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Thermodynamics question

  1. Jan 23, 2012 #1
    If in some thermodynamics system preasure [tex]P[/tex] and temperature [tex]T[/tex] are constant then Gibbs potential has minimum.

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

    Variation of [tex]G[/tex] is

    [tex]\delta G=\delta U-T \delta S+P \delta V[/tex]

    [tex]U=U(S,V)[/tex]

    [tex]\delta U=(\frac{\partial U}{\partial S})_V\delta S+(\frac{\partial U}{\partial V})_S\delta V+\frac{1}{2}(\frac{\partial^2 U}{\partial S^2})_V(\delta S)^2+\frac{1}{2}(\frac{\partial^2 U}{\partial V^2})_S(\delta V)^2+\frac{\partial^2 U}{\partial S\partial V}\delta S \delta V[/tex]

    If we use Maxwell relation we get

    [tex]\delta G=\frac{1}{2}(\frac{\partial^2 U}{\partial S^2})_V(\delta S)^2+\frac{1}{2}(\frac{\partial^2 U}{\partial V^2})_S(\delta V)^2+\frac{\partial^2 U}{\partial S\partial V}\delta S \delta V[/tex]

    and from here

    [tex]\delta G>0[/tex]

    Is it true from [tex]G[/tex] is minimum. Or [tex]\delta G=0, \delta^2 G>0[/tex]. Tnx for your answer.
     
    Last edited: Jan 23, 2012
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  3. Jan 23, 2012 #2

    Philip Wood

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    Your expansion of δU (second order Taylor expansion) seems to be missing (δS)2 and (δV)2 in the third and fourth terms.
     
  4. Jan 23, 2012 #3
    Tnx. I fixed it. Do you know answer to my question?
     
  5. Jan 23, 2012 #4

    Philip Wood

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    No. I'm still trying to follow your reasoning. Your first expression for δG should surely be

    δG = VδP - SδT

    The right hand side of your equation for δG is surely identically zero (at least to first order)?

    [Earlier mistake in signs put right following RAP's post. Sorry.]
     
    Last edited: Jan 24, 2012
  6. Jan 23, 2012 #5

    Rap

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    The correct expression for δG is derived as:

    G=U+PV-TS

    δG=δU+(PδV+VδP)-(TδS+SδT) = (TδS-PδV)+(PδV+VδP)-(TδS+SδT) = VδP-SδT

    which will be zero for a constant number of particles (N)
     
  7. Jan 28, 2012 #6
    Yes you're right. Landau says in his book. In thermodynamics equlibrium [tex]G[/tex] is minimum. Thus for any small deviation from equilibrium the change of the quontity [tex]G[/tex] musy be positive. So

    [tex]\delta U -T\delta S+P\delta V>0[/tex]

    In other words the minimum work which must be done from bring this part of body from equilibrium to any neighboring state is positive.

    Well ok. I can except that. But

    δG=δU+(PδV+VδP)-(TδS+SδT) = (TδS-PδV)+(PδV+VδP)-(TδS+SδT) = VδP-SδT

    But if I understand well in some neighboring state [tex]\delta P[/tex], and [tex]\delta T[/tex] isn't zero.
     
  8. Jan 28, 2012 #7
    Also is interesting that Landau then says

    [tex]\frac{\partial^2 U}{\partial S^2}(\delta S)^2+2\frac{\partial^2 U}{\partial S\partial V}\delta S\delta V+\frac{\partial^2 U}{\partial V^2}(\delta V)^2>0[/tex]

    If such an inequality holds for arbitrary [tex]\delta S[/tex] and [tex]\delta V[/tex] then

    [tex]\frac{\partial^2 U}{\partial S^2}>0[/tex]

    and

    [tex]\frac{\partial^2 U}{\partial S^2}\frac{\partial^2 U}{\partial V^2}-(\frac{\partial^2 U}{\partial S\partial V})^2>0[/tex]

    How he get this second inequality? I don't understand. And why he doesn't have condition

    [tex]\frac{\partial^2 U}{\partial V^2}>0[/tex]

    Thanks for your answer.
     
    Last edited: Jan 28, 2012
  9. Feb 6, 2012 #8
    You can't use eq

    [tex]\delta U=T\delta S-P\delta V[/tex]

    Second law of thermodynamics is defined by eq

    [tex]d U=TdS-Pd V[/tex]

    You can't simply change derivatives with variations. Give reference for that.
     
  10. Feb 6, 2012 #9

    Rap

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    I assumed the person was using "[itex]\delta[/itex]" incorrectly in place of "d" and didn't bother to correct it. I probably should not have done that. All system variables should use "d", but products of conjugate variables should use [itex]\delta[/itex] like [itex]\delta W=P dV[/itex]. Will that fix everything?
     
    Last edited: Feb 6, 2012
  11. Feb 6, 2012 #10
    This [tex]\delta[/tex]-s from my posts are variations, not derivatives.
     
  12. Feb 14, 2012 #11
    There are two variables that are being varied arbitrarily (del_S and del_V). So you will only need to satisfy TWO conditions. The following is one way to derive one set of two conditions:

    In the inequality (21.2) of pg 64, let x = delta_S (just for convenience;)
    Now you get a quadratic equation on the left side of the inequality (21.2) of the form:
    a x^2 + b x + c > 0

    (where a, b, c are the other terms consisting of partials etc).

    (i) For the left side of this inequality to have no roots, you want the following condition for the discriminant:

    b^2 - 4 a c < 0 (this is condition 21.4 in the book)

    (ii) and for the quadratic equation on the left side to be > 0, you need, in addition to condition (i) above, the following condition:

    a > 0 (this is condition (21.3) in the book)

    Hope that helps.

    This is just at the side (a digression): And you can similarly derive an alternative 2 conditions, if you took x = delta_V originally. In fact, one of the conditions will be the same as (i), and the other can be transformed to
    (partial_P / partial_V)_S < 0 (which is related to the isothermal compressibility in (21.6) through (16.14)). But you need to show that c_p > 0 in 16.14 in order to get 21.6 (we only know that c_v > 0 in the derivation up to 21.5); so Landau shows that c_p > c_v > 0 in pg 64-65 with the help of 16.10 and 21.6.
     
    Last edited: Feb 14, 2012
  13. Feb 15, 2012 #12

    Actually, I was wrong about the reasoning why you need two conditions. In the above inequality, it is just because that is all you need to satisfy the inequality (which is quadratic in x = delta_S). In addition, the extra condition you get from using x = delta_V can be transformed to 21.3.

    In the general case, you will have 3 conditions (see Le Chatelier's principle (eqns 22.3 - 22.4))
     
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