Stable equilibrium with retained unequal volumes

For each of two separated fixed-composition subsystems $i\in \{A,B\}$, let $O_i\subseteq D_{c_i}$ be an open dimensionless single-phase subset with physical chart ${\mathrm{\Omega }}_i={\mathrm{\Phi }}_{c_i}(O_i)$ as in Entropic temperature, pressure, and molar chemical potentials. Fix positive amounts $n_i$ with nonempty physical slices ${\mathrm{\Omega }}_{i,n_i}:=\{(U_i,V_i):(U_i,V_i,n_i)\in {\mathrm{\Omega }}_i\}$. Fix $U_0>0$ joules and $V_0>0$ cubic metres, and set $V_A=V_0$ and $V_B=2V_0$. Assume ${\mathrm{\Omega }}_{i,n_i}\subseteq (0,2U_0)\times (0,\mathrm{\infty })$ and $(U,V_i)\in {\mathrm{\Omega }}_{i,n_i}$ for every $0<U<2U_0$ and each $i$. Let each physical-coordinate entropy representative be $S_i(U_i,V_i)=a\ln (U_i\mathrm{/}{U}_0)+b\ln (V_i\mathrm{/}{V}_0)$ on its slice, where $a,b>0$ are in $\mathrm{J}{\mathrm{K}}^{-1}$. The isolated compound permits energy exchange but retains the volume and mole partitions, and its adopted total entropy is $S_A+S_B$. For total energy $2U_0$, stipulate that every split $0<U_A<2U_0$, $U_B=2U_0-U_A$ with these retained volumes and amounts is admitted, and that these are exactly the feasible endpoints. As additional model input, designate the split $U_A=U_B=U_0$ as a stable equilibrium. This designation must satisfy Stable equilibrium requires an attained global entropy maximum, which supplies a necessary maximum criterion rather than a converse. The unique entropy-maximizing energy split is $U_A=U_B=U_0$, so $T_A=T_B=U_0\mathrm{/}a$, while $p_A=b{T}_A\mathrm{/}{V}_0$ and $p_B=b{T}_B\mathrm{/}(2V_0)=p_A\mathrm{/}2$. Thus this model's stable equilibrium under energy exchange does not imply pressure equality when volume exchange is retained.