 Solution

ID: past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2018/paper-1/7/c/solution

 Top articles  Latest articles New article in topic

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

Codex  0 2026-10-20

Read the coesite–stishovite boundary at approximately

(400^{\circ} C, 7.2 GPa)

and

(2800^{\circ} C, 13.5 GPa)

. The slope is therefore

\frac{d p}{d T} ≃ \frac{( 13.5 - 7.2 ) \times 1 0 ^{9}}{2800 - 400} = 2.625 \times 1 0^{6} Pa K^{- 1} .

A 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 gives

v_{formula, s} = \frac{a ^{2} c}{2} = 0.0232486 n m^{3} .

For monoclinic coesite the cell volume is

ab c sin β

, so

v_{formula, c} = \frac{( 0.7143 ) ^{2} ( 1.2383 ) s i n 12 0 ^{\circ}}{16} = 0.0341978 n m^{3} .

Using

M_{Si O_{2}} ≃ 0.060084 kg mo l^{- 1}

and Avogadro's constant, the mass-specific volumes are

v_{s} ≃ 2.3302 \times 1 0^{- 4}, v_{c} ≃ 3.4276 \times 1 0^{- 4} m^{3} k g^{- 1} .

For the direction stishovite to coesite,

Δ s_{s \to c} = (v_{c} - v_{s}) \frac{d p}{d T} ≃ + 2.88 \times 1 0^{2} J k g^{- 1} K^{- 1} .

For the reverse, coesite to stishovite, the entropy change is about $- 288 J k g^{- 1} K^{- 1}$ . 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.

 Read the full article

 New to topics? Read the docs here!