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.© BrainMass Inc. brainmass.com June 3, 2020, 7:38 pm ad1c9bdddf
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, ...
A detailed solution is given.