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Partial derivatives (Maxwell relations) in thermodynamics

  1. Sep 26, 2010 #1
    My professor did this in lecture, and I can't figure out his logic. Can someone fill in the gaps?

    He went from:

    dS = \left( \frac{\partial S}{\partial P} \right)_T dP + \left( \frac{\partial S}{\partial T} \right)_P dT

    (which I totally understand; it just follows from the fact that [itex]S[/itex] is an exact differential) to the following:

    \left( \frac{\partial S}{\partial T} \right)_V = \left( \frac{\partial S}{\partial P} \right)_T \left( \frac{\partial P}{\partial T}\right)_V + \left( \frac{\partial S}{\partial T} \right)_P

    Where the heck does THAT come from? Anyone have any ideas?
  2. jcsd
  3. Sep 26, 2010 #2


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    The first equation tells you that [itex]S[/itex] can be written as a function of [itex]T[/itex] and [itex]P[/itex] only... i.e. [itex]S=S(P,T)[/itex]....the second equation is just a straight forward application of the chain rule to [itex]S(P,T)[/itex]
  4. Sep 26, 2010 #3
    Thanks for your response, but I'm afraid I don't understand. How can a "straightforward application of the chain rule to [itex]S(P,T)[/itex]" involve the variable [itex]V[/itex]? I would think such an application could only contain the variables [itex]S, T, P[/itex].
    Last edited: Sep 26, 2010
  5. Sep 27, 2010 #4


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    Both [itex]T[/itex] and [itex]P[/itex] may have some dependence on [itex]V[/itex].

    [tex]\left(\frac{\partial S}{\partial T}\right)_V[/tex]

    simply means the partial derivative of [itex]S[/itex], with repect to [itex]T[/itex], taken at constant [itex]V[/itex]. Whenever you calculate a partial derivative of a multivariable function, you're always holding at least one variable constant.
  6. Sep 27, 2010 #5

    George Jones

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    Actually, all the other variables are constant.
    When it come to the chain rule, often physicists are very sloppy. Start with [itex]S=S\left(P,T\right)[/itex] and suppose that pressure is a function of volume and temperature. Define

    [tex] \tilde{S} \left(V,T\right) = S \left( P\left(V,T\right) , T \right).[/tex]

    Now, use the chain rule to calculate [tex]\partial \tilde{S} / \partial T[/tex].

    Although [itex]S[/itex] and [tex]\tilde{S}[/tex] are very related, they actually are different functions.
  7. Sep 27, 2010 #6


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    You beat me to it. But let me add. The real problem (sloppiness) is that we confuse the variables with the functions e.g. x = x(t). We should really begin with entropy as a function of T, P,N and V.
    S = s(T,P,N,V) so
    [tex]dS = \frac{\partial s}{\partial P} dP + \frac{\partial s}{\partial T} dT + \frac{\partial s}{\partial V}dV + \frac{\partial s}{\partial N}dN[/tex]
    But we also have a constraint on T,P, N, and V via the physical gas law. This we can express implicitly f(P,V,N,T)=0, or solve for any of the four quantities as a function of the other three.

    [tex]P = p(V,T,N)[/tex]
    [tex]V = v(P,T,N)[/tex]
    [tex]T = t(P,V,N)[/tex]
    [tex]N = n(P,V,T)[/tex]

    Differentiate each of these to relate their partials to f's.

    We then may impose a condition such as constancy of volume or temperature, etc.

    It may be rather lengthy but I think worthwhile to go through the long derivation then practice converting to the "sloppy" form used.

    [EDIT]: Actually "we begin with" the gas law and that is incorporated into the entropy function so I got overly inclusive with the variables.
    Last edited: Sep 27, 2010
  8. Oct 10, 2011 #7
    Maybe he used Maxwell relations?
    Its a difficult to calculate.
    But you can find loads of ways to find similar relations :
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