Read interactions through virial coefficients
Derive the second virial correction from a pair potential and keep density truncation separate from uncontrolled dense-fluid extrapolation.
On this page
The question for today
What you’ll understand
- Express the equation of state as a low-density expansion with dimensional coefficients.
- Compute the classical second virial coefficient from the Mayer function.
- Distinguish exclusion diameter, particle volume, attraction, and the limits of a truncated series.
The ideal-gas equation neglects how particles exclude or attract one another. At low density these effects can be organized by the number of particles participating in a connected encounter. The virial expansion expresses pressure in powers of number density, and its second coefficient captures the leading pair correction. We will derive that coefficient from a classical pair potential, evaluate a hard-sphere example, and identify why a small-looking correction cannot license an arbitrary extrapolation toward a dense liquid or phase transition.
Expand pressure at fixed temperature
Write pressure divided by thermal energy as density plus a second coefficient times density squared and further terms. Equivalently, the compressibility factor is one plus a correction linear in density. The second coefficient has volume units and the third has volume squared units when density counts particles per cubic metre. Molar-density conventions use different units and numerical factors involving Avogadro's constant. State the density convention before substituting a tabulated coefficient.
The dilute classical regime needs both weak quantum exchange and controlled interaction corrections. Small density times thermal wavelength cubed suppresses exchange statistics. Small successive virial contributions control the interaction truncation, provided the relevant cluster integrals exist. Near a Boyle temperature the second coefficient can vanish by cancellation between attraction and repulsion while higher coefficients remain nonzero. A zero leading correction is therefore not proof that the interacting gas is exactly ideal at every density.
A pair potential determines the first correction
For particles interacting through a spherically symmetric pair potential u of separation, factor each Boltzmann pair weight into one plus a Mayer function. The function vanishes when the pair does not interact, making connected contributions identifiable. In the dilute thermodynamic limit the second virial coefficient is minus one half its integral over separation. The half removes double counting of unordered pairs; omitting it gives an excluded-volume coefficient twice too large.
Repulsive energy makes the Mayer function negative and contributes positively to the second coefficient. Attractive energy makes it positive and contributes negatively. The exponential must be retained for a strong repulsive core; expanding it in inverse temperature is not valid where potential energy diverges. An integrable short-range tail permits a finite pair integral. Unscreened long-range interactions may violate that condition and require neutralization, screening, or resummed methods rather than this simple ordinary virial formula.
Hard spheres make excluded volume precise
For hard spheres of diameter d, separations below d are forbidden and all larger separations have zero pair potential. The Mayer function is minus one inside the excluded separation sphere and zero outside. Its integral gives a second coefficient equal to half that excluded sphere volume. This is four times the physical volume of one sphere, not the entire excluded sphere volume and not the particle's own volume. Keeping diameter distinct from radius prevents another factor-of-eight error.
At sufficiently low density the positive correction raises pressure at fixed density and temperature. It describes the first pair contribution, not an exact equation of state for close packing. Attractions can lower the pressure and may make the coefficient negative, but a negative coefficient alone does not establish phase separation. The quadratic lab isolates the second-order pressure term with a chosen coefficient. It cannot derive a pair potential, calculate higher clusters, or determine convergence near a critical point.
Work it through
Evaluate a hard-sphere correction
A classical hard-sphere gas has diameter d=2.0 nm and number density n=3.0×10²³ m⁻³. Compute B₂ and the leading compressibility factor using π≈3.14. Assess the size of the retained correction.
Convert the diameter cube: d³=8.0 nm³=8.0×10⁻²⁷ m³. Multiply by 2π/3 to obtain B₂≈1.675×10⁻²⁶ m³.
The dimensionless pair correction is B₂n≈(1.675×10⁻²⁶)(3.0×10²³)=0.005024, so Zcomp≈1.005024.
The leading pressure is about 0.50% above the ideal value at the same density and temperature. This small pair correction supports a dilute interaction estimate, but the thermal wavelength must still be checked before claiming classical statistics.
B₂≈1.675×10⁻²⁶ m³ and Zcomp≈1.0050. Pair exclusion raises pressure by about half a percent in the truncated low-density description.
The living notebook
Explore the model
Interactive lab · function
Isolate the second virial term
For fresh runs Reset and set BOTH coefficient a=0.05, position x=1; then Reset and set a=0.05, x=2. Interpret x as normalized density and y as the normalized B₂n² pressure correction. Predict y=0.05 and 0.20. The parabola excludes the ideal linear term, higher virial terms, and any pair-potential calculation.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forDoubling density quadruples the pair pressure correction while only doubling the ideal pressure.
Pause & reason
Check your understanding
Your turn
Try it yourself
At a fixed temperature an interacting gas has B₂=−2.0×10⁻²⁷ m³ and density n=1.0×10²⁵ m⁻³. Estimate Zcomp to second order in pressure. What changes if density doubles, and can this truncated result establish a liquid transition?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use Zcomp=1+B₂n and distinguish a relative correction from the absolute n² pressure term.
Compare with the solution
Initially B₂n=−0.020, giving Zcomp≈0.980. Doubling density gives Zcomp≈0.960, while the absolute pair pressure correction quadruples. The negative sign indicates an attractive net pair contribution at this temperature. It does not establish condensation or the validity of truncation; higher virial coefficients and phase stability require separate analysis.
Keep the idea
Take it with you
- Virial coefficients organize low-density interaction corrections with declared density units.
- The second classical coefficient is minus one half the Mayer-function pair integral.
- A truncated virial series requires small higher contributions and cannot determine dense-fluid behavior by extrapolation.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.