    I'm still not sure I'm going to have to think about this a bit more!
     
  5. Feb 5, 2012 #4
    Is this the correct term to add to pdV to make it an exact differential; if so, how does this imply that pdV is inexact?
     
  6. Feb 5, 2012 #5

    tiny-tim

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    Yes, pdV + Vdp is an exact differential … it's exactly d(pV).

    So that means that (unless Vdp = 0, ie p is constant) pdV is not exact. :smile:
     
  7. Feb 5, 2012 #6
    Ahh yes that does make sense! I'm always bad at 'arguing', 'showing' or 'proofing' questions; thanks man.

    Cheers!
     
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