In this problem you will derive approximate formulas for the shapes of the phase boundary curves in 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.

(a) Show that in an ideal mixture of A and B, the chemical potential of species A can be written ${\mu }_A={\mu }_A°+kT\ln (1-x)$where A is the chemical potential of pure A (at the same temperature and pressure) and $x=N_B/\left(N_A+N_B\right)$. Derive a similar formula for the chemical potential of species B. Note that both formulas can be written for either the liquid phase or the gas phase.

(b) At any given temperature T, let x₁ and x_gbe the compositions of the liquid and gas phases that are in equilibrium with each other. By setting the appropriate chemical potentials equal to each other, show that x₁and x_g obey the equations =$\frac{1-x_l}{1-x_g}=e^{\Delta G_A°/RT}\text{and}\frac{x_l}{x_g}=e^{\Delta G_B°/RT}$ and where $\Delta G°$represents the change in G for the pure substance undergoing the phase change at temperature T.

(c) Over a limited range of temperatures, we can often assume that the main temperature dependence of $\Delta G°=\Delta H°-T\Delta S°$comes from the explicit T; both $\Delta H°\text{and}\Delta S°$are approximately constant. With this simplification, rewrite the results of part (b) entirely in terms of $\Delta H_A°,\Delta H_B°$ T_A, and T_B (eliminating $\Delta G\text{and}\Delta S$). Solve for x₁and x_gas functions of T.

(d) Plot your results for the nitrogen-oxygen system. The latent heats of the pure substances are$\Delta H_{{\mathrm{N}}_2}°=5570\mathrm{J}/\mathrm{mol}\text{and}\Delta H_{{\mathrm{O}}_2}°=6820\mathrm{J}/\mathrm{mol}$. Compare to the experimental diagram, Figure 5.31.

(e) Show that you can account for the shape of Figure 5.32 with suitably chosen$\Delta H°$ values. What are those values?

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:

$G-G°={\int }_{p°}^{p}VdP$

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

$V=nRT/P$

$G-G°=nRT{\int }_{p°}^p\frac{dP}{P}$

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:

$G=N\mu$

Substitute this into (1)

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

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

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

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

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

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

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

For B we get

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

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

${\mu }_g={\mu }_l$

Using (2), we have

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

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

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

But $\Delta G_A°=N_A\Delta {\mu }_A°$

$\frac{\Delta G_A°}{RT}=\ln \left(\frac{1-x_l}{1-x_g}\right)$

Exponentiation both sides

$e^{\Delta G_A°/RT}=\frac{1-x_l}{1-x_g}\left(4\right)$

Now using (3) we get

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

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

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

But $\Delta G_B°=N_B\Delta {\mu }_B°$

$\frac{\Delta G_B°}{RT}=\ln \left(\frac{x_l}{x_g}\right)$

exponentiation both sides to get:

$e^{\Delta G_B°/RT}=\frac{x_l}{x_g}\left(5\right)$

For a short range temperatures we can use $\Delta G°=\Delta H°-T\Delta S°$

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

$$\begin{gathered} e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT}=\frac{1-x_l}{1-x_g} \\ {e}^{\left(\Delta H_B°-T\Delta S_B°\right)/RT}=\frac{x_l}{x_g} \end{gathered}$$

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

$$\begin{gathered} \left(1-x_g\right)e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT}=1-x_g{e}^{\left(\Delta H_B°-T\Delta S_B°\right)/RT} \\ {x}_g\left(e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT}-e^{\left(\Delta H_B°-T\Delta S_B°\right)/RT}\right)=1-e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT} \\ {x}_g=\frac{1-e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT}}{\left(e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT}-e^{\left(\Delta H_B°-T\Delta S_B°\right)/RT}\right)} \end{gathered}$$

Substitute in second equation

$x_l=\frac{\left(1-e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT}\right)e^{\left(\Delta H_B°-T\Delta S_B°\right)/RT}}{\left(e^{\left(\Delta H_A°-T\Delta S_A°\right)/RT}-e^{\left(\Delta H_B°-T\Delta S_B°\right)/RT}\right)}$