Suppose two phases 1 and 2 of water, say ice and water, are kept in a closed container, at a fixed temperature $T$ and fixed pressure $P$? Then I have the following question:

  1. Is phase 1 in thermodynamic equilibrium with phase 2 during the first order phase transition?

  2. If yes, phase 1 must be in chemical equilibrium with phase 2. This implies $\mu_1=\mu_2$. Doesn't this imply that the rate of the transition from ice $\rightarrow$ water is same as the rate of transition from water $\rightarrow$ ice? If yes, it appears to me that, as long as phase 1 and 2 are in equilibrium with each other, the quantity of ice and water remains the same. But this obviously doesn't happen. Then where am I making the mistake?

  3. To derive the Clausius-Clapeyron equation, do we need phase 1 to be in thermodynamic equilibrium with phase 2 or is it enough to consider the full system to be in thermodynamic equilibrium at temperature $T$ and pressure $P$?

  4. Is the consequence of chemical equilibrium in case of diffusion different from that of the chemical equilibrium between two phases of a given substance?

  • 1
    $\begingroup$ Your question seems to assume the system is in thermal isolation. In that case, the total thermal energy will determine how much water is liquid and how much is ice in thermal equilibrium. In (2), you say, "But this obviously doesn't happen," by which you seem to mean that the ice will continue to melt? Only if extra thermal energy seeps into the system, i.e. if the system is not thermally isolated. $\endgroup$ – pwf Jan 16 '15 at 23:21

The answer to this starts from Gibbs-Duhem relationships that define the number of independent variables you have to completely characterize a thermodynamic systems. You can find this in textbooks or the web. I am happy to write it up later. In any case, for a closed system the number of independent variable you can pick is 2. By putting some ice and water in a closed container (rigid), you have fixed the overall density (number of molecules in the rigid volume), and in this case you have chosen temperature. Therefore, the system will come to phase equilibrium (thermal, mechanical, chemical) by adjusting how much it wants in water form and how much as ice.

This means even if you started out of equilibrium, three things will happen: 1) $T_{ice}=T_{water}$ thermal equilibrium (you already started with that) 2) $P_{ice}=P_{water}$ mechanical equilibrium 3) $\mu_{h20, ice}=\mu_{h20,water}$ chemical equilibrium. Since chemical potential is a function of T, P and composition (in this case purely H2O) these three things will adjust simultaneously to satisfy equality across the phases.

So to answer your questions:

i) You may or may not have started in equilibrium but it will settle to equilibrium, i.e., you will move the right mass split across the phases on the first-order phase transition.

ii) You are right, the phase-split stops when you reach equal chemical potentials.

iii) the complete system is in thermodynamic equilibrium only when no sub-systems are out of equilibrium. Strictly, you cannot define a unique thermodynamic state for the complete system unless all its parts are in equilibrium as well.

iv) No, chemical potential is the derivative of free energy that dictates exchange of matter between phases, reactions, etc.

  • $\begingroup$ @ Sankaran- In my example, is it possible to have arbitrary amount of phase 1 to be in equilibrium with arbitrary amount of phase 2? Does your answer imply that phase transition is possible only when the phases are out of equilibrium? $\endgroup$ – SRS Jan 17 '15 at 6:31
  • $\begingroup$ @ Sankaran- Since, $P,T$ are fixed, $\mu_1(T,P)$ and $\mu_2(T,P)$ gets fixed. Since, on the phase equilibrium line , $\g_1(T,P)=g_2(T,P)$ and $g/n=\mu$, we have $$\mu_1(T,P)=\mu_2(T,P)$$ This implies we have chemical equilibrium when we fix (T,P). Moreover, it appears that any amount of water can be in equilibrium with any amount of ice for a given (T,P) corresponding to a point on the phase equilibrium line. At (T,P) values other than those corresponding to points on the phase equilibrium line, either $\mu_1(T,P)>\mu_2(T,P)$ or $\mu_1(T,P)<\mu_2(T,P)$. $\endgroup$ – SRS Jan 17 '15 at 12:16
  • $\begingroup$ In the first case the whole substance goes into phase 1 and in the second case into the phase 2. There is no tendency to of reaching equal chemical potentials. Doesn't it contradict your first and second point? $\endgroup$ – SRS Jan 17 '15 at 12:22
  • $\begingroup$ Lets see if this explains. Chemical potential doesn't go to zero when the amount stuff in a phase is zero. Chemical potential is technically $\frac{\partial G}{\partial N}$, i.e., the differential change in free energy when a new phase nucleates which is non-zero even if the other phase is zero. In other words it is a slope. The definition $g/n=\mu$ holds only for ideal gases which neither ice or water is. So no matter where you start with $\mu_{water} >\mu_{ice}$ or the opposite the system will (don't hinder it in other ways) go to equilibrium. $\endgroup$ – Sankaran Jan 21 '15 at 23:09
  • $\begingroup$ So you can be at equilibrium in only one phase: Water above melting point (the value of $\mu_{ice}\neq 0$ at that temp). However, you can also remain in non-equilibrium if you carefully prevent nucleation. You can subcool water below its freezing point, if you do that carefully without allowing ice to nucleate. See online videos of people subcooling water in glass bottles (smooth surfaces) and degassing (removing internal nucleation sites) and when they tap the water it instantly freezes. $\endgroup$ – Sankaran Jan 21 '15 at 23:16

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