91Ó°ÊÓ

$\mathrm{β}$and liquid are in equilibrium only at temperatures below the point where the liquid is in equilibrium with infinitesimal amounts of $\mathrm{Î\pm }$ and $\mathrm{β}$ . This point is called a peritectic point.

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

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

The Gibbs free energy is given as:

$G=U-TS+PV$

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

$dG=dU-SdT+PdV$

By increasing the temperature

$\frac{∂G}{∂T}=-S$

We can see from this equation that as the temperature rises, the stability ranges of $\mathrm{Î\pm }$and $\mathrm{β}$vanish. When we lower the temperature, the stability of $\mathrm{γ}$plus the liquid appears, as shown in the preceding figure for large x. As the temperature drops, the liquid's stable range narrows until it evaporates at two locations, one of which is known as the eutectic point (at which all the liquid freezes). Only at temperatures below the red point, where the liquid is in equilibrium with tiny amounts of $\mathrm{Î\pm }$and $\mathrm{β}$ and the liquid in equilibrium. This point is known as the peritectic point.