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Thermodynamic second derivatives?

  1. Nov 2, 2017 #1
    This is for research purposes.

    I am aware that first derivatives in thermodynamics always occur (a no-brainer). Do second derivatives occur in thermodynamics commonly as well?
     
  2. jcsd
  3. Nov 2, 2017 #2

    DrDu

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    Yes, they are important to analyse stability, i.e. entropy should be maximal. In statistical thermodynamics, second derivatives of free energy gives you the mean fluctuations of e.g. energy or particle numbers.
     
  4. Nov 2, 2017 #3
    Hi, thanks for replying. Am i correct to assume that this is d2Q/dT2?

    Also, could I ask a reference for this information? Thank you very much!
     
  5. Nov 2, 2017 #4
    I mean, I need the reference for the writeup. Thank you!
     
  6. Nov 4, 2017 #5

    Mapes

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    And how! Material properties are second derivatives of thermodynamic potentials. For example, the thermal expansion coefficient is $$\alpha_V=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)=\frac{1}{V}\left(\frac{\partial^2 G}{\partial T\partial P}\right)$$ The stiffness is $$E=\left(\frac{\partial\sigma}{\partial\epsilon}\right)=\frac{1}{V}\left(\frac{\partial^2 U}{\partial\epsilon^2}\right)$$ The heat capacity is $$c=T\left(\frac{\partial S}{\partial T}\right)=-T\left(\frac{\partial^2 G}{\partial T^2}\right)$$ And so on.
     
    Last edited: Nov 5, 2017
  7. Nov 5, 2017 #6

    DoItForYourself

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    In the book "Thermodynamics foundations and applications" (E. P. Gyftopoulos, G. P. Beretta), Chapters 9 and 10 they often use the second derivative of entropy.
     
  8. Nov 5, 2017 #7
    Regarding the last equation, should there be a minus sign? dG=-SdT+VdP
     
  9. Nov 5, 2017 #8

    Mapes

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    As always, thank you Chester! Edited to fix.

    And the reason I should have caught that is that the curves of the Gibbs free energy have an increasingly negative slope with increasing temperature. And when drawn correctly, they end up at T = 0 K as a straight flat line, because the entropy and the heat capacity are zero at absolute zero.
     
  10. Nov 21, 2017 at 2:13 PM #9
  11. Nov 21, 2017 at 2:30 PM #10
    Thank you very much. I'll try and get the resource; this will be of great importance to my study :biggrin:

    For now, I'm relaxing and playing around with Laplace transforms. Thanks again!
     
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