# 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?

• It assumes that, in the gas phase, the mixture behaves as an ideal gas mixture. Apr 28, 2020 at 0:03
• Thanks @Chet Miller, it helps! So, $V_m(gas) = RT/p$, and we then change the integration variable from the total pressure $P$ to the vapor pressure $p$ by taking advantage of the fact that $dP = dp$ and replacing the gas-phase integration range accordingly. This, together with some approximation, will lead to the equation in the textbook. I see the logic now. Apr 28, 2020 at 0:12
• On second thought, it is not obvious why $dP = dp$. All we know is that the vapour pressure is some function of the ambient pressure but it is not necessarily a linear function. Apr 28, 2020 at 0:24