8.5 The van der Waals equation of state

The Lennard-Jones potential describes one pair of molecules. To predict the behaviour of a whole fluid — its pressure, its condensation, its critical point — that microscopic picture must be coarse-grained into an equation of state. The van der Waals equation does exactly this, correcting the ideal gas for the two features of the pair potential: the attractive tail and the repulsive core.

Two corrections to the ideal gas

▶ Building the equation from the pair potential Derivation

The ideal gas law $pv = RT$ (per mole, $v$ the molar volume) assumes point particles with no interactions. The pair potential supplies two corrections:

Excluded volume. The steep $1/r^{12}$ repulsion means each molecule occupies a finite volume its neighbours cannot enter. The volume available for motion is reduced from $v$ to $v - b$ , where $b$ is the excluded volume per mole.
Cohesive pressure. The $-1/r^6$ attraction pulls each molecule inward, reducing the pressure it exerts on the walls. The reduction is proportional to the density of pairs, $\propto 1/v^2$ , so the measured pressure is lowered by $a/v^2$ .

Applying both to the ideal gas law gives

\left(p + \frac{a}{v^2}\right)(v - b) \;=\; RT.

where

$p$: pressure Pa
$v$: molar volume m³/mol
$T$: temperature K
$a$: cohesion parameter (from the $1/r^6$ attraction) Pa·m⁶/mol²
$b$: excluded volume (from the $1/r^{12}$ core) m³/mol

The two phenomenological constants map directly onto the two halves of the Lennard-Jones potential: $a$ is the attraction, $b$ is the repulsion. A gas of hard spheres with no attraction would have $a=0$ ; a gas of point attractors would have $b=0$ .

a (attraction)3.00

b (excl. volume)0.333

T_c2.67

T / T_c1.000

Attraction a = 3.00 Excluded volume b = 0.333 Temperature T/T_c = 1.000

The vdW equation tweaks the ideal-gas law with two corrections: a/v² reduces pressure to account for inter-molecular attraction, and −b in the denominator reduces volume to account for excluded volume. Below T_c the isotherm develops the loop; above T_c it's monotonic. The critical point at (V_c, p_c, T_c) = (3b, a/(27b²), 8a/(27Rb)) is the only point where (∂p/∂V) and (∂²p/∂V²) both vanish.

Vary $a$ and $b$ and watch the isotherms change shape. Increasing the cohesion $a$ deepens the pressure dip that drives condensation; increasing the excluded volume $b$ shifts the whole curve to larger volumes.

The critical point and phase separation

The van der Waals isotherms change character at a single critical point, found where the isotherm has both zero slope and zero curvature, $(\partial p/\partial v)_T = (\partial^2 p/\partial v^2)_T = 0$ :

v_c = 3b, \qquad p_c = \frac{a}{27b^2}, \qquad T_c = \frac{8a}{27Rb}.

Above $T_c$ the isotherms fall monotonically and the substance is a single supercritical fluid — no distinction between liquid and gas. Below $T_c$ the isotherm develops a wiggle, and the fluid separates into coexisting liquid and vapour. The physically-observed coexistence pressure is fixed by the Maxwell equal-area construction: the horizontal tie line is drawn so that it cuts off equal areas above and below the van der Waals loop, which is the condition that the two phases have equal chemical potential.

That the same two constants $(a,b)$ set both the critical point and the coexistence curve is the strength of the model: a two-parameter caricature of the pair potential reproduces the entire qualitative topology of a fluid’s phase diagram. The one region it treats as physical but which equilibrium forbids — the rising middle branch of the loop — is where the next lesson finds the tensile limit of a liquid.