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Partition function, chemical potential and free energy

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.

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

and

μ = (∂F/∂N)_T, ...

Solution Summary

A detailed solution is given.

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