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Chain relation/ triple partial derivative rule

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    For the van der Waals equation of state, confirm the following property:

    (∂P/∂T)V (∂T/∂V)P (∂V/∂P)T = -1

    2. Relevant equations

    The van der Waals equation of state is:

    P = nRT/(v-nb) - an2/V2

    *R, n, a, b are const.

    3. The attempt at a solution

    I have come up with some partial derivatives, however, I cannot seem to figure out the algebra to make their product equal to -1. Perhaps my derivatives are incorrect?

    (∂P/∂T)V = nR/(v-nb)-1

    (∂T/∂V)P = P - an2/V2 + 2abn3/V3

    (∂V/∂P)T = 1/ (2an2/V3 - nRT/(v-nb)2)

    Any hints or ideas?
  2. jcsd
  3. Sep 4, 2011 #2
    I believe there is formula that is (∂X/∂Y) =-(F[itex]_{Y}[/itex]/F[itex]_{X}[/itex])
  4. Sep 4, 2011 #3
    Can you explain what the right side of the equality represents?
  5. Sep 4, 2011 #4
    I am not entirely sure it was just an equation in my book, I will leave it for some one else to answer because I do not want to tell you wrong
  6. Sep 5, 2011 #5
    Anyone else?
  7. Sep 5, 2011 #6


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    Show how you arrived at the last two partial derivatives. (I suggest using implicit differentiation.)
  8. Sep 6, 2011 #7
    Well, i think i figured it out. I used the reciprocal identity:

    (dx/dy)z = 1/ (dy/dx)z (should be partial derivatives here)

    to make the triple partial derivative product a double partial derivative product, and then showed it to be equal to the resulting partial derivative on the other side of the equality (which occurs when you divide -1 by one of the terms originally on the left). It worked well =)
  9. Sep 6, 2011 #8
    Sorry for the hijack but i have a similar question:

    for a recent semester test we needed to show:
    [itex]\frac{\partial{P}}{\partial{V}} \frac{\partial{V}}{\partial{T}} \frac{\partial{T}}{\partial{P}} = -1[/itex]

    i simply converted each partial into its implicit version and cancelled terms:

    [itex]\frac{\partial{P}}{\partial{V}} = \frac{-F_V}{F_P}[/itex]

    [itex]\frac{\partial{V}}{\partial{T}} = \frac{-F_T}{F_V}[/itex]

    [itex]\frac{\partial{T}}{\partial{P}} = \frac{-F_P}{F_T}[/itex]

    resulting in

    [itex]\frac{-F_V}{F_P} \frac{-F_T}{F_V} \frac{-F_P}{F_T} = -1[/itex]

    yet this was marked very clearly wrong...

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