In terms of partition function, the free energy is given by :
-NkT ln q + PV
Statistical thermodynamics connects the molecular partition function \(q\) to every macroscopic thermodynamic quantity, and the connection runs first through the Helmholtz free energy:
\(A = -NkT\ln q\)
(for distinguishable particles; an indistinguishability correction of \(N!\) is applied for a gas). Helmholtz energy appears first because the canonical ensemble is defined at constant \(N\), \(V\) and \(T\) — exactly the variables for which \(A\) is the natural potential.
The Gibbs free energy then follows from the standard thermodynamic relation
\(G = A + PV\),
which gives \(G = -NkT\ln q + PV\).
So the two candidate expressions differ precisely by the \(PV\) term, and that term is what converts the constant-volume potential into the constant-pressure one. Since laboratory chemistry is done at constant pressure, it is \(G\) that is usually meant by "the free energy", and the expression including \(PV\) is the one required.
The remaining options fail on inspection. A derivative \(\frac{d\ln q}{dT}\) appears in the expression for internal energy, \(U = NkT^{2}\left(\frac{\partial \ln q}{\partial T}\right)_V\), not free energy. And \(NkT^{2}\ln q\) has the wrong dimensions of temperature entirely.
Hence the free energy is -NkT ln q + PV.
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