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Hemholtz free energy proof

  1. Mar 8, 2014 #1
  2. jcsd
  3. Mar 8, 2014 #2

    UltrafastPED

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    Isothermal work = T is constant!
     
  4. Mar 9, 2014 #3
    yes, but you could get the equation ##dw \leq-dA## without assuming T=constant anywhere?

    I mean, from what I can see, all they do is use the 1st law of thermodynamics and the general fact that ##dQ \leq TdS##. So shouldn't the result they get, ie ##dw \leq-dA##, apply regardless if T is constant or not during the whole process?
     
  5. Mar 9, 2014 #4

    DrDu

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    I also don't see why constant temperature should be necessary.
     
  6. Mar 9, 2014 #5

    stevendaryl

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    Well, the differential equality for Helmholtz free energy is:

    [itex]dA = -S dT -p dV = -S dT - dW[/itex]

    It seems to me that if [itex]T[/itex] is not constant, then the term [itex]-S dT[/itex] could be either positive or negative, depending on whether the temperature is increasing or decreasing.
     
  7. Mar 9, 2014 #6

    DrDu

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    You are right. The point is that in general, dA =dU -dTS. This only reduces to dA=dU-TdS if T is constant.
     
  8. Mar 10, 2014 #7
    Ahh, but of course. thank you.
     
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