91Ó°ÊÓ

Diagrams such as Figures 5.31 and 5.32, assuming that both phases behave as ideal mixtures. For definiteness, suppose that the phases are liquid and gas.

The Gibbs free energy is given by:

$G=U-TS+PV$

For a single component ideal gas system at constant temperature, the change in Gibbs free energy with pressure is:

$G-G°={∫}_{p°}^{p}VdP$

where,

$p°$is the pressure of the pure Substance

$G°$is the Gibbs free energy of the pure substance.

Substituting from the ideal gas law, we get

$V=nRT/P$

$G-G°=nRT{∫}_{p°}^p\frac{dP}{P}$

Integrating it,

role="math" localid="1647032806277" $$\begin{gathered} G-G°=nRT\left[\ln \right(P){]}_{p°}^p \\ G-G°=nRT\ln \left(\frac{p}{p°}\right) \\ G=G°+nRT\ln \left(\frac{p}{p°}\right)(1) \end{gathered}$$

In terms of chemical potential, the Gibbs free energy is given by:

$G=N\mathrm{μ}$

Substitute this into (1)

$$\begin{gathered} N\mathrm{μ}=N\mathrm{μ}°+NkT\ln \left(\frac{p}{p°}\right) \\ \mathrm{μ}=\mathrm{μ}°+kT\ln \left(\frac{p}{p°}\right) \end{gathered}$$

Let's say we have substance A, and its chemical potential is as follows:

${\mathrm{μ}}_A={\mathrm{μ}}_A°+kT\ln \left(\frac{p}{p°}\right)$

When another substance B is added to A, the chemical potential of A in the mixture is calculated as follows:

${\mathrm{μ}}_A={\mathrm{μ}}_A°+kT\ln \left(\frac{p_A}{p°}\right)$

Let the fraction of the substance B be r thus the fraction of substance A is 1- a, the pressure can be written as:

$p_A=(1-x)p°$

Now, we get

${\mathrm{μ}}_A={\mathrm{μ}}_A°+kT\ln (1-x)\left(2\right)$

For B we get

${\mathrm{μ}}_B={\mathrm{μ}}_B°+kT\ln \left(x\right)\left(3\right)$

Let x₁ and x_g be the equilibrium compositions of the liquid and gas phases, respectively, and equating the chemical potentials yields:

${\mathrm{μ}}_g={\mathrm{μ}}_l$

Using (2), we have

$$\begin{gathered} {\mathrm{μ}}_g°+kT\ln \left(1-x_g\right)={\mathrm{μ}}_l°+kT\ln \left(1-x_l\right) \\ \frac{{\mathrm{μ}}_g°-{\mathrm{μ}}_l°}{kT}=\ln \left(1-x_l\right)-\ln \left(1-x_g\right) \end{gathered}$$

For 1 mole, we have $N_{A}k=R$

$$\begin{gathered} \frac{N_A\left({\mathrm{μ}}_g°-{\mathrm{μ}}_l°\right)}{RT}=\ln \left(\frac{1-x_l}{1-x_g}\right) \\ \frac{N_A\mathrm{Δ}{\mathrm{μ}}_A°}{RT}=\ln \left(\frac{1-x_l}{1-x_g}\right) \end{gathered}$$

But $\mathrm{Δ}{G}_A°=N_A\mathrm{Δ}{\mathrm{μ}}_A°$

$\frac{\mathrm{Δ}{G}_A°}{RT}=\ln \left(\frac{1-x_l}{1-x_g}\right)$

Exponentiation both sides

$e^{\mathrm{Δ}{G}_A°/RT}=\frac{1-x_l}{1-x_g}\left(4\right)$

Now using (3) we get

$$\begin{gathered} {\mathrm{μ}}_g°+kT\ln \left(x_g\right)={\mathrm{μ}}_l°+kT\ln \left(x_l\right) \\ \frac{{\mathrm{μ}}_g°-{\mathrm{μ}}_l°}{kT}=\ln \left(x_l\right)-\ln \left(x_g\right) \end{gathered}$$

For 1 mole we have $N_{B}k=R$

$$\begin{gathered} \frac{N_B\left({\mathrm{μ}}_g°-{\mathrm{μ}}_l°\right)}{RT}=\ln \left(\frac{x_l}{x_g}\right) \\ \frac{N_B\mathrm{Δ}{\mathrm{μ}}_A°}{RT}=\ln \left(\frac{x_l}{x_g}\right) \end{gathered}$$

But $\mathrm{Δ}{G}_B°=N_B\mathrm{Δ}{\mathrm{μ}}_B°$

$\frac{\mathrm{Δ}{G}_B°}{RT}=\ln \left(\frac{x_l}{x_g}\right)$

exponentiation both sides to get:

$e^{\mathrm{Δ}{G}_B°/RT}=\frac{x_l}{x_g}\left(5\right)$

For a short range temperatures we can use $\mathrm{Δ}G°=\mathrm{Δ}H°-T\mathrm{Δ}S°$

So from (4) and (5), we get

$$\begin{gathered} e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT}=\frac{1-x_l}{1-x_g} \\ {e}^{\left(\mathrm{Δ}{H}_B°-T\mathrm{Δ}{S}_B°\right)/RT}=\frac{x_l}{x_g} \end{gathered}$$

To eliminate it, we must solve these equations for and x₁and x_g, substituting from the second equation into the first:

$$\begin{gathered} \left(1-x_g\right)e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT}=1-x_g{e}^{\left(\mathrm{Δ}{H}_B°-T\mathrm{Δ}{S}_B°\right)/RT} \\ {x}_g\left(e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT}-e^{\left(\mathrm{Δ}{H}_B°-T\mathrm{Δ}{S}_B°\right)/RT}\right)=1-e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT} \\ {x}_g=\frac{1-e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT}}{\left(e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT}-e^{\left(\mathrm{Δ}{H}_B°-T\mathrm{Δ}{S}_B°\right)/RT}\right)} \end{gathered}$$

Substitute in second equation

$x_l=\frac{\left(1-e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT}\right)e^{\left(\mathrm{Δ}{H}_B°-T\mathrm{Δ}{S}_B°\right)/RT}}{\left(e^{\left(\mathrm{Δ}{H}_A°-T\mathrm{Δ}{S}_A°\right)/RT}-e^{\left(\mathrm{Δ}{H}_B°-T\mathrm{Δ}{S}_B°\right)/RT}\right)}$