# Chemical potential of mixtures of ideal gases and the Gibbs-Duhem equation

As is well known for a constant composition system, $\mu$ (symbolizing chemical potential) is equal to the molar Gibbs energy. Using $$\mathrm{d}G = V\mathrm{d}p-S\mathrm{d}T,$$ we can write $$\mu = \mu^{0} + RT\ln P.$$ But for mixtures, $$\mathrm{d}G = V\mathrm{d}p - S\mathrm{d}t + \mu_{\ce{A}} \mathrm{d}n_{\ce{A}} + \mu_{\ce{B}} \mathrm{d}n_{\ce{B}}$$ Can the original formula for chemical potential in the case of mixtures?

Moreover, using the Gibbs-Duhem equation, $$n_{\ce{A}} \mathrm{d}\mu_{\ce{A}} + n_{\ce{B}} \mathrm{d}\mu_{\ce{B}} = 0,$$ which means that $\mu_{\ce{A}}$ changes upon adding B to the mixture, even when the partial pressure of A is constant. This contradicts the formula.

• Who says that the partial pressure of A is constant when you add B to the mixture at constant pressure? – Chet Miller Apr 1 '18 at 15:25
• I meant adding b at constant volume of the mixture – DHYEY Apr 8 '18 at 5:06

The chemical potential of a the two species in your ideal gas mixture are $$\mu_A=\mu^0_A(T)+RT\ln\left(P\frac{n_A}{(n_A+n_B)}\right)$$ $$\mu_B=\mu^0_B(T)+RT\ln\left(P\frac{n_B}{(n_A+n_B)}\right)$$ and the total free energy of the mixture is: $$G=n_A\mu_A+n_B\mu_B$$So, the chemical potential of A depends on the number of moles of B and the chemical potential of B depends on the number of moles of A. Do these equations satisfy the Gibbs-Duhem equation? Try them out and see.