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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 $$\mu_1 =\mu_2 =\mu_3~?$$

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    $\begingroup$ 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. $\endgroup$
    – rmhleo
    Oct 11, 2014 at 11:09
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    $\begingroup$ 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. $\endgroup$
    – hyportnex
    Oct 11, 2014 at 12:45
  • $\begingroup$ @user31748 your answer is convincing. Thanks! $\endgroup$
    – levitt
    Oct 11, 2014 at 13:45
  • $\begingroup$ Does the fact that the 3 phases are in equilibrium help? $\endgroup$ Jul 30, 2021 at 18:08

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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 $$\begin{align} \mu_2 &= \mu_1 \\ \mu_3 &= \mu_1 \end{align}$$

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