Questions: Phase Transitions and Equilibrium Phase Diagrams

Question 1

At the liquid-gas coexistence curve (boiling point), both liquid and gas phases coexist. What determines the fraction of the system that is in the gas phase?

• A. The temperature alone — the fraction is fixed at each temperature
• B. The pressure alone — higher pressure means more liquid
• C. The total volume of the system — the fraction adjusts to satisfy volume conservation at equal Gibbs free energy
• D. Random fluctuations — the system constantly shifts between all-liquid and all-gas

Answer: (C)

At the coexistence point, G has two equal-depth minima (liquid and gas). The actual fraction in each phase is determined by the lever rule: the total volume of the system must be conserved as it partitions between the two phases at their respective specific volumes. A container with more total volume at fixed temperature will have more gas; less volume means more liquid. Temperature alone doesn't fix the fraction — it only fixes the coexistence condition (equal G). This is why a pressure cooker at its rated pressure contains liquid water as long as liquid remains.

Question 2

Why does increasing pressure lower the melting point of ice, while increasing pressure raises the boiling point of water?

• A. Ice is less dense than liquid water, so ΔV < 0 for melting; liquid water is less dense than steam, so ΔV > 0 for vaporization
• B. The Clausius-Clapeyron equation only applies to liquid-gas transitions
• C. Ice and water have the same density, so the slope is governed by latent heat alone
• D. The anomaly arises because ice has lower entropy than water

Answer: (A)

The Clausius-Clapeyron equation dP/dT = ΔS/ΔV = L/(TΔV) shows that the sign of the coexistence curve slope depends on the sign of ΔV. For the solid-liquid transition of water, ice is less dense than liquid water, so melting increases density and ΔV = V_liquid − V_ice < 0. This gives a negative slope: increasing pressure lowers the melting point. For liquid-gas, gas is far less dense than liquid (ΔV > 0), giving a positive slope: increasing pressure raises the boiling point. This is not an anomaly requiring special explanation — it follows directly from the equation and the known density ordering.

Question 3

A second-order phase transition involves a discontinuous jump in the order parameter (e.g., magnetization drops abruptly to zero at the Curie temperature).

• T. True
• F. False

Answer: False

This is the defining difference between first-order and second-order transitions. In a second-order (continuous) transition, the order parameter decreases *continuously* to zero as the critical temperature is approached — no abrupt jump, no latent heat, no phase coexistence. What diverges at a second-order transition is not the order parameter but its *susceptibility* (response to external fields) and the correlation length. A first-order transition like melting does show a discontinuous jump in density and entropy.

Question 4

Near a second-order critical point, both the susceptibility and the correlation length diverge, even though the order parameter itself goes smoothly to zero.

• T. True
• F. False

Answer: True

This is precisely what characterizes a second-order transition and distinguishes it from smooth crossovers. As T → T_c, the free energy curvature flattens — the system becomes infinitely responsive to small perturbations (divergent susceptibility). The correlation length — the scale over which fluctuations are correlated — also diverges, meaning fluctuations occur on all length scales simultaneously. This scale-invariance at the critical point is responsible for critical opalescence: light scatters off density fluctuations at all scales, making the fluid appear milky.

Question 5

Explain how the Clausius-Clapeyron equation is derived, and what physical insight does it capture about coexistence curves?

Model answer: Along a coexistence curve, both phases have equal Gibbs free energy: G_A = G_B. As T and P vary together along the curve, dG_A = dG_B, which gives −S_A dT + V_A dP = −S_B dT + V_B dP. Rearranging: dP/dT = (S_A − S_B)/(V_A − V_B) = ΔS/ΔV = L/(TΔV). The equation captures that phase boundaries in P-T space are not arbitrary lines but are determined by the competition between entropy gain and volume change accompanying the transition.

The derivation uses only the condition of equal Gibbs free energy along the coexistence curve — no detailed molecular model is needed. The result encodes two key insights: the slope of the coexistence curve depends on the ratio of latent heat to volume change, and the sign of ΔV determines whether increased pressure favors the denser phase (raises its transition temperature) or the less dense phase (lowers it). Water's anomalous negative melting-point slope is explained entirely by the fact that ice is less dense than liquid water.