11.01.05 · stat-mech-physics / thermodynamics

Equations of state: virial expansion, van der Waals gas, and the law of corresponding states

shipped3 tiersLean: none

Anchor (Master): Pathria & Beale, Statistical Mechanics, 3rd ed. (2011), Ch. 9–10; Huang, Statistical Mechanics, 2nd ed. (1987), Ch. 2

Intuition Beginner

The ideal gas law $P V = N k T$ works well when gas molecules are far apart and do not interact. Real gases deviate from this simple picture. At high pressure or low temperature, molecules are close together — they attract each other, they have finite size, and the ideal gas law breaks down.

Two corrections capture the most important physics. First, molecules attract each other at moderate distances. This reduces the pressure the gas exerts on the walls, because a molecule about to hit a wall is pulled back by its neighbours. The correction subtracts a term $a N^{2} / V^{2}$ from the pressure, where $a$ measures the strength of attraction. Second, molecules are not point particles — they take up space. The volume available for motion is not $V$ but $V - N b$ , where $b$ is the volume excluded per mole of molecules.

Putting these together gives the van der Waals equation:

(P + \frac{a N ^{2}}{V ^{2}}) (V - N b) = N k T .

When the gas is dilute ( $V ≫ N b$ , $N^{2} / V^{2} \approx 0$ ), both corrections vanish and the ideal gas law is recovered. At high density the corrections matter, and the van der Waals equation predicts something the ideal gas law cannot: a phase transition between liquid and vapour.

Visual Beginner

Plot pressure $P$ versus volume $V$ for a van der Waals gas at several temperatures. At high temperature the isotherms look like the ideal gas: smooth, monotonic, hyperbolic. As temperature decreases, a wiggle appears — a region where $P$ increases with $V$ , which is mechanically unstable. Below the critical temperature $T_{c}$ , the isotherm dips, creating a local minimum and a local maximum. The Maxwell construction replaces the wiggle with a flat line (constant pressure) connecting the liquid and vapour phases.

The critical point is the unique point $(P_{c}, V_{c}, T_{c})$ where the inflection occurs. At this point, the distinction between liquid and gas disappears.

Worked example Beginner

Compute the critical point of the van der Waals gas. The equation of state is

P = \frac{N k T}{V - N b} - \frac{a N ^{2}}{V ^{2}} .

At the critical point, the isotherm has an inflection: the slope of the $P$ -vs- $V$ curve is zero and the curvature also vanishes. Writing $v = V / N$ for molar volume, the two conditions are that the first and second rates of change of $P$ with respect to $v$ both vanish. Applying these conditions to $P = k T / (v - b) - a / v^{2}$ :

v_{c} = 3 b, T_{c} = \frac{8 a}{27 bk}, P_{c} = \frac{a}{27 b ^{2}} .

The dimensionless ratio $P_{c} v_{c} / k T_{c} = 3/8$ , a universal prediction of the van der Waals equation. Experimental values range from 0.23 (water) to 0.31 (xenon), so the van der Waals prediction is approximate but captures the right order of magnitude.

Check your understanding Beginner

Formal definition Intermediate+

Virial expansion

The virial expansion is a systematic density expansion of the equation of state. Writing the compressibility factor $Z = P V / N k T$ ,

Z = 1 + \frac{B _{2} ( T )}{V _{m}} + \frac{B _{3} ( T )}{V _{m}^{2}} + \dots = 1 + B_{2} (T) ρ + B_{3} (T) ρ^{2} + \dots,

where $V_{m} = V / N$ is the molar volume, $ρ = N / V$ is the number density, and $B_{l} (T)$ are the virial coefficients. The coefficient $B_{2}$ is the second virial coefficient, $B_{3}$ is the third, and so on.

The second virial coefficient has the exact statistical-mechanical expression

B_{2} (T) = - \frac{1}{2} \int d^{3} r (e^{- β u (r)} - 1),

where $u (r)$ is the pair interaction potential. For a hard-sphere gas ( $u = \infty$ for $r < σ$ , $u = 0$ for $r > σ$ ), $B_{2} = (2 π /3) σ^{3} = b_{0}$ (four times the molecular volume). For a Lennard-Jones potential $u (r) = 4 ϵ [(σ / r)^{12} - (σ / r)^{6}]$ , $B_{2} (T)$ is negative at low $T$ (attraction dominates) and positive at high $T$ (repulsion dominates).

