# Grand free energy as force towards equilibrium

My book on thermal physics shows that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), the grand free energy, $$\Phi$$, tends to decrease. The definition for the grand free energy is $$\Phi\equiv U-TS-\mu N.$$ Now, they wrote the entropy of the reservoir as: $$S_R=1/T[U_T-U+P(V_T-V)-\mu(N_T-N)],$$ where subscript $$R$$ stands for reservoir, and subscript $$T$$ stands for total. I can see that that they’re using the thermodynamic identity, $$dU=TdS-PdV+\mu dN,$$ but I don’t see why the pressure is assumed fixed here. Thermal equilibrium implies constant and equal $$T$$, and diffusive equilibrium implies constant and equal $$\mu$$. How can we conclude that $$P$$ must be constant too? As far as I know, we don’t have mechanical equilibrium.

I have three guesses:

1) We are not in mechanical equilibrium, and it's just the reservoir's pressure that can be considered fixed (while the system's pressure may vary.

2) Somehow thermal and diffusive equilibrium also imply mechanical equilibrium, so the pressure of the system and its surroundings (=reservoir) are always equal.

3) Use the thermodynamic identity for the grand potential, which yields the following partial derivative: $$-\left(\frac{\partial\Phi}{\partial V}\right)_{T,\mu}=P.$$ If would seem then that this partial derivative is the same for the reservoir, if our system transforms during diffusive and thermal equilibrium. Hm, I'm still confused.

Which one is correct?

• You don't need to include a specific edit history in your question. An edit history is available for those who are interested :) I have edited the question to make it a cohesive question. Please make sure I didn't change anything that shouldn't have been changed – Aaron Stevens Dec 19 '18 at 2:45

The grand potential is a minimum for fixed $$\mu, V, T$$. Energy and particles are allowed to exchange with the reservoir, but the system volume $$V$$ is fixed. Therefore, as you surmise, there is no condition of mechanical equilibrium. In working through the conditions for equilibrium, any $$dV$$ or $$\Delta V$$ terms are zero by definition. So there is no need to consider the pressure in the system, or in the reservoir, when discussing the minimum of $$\Phi$$.
Of course, at equilibrium, assuming homogeneous systems in the thermodynamic limit, they will be given by the same equation of state $$P=P(\mu,T)$$. So the system pressure and the reservoir pressure will be equal, at equilibrium.
As a side note, the grand ensemble is quite useful for discussing cases which are trickier than the standard ones: when the system is small, for instance, or has some inhomogeneous features such as walls (so we can discuss capillary condensation). As long as the system may be considered in equilibrium with a reservoir, with respect to exchanging particles and energy, the state point is defined by the parameters $$\mu$$ and $$T$$ which characterize the reservoir.
• I don't really see why $P=P(\mu,T)$, but I can see that if we have fixed $V$ and $T$, and fixed $N$ I suppose, since we're in diffusive equlibrium, than the equation of state which relates $P,V,T$ (and $N$?) will fix $P$. – Sha Vuklia Dec 19 '18 at 20:17
• Thermodynamic equations of state are not restricted to relations between $P,V,T$ and $N$. Most of the thermodynamic relations that you will see are based on converting from one set of variables to another. Having specified $\mu$ and $T$, all other intensive thermodynamic variables are defined. For example, $\rho$ and $P$ may each be expressed as a function of those two. Usually, the function is not known (except for a few exactly soluble models), but that doesn't matter. – user197851 Dec 19 '18 at 20:35