Vapour pressure of a liquid subjected to pressure I am following the derivation by Atkins in section 4B(c) (Physical transformations of pure substances; thermodynamic aspects of phase transitions), in which the authors derive the equation ${p = p_* * exp[V_m(liquid)* ΔP/RT]}$. The derivation starts by noting that chemical potentials of vapour and liquid are equal at equilibrium, $μ_{liquid} = μ_{gas}$, and follows by integration over the pressure to calculate the change in chemical potentials due to the added external pressure. 

What is unclear to me is why the integration for the liquid phase is taken over the ambient pressure while the integration for the gas phase is taken over the vapour (NOT ambient) pressure. It seems counterintuitive to me because I would expect that $μ_{gas}$ is a function of the ambient pressure and therefore should be integrated over the ambient pressure. Am I missing something?

EDIT: I guess the question could be rephrased: in the definition of molar volume for gas, $V_m=(\frac{\partial \mu} {\partial P})_{T,n}$, is $P$ the partial or ambient pressure? More generally, does $\mu_{gas}(T,P)$ depend on the partial or ambient pressure?
Here's how it works.  The fugacity of the liquid at temperature T and total pressure P is equal to the equilibrium vapor pressure at T times the Poynting correction (assuming that the vapor phase behaves as an ideal gas mixture).  The fugacity of the same species in the vapor is equal to the partial pressure of the substance in the gas phase.  The vapor and liquid are at equilibrium at temperature T and total pressure P if the fugacity of the substance in the liquid is equal to the fugacity of the substance in the vapor phase.