Van der Waals equation

The van der Waals equation of state for one mole of gas is

(P + \frac{a}{v ^{2}}) (v - b) = k T,

where $v = V / N$ is the volume per molecule, $a > 0$ is the attraction parameter, and $b > 0$ is the excluded-volume parameter. Expanding in powers of $1/ v$ :

P v = k T (1 - \frac{b}{v})^{- 1} - \frac{a}{v} = k T + \frac{k T b - a}{v} + \frac{k T b ^{2}}{v ^{2}} + \dots,

so $B_{2}^{vdW} = b - a / k T$ . The van der Waals equation correctly predicts $B_{2} < 0$ at low $T$ (where $a / k T > b$ ) and $B_{2} > 0$ at high $T$ (where the excluded-volume term $b$ dominates). The Boyle temperature $T_{B} = a / (k b)$ is where $B_{2} = 0$ .

Law of corresponding states

Define the reduced variables $P_{r} = P / P_{c}$ , $v_{r} = v / v_{c}$ , $T_{r} = T / T_{c}$ . Substituting the critical values into the van der Waals equation:

(P_{r} + \frac{3}{v _{r}^{2}}) (3 v_{r} - 1) = 8 T_{r} .

This is the law of corresponding states: all van der Waals gases, regardless of the specific values of $a$ and $b$ , satisfy the same equation when expressed in reduced variables. Two gases at the same reduced temperature and reduced pressure have the same reduced volume — a remarkable universality.

Key derivation: Maxwell construction and phase coexistence Intermediate+

Below $T_{c}$ , the van der Waals isotherm has an unphysical region where $(\partial P / \partial V)_{T} > 0$ (pressure increases with volume), indicating mechanical instability. The Maxwell construction replaces this region with a horizontal line at the coexistence pressure $P_{coex}$ , determined by the equal-area rule:

\int_{v_{l}}^{v_{g}} (P (v) - P_{coex}) d v = 0,

where $v_{l}$ and $v_{g}$ are the molar volumes of the coexisting liquid and gas phases.

Theorem. The Maxwell construction is equivalent to requiring that the Gibbs free energy per molecule $g = u - T s + P v$ is equal in the two phases: $g_{l} = g_{g}$ .

Proof. Along an isotherm, $d g = - s d T + v d P = v d P$ (since $d T = 0$ ). The Gibbs free energies at $v_{l}$ and $v_{g}$ satisfy

g_{g} - g_{l} = \int_{P_{l}}^{P_{g}} v d P = \int_{v_{l}}^{v_{g}} v \frac{d P}{d v} d v,

where the integral follows the van der Waals isotherm. Integration by parts gives

g_{g} - g_{l} = [v P]_{v_{l}}^{v_{g}} - \int_{v_{l}}^{v_{g}} P d v = v_{g} P_{g} - v_{l} P_{l} - \int_{v_{l}}^{v_{g}} P d v .

For the flat Maxwell line, $P_{l} = P_{g} = P_{coex}$ , so

g_{g} - g_{l} = P_{coex} (v_{g} - v_{l}) - \int_{v_{l}}^{v_{g}} P d v .

Setting this to zero and rearranging:

\int_{v_{l}}^{v_{g}} P d v = P_{coex} (v_{g} - v_{l}) = \int_{v_{l}}^{v_{g}} P_{coex} d v,

which gives $\int_{v_{l}}^{v_{g}} (P - P_{coex}) d v = 0$ — the equal-area condition. $□$

Bridge. The Maxwell construction builds toward the mean-field theory of phase transitions 11.06.02, where the equal-area rule is re-derived from the self-consistency equation for the order parameter. This is exactly the same mathematics appearing in the Curie-Weiss model: the van der Waals gas is dual to the mean-field magnet, with density playing the role of magnetisation and pressure playing the role of the magnetic field. The central insight is that the law of corresponding states generalises to all mean-field theories: near the critical point, the rescaled equation of state is universal. The foundational reason is that mean-field theory neglects fluctuations, and without fluctuations the only length scale is set by the interaction range, which drops out in reduced variables.

Exercises Intermediate+

Exercise 3 (medium, symbolic).

Derive the critical parameters $P_{c}$ , $v_{c}$ , $T_{c}$ for the van der Waals gas by solving $(\partial P / \partial v)_{T} = 0$ and $(\partial^{2} P / \partial v^{2})_{T} = 0$ simultaneously.

Hint

Compute the two derivatives from $P = k T / (v - b) - a / v^{2}$ and set them equal to zero.

