Fugacity and Real Gases

Introduction

Real gas molecules attract and repel one another, so their molar volume is not exactly RT/p. The ideal-gas chemical-potential formula then fails if ordinary pressure is inserted without correction. Fugacity preserves the useful logarithmic form by replacing pressure with an effective pressure defined through the actual chemical potential. The fugacity coefficient measures how far that effective pressure differs from the measured one.

Core explanation

For a pure real gas at temperature T and pressure p, define fugacity f by μ(T,p) = μ° ig(T) + RT ln(f/p°), where μ° ig uses a consistent ideal-gas standard-state convention at p°. The fugacity has pressure units, and f/p° is dimensionless. Define φ = f/p. In the zero-pressure limit gas molecules are far apart and ideal behaviour is recovered, so φ → 1 and f → p. At finite pressure, φ can differ from one; it is not generally a constant for a substance.

The difference between real and ideal chemical potentials at the same T and p is μ real − μ ig = RT ln φ. If φ < 1, the real-gas μ lies below the ideal-gas reference at that p; attractions commonly contribute to this behaviour in some ranges. If φ > 1, the real-gas μ lies above. This simple interpretation is local to the stated conditions; real gases can change behaviour with pressure and temperature.

The pressure dependence connects φ to measurable volume deviations. At constant T, dμ real = V m dp and dμ ig = (RT/p)dp. Subtracting gives RT dln φ = (V m − RT/p)dp. With compressibility factor Z = pV m/(RT), the equation becomes dln φ = (Z − 1)dp/p. Integrating from the dilute limit, where φ → 1, gives ln φ = ∫₀ᵖ [Z(p′) − 1]dp′/p′ at fixed T, provided the data and limiting behaviour are appropriate. This shows that fugacity is anchored in an equation of state rather than being an arbitrary correction factor.

For mixtures, a component has a species fugacity or partial fugacity and a coefficient relative to its partial pressure under a specified convention. Replacing every p i by the total mixture pressure would be incorrect. Accurate high-pressure reaction and phase-equilibrium calculations use fugacities or activities consistently in place of ideal partial-pressure ratios.

Step-by-step reasoning

Check whether ideal-gas behaviour is justified. If not, obtain a fugacity coefficient from data or an equation of state at the stated T, p and composition. Calculate f = φp for a pure gas, then use μ = μ° ig + RT ln(f/p°). To compare with the ideal model at the same pressure, calculate RT ln φ. State the reference convention; do not assume φ equals the compressibility factor Z.

Visual explanation

Plot pressure along a horizontal axis and draw a straight line f = p for the ideal gas. A real-gas f curve may dip below or rise above that line. At one pressure draw a vertical distance between f and p, labelled by φ = f/p. A second plot of ln φ against p starts at zero in the dilute limit.

Real-world analogy

An advertised ticket price may differ from the effective cost after fees and discounts; an effective price can preserve a familiar decision rule while including corrections. Fugacity similarly acts as an effective pressure in a logarithmic chemical-potential equation. Unlike a commercial price, it is rigorously defined through thermodynamics and approaches actual pressure in the dilute limit.

Real-world example

High-pressure gas processing cannot always use simple partial pressures to predict equilibrium accurately. If measured or modelled φ for a pure gas at a condition is 0.80, its fugacity is 0.80p, and its chemical potential differs from the ideal-gas value at the same p by RT ln 0.80. Engineers use such corrections when estimating gas absorption and reaction equilibria.

Why?

The ideal expression uses pressure because ideal V m = RT/p integrates to a logarithm. A real gas has a different V m, but its chemical potential still has a well-defined pressure dependence. Defining f through the logarithm packages the difference into a quantity with pressure units, so familiar equilibrium equations can retain their form while using physically corrected activities.

Common misconception

Fugacity is not a new mechanical pressure read by an ordinary gauge. The gauge reads p; f is inferred from the chemical potential or an equation of state. Also, φ is not the same as Z. They are related by an integral over pressure, so a single Z value at one pressure does not generally give φ by simple substitution.

Worked example

Suppose a pure real gas at 300 K and 10.0 bar has φ = 0.85 under a specified model. Its fugacity is f = φp = 8.5 bar. Relative to an ideal gas at the same T and p, Δμ = RT ln φ = (8.314)(300)ln(0.85) J mol⁻¹ ≈ −405 J mol⁻¹. The negative correction does not mean the measured pressure is 8.5 bar; it remains 10.0 bar.

Quick check

1. What limit must φ approach as a pure gas becomes very dilute? Answer: φ → 1 as p → 0, so f/p → 1 and fugacity approaches ordinary pressure. This anchors the real-gas convention to the ideal-gas limit.

Exam focus

Write f = φp and use the dimensionless ratio f/p° inside a logarithm. State T and p for any quoted φ. Compare real and ideal chemical potentials with RT ln φ, not RTφ. Keep compressibility factor Z distinct and use its pressure integral only when asked to derive φ from an equation of state.

Advanced insight

The integral ln φ = ∫₀ᵖ(Z − 1)dp′/p′ reveals that fugacity depends on the entire path of equation-of-state deviations from the dilute limit to p. This matters near phase changes, where a simplistic local correction may be unreliable. A consistent thermodynamic model must also give the same chemical potential by integrating along any reversible path between states.

Summary

Fugacity f replaces p in the chemical-potential logarithm for a real gas, and φ = f/p measures the departure from ideality. In the dilute limit φ tends to one. The correction to ideal-gas μ at the same p is RT ln φ, and the coefficient can be related to measured compressibility through a pressure integral.

Practice questions

1. A gas at 5.0 bar has φ = 1.10. Find its fugacity and the sign of its μ correction relative to an ideal gas at that pressure. Answer: f = 1.10 × 5.0 = 5.5 bar. Since ln 1.10 > 0, μ real − μ ig = RT ln 1.10 is positive. 2. If Z = 1 at one isolated pressure, must φ = 1 there? Answer: No. ln φ is an integral of (Z − 1)/p over the range from dilute gas to that pressure. Deviations elsewhere along the path may give φ different from one even if Z equals one at the endpoint. 3. Why must f/p° rather than f alone appear inside the logarithm? Answer: A logarithm needs a dimensionless argument. Dividing fugacity by standard pressure gives a unitless activity and makes the reference chemical potential explicit.