 Solution

ID: past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2015/paper-1/7/a/solution

At two-phase equilibrium, the endpoints of a common tangent to the two molar free-energy curves give the coexisting compositions. They are not the separate curve minima. Reading the graph gives approximately

x_{B, α} ≃ 0.02

and

x_{B, m} ≃ 0.64

, with a few atomic-percent graphical uncertainty.

Let

f_{α}

be the fraction of atoms in the precipitate thermodynamic phase. Conservation of the overall B fraction gives

0.55 = f_{α} x_{B, α} + (1 - f_{α}) x_{B, m},

and hence

f_{α} = \frac{x _{B, m} - 0.55}{x _{B, m} - x _{B, α}} ≃ \frac{0.64 - 0.55}{0.64 - 0.02} ≃ 0.145 \approx 15%.

About85percent remains in the B-richer matrix. Smooth readings of the original curves put the result near14–15percent; the graph does not justify many significant figures. The result is an atomic or molar thermodynamic phase fraction. Its identification with a volume fraction in the next estimate requires comparable atomic volumes of the two thermodynamic phases.

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