# Boltzman distribution for chemical potentials

I read that if we have a system with two co-existing phases with chemical potentials respectively $\mu_1$ and $\mu_2$ then, at the equilibrium, the concentrations $X_1$ and $X_2$ are related by the relation:

$X_1=X_2e^{[\frac{-(\mu_1-\mu_2)}{kT}]}$

Is that true? How can I derivate this expression?

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I could copy the argument here, but I think there is a book that is so good that I will recommend it without actually giving an answer. Look at Landau's Statistical Physics, part I (volume 5), chapter 9. There, you will find the argument and the conditions under which this statement is valid. And, if you have time, read the whole book. Every minute spent will pay off. – Rafael Jan 26 '11 at 19:09
Certainly Landau's book is excellent. Depending on the reader's background and level of expertise, though, some other, gentler books might be appropriate. The last time I taught undergraduate statistical mechanics, I used the book An Introduction to Thermal Physics by Daniel Schroeder. I think it's quite good at deriving rules like these in a clear, accessible way. – Ted Bunn Jan 26 '11 at 19:31
Just to clarify: I assume that you mean that you are adding a very small amount of an impurity that was not originally in the two coexisting phases. What you say can not be true of the original ingredients, since by definition they have the same chemical potential in the two coexisting phases. For example, if you are talking about phase coexistence between oil and water, then your formula can't apply to the concentrations of oil in the two phases, because the chemical potential of oil is the same in both phases! – Greg P Jan 26 '11 at 19:39

It is just the Boltzmann distribution, with the energy being given by the chemical potential.

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