The chemical potentials of substance A in the liquid and gas phases depend on the concentration c^{(g)}_A in the gas phase and c^{(\ell)}_A in the liquid phase according to the ideal mixture mixing law (8.68),
μ_A (T, p, c_A) = μ_A (T, p) + R T \ln (c_A) (8.68)
\mu ^{(\ell )}_A \Bigl(T , p ,c^{(\ell )}_A\Bigr) = \mu ^{(\ell )}_A (T,p) + R T \ln \Bigl(c^{(\ell )}_A\Bigr) .
\mu ^{(g)}_A \Bigl(T , p ,c^{(g)}_A\Bigr) = \mu ^{(g)}_A (T,p) + R T \ln \Bigl(c^{(g)}_A\Bigr) .
Here, \mu ^{(\ell )}_A (T,p) and \mu ^{(g)}_A (T,p) are the chemical potential of the pure substance in the liquid and gas phases. When a concentration appears in the argument, then the substance is part of a mixture. Since the problem refers to the saturation pressure p^\circ _A, we want to introduce it in the relations above. For the gaseous phase, we simply apply relation (8.58) and write,
μ (T, p) = μ (T, p_0) + R T \ln \Bigl(\frac{p}{p_0}\Bigr) . (8.58)
\mu ^{(g)}_A \Bigl(T , p ,c^{(g)}_A\Bigr) = \mu ^{(g)}_A (T,p^\circ _A) + R T \ln \Bigl(\frac{p}{p^\circ _A}\Bigr) + R T \ln \Bigl(c^{(g)}_A\Bigr) .
For the liquid phase, we turn to relation (8.85), that was established for incompressible liquids, and write,
\mu (T,P) – \mu (T,p_{ext}) = \int_{p_{ext}}^{p}{\frac{\partial \mu }{\partial p}dp } = \nu \int_{p_{ext}}^{p}{dp} = \nu (p-p_{ext}) . (8.85)
\mu ^{(\ell )}_A \Bigl(T , p ,c^{(\ell )}_A\Bigr) = \mu ^{(\ell )}_A (T, p^\circ _A) + (p-p^\circ _A) \nu ^{(\ell )}_A + R T \ln \Bigl(c^{(\ell )}_A\Bigr) .
The equilibrium condition for substance A in the mixture reads,
\mu ^{(g)}_A \Bigl(T , p ,c^{(g)}_A\Bigr) = \mu ^{(\ell )}_A \Bigl(T , p ,c^{(\ell )}_A\Bigr) .
The saturation pressure p^\circ _A is defined by the equilibrium between the liquid and the gas phases of the pure substance, which is characterised by,
\mu ^{(g)}_A (T,p^\circ _A) = \mu ^{(\ell )}_A (T, p^\circ _A) .
Therefore, the equality of the chemical potentials of the substance A in the gas and liquid phases yields,
R T \ln \Bigl(\frac{p}{p^\circ _A}\Bigr) + R T\ln \bigl(c^{(g)}_A\bigr) = (p-p^\circ _A)\nu ^{(\ell )}_A + R T \ln \bigl(c^{(\ell )}_A\bigr) .
which can be recast as,
R T \ln \Bigl(\frac{p c^{(g)}_A }{p^\circ _A c^{(\ell )}_A}\Bigr) = (p-p^\circ _A)\nu ^{(\ell )}_A .
According to relation (8.67), the partial pressure of substance A in the gaseous phase is p_A = p c^{(g)}_A , and according to relation (8.89),
\frac{p_A}{p} = \frac{N_A}{\sum\limits_{B=1}^{r}{N_B} } =c_A (8.67)
p − p_{ext} = \frac{N_s RT}{V} (8.89)
(p − p^\circ _A) ν^{(g)}_A = R T
which implies that,
\ln \Biggl(\frac{p_A}{p^\circ _A c^{(\ell )}_A}\Biggr) = \frac{\nu ^{(\ell )}_A}{\nu ^{(g)}_A}
Since the molar volume in the liquid phase is negligible compared to the molar volumes in the gas phase, i.e. \nu ^{(\ell )}_A \ll \nu ^{(g)}_A
\ln \Biggl(\frac{p_A}{p^\circ _A c^{(\ell )}_A}\Biggr) \simeq 0
Thus, we recover Raoult’s law,
p_A = p^\circ _A c^{(\ell )}_A .