Question

Figure 5.35 (left) shows the free energy curves at one particular temperature for a two-component system that has three possible solid phases (crystal structures), one of essentially pure A, one of essentially pure B, and one of intermediate composition. Draw tangent lines to determine which phases are present at which values of x. To determine qualitatively what happens at other temperatures, you can simply shift the liquid free energy curve up or down (since the entropy of the liquid is larger than that of any solid). Do so, and construct a qualitative phase diagram for this system. You should find two eutectic points. Examples of systems with this behaviour include water + ethylene glycol and tin - magnesium.

Short answer

Starting at x=0 on the left, the $\alpha$phase and liquid phase are stable, then the liquid phase is stable, then the $\beta$ phase and liquid phase are stable, then a small range of x where only the $\beta$ phase is stable, and ultimately just the liquid phase is stable.

Step-by-step solution

Given information

The free energy curves at one particular temperature for a two-component system that has three possible solid phases (crystal structures), one of essentially pure A, one of essentially pure B, and one of intermediate composition.

Step 2: 

Consider the Gibbs free energy graph below, which shows a system with three solid phases: $\alpha ,\beta \text{and}\gamma$. One is a pure A substance, one is a pure B substance, and one is a mixture of the two.

Explanation 

First, as shown in the diagram, we draw three tangent lines from left to right: first, the $\alpha$phase plus the liquid phase are stable, then the liquid phase is stable, then the $\beta$phase plus the liquid phase are stable, then a narrow range of x where only the $\beta$phase is stable, and finally only the liquid phase is stable.

Calculations

The Gibbs free energy is given by

$G=U-TS+PV$

At constant entropy and constant pressure, differentiate the Gibbs free energy to get:

role="math" localid="1647065790638" $dG=dU-SdT+PdV$

By increasing the temperature,

$\frac{\partial G}{\partial T}=-S$

We can see from this equation that as the temperature rises, the stability ranges of $\alpha$and $\beta$ vanish. When we lower the temperature, the stability of $\gamma$ plus the liquid appears, as seen in the previous figure for large x. As the temperature drops, the liquid's stable range narrows until it vanishes at two locations, which are known as the eutectic points (where all the liquid freezes).