# Chemical potentials at triple point

How to prove that chemical potential of each phase at triple point are equal?

My attempt: At phase transition, $$P$$ and $$T$$ are constant and using $$dG=0$$(Gibbs free energy at minimum) at equilibrium gives $$\mu_1dN_1 + \mu_2dN_2 + \mu_3dN_3= 0$$ $$dN_1 + dN_2 + dN_3 =0$$ if we assume no additional substance is being added. Using this, we'll have $$(\mu_2 - \mu_1)dN_2 + (\mu_3 - \mu_1)dN_3 =0).$$

I am stuck here. How to prove from here that $$\mu_1 =\mu_2 =\mu_3~?$$

• I am not sure that you can prove that. The fact that the phases coexist in stability for $P$ and $T$ invariable, means that the Gibbs potential (function of $P$,$T$,($\mu_1,\mu_2,\mu_3$) does not change, as you shown. And it only means the migration of particles between phases is equilibrated, thus the energies related to this migration ($\mu_1,\mu_2,\mu_3$) are equal. In other words, if one of them were lower, all the particles would go to this phase and the system would collapse into only one phase. Oct 11, 2014 at 11:09
• Mass conservation will only get the ratio of the potentials but if you move an infinitesimal amount away from the triple point onto one of the three 2-phase coexistence curves, say $ik$ you will have $\mu_i=\mu_k$. So if you assume continuity of the chemical potential their equality is assured at the triple point. Oct 11, 2014 at 12:45
• @user31748 your answer is convincing. Thanks! Oct 11, 2014 at 13:45
• Does the fact that the 3 phases are in equilibrium help? Jul 30, 2021 at 18:08

The answer depends on the conceptual understanding of the formulas, more than on algebraic manipulations. The final formula, $$(\mu_2 - \mu_1)dN_2 + (\mu_3 - \mu_1)dN_3 =0)$$ should hold for all possible changes of the two independent variables $$N_2$$ and $$N_3$$. This implies \begin{align} \mu_2 &= \mu_1 \\ \mu_3 &= \mu_1 \end{align}