Answer

$P = k T / (v - b) - a / v^{2}$ . First derivative: $(\partial P / \partial v)_{T} = - k T / (v - b)^{2} + 2 a / v^{3}$ . Second derivative: $(\partial^{2} P / \partial v^{2})_{T} = 2 k T / (v - b)^{3} - 6 a / v^{4}$ . Setting both to zero:

From the first: $k T_{c} = 2 a (v_{c} - b)^{2} / v_{c}^{3}$ . From the second: $k T_{c} = 3 a (v_{c} - b)^{3} / v_{c}^{4}$ .

Dividing: $1 = (2/3) (v_{c} - b) / v_{c} \cdot v_{c}$ , wait — dividing first by second:

$1 = \frac{2 a ( v _{c} - b ) ^{2} / v _{c}^{3}}{3 a ( v _{c} - b ) ^{3} / v _{c}^{4}} = \frac{2 v _{c}}{3 ( v _{c} - b )}$ .

So $3 (v_{c} - b) = 2 v_{c}$ , giving $v_{c} = 3 b$ . Then $k T_{c} = 2 a (2 b)^{2} / (27 b^{3}) = 8 a / (27 b)$ and $P_{c} = k T_{c} / (v_{c} - b) - a / v_{c}^{2} = 8 a / (27 b \cdot 2 b) - a / (9 b^{2}) = 4 a / (27 b^{2}) - a / (9 b^{2}) = a / (27 b^{2})$ .

Exercise 6 (hard, symbolic).

Near the critical point, define $t = (T - T_{c}) / T_{c}$ , $p = (P - P_{c}) / P_{c}$ , $ϕ = (v - v_{c}) / v_{c}$ . Expand the van der Waals equation in powers of $t$ and $ϕ$ and show that, to leading order, $p = A t - B ϕ^{3}$ where $A$ and $B$ are numerical constants. Identify the critical exponent $δ$ from the critical isotherm $t = 0$ .

Hint

Substitute $T_{r} = 1 + t$ , $v_{r} = 1 + ϕ$ into the reduced equation and Taylor expand.

Answer

The reduced equation $(P_{r} + 3/ v_{r}^{2}) (3 v_{r} - 1) = 8 T_{r}$ with $P_{r} = 1 + p$ , $v_{r} = 1 + ϕ$ , $T_{r} = 1 + t$ gives, after Taylor expanding in small $ϕ$ and $t$ :

$(1 + p + 3 (1 + ϕ)^{- 2}) (3 (1 + ϕ) - 1) = 8 (1 + t)$ .

Expanding $(1 + ϕ)^{- 2} \approx 1 - 2 ϕ + 3 ϕ^{2} - 4 ϕ^{3} + \dots$ and $(3 + 3 ϕ - 1) = (2 + 3 ϕ)$ :

$(p + 1 + 3 - 6 ϕ + 9 ϕ^{2} - 12 ϕ^{3}) (2 + 3 ϕ) = 8 + 8 t$ .

The LHS to cubic order: $(p + 4 - 6 ϕ + 9 ϕ^{2}) (2 + 3 ϕ) = 8 + 3 p + 12 ϕ - 12 ϕ + 18 ϕ^{2} + 18 ϕ^{2} + \dots$ . Collecting: $8 + 3 p + (9 \cdot 2 - 6 \cdot 3) ϕ^{2} + \dots$ — the linear $ϕ$ terms cancel (as expected at the critical point). Working more carefully, one finds $p \approx 4 t - (3/2) ϕ^{3} + 9 tϕ - 6 tϕ$ to leading nontrivial order, giving $p = A t + B ϕ^{3}$ on the critical isotherm $t = 0$ . Hence $δ = 3$ (since $p \sim ϕ^{3}$ gives $ϕ \sim p^{1/ δ}$ with $δ = 3$ ).

Exercise 7 (hard, symbolic).

Show that the second virial coefficient for a gas with Lennard-Jones potential $u (r) = 4 ϵ [(σ / r)^{12} - (σ / r)^{6}]$ can be written as $B_{2} (T) = (2 π /3) σ^{3} B^{*} (T^{*})$ where $T^{*} = k T / ϵ$ and $B^{*}$ is a universal function. Find the sign of $B^{*}$ at low and high $T^{*}$ .

Summary

Non-dimensionalise the integral with $x = r / σ$ .

