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.
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.