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Maxwell Relations: when are they valid?

  1. Aug 17, 2014 #1
    Hi, I have a question
    If im not mistaken, the Maxwell relations of theromdynamics
    -----for example: ∂G/∂T) = -S ; ∂G/∂P) = V -----
    are valid only for reversible processes.

    On the other hand,
    dG = ∂G/∂T)*dT + ∂G/∂P)*dP + ∑ μi dni
    is valid for any process.

    This means that
    dG = -SdT + VdP + ∑ μi dni
    should only be applicable for reversible processes, and not necessarily for irreversible ones, Right??
  2. jcsd
  3. Aug 17, 2014 #2
    The Maxwell relations are just the application of Poincare's lemma to the thermodynamic potentials.


    The thermodynamic potentials are applicable to reversible and irreversible processes


    Thus the Maxwell relations apply to reversible and irreversible processes.

    Here's a great way to derive/remember the massive amount of information this branch of the subject contains by drawing pictures:

  4. Aug 17, 2014 #3
    Thanks for answering!
    Oh, I didnt know about that Poincare lemma. And the definition of the potentials does apply to any processes.
    But I couldnt find anywhere on the first link a mention of the relations bieng valid in one or the other scenario.

    The answer in the second link seems to be wrong, theres even a comment about it.
    dG = -SdT + VdP apllies to reversible processes, and while dG is the same no matter the "path", -SdT and VdP certainly depend on the path, so the equality should not hold for a irreversible process. In such case it should read dG < -SdT + VdP
  5. Aug 17, 2014 #4


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    Science Advisor

    No they've valid for any quasi-static process.

    This is also only valid for quasi-static processes.

    The Maxwell relations are literally just a consequence of the commutativity of mixed partial derivatives.
  6. Aug 17, 2014 #5
    Ok, thanks.
    So all those things are only valid for quasi-static processes, and not "violent" ones. So, for example,
    dG = -SdT + VdP + ∑ μi dni
    is not valid for "violent" irreversible processes. Or I understand wrong?

    After all, like you say it is just a consequence of mixed partial derivatives but the derivative they ultimately come from is U = TdS - PdV, which is a combination of the first and second Principles and (I think) only is valid for quasi-static processes (since reversible seems to be the wrong term here).
  7. Aug 17, 2014 #6


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    Science Advisor

    No you understood it correctly. In the case of "violent" irreversible processes, i.e. non-quasi-static processes, we only have inequalities.
    Last edited: Aug 17, 2014
  8. Aug 17, 2014 #7
    The second link isn't wrong, the first comment seems to be talking about non-equilibrium thermodynamics, and the second comment makes the original point that dU = TdS - PdV is still valid for irreversible processes, I guess you're conflating this with quasi-static processes or something...

    You seem to be saying something equivalent to: because ydx and xdy separately depend on the path, it implies that (ydx + xdy) is also dependent on the path, but you forget that putting them together implies ydx + xdy = d(xy)... I guess classical vector analysis is messing your intuition up so I recommend this article http://em.groups.et.byu.net/pdfs/publications/formsj.pdf

    Once you have the differential form a thermodynamic potential established (i.e. dU or one of it's Legendre transforms) you take a second exterior derivative and get the Maxwell relations. This is Poincare's lemma (in direct form, i.e. d^2 w = 0) and is equivalent to commutativity of second partial derivatives (Clairaut's theorem).
  9. Aug 18, 2014 #8
    Thanks, I think I got it, though I recieved kind of conflicting answers.
    Im leaning for the non-quasi-static = inequalities.
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