= 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 therefore
$$
\frac{dp}{dT}\simeq\frac{(13.5-7.2)\times10^9}{2800-400}
=2.625\times10^6\,\mathrm{Pa\,K^{-1}}.
$$
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
$$
v_{{\rm formula},s}=\frac{a^2c}{2}
=0.0232486\,\mathrm{nm^3}.
$$
For monoclinic <coesite> the cell volume is $abc\sin\beta$, so
$$
v_{{\rm formula},c}=\frac{(0.7143)^2(1.2383)\sin120^\circ}{16}
=0.0341978\,\mathrm{nm^3}.
$$
Using $M_{\mathrm{SiO_2}}\simeq0.060084\,\mathrm{kg\,mol^{-1}}$ and Avogadro's constant, the mass-specific volumes are
$$
v_s\simeq2.3302\times10^{-4},\qquad
v_c\simeq3.4276\times10^{-4}\quad\mathrm{m^3\,kg^{-1}}.
$$
For the direction \b[<stishovite> to <coesite>],
$$
\boxed{\Delta s_{s\to c}=(v_c-v_s)\frac{dp}{dT}
\simeq+2.88\times10^2\,\mathrm{J\,kg^{-1}\,K^{-1}}.}
$$
For the reverse, <coesite> to <stishovite>, the <entropy> change is \b[about $-288\,\mathrm{J\,kg^{-1}\,K^{-1}}$]. <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.
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