# Proof of equality of chemical potentials and temperature for diphasic system

My question is the following :

I have a system at (T,V,N) which is composed of two phases : (T1,V1,N1) and (T2,V2,N2).

Initially I wanted to proove that $\mu_1 = \mu_2$, but I had troubles.

To proove it i use the fact that $F=F_1+F_2$ must be minimised at equilibrium because we are in (T,V,N).

So we have :

$dF=dF_1+dF_2$.

I can write : $V=V_1+V_2$, $N=N_1+N_2$ and s N and V are fixed, $dV_2=-dV_1$, $dN_2=-dN_1$.

So :

$$dF=(-P_1+P_2)dV_1+(\mu_1-\mu_2)dN_1-S_1dT_1-S_2dT_2=0$$

Then I wanted to say "well, $N_1$, $V_1$, $T$ are independant variables so I have to cancel the terms in frond of $dN_1$, $dV_1$ and I would have $P_1=P_2$ and $\mu_1=\mu_2$. But there is these terms in $dT_1$ and $dT_2$ that I don't know how to replace by $dT$...

Since in your case temperature does not remain constant, there is no point in trying to minimize $F$. If the reaction occurs in an isolated container, then total internal energy of the combined system, $U=U_1+U_2$, remains constant. If this is the case then it proper to maximize entropy, $S$, which is a function of $U,V,N$ ($F$ is obtained by Legendre transform of this fundamental relation).
$dS=(\frac{1}{T_1}-\frac{1}{T_2})dU_1+(\frac{p_1}{T_1}-\frac{p_2}{T_2})dV_1+(-\frac{\mu_1}{T_1}+\frac{\mu_2}{T_2})dN_1=0$, at equilibrium.
• @user3183950 You see that in entropy formulation each variation is independent, so you may equate each coefficient to zero, which yields $T_1=T_2$ at equilibrium, from which follows $\mu_1=\mu_2$ at equilibrium.