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.