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How they derived this eqn?

  1. Mar 9, 2007 #1
    In the lecture notes
    http://webraft.its.unimelb.edu.au/640322/pub/notes/lectures_common/lecture3.pdf [Broken]

    How did they derive the equation on the line above "L.H.S bracket is not a function of V thus"?

    I can see that (d/dv)(ds)= "that equation except with the 'equals sign' replaced by '+' "

    (dU/dV)=0 because it is an ideal gas.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 9, 2007 #2


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    The expression for dS has three terms so we can write it as

    dS = AdT + BdV + CdV

    The condition for dS being an exact differential is

    d(A)/dV = d(B+C)/dT ( partial differentiation)

    But dB/dT is zero because B is held at fixed T, so

    d(A)/dV = d(C)/dT

    which is the equation you asked about.
    Last edited: Mar 9, 2007
  4. Mar 10, 2007 #3
    I don't understand why the condition on dS in order for it to be an exact differential. The reasoning for the exact differential on the notes I don't follow either.
  5. Mar 10, 2007 #4


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    I'm not clear what's troubling you.

    dS has to be an exact differential for physical reasons. It just means there are no other sources of variation in S but these.

    The little equation in brackets ( 'dz = ...') is a theorem, which says that if a total differential is the sum of two sources, then the sources have a special relationship. Just accept it for now. It's used to get the next line.
  6. Mar 11, 2007 #5

    I see now but I seem to be more interested to see why the exact differential demands this form. You want dS to be an exact differential because S is a function of two variables and is a state function which behaves just like a perfect mathematical function of two variables.
  7. Mar 12, 2007 #6


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    That's how I see it, but talk to other people about this, it might be important.
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