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Statistical Mechanics to Thermo for Isothermal Isobaric Ensemble

  1. Mar 4, 2006 #1

    JM1

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    I am trying to make the connection from statistical mechanics to thermodynamics for the isothermal isobaric ensemble. Partition function = (sum of)exp(-BEj-gamma*Vj).

    I have followed T.L. Hill [Statistical Mechanics, p. 67] but can not understand how he justifies dE=(sum of)EdP, rather than (sum of) EdP +(sum of)PdE. This makes it easy, but I think (sum of)PdE is not zero. He doesn't make this simplification for the canonical or grand canonical ensembles.

    However, since Ej = Ej(P, N) this will introduce an unpleasant (dE/dP)dP term (partial derivative with N held constant) which I do not know what to do with. I can get the TdS and PdV terms but am stuck with what this nasty extra term means. Any ideas would be appreciated.
     
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  3. Mar 5, 2006 #2

    Physics Monkey

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    I don't have Hill's book, so I can't directly help you unless you can be more specific about what Hill is trying to do. I have provided a brief sketch of one way to obtain the Gibbs ensemble from the canonical ensemble so perhaps you will find that helpful. If not please feel to try and help me understand your question a bit more. Also, I'm sure someone else will come along eventually who has the book and can perhaps be more helpful.

    Given the partition function in the canonical ensemble [tex] Z(T,V,N) [/tex] you can get the partition function in the Gibbs ensemble by doing a Laplace transform [tex] \mathcal{Z} = \int^\infty_0 e^{- \beta P V} Z(T,V,N) [/tex]. This just follows straight from the definition. The Helmholtz free energy is [tex] F = - kT \ln{Z} = E - TS [/tex]. I assume you've already convinced yourself of this formula. In order to evaluate the integral in the thermodynamic limit you need to find the largest value of the integrand; this term will dominate the whole integral.

    The [tex] V [/tex] where the integrand takes its largest value is given by solving [tex] P = -\frac{\partial F}{\partial V} [/tex] which is the usual formula for pressure in the canonical ensemble. You can solve for [tex] V [/tex] in terms of [tex] P [/tex] so that the integral is given (keeping the largest term in the thermodynamic limit) as [tex] \mathcal{Z}(T,P,N) = e^{- \beta P V(P) - \beta F(T,V(P),N)} [/tex]. The Gibbs free energy is then [tex] G = - k T \ln{\mathcal{Z}} = P V + F [/tex] which is the usual thermodynamic formula.

    Hope this helps some.
     
    Last edited: Mar 5, 2006
  4. Mar 5, 2006 #3

    JM1

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    Thanks,
    I can get the partition function; but I can't derive the formula for the Gibbs function. I can get the Helmholtz function from the canonical ensemble partition function by comparing the terms in dE with the thermodynamic terms in dU while holding N constant.

    dE=sum (EndPn) +sum (PndEn) Then I substituted the term for En from the partition function and dEn from the fact that En=En(V,N) but N is held constant. This gives me terms that can be compared to dU=TdS-PdV.

    However, when I try to do the same with the isothermal isobaric ensemble the sum (PndEn) term gives me problems as En = En(P,N), so dEn = (dEn/dP)dP, where this is a partial derivative with N constant. I don't know what to do with this term. Hill gets around this problem by saying dEn = sum of (EndPn) which I can't justify. I know G = -kTln Z, but I can't derive it. Thanks,
     
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