2
$\begingroup$

I've read from a few places like Kjellander, R. (2019). Statistical Mechanics of Liquids and Solutions: Intermolecular Forces, Structure and Surface Interactions Volume I. p.83. and this Physics Stack Exchange post that (quote from Kjellander):

The grand potential of a system always decreases for a spontaneous process at constant temperature, volume and chemical potential.

My confusion with this statement is that; since the grand potential $\Phi$, is a function of $T$, $V$, and $\mu$, if $T$, $V$, and $\mu$ are fixed, then it seems like $\Phi(T, V, \mu)$ should also be fixed, so what is there to minimize?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

The spontaneous process that the statement is referring to is meant to bring the system from some unstable state to a stable equilibrium state. One function $\Phi$ describes only one of these states, usually the stable equilibrium state with constant chemical potential throughout. The other state may not even have the same chemical potential throughout, it can consist of several subsystems where chemical potential has different values.

For example, if in the system there are two phases in a an unstable ratio (e.g. too much liquid and too little vapor, for the given $T,V$), then chemical potential can be different in the liquid and in the vapor phase. Let the grand potential per unit volume be given by function $\phi_{liq}(T,\mu_{liq})$ and for vapor $\phi_{vap}(T,\mu_{vap})$.

There may be a state where we have non-equilibrium volumes and chemical potentials of liquid and vapor, e.g. too much liquid and too little vapor. The grand potential in this non-equilibrium state has value

$$ \Phi_1 = V_{liq,1}\phi_{liq}(T,\mu_{liq,1}) + (V-V_{liq,1})\phi_{vap}(T,\mu_{vap,1}). $$

The system then may evolve towards a stable state where liquid volume goes from $V_{liq,1}$ to $V_{liq,2}$ (some liquid will evaporate), both phases have the same temperature $T$, and the same chemical potential $\mu_2$ . The grand potential of the system in the equilibrium state will be

$$ \Phi_2 = V_{liq,2}\phi_{liq}(T,V,\mu_2) + (V-V_{liq,2})\phi_{vap}(T,V,\mu_2). $$

The decrease of grand potential means that grand potential of the later equilibrium state is lower or same than the grand potential of the original non-equilibrium state:

$$ \Phi_2 \leq \Phi_1. $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ I think my main confusion I had at the time was I thought the grand potential $\Phi$ is uniquely defined by the total system's temp $T$, volume $V$, and chem potential $\mu$. Whereas it depends also on the "sub-system" configuration, e.g. $V_{liq}$ in your example. Or, in another (contrived) example: an enclosed box with a water droplet (at constant temp. and equilibrium vapor pressure), the contact angle the drop makes with the side walls will minimise the grand potential. $\endgroup$
    – eugenhu
    Commented Jan 18, 2023 at 14:09
  • $\begingroup$ I think $\Phi$ can be determined by those few variables, but only when we are sure the system is in the unique stable equilibrium. If there are multiple equilibria for given $T,V,\mu$, or the system is in a non-equilibrium state, then $\Phi$ either depends on details (how much liquid, vapor), or may not be even defined at all (violent turbulent motion etc.) $\endgroup$
    – Ján Lalinský
    Commented Jan 18, 2023 at 15:54
0
$\begingroup$

Imagine a bottom of water，although $p$, $T$, $\mu$ are constant for both liquid and gas contents，however，some liquid may convert to gas and the total grand potential changes.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for answering. I'm not quite sure I understand your answer and example though. Do you mind rephrasing and expanding on some of the details? I'm still learning the topic. $\endgroup$
    – eugenhu
    Commented Sep 9, 2019 at 16:28
  • $\begingroup$ @eugenhu actually grand potential eqauls $/mu/$N，as you can see，it relies on both partical number and chemical potential，so it wont be fixed if particles change type $\endgroup$
    – Andrew
    Commented Sep 10, 2019 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.