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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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
and
μ = (∂F/∂N)_T, ...
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