Temperature and Pressure Dependence of Chemical Potential

Introduction

Phase stability can change when temperature or pressure changes because each phase's chemical potential responds differently. A pure phase has a particularly useful differential: dμ = −S m dT + V m dp. Its slopes explain why heating usually favours vapour over liquid and liquid over solid, while pressure often favours the denser phase. The same differential is the starting point for calculating phase-boundary slopes.

Core explanation

For a pure one-component phase, μ = G m. From the Gibbs differential dG = −S dT + V dp at fixed amount, dividing by amount gives dμ = −S m dT + V m dp. Thus (∂μ/∂T) p = −S m and (∂μ/∂p) T = V m. Molar entropy and volume are positive for ordinary stable phases, so μ generally falls as T rises at fixed p and rises as p rises at fixed T. The rates of change differ among phases.

At a fixed pressure, vapour usually has much greater molar entropy than liquid, so its μ versus T curve has a more negative slope. Liquid usually has greater entropy than solid. Starting at a temperature where solid is stable, warming may make liquid's μ cross below solid's at melting. Warming further may make vapour's μ cross below liquid's at boiling. At each crossing, the two phases have equal μ. This picture explains the order of transitions without claiming that every substance has every phase under all pressures.

At fixed temperature, the phase with larger molar volume has a steeper positive μ versus p slope. Gas has much larger V m than liquid, so compressing a gas raises its μ quickly relative to liquid and can favour condensation. For many substances, liquid has larger V m than solid, so increasing pressure favours the solid. Water is a famous exception near ordinary conditions: ice has larger molar volume than liquid water, so raising pressure can favour liquid relative to ice and lower the melting temperature over that region.

These slopes are local. S m and V m themselves can change with T and p, so a straight-line extrapolation over a wide range may fail. At a coexistence boundary, differentiating μ^α = μ^β along the boundary gives −S m^α dT + V m^α dp = −S m^β dT + V m^β dp. Rearranged, dp/dT = ΔS m/ΔV m = ΔH m/(TΔV m) for a first-order transition, the Clapeyron relation developed next.

Step-by-step reasoning

Identify a pure phase and write dμ = −S m dT + V m dp. For a temperature change at constant pressure, compare molar entropies; the higher-entropy phase's μ falls faster. For a pressure change at constant temperature, compare molar volumes; the larger-volume phase's μ rises faster. Set two phase μ values equal to locate coexistence, and use local slopes rather than extrapolating a constant S m or V m indefinitely.

Visual explanation

On a μ-versus-T graph at fixed pressure, draw solid, liquid and vapour lines with increasingly steep negative slopes; their crossings indicate melting and boiling. On a μ-versus-p graph at fixed T, draw gas with a much steeper positive slope than liquid. Label each slope directly as −S m or V m.

Real-world analogy

Two runners may start at different heights and descend at different rates. The one with the steeper descent can cross below the other later, changing who is lower. Phase chemical potentials similarly cross as T changes because their entropy-related slopes differ. The analogy describes curve crossing, not physical running or a claim that entropy itself is height.

Real-world example

Increasing pressure at fixed temperature can condense a vapour because its large molar volume makes its μ rise more rapidly than the liquid's. Refrigeration and gas-liquefaction processes exploit pressure and temperature changes, although practical devices must also account for heat exchange and nonideal behaviour.

Why?

Temperature and pressure are conjugate to entropy and volume in the Gibbs differential. A phase with greater entropy gains a larger μ reduction when heated; a phase with greater volume pays a larger μ increase when compressed. Comparing these responses with the lowest-μ stability criterion explains the direction of many phase transitions.

Common misconception

Heating does not lower every phase's μ by the same amount, so equal μ at one temperature does not persist at all temperatures. Pressure does not invariably favour solid: it favours the phase with smaller molar volume, which may be liquid for water near its melting line. Also, −S m and V m are local derivatives, not constant slopes over arbitrary distances.

Worked example

Near a fixed pressure, suppose liquid and vapour have equal μ at T₀ and molar entropies S m,liquid = 70 and S m,vapour = 170 J mol⁻¹ K⁻¹. For a small 1.0 K temperature rise at the same pressure, μ liquid falls by about 70 J mol⁻¹, while μ vapour falls by about 170 J mol⁻¹. Vapour becomes lower by roughly 100 J mol⁻¹ and is favoured. This first-order estimate assumes the molar entropies change little over the one-degree interval.

Quick check

1. At fixed T, which pure phase's chemical potential rises faster with pressure: one with V m = 1 L mol⁻¹ or one with V m = 20 L mol⁻¹? Answer: The 20 L mol⁻¹ phase, because (∂μ/∂p) T = V m. The numerical unit conversion would be needed for an energy change, but the slope comparison is immediate.

Exam focus

Use dμ = −S m dT + V m dp with signs intact. For stability, compare μ values after the change, using entropy or volume differences to determine which curve moves faster. State that pressure favours smaller V m rather than naming a particular phase automatically. Keep slope reasoning local unless heat capacities and compressibilities are supplied.

Advanced insight

The differential equality along a phase boundary leads to the Clapeyron equation. Its numerator ΔS m equals ΔH m/T at a reversible first-order transition, and its denominator ΔV m can be positive or negative. This explains why most melting boundaries slope upward in a p–T diagram while water's familiar ice–liquid boundary slopes downward over the ordinary-pressure range.

Summary

For a pure phase, dμ = −S m dT + V m dp. Higher-entropy phases gain more stability on heating; smaller-volume phases tend to gain relative stability on compression. Crossings of phase chemical-potential curves mark coexistence. These local slopes lead directly to the Clapeyron phase-boundary equation.

Practice questions

1. At fixed p, phase A has S m = 40 and phase B has S m = 90 J mol⁻¹ K⁻¹. Which μ falls faster when temperature rises? Answer: Phase B's μ falls faster because its temperature slope is −90 rather than −40 J mol⁻¹ K⁻¹. A sufficiently large change could alter which phase has lower μ. 2. At fixed T, liquid has smaller molar volume than vapour. What does increasing pressure do to their relative μ values? Answer: Vapour μ rises faster because its V m is larger, so pressure tends to favour liquid relative to vapour, if both phases are accessible. 3. Why can water's melting line have a negative dp/dT slope? Answer: Melting increases entropy, so ΔS m > 0, but liquid water has smaller molar volume than ice near ordinary conditions, so ΔV m < 0. Their ratio ΔS m/ΔV m is negative.