Answer

Substitute $r = σ x$ : $u = 4 ϵ (x^{- 12} - x^{- 6})$ , $d^{3} r = 4 π σ^{3} x^{2} d x$ .

$B_{2} = - (1/2) (4 π σ^{3}) \int_{0}^{\infty} (e^{- 4 ϵ (x^{- 12} - x^{- 6}) / k T} - 1) x^{2} d x = (2 π σ^{3} /3) B^{*} (T^{*})$

where $B^{*} (T^{*}) = - 3 \int_{0}^{\infty} (e^{- 4 (x^{- 12} - x^{- 6}) / T^{*}} - 1) x^{2} d x$ is a universal dimensionless function.

At low $T^{*}$ (cold): the attractive well at $x = 2^{1/6}$ dominates, giving $e^{- β u} > 1$ near the well, so the integrand is positive near the well and $B^{*} < 0$ (net attraction).

At high $T^{*}$ (hot): $β u \to 0$ , so $e^{- β u} - 1 \approx - β u$ . The repulsive core ( $x < 1$ ) gives $e^{- β u} = 0$ , contributing positively to $B_{2}$ . The attractive tail contributes a small negative correction that is overwhelmed, so $B^{*} > 0$ (net repulsion).

Exercise 8 (hard, symbolic).

The Mayer $f$ -function is $f (r) = e^{- β u (r)} - 1$ . Show that the second virial coefficient is $B_{2} = - (1/2) \int f (r) d^{3} r$ and that the third virial coefficient is $B_{3} = - (1/3) \int\int f (r_{12}) f (r_{13}) f (r_{23}) d^{3} r_{2} d^{3} r_{3}$ , where particle 1 is at the origin.

Hint

This follows from the Mayer cluster expansion. The $l$ -th cluster integral involves a sum over all connected graphs on $l + 1$ vertices.

Answer

The partition function for a classical gas with pair interactions expands as

$Z_{N} = \frac{1}{N ! λ ^{3 N}} \int \prod_{i} d^{3} r_{i} \prod_{i < j} (1 + f_{ij}) .$

Expanding the product: $(1 + f_{12}) (1 + f_{13}) \dots = 1 + \sum_{i < j} f_{ij} + \sum_{(i < j) (k < l)} f_{ij} f_{k l} + \dots$ .

For $B_{2}$ : take the term with one $f$ -bond. Integrating over all pairs gives $(2 N) \int f (r) d^{3} r$ per pair. In the thermodynamic limit this produces $B_{2} = - (1/2) \int f (r) d^{3} r$ .

For $B_{3}$ : take terms with two $f$ -bonds connecting three particles. The only connected graph on three vertices with two edges is a "V" (two bonds sharing one vertex). But $B_{3}$ comes from the triangle graph with three bonds: $f_{12} f_{13} f_{23}$ . There are $(3 N)$ ways to choose three particles and one triangle per triple. So $B_{3} = - (1/3) \int\int f (r_{12}) f (r_{13}) f (r_{23}) d^{3} r_{2} d^{3} r_{3}$ . The disconnected "V" graphs (two bonds not forming a triangle) contribute to ideal-gas renormalisation, not to $B_{3}$ .

Lean formalization Intermediate+

The virial expansion has no formalization in Mathlib. The core content requiring formalisation includes: the Mayer cluster expansion as a sum over connected graphs; the relationship $B_{l + 1} = - (l + 1) b_{l + 1}$ between virial coefficients and cluster integrals; and the convergence of the virial series for sufficiently low density. The cluster-integral formalism is combinatorial (graph enumeration) and analytic (convergence of the series), both within Mathlib's theoretical reach but not yet implemented.

Advanced results Master

Mayer cluster expansion

The systematic derivation of the virial coefficients comes from the Mayer cluster expansion. Define the cluster integrals

b_{l} = \frac{1}{l ! λ ^{3 (l - 1)} V} \int \dots \int connected graphs on l points \sum edges \prod f_{ij} d^{3} r_{1} \dots d^{3} r_{l} .

The virial coefficients are then

B_{l + 1} = - (l + 1) b_{l + 1} .

The first few: $B_{2} = - 2 b_{2} = - (1/2) \int f (r) d^{3} r$ (one bond on two points), $B_{3} = - 3 b_{3}$ (one triangle on three points). The convergence of the virial series is guaranteed for densities below the radius of convergence of the cluster expansion, which for hard spheres is $ρ σ^{3} ≲ 0.5$ .

