Past exam of the materials science course of the University of Cambridge / 2018 / Paper 1 / 7 / c / Solution

Read the coesite–stishovite boundary at approximately $(400{}^{\circ }\mathrm{C},7.2\mathrm{GPa})$ and $(2800{}^{\circ }\mathrm{C},13.5\mathrm{GPa})$. The slope is thereforeA Celsius temperature difference equals the corresponding kelvin difference. The graph does not justify treating the endpoints as exact measurements.

For stishovite, the tetragonal cell volume divided by two givesFor monoclinic coesite the cell volume is $abc\sin \beta$, soUsing $M_{\mathrm{Si}{\mathrm{O}}_2}\simeq 0.060084\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ and Avogadro's constant, the mass-specific volumes areFor the direction stishovite to coesite,For the reverse, coesite to stishovite, the entropy change is about $-288\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Stishovite is the denser phase: a positive boundary slope and its smaller volume imply a lower entropy on the high-pressure side. Stating the direction prevents an otherwise avoidable sign ambiguity.