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Thermodynamic Identities Proof - Gibbs and Helmholtz

  1. Feb 25, 2009 #1

    TFM

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    1. The problem statement, all variables and given/known data

    Use the definitions of Helmholtz free energy, F, and Gibbs free energy, H, together with the thermodynamic identity, to show that

    [tex] S = -(\frac{\partial F}{\partial T})_V [/tex] and [tex] S = -(\frac{\partial F}{\partial T})_P [/tex]

    Then use those definitions again, and the expressions above, to show that

    [tex] U = -T^2 \left(\frac{\partial (F/T)}{\partial T} \right)_V [/tex] and [tex] H = -T^2 \left(\frac{\partial (G/T)}{\partial T} \right)_P [/tex]

    These equations are known as the Gibbs-Helmholtz equations and are useful in chemical thermodynamics.

    2. Relevant equations

    Helmholtz:

    [tex] F = U - TS [/tex]

    Gibbs:

    [tex] G = H - TS = U - TS + PV [/tex]

    3. The attempt at a solution

    I have succesfully done the first part (showing S = F/T...) But I am stuck on the second.

    I asked the ;ectuirer, whop said I should first work backwards, and do df/DT, and insert into the equation. This has given me?

    [tex] F(T) = V - TS [/tex]

    [tex] \frac{\partial F}{\partial T} = -S [/tex]

    insert:

    [tex] U = -T^2 \left(\frac{-S}{\partial T} \right)_V [/tex]

    now the thermodynamic identity is:

    dU = Tds - pdv

    I do I get rid of the squared part?

    because if I just differentiate T squared, I get 2T,

    so this could leave me with:

    [tex] U = -2T \left(-S} \right)_V [/tex]

    but this doesn't make the S a ds? and I have a factor of 2?

    Any ideas?

    Many thanks,

    TFM
     
  2. jcsd
  3. Feb 25, 2009 #2

    Mapes

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    I would use the chain rule to evaluate

    [tex]\left(\frac{\partial (F/T)}{\partial T}\right)_V[/tex]

    which will give you two terms. Then plug in [itex]F=U-TS[/itex], which leads to the desired first expression. The same approach works for the H-G pair.
     
  4. Feb 26, 2009 #3

    TFM

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    Okay, so using chain rule:

    [tex] \frac{\partial \frac{F(T)}{T}}{dT} [/tex]

    Would the quotient rule not be best:

    [tex] u = F(T), v = T, du/DT = F'(t), dv/dT = 1 [/tex]

    giving:

    [tex] \frac{TF'(T) - 1F(T)}{T^2} [/tex]

    Plug in F = U - TS, F' = -S


    [tex] \frac{T(-S) - U - TS}{T^2} [/tex]

    [tex] \frac{-TS - U - TS}{T^2} [/tex]

    [tex] \frac{-2TS - U}{T^2} [/tex]

    Reinsert:

    [tex] U = -T^2 \left(\frac{-2TS - U}{T^2} \right)_V [/tex]

    This cancels to:


    [tex] U = -2TS - U [/tex]

    [tex] 2U = -2TS [/tex]

    [tex] U = -TS [/tex]

    I don't think this is right...?
     
  5. Feb 26, 2009 #4
    Try this approach,

    First divide F by T. Then differentiate the answer. Then plug in the thermodynamic identity for dU. Rearrange, and let N and V be constant. This should give you the right answer.
     
  6. Feb 26, 2009 #5

    TFM

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    So when you say divide F by T, would thuis be F/T, or do we use F = U - TS, giving U/T - S?
     
  7. Feb 26, 2009 #6
    You want both of those.
     
  8. Feb 26, 2009 #7

    TFM

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    Okay, so

    [tex] U = -T^2 \left(\frac{\partial (F/T)}{\partial T} \right)_V [/tex]

    F/T = U/T - S

    put into equation:

    [tex] U = -T^2 \left(\frac{\partial (\fracd{U}{T} - S)}{\partial T} \right)_V [/tex]

    differentiating this gives:

    [tex] U = -T^2 \left(U)_V [/tex]

    ?
     
  9. Feb 26, 2009 #8
    What you want is this;

    [tex]\frac{\partial (F/T)}{\partial T}=\frac{\partial (\frac{U}{T} - S)}{\partial T}[/tex]

    Then think what is d(U/T)/dT?
     
    Last edited: Feb 26, 2009
  10. Feb 26, 2009 #9

    Mapes

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    You made an arithmetic error in post #3. You're almost there.
     
  11. Feb 26, 2009 #10
    That is strange we posted at exactly the same time. Are you my long lost twin brother? :wink:
     
  12. Feb 26, 2009 #11

    TFM

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    Would it be U/T^2 ?
     
  13. Feb 26, 2009 #12
    That wold be part of it, although I think it should be negative U/T^2. Use the quotient rule and remember that you can't actually differentiate U so you have to leave it as dU/dT.
     
  14. Feb 26, 2009 #13

    TFM

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    Okay, so using the quotient rule:

    U = U (:rolleyes:)
    V = T

    du/dt = du/dt

    dvdt = 1

    thus:

    [tex] = \frac{U - Tdu/dT}{T^2} [/tex]

    Look okay now?
     
  15. Feb 26, 2009 #14
    YOur quotient rule is back to front I believe. It should be (the following are for the general case);

    [tex](\frac{u}{v})'=\frac{u'*v-u*v'}{v^{2}}[/tex] Note where the primes are, differentiate top times bottom, then take away the derivative of the bottom times the top and divide the whole lot by the bottom squared.

    Once you've changed that let dU=identity and the rest is just rearranging.

    Edit: I don't mean dU=1, I just mean the thermodynamic identity for dU, which you must have used earlier.
     
    Last edited: Feb 26, 2009
  16. Feb 26, 2009 #15

    TFM

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    Okay, so it should have been:

    [tex] \frac{dU/dT T - U}{T^2} [/tex]

    so now the thermodynamic identity:

    dU = Tds - pdv

    so now I have to rearrange

    [tex] \frac{dU/dT T - U}{T^2} [/tex]

    into dU = , then replace du with the thermo identity, and solve?
     
  17. Feb 26, 2009 #16
    It would be easier if you just substituted for dU in the differential eqution and solved. Don't forget the -dS/dT term from earlier when you put everything back together.
     
  18. Feb 26, 2009 #17

    TFM

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    so,

    dU = Tds + pdV

    sub into:

    [tex] \frac{dU/dT T - U}{T^2} [/tex]


    [tex] \frac{(Tds + pdV/dT) T - U}{T^2} [/tex]

    this will give:

    [tex] \frac{T^2 ds/dT + pT dV/dT - U}{T^2} [/tex]

    ???
     
  19. Feb 26, 2009 #18
    Thats right. Now all you have to do is simplify and you will have the right answer.
     
  20. Feb 26, 2009 #19

    TFM

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    Okay so:

    [tex] \frac{T^2 ds/dT + pT dV/dT - U}{T^2} [/tex]

    break it down into:

    [tex] \frac{T^2 ds/dT}{T^2} + \frac{PT dV/dT}{T^2} - \frac{u}{T^2} [/tex]

    [tex] \frac{ds}{dT} + \frac{P dV/dT}{T} - \frac{u}{T^2} [/tex]

    [tex] \frac{ds}{dT} + \frac{P dV/dT}{T} = \frac{u}{T^2} [/tex]

    Okay so far?
     
  21. Feb 26, 2009 #20
    Not sure about the equals sign here, don't forget this is part of the rhs of the equation,

    d(F/T)/dT=d/dT(U/T-S)
     
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