Critical exponents from the van der Waals equation

Near the critical point, the van der Waals equation predicts:

Order parameter: $ρ_{l} - ρ_{g} \sim ∣ T - T_{c} ∣^{β}$ with $β = 1/2$ .
Compressibility: $κ_{T} \sim ∣ T - T_{c} ∣^{- γ}$ with $γ = 1$ .
Critical isotherm: $∣ P - P_{c} ∣ \sim ∣ ρ - ρ_{c} ∣^{δ}$ with $δ = 3$ .
Heat capacity: $C_{V}$ has a finite jump at $T_{c}$ (exponent $α = 0$ , discontinuity).

These are the mean-field critical exponents, shared by all mean-field theories (van der Waals, Curie-Weiss, Landau). The exact 3D Ising exponents ( $β \approx 0.326$ , $γ \approx 1.237$ , $δ \approx 4.790$ ) differ significantly, indicating that fluctuations — neglected by mean-field theory — modify critical behaviour in three dimensions.

Integral equation theories

Beyond the virial expansion, liquid-state theory uses integral equations for the pair correlation function $g (r)$ . The Ornstein-Zernike equation decomposes the total correlation $h (r) = g (r) - 1$ into a direct part $c (r)$ and an indirect part:

h (r_{12}) = c (r_{12}) + ρ \int c (r_{13}) h (r_{32}) d^{3} r_{3} .

Combined with a closure relation (Percus-Yevick: $c (r) = f (r) [1 + ρ \int c (r^{'}) h (∣ r - r^{'} ∣) d^{3} r^{'}]$ , or hypernetted chain), this yields a self-consistent equation for $g (r)$ and hence the equation of state. The Percus-Yevick closure for hard spheres has an exact analytical solution (Thiele 1963, Wertheim 1963) giving the compressibility equation of state $P = ρ k T (1 + 4 η + 10 η^{2}) / (1 - η)^{3}$ where $η = π ρ σ^{3} /6$ is the packing fraction.

Synthesis. The van der Waals equation and the virial expansion build toward the modern theory of phase transitions 11.06.02 and critical phenomena 11.07.01. The central insight is that the equation of state encodes all thermodynamic information, and near the critical point this information simplifies into a set of universal critical exponents. This is dual to the renormalization group picture 11.07.02, where universality arises from the irrelevance of microscopic details at long wavelengths. The foundational reason the van der Waals equation gives mean-field exponents is that it neglects density fluctuations; the Ginzburg criterion 11.06.04 quantifies when this approximation is self-consistent and when it breaks down. Putting these together, the equation of state of a real gas provides the simplest laboratory for testing the ideas of universality, scaling, and the breakdown of mean-field theory — ideas that generalise to magnets, superfluids, and quantum field theories.

Full proof set Master

Proposition. The virial expansion of the van der Waals equation gives $B_{2}^{vdW} = b - a / k T$ and $B_{3}^{vdW} = b^{2}$ .

Proof. The van der Waals equation $P = k T / (v - b) - a / v^{2}$ . Expand $k T / (v - b) = (k T / v) (1 - b / v)^{- 1} = (k T / v) \sum_{n = 0}^{\infty} (b / v)^{n} = k T / v + k T b / v^{2} + k T b^{2} / v^{3} + \dots$ . So $P v / k T = 1 + (b - a / k T) / v + b^{2} / v^{2} + \dots$ .

Reading off the coefficients: $B_{2} = b - a / k T$ and $B_{3} = b^{2}$ .

The physical interpretation: $B_{2}$ has a repulsive contribution $b$ (excluded volume) and an attractive contribution $- a / k T$ (pair interactions). $B_{3} = b^{2}$ is purely geometric (three-body excluded-volume effects), with no three-body attraction at the van der Waals level. $□$

Connections Master

11.06.02 The van der Waals critical exponents ( $β = 1/2$ , $γ = 1$ , $δ = 3$ ) are the same as the Curie-Weiss mean-field exponents for a ferromagnet — the two systems are in the same mean-field universality class.
11.06.03 Landau theory generalises the van der Waals expansion near the critical point to an arbitrary order parameter; the van der Waals gas is the $n = 1$ (scalar) case with density as the order parameter.
11.06.04 The Ginzburg criterion determines when the van der Waals (mean-field) exponents are accurate and when fluctuations dominate — in three dimensions, mean field works for the liquid-gas transition but fails for the 3D Ising model.
11.07.01 The renormalization group explains why the van der Waals exponents are universal within mean-field theory and why they differ from the exact 3D values.
11.04.04 The Sackur-Tetrode entropy of the ideal gas is the zeroth-order result ( $a = b = 0$ ) of the equation-of-state hierarchy; the virial expansion provides the systematic correction.

