The Al2SiO5 system has three polymorphs: andalusite (low P, moderate T), kyanite (high P), and sillimanite (high T). A metamorphic rock contains kyanite partially replaced by sillimanite. What does this indicate about the rock's thermal history?
AThe rock cooled rapidly from magmatic temperatures
BThe rock experienced increasing temperature at relatively high pressure, crossing the kyanite-sillimanite phase boundary during prograde metamorphism
CThe rock was weathered at the surface
DKyanite transformed to sillimanite during decompression at constant temperature
The kyanite-sillimanite boundary has a positive slope in P-T space -- crossing it requires temperature increase at pressures where kyanite is initially stable (above the andalusite field). The partial replacement texture records the reaction in progress, indicating the rock was heating. This mineral transition is diagnostic of upper amphibolite to granulite facies metamorphism and is a classic example of using phase diagrams to interpret metamorphic history.
Question 2 True / False
A univariant reaction boundary on a P-T phase diagram has zero width because the reaction occurs at a single precise temperature for any given pressure.
TTrue
FFalse
Answer: True
For a univariant reaction in a pure system (one degree of freedom along the boundary by the phase rule), at any given pressure there is exactly one temperature where the reactant and product assemblages coexist in equilibrium. Above that temperature, products are stable; below, reactants are stable. The boundary is a line, not a band. In natural systems with solid solutions, reactions may occur over a narrow T-P interval, but the fundamental thermodynamic constraint makes the boundary a sharp line in end-member systems.
Question 3 Short Answer
Explain how the Clausius-Clapeyron equation predicts the slope of the graphite-diamond phase boundary and what this slope tells us about diamond formation.
Think about your answer, then reveal below.
Model answer: The Clausius-Clapeyron slope dP/dT = delta-S/delta-V. Diamond has lower molar volume than graphite (denser packing) so delta-V (graphite to diamond) is negative. The entropy change is small and slightly negative (diamond is more ordered). The ratio gives a steep positive slope in P-T space, meaning diamond stability requires very high pressure (>4 GPa) regardless of temperature. This tells us diamonds form only at depths exceeding ~150 km in the mantle, where pressures are sufficient. The steep slope means temperature variations matter much less than pressure for crossing this boundary.
The negative delta-V dominates: converting to the denser phase requires high pressure. The steep positive slope means diamonds are a pressure indicator -- they require mantle depths, not just high temperatures.