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A Thermodynamics Pathria Eq.4.5.9

  1. Oct 9, 2018 #1
    I'm confused about the mathematics that led to the equation 4.5.9.

    upload_2018-10-9_16-19-51.png

    Specifically, I'm confused about what the variables that describe U are.

    From the equation

    upload_2018-10-9_16-26-44.png

    I think temperature T(through beta), chemical potential (through alpha), V (through E_s) and N (through... restriction on the summation?) defines the (average) internal energy U,

    but how should I do the differentiation? What I ended up was that the chemical potential dependence can be changed to the dependence on the fugacity z so that

    upload_2018-10-9_16-29-42.png

    but it doesn't seem to be exactly eq.4.5.9 to me.
     
  2. jcsd
  3. Oct 9, 2018 #2

    Charles Link

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    I have the answer. It's going to take me a few minutes to write out the partial derivatives in Latex... ## \\ ## ## z=z(T, N,V) ##, so that ## U(T,z,V)=U(T,N,V) ## . ## \\ ## Three parameters are need to specify ## U ##, not 4 . Any 3 of the 4, (##z,T,N,V ##), can be chosen.## \\ ## Now using the first expression for ## U ##, ## \\ ## ## dU=(\frac{\partial{U}}{\partial{T}})_{z,V} \, dT+(\frac{\partial{U}}{\partial{z}})_{V,T} \, dz+(\frac{\partial{U}}{\partial{V}})_{z,T} \, dV ##. ## \\ ## Taking the second expression: ## \\ ## ## dU=(\frac{\partial{U}}{\partial{T}})_{N,V} \, dT+(\frac{\partial{U}}{\partial{N}})_{V,T} \, dN+(\frac{\partial{U}}{\partial{V}})_{N,T} \, dV ##. ## \\ ## Now, set ## dz=0 ## and ## dV=0 ##, and set the two ## dU's ## equal. Take a partial w.r.t. ## T ## at constant ## z,V ##, (we set ## dz ## and ## dV=0 ## ), =divide through by ## dT ## everywhere, and the result is the equation 4.5.9. ## \\ ## @Chestermiller This is a good thermodynamics problem that the OP has. Might you have anything to add?
     
    Last edited: Oct 9, 2018
  4. Oct 10, 2018 #3

    Charles Link

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    @IcedCoffee You have a very interesting question, but so far you haven't returned to see the answer.
     
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