Question 8.10: A container at a pressure p and temperature T contains two s...

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)

which implies that,

Since the molar volume in the liquid phase is negligible compared to the molar volumes in the gas phase, i.e. $\nu ^{(\ell )}_A \ll \nu ^{(g)}_A$

Thus, we recover Raoult’s law,

$p_A = p^\circ _A c^{(\ell )}_A$.