Real Gases and the Virial Expansion

Introduction

Real gas particles occupy space and exert forces on one another. At low density, the ideal-gas law is a useful first term, and interactions appear as systematic corrections. The virial expansion organises pressure by density. Its second coefficient captures the leading effect of pair interactions and offers a bridge from molecular forces to measured gas pressure.

Core explanation

Let ρ = N/V be number density. Write the compressibility factor Z = p/(ρkT) as Z = 1 + B₂(T)ρ + B₃(T)ρ² + … . The coefficient B₂ has units of volume per particle under this convention; if molar density c = n/V is used with p/(cRT), a molar B₂ has units of volume per mole and differs by Avogadro scaling. The expansion is most reliable at sufficiently low density, where higher powers become small. The ideal gas has B₂ = B₃ = … = 0 in its point-particle noninteracting model.

For a spherically symmetric classical pair potential u(r), the leading coefficient is B₂(T) = −2π∫₀∞[exp(−u(r)/kT) − 1]r²dr under the number-density convention. Repulsive regions with large positive u make the bracket negative and contribute positively to B₂. Attractive regions with negative u make the bracket positive and contribute negatively. Their competition can make B₂ positive or negative as temperature changes. At a Boyle temperature, B₂ = 0; the gas may look ideal to first order in density, but higher virial coefficients need not vanish.

For hard spheres of collision diameter d, u(r) is infinite for r < d and zero beyond. The integral then gives B₂ = 2πd³/3, a positive excluded-volume correction. This predicts Z > 1 at low density because finite-size repulsion raises pressure relative to the ideal point-particle law. Attractive forces can lower pressure relative to ideal, giving Z < 1 over some range, because particles pull one another back from the walls in a simple mechanical picture.

The virial coefficients depend on temperature and the interaction model. Fitting measured p–V–T data can provide coefficients, and comparing them with pair-potential predictions tests molecular-force assumptions. At higher density, three-body correlations and higher-order terms matter; truncating after B₂ may become inaccurate even if a plot looks nearly linear over a narrow range.

Step-by-step reasoning

Choose number density or molar density and write the corresponding Z definition and coefficient units. At low density, estimate Z ≈ 1 + B₂ρ. Use the sign of B₂ to predict pressure above or below ideal at the same ρ,T. If a pair potential is supplied, identify repulsive and attractive regions in the integral. Check that neglected B₃ρ² is plausibly small before interpreting a two-term result.

Visual explanation

Plot Z against ρ near zero. All gas curves begin at Z = 1; their initial slopes are B₂. Draw one positive-slope curve for repulsion-dominated conditions and one negative-slope curve for attraction-dominated conditions. Beside it sketch a pair potential with a steep repulsive core and a shallow attractive well.

Real-world analogy

A crowd in a corridor may exert more pressure on the walls if people cannot overlap, while attractive groups may cluster and hit the walls less often. The analogy hints at excluded volume and attractions. Molecular pressure is calculated from statistical forces and collisions, so the virial coefficient is more precise than the crowd story.

Real-world example

Low-pressure gas data often show a linear first departure of Z from one as density grows. Measuring that slope at several temperatures reveals how attraction and excluded-volume effects balance. A temperature at which the slope is near zero does not mean gas molecules cease interacting; opposing contributions can cancel in B₂.

Why?

At very low density, interactions occur primarily in isolated pairs. Their statistical effect is therefore proportional to the chance of finding another particle, which grows with density. The pair-potential integral weighs each separation by both available spherical volume and a Boltzmann factor, converting molecular attraction and repulsion into the first nonideal pressure correction.

Common misconception

Z = 1 at one finite-density condition does not prove an ideal gas. Virial terms can cancel, and other properties may remain nonideal. Another mistake is to confuse B₂'s units under number-density and molar-density expansions. The sign of B₂ can change with T without any change in the identity of the molecules.

Worked example

At one T, suppose B₂ = −0.020 nm³ per particle and ρ = 2.0 nm⁻³, with higher terms negligible for illustration. Then Z ≈ 1 + B₂ρ = 1 − 0.040 = 0.960. Pressure is about 4% below the ideal prediction ρkT at the same number density. If density doubles, the two-term model predicts Z ≈ 0.920, but higher-density validity must be checked before accepting that value.

Quick check

1. What sign of B₂ results for an ideal hard-sphere gas at low density? Answer: Positive. Excluded volume is repulsive, and for sphere diameter d the classical number-density coefficient is B₂ = 2πd³/3.

Exam focus

Define Z and density before quoting B₂. Check that B₂ρ is dimensionless. State low-density truncation and avoid claiming that B₂ = 0 removes all interactions. For a pair potential, treat repulsive and attractive contributions with the correct sign inside the integral.

Advanced insight

The second virial coefficient is determined by two-body physics, but B₃ and higher coefficients involve more complex clusters and correlations. Quantum gases can have exchange contributions to virial coefficients even without a classical pair potential. The classical integral is therefore tied to its interaction and temperature regime.

Summary

The virial expansion Z = 1 + B₂ρ + B₃ρ² + … measures low-density departures from ideal pressure. B₂ is related to pair forces: repulsion contributes positively and attraction negatively. Hard spheres give B₂ = 2πd³/3. A zero B₂ cancels only the leading correction, not all nonideality.

Practice questions

1. With B₂ = +0.030 L mol⁻¹ and molar density c = 0.20 mol L⁻¹, estimate Z to first order. Answer: B₂c = 0.0060, so Z ≈ 1.006. The molar units cancel correctly. 2. A measured low-density Z curve has negative initial slope versus number density. Which effect likely dominates B₂? Answer: Attractive pair interactions dominate the leading correction under those conditions, making B₂ negative. Repulsion may still be present but outweighed in the integral. 3. What does B₂ = 0 at a Boyle temperature say about B₃? Answer: Nothing forces B₃ to vanish. The gas is ideal only to first order in density; higher-density terms may still produce deviations.