thermodynamics - Deriving Enthalpy from Statistical Mechanics

One can derive all the numerous thermodynamic potentials (Helmholtz, Gibbs, Grand, Enthalpy) by Legendre transformations, but I'm interested in seeing each from Stat Mech instead (ie taking the log of a partition function). I can do this easily for all but the Enthalpy, which has me stumped.

Easy example: Deriving Helmholtz

The Helmholtz energy can be derived by putting our system $S$ in thermal contact with a heat bath $B$ at temperature $T$ and maximizing the total entropy:

$$S_\mathrm{tot}/k_b=\log\sum_{E_S}\Omega_B(E_\mathrm{tot}-E_S)\Omega_S(E_S)$$

If the heat bath is big enough to maintain constant temperature $1/T=\left(\frac{dS_B}{dE_B}\right)_{V_B}$, then we can say $S_B\approx -E_S/T+S_0$ for small $E_S$. Then $\Omega_B\approx \Omega_0 e^{-\beta E_S}$ where $S_0=k_b\log\Omega_0$, $\beta=1/k_bT$, so

$$S_\mathrm{tot}/k_b=\log\sum_{E_S}\Omega_0e^{-\beta E_S}\Omega_S(E_S)=S_0/k_b+\log Z_S$$ where $Z_S=\sum_{E_S}e^{-\beta E_S}\Omega_S(E_S)$. So maximizing the total entropy is just maximizing $Z_S$. If we define the Helmholtz Free Energy $A_S$ as $-\beta A_S=\log Z_S$, and use the Shannon entropy $S/k_b=-\sum p\log p $, we see

$$S_S/k_b=-\sum \frac{\Omega_S(E_S)e^{-\beta E_S}}{Z}\log \frac{\Omega_S(E_S)e^{-\beta E_S}}{Z}$$ $$S_S/k_b=\frac{\beta}{Z}\sum \Omega_S(E_S)E_S e^{-\beta E_S} + \log Z$$ $$S_S/k_b=-\frac{\beta}{Z}\partial_\beta Z + \log Z$$ $$S_S/k_b=\beta \langle E_S\rangle -\beta A_S$$ $$A_S=\langle E_S\rangle -TS_S$$

The other thermodynamic potentials at fixed $T$ are similarly easy to derive.

But now try deriving Enthalpy

The same procedure does not work for enthalpy, because,

$$S_\mathrm{tot}/k_b=\log\sum_{V_S}\Omega_B(V_\mathrm{tot}-V_S,E_{B0}+pV_S)\Omega_S(V_S,E_{S0}-pV_S)$$

...if the bath is big enough to maintain a constant temperature, then its total entropy is constant as a function of $V_S$. That is, if $V_S$ increases, the bath entropy decreases due to less volume, but the bath energy increases by the same amount due to increased energy. So, to first-order, $\Omega_B$ is constant and the total entropy splits into a sum of both individual subsystem entropies.

Is there a way to derive the enthalpy from stat mech considerations, as there is for the other potentials, rather than by Legendre transforming the energy?

By "derive the enthalpy", I mean "derive that the quantity which should be minimized in equilibrium is given by $H=\langle E \rangle+p\langle V \rangle$."