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    Partition function, chemical potential and free energy

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    Derive equations: S = -(∂F/∂T)_V, N = Nk [ln (V/N_vQ) + 5/2] - ∂F_int/∂T and μ = (∂F/∂N)_T, V = -kT ln (VZ_int/N_vQ) for the entropy and chemical potential of an ideal gas.

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    Solution Preview

    We can derive the expressions from the fundamental law of thermodynamics:

    dU = T dS -PdV +μ dN

    If you write.

    T dS = d(TS) - S dT

    and bring the d(TS) to the left hand side, you get:

    dF = -S dT - PdV + μ dN

    From this equation you read off that:

    S = -(∂F/∂T)_V, N


    μ = (∂F/∂N)_T, ...

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