Chemical potential of mixtures
It's much simpler than all that.

Pure gas
$$\mu_g(T,P)=\mu^0(T)+RT\ln P$$
where $\mu^0(T)$ is the molar free energy of the pure gas at temperature T and pressure 1 Bar, and P is the pressure expressed in bars.

Pure liquid

If $P_{sat}(T)$ is the equilibrium vapor pressure at temperature T, then, at pressure $P_{sat}(T)$, we know that the free energy of the liquid is equal to the free energy of the vapor.  So, $$\mu_l(T,P_{sat})=\mu^0(T)+RT\ln P_{sat}(T)$$If we neglect the effect of pressure on the chemical potential of the liquid (because the specific volume is very small), then, for a pure liquid at temperature T and pressure P:$$\mu_l(T,P)=\mu^0(T)+RT\ln P_{sat}(T)$$

Ideal gas mixture

For species A in an ideal gas mixture, the chemical potential of species A is the same as the chemical potential of the pure species, except evaluated at the partial pressure of A in the gas mixture:$$\mu_{Ag}(T,P,y_A)=\mu^0_A(T)+RT\ln p_A=\mu^0_A(T)+RT\ln (Py)_A=\mu_A(T,P)+RT\ln y_A$$
where $p_A$ is the partial pressure of A (bars), $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, P is the total pressure of the mixture (bars), $\mu_A(T,P)$ is the free energy of pure A at the temperature and  pressure of the mixture, and $y_A$ is the mole fraction of A in the mixture.  So the chemical potential of A in the mixture is equal to the chemical potential of pure A at the same temperature and pressure as the mixture plus the correction term for mixing $RT\ln y_A$.

Ideal liquid mixture

By analogy to an ideal gas mixture, the chemical potential of species A in an ideal liquid solution is given by:

$$\mu_{Al}(T,P,x_A)=\mu_{Al}(T,P)+RT\ln x_A=\mu^0_A(T)+RT\ln P_{A,sat}(T)+RT\ln x_A$$
where $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, $P_{A,sat}(T)$ is the saturation vapor pressure of species A at temperature T (bars), and $x_A$ is the mole fraction of species A in the liquid.