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In explaining phase equilibrium, my textbook gives an example of two phases ($\alpha$ and $\beta$) of the same species coexisting together and isolated from the rest of the world. With the total volume, energy, and a number of particles of the combined two "systems" phases being constant, the total entropy is maximized according to the 2nd law:

$$ds_{total}=ds_{\alpha}+s_{\beta}=0$$

Where the subscripts indicate the phase, then using the fundamental equation we have:

$$\frac{1}{T_{\alpha}}dE_{\alpha}+\frac{P_{\alpha}}{T_{\alpha}}dV_{\alpha}-\frac{\mu_{\alpha}}{T_{\alpha}}dN_{\alpha}+\frac{1}{T_{\beta}}dE_{\beta}+\frac{P_{\beta}}{T_{\beta}}dV_{\beta}-\frac{\mu_{\beta}}{T_{\beta}} dN_{\beta}=0$$

With the total energy, volume, and number of particles being constant, we have $dV_{\beta}=-dV_{\alpha}$, $dE_{\beta}=-dE_{\alpha}$ and $dN_{\beta}=-dN_{\alpha}$ so the entropy maximization gives : $$(\frac{1}{T_{\alpha}}-\frac{1}{T_{\beta}})dE_{\alpha}+(\frac{P_{\alpha}}{T_{\alpha}}-\frac{P_{\beta}}{T_{\beta}})dV_{\alpha}-(\frac{\mu_{\alpha}}{T_{\alpha}} -\frac{\mu_{\beta}}{T_{\beta}})dN_{\alpha}=0$$

Then, the book claims that the pressures, temperatures, and chemical potentials of the two phases are equal because $dE_{\alpha}$, $dV_{\alpha}$, and $dN_{\alpha}$ vary independently. How can the book assume these vary independently? when one phase gains a particle it gains more volume while the converse is true for the other phase. Also, the book goes on and applies Gibbs and Helmholtz free energies for each phase. Aren't the thermodynamic potentials only applicable for systems in contact to a bath?

Textbook: Thermodynamics and Statistical Mechanics: An Integrated Approach by M. Scott Shell

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3 Answers 3

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In fact, they are not always independently variable but when they are independently variable the equilibrium condition is as stated. When they are not independently variable then you need to set up the auxiliary condition and vary that with the one obtained for $\delta s=0$ for an isolated system or the thermodynamic potential corresponding to the reservoir attached to the system.

A simple example is a spherical bubble (b) in equilibrium with its vapor (v) together in a rigid container; now the internal energy depends both on the bubble's volume via internal pressure and its surface area via surface tension. Because of the geometric relationship $V=\frac{4 \pi}{3}r^3$ and $A=4\pi r^2$, you also have $dV=4\pi r^2dr$ and $dA=8\pi rdr$, or $dV=\frac{r}{2}dA$. So when you write $dE=TdS-pdV+\gamma dA$ as the energy of the bubble the infinitesimals $dV$ and $dA$ are not independently variable. Now in thermal equilibrium $T_1=T_2$ and you can write that $-(p_b-p_v)\delta V +(\gamma_b-0)\delta A = 0$, (assume $\gamma_v=0$), from which you get Kelvin's law $p_b-p_v=\frac{2\gamma_b}{r}$. (You can also get some very interesting results in electrochemistry establishing equilibrium relationships between voltages and diffusivity and, some such.)

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From a mathematical point of view, $dE_{\alpha}$, $dV_{\alpha}$, and $dN_{\alpha}$ may vary independently because are independent variables. This is also true from the physical point of view. It is enough to consider the possibility that there is a separation wall between the two subsystems such that it is possible to fix, in turn, energy, volume, and the number of particles of one subsystem. This is always possible by using adiabatic/diathermal, rigid/movable, or permeable/not-permeable walls.

Let's consider the case of a rigid but diathermal and permeable wall. In such a case, $dV_{\alpha}=0$, therefore the extremum conditions becomes

$$\left(\frac{1}{T_{\alpha}}-\frac{1}{T_{\beta}}\right)dE_{\alpha}-\left(\frac{\mu_{\alpha}}{T_{\alpha}} -\frac{\mu_{\beta}}{T_{\beta}}\right)dN_{\alpha}=0. $$ This equation may be satisfied in the presence of non-zero differences in the parentheses only if there is a relation between $dE_{\alpha}$ and $dN_{\alpha}$. If such variables may vary independently, the only way to get the left-hand side equal to zero is by vanishing of both terms in the parentheses.

Regarding the relationship between thermodynamic potentials and thermal or pressure baths, to answer, it would be necessary to know how the author of the textbook applies Gibbs and Helmholtz free energies

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  • $\begingroup$ I added the textbook name in my original post. I am still having a hard imagining a case where $dV_{\alpha}=0$ and $dN_{\alpha}\not\equiv 0$. Exchanging particles while the volume of both phases remains constant is contradictory. Unless the particles are point particles? $\endgroup$
    – Abe
    Nov 26, 2022 at 0:05
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How can the book assume these vary independently? when one phase gains a particle it gains more volume while the converse is true for the other phase.

In what you are describing you have choosen not to vary these variables independently. The point however is that you can, because, as GiorgioP says, these are independent variables. Here is how you can do that in a thought experiment:

Divide your box in two parts, one that contains phase $\alpha$ and the other phase $\beta$. Make sure the walls of each compartment are rigid. Then move a particle from one phase to the other: the transfer is under constant volume. What if the energy in the two parts changed during the transfer? Add/remove heat as needed from each part to restore its energy. The net result is a transfer of particles at constant $V$ and constant $E$.

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