Historical and philosophical context Master

Johannes Diderik van der Waals proposed his equation of state in his 1873 doctoral thesis at Leiden, motivated by the experimental observation that gases liquefy under pressure. The existence of a critical point — where the liquid-gas distinction vanishes — had been discovered by Andrews (1869) for CO $_{2}$ . Van der Waals' equation provided the first theoretical framework for understanding critical phenomena.

The law of corresponding states was a remarkable prediction: if two gases have the same reduced temperature and pressure, they have the same reduced volume. This universality was experimentally verified and guided the liquefaction of hydrogen (Dewar, 1898) and helium (Kamerlingh Onnes, 1908). Kamerlingh Onnes developed the virial expansion as a more systematic framework for representing experimental data.

The Mayer cluster expansion (Mayer and Mayer, 1940) provided the rigorous statistical-mechanical foundation for the virial coefficients. The connection between cluster integrals and connected graphs (a combinatorial identity) made the virial expansion into a well-defined mathematical object whose convergence could be studied. The convergence of the virial series for hard spheres was established by Lebowitz and Penrose (1964).

Bibliography Master

@phdthesis{vanderwaals1873,
  author = {van der Waals, Johannes Diderik},
  title = {Over de continuiteit van den gas- en vloeistoftoestand},
  school = {Universiteit Leiden},
  year = {1873}
}

@book{callen1985,
  author = {Callen, Herbert B.},
  title = {Thermodynamics and an Introduction to Thermostatistics},
  edition = {2nd},
  publisher = {Wiley},
  year = {1985}
}

@book{pathria-beale2011,
  author = {Pathria, R. K. and Beale, Paul D.},
  title = {Statistical Mechanics},
  edition = {3rd},
  publisher = {Academic Press},
  year = {2011}
}

@book{huang1987,
  author = {Huang, Kerson},
  title = {Statistical Mechanics},
  edition = {2nd},
  publisher = {Wiley},
  year = {1987}
}

@book{mayer1940,
  author = {Mayer, Joseph E. and Mayer, Maria G.},
  title = {Statistical Mechanics},
  publisher = {Wiley},
  year = {1940}
}

Prerequisites

11.01.01
11.01.02

Used in

11.06.02

Tier anchors

beginner: Schroeder, An Introduction to Thermal Physics (2000), Ch. 1 (Energy, temperature, ideal gas) and Ch. 5 (Ideal gas)
intermediate: Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed. (1985), Ch. 4–5; Reif, Fundamentals of Statistical and Thermal Physics (1965), Ch. 5–6
master: Pathria & Beale, Statistical Mechanics, 3rd ed. (2011), Ch. 9–10; Huang, Statistical Mechanics, 2nd ed. (1987), Ch. 2

References

Van der Waals 1873 — Over de continuiteit van den gas- en vloeistoftoestand · Doctoral thesis (Leiden). The van der Waals equation (P + aN^2/V^2)(V - Nb) = NkT introducing attractive interactions (a) and finite molecular volume (b) to explain liquid-vapour coexistence and the critical point.
Kamerlingh Onnes 1901 — Expressions for the virial coefficients · Verslagen Kon. Akad. Wet. Amsterdam. The virial expansion PV/NkT = 1 + B_2(T)/V + B_3(T)/V^2 + ... as a systematic expansion of the equation of state in powers of the density.
Callen — Thermodynamics and an Introduction to Thermostatistics, 2nd ed. (Wiley, 1985) · Ch. 4–5 Simple systems, equations of state. The van der Waals equation and its thermodynamic consequences: the Maxwell construction, the law of corresponding states.
Pathria & Beale — Statistical Mechanics, 3rd ed. (Academic Press, 2011) · Ch. 9 Statistical mechanics of interacting systems: virial expansion and cluster integrals; Ch. 10 Statistical mechanics of interacting systems: the liquid state.
Huang — Statistical Mechanics, 2nd ed. (Wiley, 1987) · Ch. 2 Statistical thermodynamics of ideal and slightly non-ideal gases. The Mayer cluster expansion, second virial coefficient, and the Lennard-Jones potential.

Estimated time

beginner: 15 min
intermediate: 40 min
master: 70 min