Phase Coexistence as Equality of Chemical Potential

Introduction

Liquid water and its vapour can share a container at equilibrium even though their densities, entropies and molecular arrangements differ. The thermodynamic requirement is not equality of those properties. It is equality of the chemical potential of water in the two phases, together with common temperature and pressure under ordinary mechanical and thermal equilibrium. This condition ties microscopic particle exchange to the geometry of a phase diagram.

Core explanation

Take a pure substance in phases α and β at fixed T and p. Transfer a small positive dn from α to β while conserving total amount. The combined Gibbs-energy change is dG = (μ β − μ α)dn. If μ β < μ α, the transfer lowers G and β is favoured. At coexistence μ α(T,p) = μ β(T,p), so the first-order transfer cost vanishes. This equality is a necessary equilibrium condition when transfer is possible, independent of how much material sits in each phase.

For a one-component system, each phase has μ = G m. Plotting μ for solid, liquid and gas at a fixed pressure, the lowest curve is stable and crossings mark transitions. In a p–T phase diagram, the condition μ α(T,p) = μ β(T,p) traces a line. Three-phase coexistence requires equality of three μ values and occurs at an isolated triple point for the usual one-component phase diagram. Away from coexistence, one phase has the lower μ, though another may persist temporarily through nucleation barriers.

Statistical mechanics interprets μ as an incremental free-energy cost of adding a particle. At coexistence, moving a particle from one phase to the other has no net free-energy preference after accounting for energy and entropy. Microscopic evaporation and condensation continue; their average rates balance. Different densities can coexist because the free-energy minimum may support two distinct structures at the same μ, T and p.

For mixtures, phase equilibrium requires μ i^α = μ i^β for every component i that can exchange between phases. The compositions of α and β usually differ. For example, a binary liquid and vapour can have different mole fractions while each component's chemical potential matches across the boundary. Equal total Gibbs energies or equal average compositions are not substitutes for these componentwise conditions.

Step-by-step reasoning

State common T and p and identify which species can cross. Write one positive and one negative amount change for the transfer and calculate dG. Set its coefficient to zero for coexistence. For pure phases, compare μ curves to determine stability. For mixtures, impose equality for each transferable component and solve with material balances to find compositions and phase amounts.

Visual explanation

Draw two boxes labelled liquid and vapour with double arrows of molecules crossing. Put μ l and μ v beside them. Under the drawing plot μ l(T) and μ v(T) crossing at the boiling temperature for one pressure. On the p–T diagram, extend the equality into a coexistence curve rather than one isolated temperature.

Real-world analogy

Two connected reservoirs can have different shapes and water depths yet remain in balance when the relevant potential for moving a small amount is equal. Phase equilibrium likewise balances the transfer cost, not visual similarity of phases. The analogy is limited because phase chemical potential includes entropy and interactions, not just gravitational height.

Real-world example

In a sealed vessel containing liquid and vapour of one pure substance, increasing temperature changes both phase μ values and the equilibrium vapour pressure needed for equality. At a temperature below the critical point, the system can retain both phases if overall amount and volume allow. A vapour-pressure curve is therefore a record of chemical-potential equality across temperatures.

Why?

At fixed T and p, Gibbs energy must be at a minimum with respect to allowed transfers. The derivative of G with respect to transfer amount is the difference in μ. A nonzero difference gives a downhill direction; equality removes that first-order drive. The condition is local, while global stability must also rule out a still lower free-energy phase.

Common misconception

Equal chemical potentials do not require equal phase amounts or densities. A cup can contain much more liquid than vapour while both coexist. Nor does an observed higher-μ phase prove equilibrium; metastable phases can persist until nucleation occurs. For mixtures, equality must be checked species by species, not by comparing one averaged μ.

Worked example

At one T,p, suppose μ liquid = −8.00 kJ mol⁻¹ and μ vapour = −7.80 kJ mol⁻¹ on a common reference. Converting 0.010 mol of vapour to liquid changes G by (−8.00 − (−7.80))(0.010) = −0.0020 kJ = −2.0 J, so liquid is favoured locally. At a different pressure where both μ equal −7.90 kJ mol⁻¹, the first-order transfer change is zero and they may coexist.

Quick check

1. Can a pure liquid and its vapour coexist at equilibrium with different molar volumes? Answer: Yes. Coexistence requires equal chemical potentials at common T and p, not equal molar volumes. Their different volumes shape the slope of the coexistence curve.

Exam focus

Write dG = (μ target − μ source)dn with consistent signs. State equality of T and p as well as μ for ordinary phase equilibrium. Distinguish necessary local equality from global stability and from kinetic metastability. For mixtures, use componentwise μ equalities and separate material balances.

Advanced insight

At a finite system's first-order transition, fluctuations may switch between phase-like configurations, producing a bimodal distribution. In the thermodynamic limit, interfacial costs and free-energy densities determine how coexistence appears. Equality of bulk μ remains central, while surface tension affects nucleation and finite-size behaviour.

Summary

Phase coexistence follows from zero Gibbs-energy cost for material transfer: μ α = μ β for a pure substance at common T,p. The lowest μ phase is stable; equality traces a phase boundary. Mixtures require each transferable component's μ to match across phases, even when phase compositions and densities differ.

Practice questions

1. If μ solid < μ liquid at a stated T,p, which phase is stable and what is the sign for liquid → solid transfer? Answer: Solid is stable. The transfer has dG = (μ solid − μ liquid)dn < 0 for positive dn, so it lowers Gibbs energy. 2. Why can evaporation and condensation both continue at liquid–vapour equilibrium? Answer: Equality of μ removes net thermodynamic preference, but molecules still cross microscopically in both directions. Their average rates balance, giving no net phase change. 3. A binary liquid and vapour have different compositions. What must match for equilibrium? Answer: Each transferable component's chemical potential must match between liquid and vapour at common T,p: μ A^l = μ A^v and μ B^l = μ B^v.