Questions: Phase Diagrams and Clausius-Clapeyron Equation

Question 1

Water's solid-liquid phase boundary has a negative slope on a P-T diagram — it tilts left as pressure increases. What physical property of water causes this anomaly?

• A. Water has an unusually high enthalpy of vaporization compared to other small molecules
• B. Ice is less dense than liquid water, so increased pressure destabilizes the solid and favors melting
• C. Hydrogen bonds in liquid water are stronger than in ice, making the liquid thermodynamically preferred
• D. The critical point for water occurs at an exceptionally high temperature and pressure

Answer: (B)

For most substances the solid is denser than the liquid, so increasing pressure favors the solid phase and the solid-liquid line slopes right (positive slope). Water is anomalous because ice is less dense than liquid water — pressure therefore favors the denser liquid, and the melting point actually decreases as pressure rises, producing a negative slope. This is why ice melts under the blade of an ice skate.

Question 2

A substance has a large positive enthalpy of vaporization (ΔH_vap). Compared to a substance with a small ΔH_vap, how will its vapor pressure respond to a given temperature increase?

• A. Vapor pressure will change very little because the large enthalpy barrier slows evaporation
• B. Vapor pressure will decrease because high ΔH_vap stabilizes the liquid phase
• C. Vapor pressure will increase more steeply because the large ΔH_vap amplifies the temperature dependence
• D. Vapor pressure is independent of ΔH_vap; only molar mass determines how it changes with temperature

Answer: (C)

From the integrated Clausius-Clapeyron equation, ln(P₂/P₁) = −ΔH_vap/R × (1/T₂ − 1/T₁), a larger ΔH_vap multiplies the right-hand side, so the same temperature change produces a larger change in ln(P). High-ΔH_vap substances have vapor pressure curves that rise steeply with temperature. The common misconception is confusing 'hard to boil' (high boiling point) with 'vapor pressure changes slowly.'

Question 3

At the triple point of a substance, all three phases — solid, liquid, and gas — coexist in thermodynamic equilibrium simultaneously.

• T. True
• F. False

Answer: True

The triple point is the unique temperature-pressure combination where the free energies of all three phases are equal, so all three phases can coexist. For water this is 0.01°C and 611 Pa. It is the only point where the solid-liquid, liquid-gas, and solid-gas boundary lines all meet.

Question 4

Above the critical temperature, applying sufficient pressure to a gas will eventually condense it into a distinct liquid phase.

• T. True
• F. False

Answer: False

Above the critical temperature, the liquid-gas phase boundary no longer exists — the distinction between liquid and gas disappears entirely. Compressing a gas above its critical temperature produces a supercritical fluid, which has properties intermediate between liquid and gas, but no phase transition occurs. You can only condense a gas into a liquid if the temperature is below the critical temperature.

Question 5

Why does water boil at a lower temperature in Denver than at sea level, and how does the Clausius-Clapeyron equation explain this quantitatively?

Model answer: Denver's higher altitude means lower atmospheric pressure. The boiling point is the temperature at which vapor pressure equals atmospheric pressure, so lower pressure means the vapor pressure threshold is reached at a lower temperature. The Clausius-Clapeyron equation, ln(P₂/P₁) = −ΔH_vap/R × (1/T₂ − 1/T₁), lets you calculate the new boiling point from the known boiling point at sea level, ΔH_vap, and the reduced pressure at altitude.

This connects the abstract equation to a concrete, observable phenomenon. The equation shows that pressure and boiling temperature are locked together via ΔH_vap: lower P means lower T_boil. This is why pressure cookers work in reverse — elevated pressure raises the boiling point above 100°C, cooking food faster.