Chapter 15, Problem 15.36P

(a)

Interpretation Introduction

Interpretation:

To prove $\frac{G^E}{RT}=\frac{F_R\left(x\right)}{RT}$ .

Concept Introduction:

From equation of Gibbs Energy and related properties $S^E=-{\left(\frac{\partial G^E}{\partial T}\right)}_{P,x}$ .

For regular solution $S^E=0$everywhere.

(b)

Interpretation Introduction

Interpretation:

To prove $\frac{G^E}{RT}=F_A\left(x\right)$ .

Concept Introduction:

From equation of Excess Property Relations $\frac{H^E}{RT}=-T{\left(\frac{\partial \left(G^E/RT\right)}{\partial T}\right)}_{P,x}$ .

For a thermal solution $H^E=0$ everywhere.

(c)

Interpretation Introduction

Interpretation:

To interpret the implications to the equations $\frac{G^E}{RT}=\frac{\alpha }{RT}{x}_1{x}_2$ and $\frac{G^E}{RT}=\beta x_1{x}_2$ with respect to the shapes of predicted solubility diagrams of LLE.

Concept Introduction:

The simplest equation to calculate $\frac{G^E}{RT}$ is $\frac{G^E}{RT}=A\left(T\right)x_1{x}_2$and can be written in two different forms.

The regular solution is $\frac{G^E}{RT}=\frac{\alpha }{RT}{x}_1{x}_2$   --- $\left(A\right)$

The a thermal solution is $\frac{G^E}{RT}=\beta x_1{x}_2$ .  --- $\left(B\right)$

Here $\alpha ,\beta$ are constants.