Is the Gibbs free energy not/less important for canonical ensembles? If so, why? From canonical ensembles, we find that the Helmholtz free energy $F=U-TS$ is related to the canonical partition function as $$F=U-TS=-k_BT\ln Z$$ where $Z$ is the canonical partition function. Therefore, the Gibbs free energy $G$ is given by $$G=F+PV=-k_BT\ln Z+\frac{1}{\beta}V\frac{\partial}{\partial V}(\ln Z)$$ where the expression of pressure $P$ has been used. But textbooks typically mention Helmholtz free energy $F$ but not Gibbs free energy in the context of canonical ensembles. Is Gibbs free energy not as useful as Helmholtz free energy for canonical ensembles? If so, why?
 A: $G$ and $F$ are the thermodynamic potentials of different ensembles, the one for $G$ has, as far as I know, no commonly accepted name. The one for $F$ is the canonical ensemble. 
The bath in the ensemble corresponding to $G$ can exchange volume and energy with the system and is characterized by two intense parameters the system gets into equilibrium with: temperature and pressure. 
For a system characterized by the common parameters $S$, $N$, $V$ there are 8 ensembles depending on which are considered fixed or exchanged with the bath. The bath is then characterized by the corresponding intensive parameters ($T$, $\mu$ and $p$).
The thermodynamic potentials that are extremal in the equilibrium of those ensembles are related by Legendre transformations that change the variables from a quantity to the derivative of the function with respect to that quantity (yes, the same Legendre transformation as in Hamiltonian mechanics). The potential of the microcanoical ensemble is the energy $E(S, V, N)$ expressed in terms of the natural variables. From there we get to the Free energy by a Legendre transform $F = E - TS$, where $T = \partial_S E$ and $S$ must be eliminated by calculating and substituting $S(T)$ to express $F$ in terms of its so called natural variables.
As a general thermodynamic result the potential corresponding to the ensemble with the bath parameters $(T, \mu, p)$ is zero. This gives the Gibbs-Duhem relation:
$$ S \, dT - V\,dp + N\,d\mu = 0 $$
If we introduce other quantities describing the overall system (e.g. magnetization) there are even more ensembles and we again can change between extensive and intensive parameters by Legendre transformation.
