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How they got this Identity

  1. Mar 18, 2007 #1
    1. The problem statement, all variables and given/known data
    If dU=TdS-PdV then
    (dS/dV)T=(dP/dT)V

    the T and V at the end means that T and V are constant
    How did they get this identity? It came from a thermodynamics hence for their notations.

    I have tried ways like rearranging but it dosen't seem to work. I think it has something to do with exact differentials.
     
  2. jcsd
  3. Mar 18, 2007 #2

    Hootenanny

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    Are you familiar with the Helmholtz function?
     
  4. Mar 18, 2007 #3
    by Clairaut's theorem,
    let [itex]z=z(x,y)[/tex]
    [tex]\left ( \frac{\partial^2 z}{\partial x\partial y}\right )=\left ( \frac{\partial^2 z}{\partial y\partial x}\right )[/tex]

    or more explicitly, using "thermodynamic notations":
    [tex]\left [\frac{\partial}{\partial y}\left ( \frac{\partial z}{\partial x}\right )_y\right ]_x=\left [\frac{\partial}{\partial x}\left ( \frac{\partial z}{\partial y}\right )_x\right ]_y[/tex]

    then look at:
    [tex]F=U-TS[/tex]

    [tex]dF=-SdT-PdV[/tex]
    so F is a function of T and V, ie. [itex]F=F(T,V)[/tex] (when N is treated as a constant)
     
    Last edited: Mar 18, 2007
  5. Mar 18, 2007 #4
    I see. The Clairaut's theorem is a formal way of stating exact differentials isn't it?

    Helmholtz function is F=U-TS and sub dU=TdS-PdV and apply Clairaut's theorem.

    Using F=U-TS was the important bit. The sheet didn't mention Helmoholtz anywhere so it would have been hard to know what to use in order to derive the relation. Wouldn't you say?
     
  6. Mar 18, 2007 #5
    not really.. hehe, looking at the derivatives you posted. One easily notice that the variables involved are V and T.

    so you should know the one thermodynamic potential that is a function of V and T. (which is F)

    in general, you have
    U(S,V)
    H(S,P)
    F(T,V)
    G(T,P)

    H=U+PV
    F=U-TS
    G=U-TS+PV

    so when you see an identity, check what variables it involves and use the appropriate thermodynamic functions.
     
    Last edited: Mar 18, 2007
  7. Mar 18, 2007 #6

    robphy

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    There's a neat calculation involving differential forms.
    [tex]0=d(dU)=d(TdS-PdV)=dT \wedge dS+Td(dS) - dP\wedge dV-Pd(dV)=dT\wedge dS-dP\wedge dV[/tex]
    Then write
    [tex]dS=\left(\frac{\partial S}{\partial T}\right)_V dT + \left(\frac{\partial S}{\partial V}\right)_T dV [/tex]
    and
    [tex]dP=\left(\frac{\partial P}{\partial T}\right)_V dT + \left(\frac{\partial P}{\partial V}\right)_T dV [/tex].
    Note that [tex]dT\wedge dT=0[/tex] and [tex]dT\wedge dV= - dV\wedge dT[/tex].
     
  8. Mar 18, 2007 #7
    OK. But you could interchange these functions couldn't you to get them into different variables intermingled together.

    What is the upside down V in the equations?
     
  9. Mar 18, 2007 #8

    robphy

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  10. Mar 18, 2007 #9
    the notation looks insane.

    just one question, why does d(dU)=0??

    and I am very confused by the info from mathworld...

    so
    [tex]dT\wedge dS[/tex]

    equals what in terms of partial derivatives???

    is it like a vector cross product or something like that???
    suppose a=a(x,y), b=b(x,y)

    what would
    [tex]da\wedge db[/tex]
    be?
     
    Last edited: Mar 18, 2007
  11. Mar 19, 2007 #10

    cristo

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    This comes from the definition of the exterior derivative
    Plug dS and dP into the first equation in robphy's post.
    Look up the exterior derivative above.
     
  12. Mar 19, 2007 #11
    wow.. thanks for the info. differential forms are awesome! time for me to pick up a book on such topic.
     
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