A Statistical View of Phase Equilibrium

Introduction

Phase diagrams can be built from molecular statistics because equilibrium seeks the state with appropriate minimum free energy. The partition function counts accessible microscopic states, and its logarithm gives a thermodynamic potential. Comparing chemical potentials of liquid and vapour then yields the coexistence condition. This chain connects molecular energy levels with an observable vapour pressure, while reminding us that knowing only gas molecules is insufficient to calculate the liquid's properties.

Core explanation

For N indistinguishable, weakly interacting ideal-gas molecules at fixed N, volume V and temperature T, the approximate canonical partition function is Q N=q^N/N!, where q is the single-molecule partition function. Translational and internal parts factor in a suitable ideal model: q=(V/Λ³)q int, with Λ=h/(2πmkT)^(1/2). Here q int sums electronic, rotational and vibrational Boltzmann weights under consistent reference energies. The factorial accounts for indistinguishability; omitting it leads to an unphysical entropy and an incorrect N dependence.

The Helmholtz free energy is A=−kT ln Q N. Using Stirling's approximation ln N!≈N ln N−N for large N, A≈−NkT[ln(q/N)+1]. Holding T and V fixed, chemical potential is μ=(∂A/∂N) (T,V)≈−kT ln(q/N), provided the ideal single-molecule q does not depend on N. Substituting q gives μ g≈kT ln(nΛ³/q int), where n=N/V is number density. For an ideal gas n=P/(kT), so at fixed T a pressure increase changes μ g by kT ln(P₂/P₁) per molecule, or RT ln(P₂/P₁) per mole.

For a pure liquid and its vapour at equilibrium, μ l(T,P sat)=μ g(T,P sat). This equality determines the saturation pressure in principle. At pressures where the vapour behaves nearly ideally and liquid pressure dependence is modest over the relevant range, one often writes μ g(T,P)=μ g°(T)+RT ln(P/P°). Then RT ln(P sat/P°)=μ l(T,P sat)−μ g°(T). The liquid chemical potential must be known or modelled. A gas partition function alone cannot provide a numerical vapour pressure, because the cohesive energies and configurational entropy of the liquid are crucial.

The temperature dependence of the equality leads to the Clapeyron relation. Differentiating equal chemical potentials along coexistence and using dμ=−s m dT+v m dP gives (v g−v l)dP=(s g−s l)dT. Therefore dP/dT=Δs/Δv=Δh/(TΔv). The statistical and macroscopic approaches are consistent: partition functions produce μ and thermodynamic derivatives, while Clapeyron expresses how equality persists as temperature changes.

For a nonideal vapour, pressure in the logarithmic expression is replaced by fugacity in a suitable convention. For mixtures, each component's liquid activity and vapour fugacity enter separate chemical-potential equalities. Near a critical point, ideal-gas and simple liquid approximations fail more strongly because density fluctuations and interactions matter. A molecular model with intermolecular forces is required to predict the full coexistence curve.

The reference energy in q int can shift absolute chemical potentials, but physically consistent phase comparisons include matching reference conventions. Observable coexistence conditions cannot depend on arbitrary zero choices. This is why one should not compare a gas μ computed from one energy zero with an unrelated liquid μ from another without accounting for the offset.

Step-by-step reasoning

Write Q N=q^N/N! for the ideal gas and take A=−kT ln Q N. Differentiate with respect to N to obtain μ g. Express density in terms of gas pressure, then set μ g=μ l at saturation. Finally, differentiate that equality along the coexistence line to recover the Clapeyron slope and identify where ideal approximations enter.

Visual explanation

Picture two μ-versus-pressure curves at one temperature. The vapour μ rises logarithmically with pressure in its ideal region, while the liquid curve changes more gently because of its smaller molar volume. Their intersection identifies saturation pressure. Repeating at nearby temperatures traces the vapour-pressure line.

Real-world analogy

Two routes with equal total cost mark a choice boundary. Counting microscopic states affects each route's free-energy “cost,” and the equality point shifts when temperature or pressure changes. The analogy helps explain coexistence, but chemical potential is a precisely defined partial molar quantity rather than a subjective price.

Real-world example

A sealed vial containing pure liquid solvent and headspace vapour reaches a saturation pressure at fixed temperature if enough liquid remains. Increasing temperature usually increases that pressure. Molecular motions and liquid cohesion determine the numerical value; phase equilibrium imposes equality of chemical potentials between the liquid and vapour.

Why?

Why does gas μ increase with pressure? Compressing an ideal gas reduces translational volume per molecule, lowering the number of accessible positional states at the same temperature. The entropy contribution to free energy changes, giving μ g(T,P₂)−μ g(T,P₁)=RT ln(P₂/P₁) per mole.

Common misconception

A large gas partition function does not by itself say the gas is always favoured. Phase stability compares complete chemical potentials at the same T and pressure, including energetic and entropic contributions in both phases. Another mistake is to use ideal-gas formulas close to critical conditions where intermolecular interactions are essential.

Worked example

At a fixed temperature, an ideal vapour's pressure doubles from P₁ to P₂=2P₁. Its molar chemical potential changes by Δμ=RT ln 2. At 300 K, this is approximately (8.314 J mol⁻¹ K⁻¹)(300 K)(0.693)=1.73 kJ mol⁻¹. If it initially matched a liquid μ at P₁, the doubled-pressure vapour no longer automatically matches; the liquid's pressure response and phase stability must be reconsidered.

Quick check

1. What equality defines pure liquid–vapour saturation at a chosen temperature? Answer: The liquid and vapour chemical potentials are equal at the saturation pressure: μ l(T,P sat)=μ g(T,P sat).

Exam focus

Show Q N=q^N/N!, A=−kT ln Q N and μ=(∂A/∂N) before quoting an ideal-gas logarithm. State the ideal and large-N assumptions. Explain that a liquid free-energy model or measurement is needed to calculate saturation pressure numerically.

Advanced insight

Intermolecular potentials can be included through configurational integrals or simulation to obtain liquid and vapour free energies consistently. Equality of chemical potentials and pressure then determines coexistence. Near criticality, long-range correlations change the form of fluctuations, so a simple independent-molecule partition product cannot reproduce the critical endpoint.

Summary

For an ideal gas, Q N=q^N/N! leads to μ g≈kT ln(nΛ³/q int). Pure liquid–vapour equilibrium requires μ l=μ g, defining saturation pressure. Differentiating that equality yields Clapeyron. Numerical predictions need a liquid model as well as the gas statistics, and nonideal systems require fugacity or richer molecular descriptions.

Practice questions

1. Why does the ideal-gas canonical partition function include N!? Answer: It corrects counting for indistinguishable molecules and produces appropriate extensive thermodynamics and N dependence. 2. At fixed T, how does ideal-gas molar μ change when pressure triples? Answer: It increases by RT ln 3 under the ideal-gas approximation. 3. Can a gas-only partition function determine a liquid's vapour pressure? Answer: No. Saturation requires equality with the liquid chemical potential, which depends on liquid cohesion and configurational states. 4. Which differential identity links chemical-potential equality to Clapeyron? Answer: dμ=−s m dT+v m dP for each pure phase along the coexistence line.