Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

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

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

    [tex]
    \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
    [/tex]

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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    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 :
    G(T,P)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook