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Partial Derivative of U

  1. Aug 26, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is attached in the picture.


    3. The attempt at a solution

    I'm aware that:

    dU = T dS - P dV

    ∫ dU = ∫ (T) dS - ∫ P dV

    Are they assuming that T, P are constant so

    U = TS - PV

    ∂U/∂X = T (∂S/∂X) - P (∂V/∂X)


    Or is there a way to directly change dU to ∂U?
     

    Attached Files:

  2. jcsd
  3. Aug 26, 2012 #2
    dU is a differential. The first differential is invariant, so if you know what it looks like in terms of dS and dV, you can express it in terms of dX and dY if you express the former via the latter. That's exactly what they do.
     
  4. Aug 27, 2012 #3
    I'm not sure what you mean...My question is how did they "convert"

    dU = T dS - P dV

    into

    (∂U/∂X) = T (∂S/∂X) - P (∂V/∂X) ?

    Are you allowed to simply replace each 'd' by '∂' ?
     
  5. Aug 27, 2012 #4
    The first differential is invariant. That means that for a function U(S, V), dU = TdS - PdV, if S and T are functions of (X, Y), with their differentials dS = idX + jdY, dV = kdX + ldY, we can simply plug these dS into dV into the original equation and we will get dU = T(idX + jdY) - P(kdX + ldY) = (Ti - Pk)dX + (Tj - Pl)dY.

    Now, it is known that if we have a differential dU = AdX + BdY, then A is the partial derivative with respect to X, and B is the partial derivative with respect to Y. That is also true for i, j, k, l above.

    Now combine all of this.
     
  6. Aug 27, 2012 #5
    Got it! Thanks!
     
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