Questions: Phase Equilibrium and Coexistence Conditions

Question 1

Two phases are in contact at the same temperature and pressure. What additional condition must be satisfied for them to coexist stably at equilibrium?

• A. The two phases must have equal entropy per particle
• B. The two phases must have equal chemical potential
• C. The latent heat of the transition between them must be zero
• D. The total Gibbs free energy of the system must equal zero

Answer: (B)

Thermodynamic equilibrium requires three conditions between coexisting phases: equal temperature (no heat flow), equal pressure (no mechanical work), and equal chemical potential (no particle transfer). Equal T and P are necessary but not sufficient — it is the chemical potential equality μ_α(T,P) = μ_β(T,P) that selects which (T,P) combinations allow coexistence. When μ differs between phases, particles migrate from the high-μ phase to the low-μ phase until equality is restored or one phase disappears. The phase diagram is a map of the (T,P) locus where these chemical potentials are equal.

Question 2

A student says: 'I can tell which phase is stable at a given temperature and pressure by finding which phase has higher entropy — nature maximizes entropy.' What is wrong with this reasoning?

• A. At fixed temperature and pressure, the stable phase is the one with lower Gibbs free energy (lower chemical potential per particle) — entropy maximization applies to isolated systems, not systems at fixed T and P
• B. Entropy is completely irrelevant to phase stability
• C. Nature always prefers the solid phase because it has the lowest entropy
• D. Only temperature determines phase stability; pressure is irrelevant

Answer: (A)

The entropy maximization principle applies to isolated systems (fixed U, V, N). At fixed temperature and pressure — the typical experimental condition — the relevant criterion is minimization of the Gibbs free energy G = H − TS. The phase with lower G (equivalently, lower chemical potential μ = G/N) is thermodynamically stable. This correctly incorporates both energetic and entropic contributions: at high T, the entropy term −TS dominates and favors disordered phases (liquid, gas); at low T, the enthalpy H dominates and favors ordered phases (solid). Entropy alone would always favor the gas phase.

Question 3

The solid-liquid phase boundary for water slopes negatively in the P-T diagram, meaning that applying pressure at constant temperature can cause ice to melt.

• T. True
• F. False

Answer: True

The Clausius-Clapeyron equation gives dP/dT = L/(TΔV), where ΔV = V_liquid − V_solid. For water, ice is less dense than liquid water, so V_liquid < V_solid and ΔV < 0. With L > 0 (melting absorbs heat) and T > 0, the slope dP/dT is negative. This means the melting point decreases as pressure increases — apply enough pressure at a temperature just below 0°C and the ice will melt. This anomaly (water expanding on freezing) is rare among substances and has practical consequences including ice skating.

Question 4

Above the critical point in a substance's phase diagram, there is still a phase boundary separating the liquid from the gas phase.

• T. True
• F. False

Answer: False

Above the critical point (Tc, Pc), the distinction between liquid and gas disappears entirely. The latent heat goes to zero and the density difference ΔV → 0 as the critical point is approached. Above it, the substance exists as a supercritical fluid — a single phase that can be continuously converted from liquid-like to gas-like density without crossing any phase boundary. The liquid-gas boundary in the P-T diagram terminates at the critical point; it does not extend beyond it. This means you can take liquid water, heat it above Tc at high pressure, reduce pressure, and end up as steam — never crossing a phase boundary.

Question 5

Why is equal chemical potential — rather than just equal temperature and pressure — the correct condition for phase coexistence, and how does this directly lead to the Clausius-Clapeyron equation?

Model answer: Chemical potential μ is the free energy cost of adding one particle at fixed T and P. Any imbalance in μ between phases drives particle flow from the high-μ phase to the low-μ phase, just as a temperature difference drives heat flow. At equilibrium, T, P, and μ must all be equal between coexisting phases. The phase diagram traces the (T,P) locus where μ_α(T,P) = μ_β(T,P). The Clausius-Clapeyron equation follows directly: as T changes along this coexistence curve, P must adjust to maintain μ_α = μ_β. Using dμ = −s dT + v dP (from the Gibbs-Duhem relation), setting dμ_α = dμ_β gives (−s_α + v_α dP/dT) dT = (−s_β + v_β dP/dT) dT, which rearranges to dP/dT = (s_β − s_α)/(v_β − v_α) = ΔS/ΔV = L/(TΔV), where L = TΔS is the latent heat. The slope of every phase boundary is determined by the latent heat and the volume change at that transition.