Consider a system formed by two phases, liquid and gas. The composition of the liquid phase is given by the mole fractions $x_i$ and that of the gas phase by $x_{iv}$. The partial pressures exerted by the components in the gas phase are $P_i$.

Once equilibrium is reached, the chemical potential of each component in the liquid and vapor phases equalize. $$\mu_{i,v}=\mu_{i,l}$$ We write the chemical potential of a component $i$ in the vapor phase, assuming that it behaves ideally.

$$\mu_{i,v}=\mu_{i,v}^{0}+RTln\frac{P_i}{P^0}$$

The chemical potential of a component $i$ in the liquid, assuming ideal solution, it will be given by:

$$\mu_{i,l}=\mu_{i(l)}^{\ast}+RTlnx_i$$

Equating both chemical potentials

$$\mu_{i,v}^{0}+RTln\frac{P_i}{P^0}=\mu_{i,l}^{\ast}+RTlnx_i \label{130}$$

We now consider the equilibrium of the pure liquid with its vapor

$$\mu_{i,v}^{\ast}(T,P_{i}^{\ast})=\mu_{i,l}^{\ast} (T,P_{i}^{\ast})$$

Since the chemical potential of pure $i$ in the vapor phase is given by:

$$\mu_{i,v}^{\ast}(T,P_{i}^{\ast})=\mu_{i,v}^{0}(T)+RTln\frac{P_{i}^{\ast}}{P^0}$$

Substituting

$$\mu_{i,v}^{0}(T)+RTln\frac{P_{i}^{\ast}}{P^0}=\mu_{i ,l}^{\ast}(T,P_{i}^{\ast})\label{131}$$

Since chemical potential changes little with pressure for liquids, $\mu_{i,l} ^{\ast}(T,P_{i}^{\ast})=\mu_{i,l}^{\ast}(T,P)$, substituting  (\ref{131}) into equation (\ref{130})

$$\cancel{\mu_{i,v}^{0}}+RTln\frac{P_i}{P^0}=\cancel{\mu_{i,v}^{ 0}}+RTln\frac{P_{i}^{\ast}}{P^0}+RTlnx_i$$

Simplifying the normal potentials in the gas phase and the RT we get:

$$ln\frac{P_i}{P^0}-ln\frac{P_{i}^{\ast}}{P_0}=lnx_i$$

Applying properties of natural logarithms

$$ln\frac{P_i} {P_{i}^{\ast}}=lnx_i$$

Taking the inverse of the Neperian and solving for the pressure, gives us:

$$P_i=x_iP_{i}^{\ast}$$

This last equation is known as Raoult's law

$P_i$: Vapor pressure of component i.

$P_{i}^{\ast}$: Vapor pressure of pure component i.

$x_i$: Mole fraction of component i in the solution.