0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Who says that the partial pressure of A is constant when you add B to the mixture at constant pressure? $\endgroup$ Apr 1, 2018 at 15:25
  • $\begingroup$ I meant adding b at constant volume of the mixture $\endgroup$
    – DHYEY
    Apr 8, 2018 at 5:06

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
11
  • $\begingroup$ I also had one more aspect in my question as to how can we arrive at this equation for mixtures. because deriving it is straightforward for one component systems but not for mixtures. $\endgroup$
    – DHYEY
    Apr 8, 2018 at 5:44
  • $\begingroup$ I don't quite understand what you are asking here. Are you asking about your third equation, and where it comes from? $\endgroup$ Apr 8, 2018 at 12:12
  • $\begingroup$ no i mean the logarithmic dependence of chemical potential... how do you reach to that in the case of mixtures.. $\endgroup$
    – DHYEY
    Apr 8, 2018 at 17:02
  • $\begingroup$ I still need a little more clarification. Are you asking how I know that the chemical potential of a species in an ideal gas mixture is determined by the log of its partial pressure in the mixture? Are you familiar with the change in entropy between an ideal gas mixture at T and P, and the same amounts of pure species, each at P and T? $\endgroup$ Apr 8, 2018 at 18:21
  • $\begingroup$ I just need a simple derivation of the explicit expression of u if possible from the expression of G=VdP-SdT+u(a)dn(a)+u(b)dn(b).. $\endgroup$
    – DHYEY
    Apr 10, 2018 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.