 Solution

ID: past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2015/paper-1/6/d/solution

Use the same molar amount for both thermodynamic phases. A hexagonal cell built from two basal lattice vectors at120degrees has volume

(3 /2) a^{2} c

, not the volume of a three-times-larger full hexagonal prism. With four atoms per stated graphite cell,

V_{m, g} = \frac{N _{A}}{4} \frac{3}{2} (0.246 \times 1 0^{- 9})^{2} (0.671 \times 1 0^{- 9}) ≃ 5.2944 \times 1 0^{- 6} m^{3} mo l^{- 1} .

Diamond has eight atoms per conventional cubic cell, so

V_{m, d} = \frac{N _{A}}{8} (0.357 \times 1 0^{- 9})^{3} ≃ 3.4250 \times 1 0^{- 6} m^{3} mo l^{- 1} .

For graphite transforming to diamond,

Δ V_{m} ≃ - 1.8693 \times 1 0^{- 6} m^{3} mo l^{- 1}, Δ S_{m} = 2.45 - 5.69 = - 3.24 J K^{- 1} mo l^{- 1} .

Therefore

\frac{d P}{d T} ≃ 1.73 \times 1 0^{6} Pa K^{- 1} = 1.73 MPa K^{- 1} .

The positive slope reflects diamond's smaller entropy and volume. Higher temperature favors the higher-entropy graphite thermodynamic phase, requiring greater pressure to compensate in favor of the smaller-volume diamond thermodynamic phase.

This estimate treats the supplied ambient molar entropies and volumes as applicable near coexistence and neglects their changes through thermal expansion, compressibility and heat capacity. It gives an approximate gradient, not the transition pressure or a complete high-pressure thermodynamic phase boundary; an intercept would require an additional free-energy datum.

 New to topics? Read the docs here!