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Application of Maxwell's Relations

  1. Nov 7, 2016 #1
    1. The problem statement, all variables and given/known data
    This is question 2.18 from Bowley and Sanchez, "Introductory Statistical Mechanics" .

    Show with the help of Maxwell's Relations that
    $$T dS = C_v dT + T (\frac{\partial P}{\partial T})_V dV$$
    and
    $$TdS = C_p dT - T( \frac{\partial V}{\partial T})_P dP.$$

    Then, prove that
    $$(\frac{\partial U}{\partial V})_T = T (\frac{\partial P}{\partial T})_V - P$$
    2. Relevant equations
    The four common Maxwell's Relations. They can be found here: https://en.wikipedia.org/wiki/Maxwell_relations.

    3. The attempt at a solution
    The issue for me is the first part. I have managed to prove that the second is true.
    My attempt thus far:
    Using the fact that ##dU = TdS - PdV##, we can rewrite the equation as $$TdS = dU + PdV.$$
    We can then multiply and divide dU by dT to get $$TdS = (\frac{\partial U}{\partial T})_V dT + PdV.$$ However, ##\frac{\partial U}{\partial T}_V = C_v##, and so we have $$TdS = C_v dT + PdV.$$
    This then leaves me with the issue of how to convert the term PdV ##\to T (\frac{\partial P}{\partial T})_V dV##. I was considering using the relation ##\frac{\partial P}{\partial T} = \frac{\partial S}{\partial V}##, but I'm not sure how to get that to work. I'll keep working at it, but any help is appreciated!
     
  2. jcsd
  3. Nov 8, 2016 #2
    Well, this is certainly not true! Recall that given a multivariable function ##f(x,y)##, we have
    [tex]df(x,y) = \left(\frac{\partial f}{\partial x}\right)_{y} dx + \left(\frac{\partial f}{\partial y}\right)_{x} dy [/tex]
     
  4. Nov 8, 2016 #3
    I solved it!
    Consider the fact that $$ dU = (\frac{\partial U}{\partial S}) dS + (\frac{\partial U}P\partial V}) dV.$$ Then since ##dU = TdS - PdV##, we can see that ##\frac{\partial U}{\partial S} = T##, or ##\frac{\partial S}{\partial U} = \frac{1}{T}##. Now, consider a function ##S=S(U,V)##.
    $$dS = (\frac{\partial S}{\partial U}) dU_V + (\frac{\partial S}{\partial V}) dV$$
    Using one of the Maxwell relations, we know that ##\frac{\partial S}{\partial V} = \frac{\partial P}{\partial T}##. We also know that ##dU_V = C_V dT##. Therefore, we have
    $$dS = \frac{1}{T} C_V dT + (\frac{\partial P}{\partial T}) dV$$
    and multiplying through by T gives
    $$TdS = C_V dT + T(\frac{\partial P}{\partial T}) dV$$

    A similar process using S(U,P) will give the other. Thank you!
     
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