The van der Waals Equation of State

A balloon of carbon dioxide, chilled and squeezed hard enough, stops behaving like a gas at all. It condenses into a liquid, something the ideal gas law insists can never happen. Real molecules take up space and pull on their neighbors, two facts the ideal gas equation quietly ignores. Johannes Diderik van der Waals fixed this in his 1873 doctoral thesis with a single, elegant modification of $pV = nRT$ . His equation captures both molecular size and intermolecular attraction using just two constants. It remains the simplest model that predicts condensation, critical points, and the shape of real isotherms. This article derives the van der Waals equation from physical reasoning, not just curve-fitting. You will see where $a$ and $b$ come from. You will learn how to extract them from measurable data. You will work two full numeric examples. Finally, you will see where the model breaks down and why it still matters.

Key takeaways

The van der Waals equation is $\left(p + \dfrac{an^2}{V^2}\right)(V - nb) = nRT$ , a two-parameter correction to the ideal gas law.
$b$ (in L/mol) accounts for the finite volume excluded by molecules themselves; $V$ is replaced by $V - nb$ .
$a$ (in L²·atm/mol²) accounts for attractive forces that lower the pressure a real gas exerts on its walls.
The equation predicts liquid–gas coexistence, a critical point, and the loop-shaped isotherms seen below $T_c$ .
$a$ and $b$ can be found from critical constants: $a = \dfrac{27R^2T_c^2}{64p_c}$ and $b = \dfrac{RT_c}{8p_c}$ .
Key idea: real gas deviations from ideality arise from finite molecular size (repulsion at short range) and mutual attraction (at longer range), and the van der Waals equation builds both into pressure and volume terms separately.

The core insight

Real molecules exclude volume to each other and pull each other inward, so the van der Waals equation shrinks the available volume by $nb$ and boosts the effective pressure by $an^2/V^2$ , turning the ideal gas law into a model that can condense, boil, and have a critical point.

The physical picture

The ideal gas law treats molecules as point particles with no interactions. Real molecules violate both assumptions. At short range, electron clouds repel strongly; molecules behave like small hard spheres that cannot overlap. At intermediate range, induced-dipole and dispersion forces pull molecules together — this is the attraction responsible for condensation. Van der Waals repaired the ideal gas law by patching both terms in $pV = nRT$ . First, the volume available for a molecule to move in is not the container volume $V$ , but $V$ minus the volume physically occupied by all the other molecules. He wrote this excluded volume as $nb$ , where $b$ is a molar parameter characteristic of molecular size. Second, a molecule about to strike the wall is pulled backward by its neighbors, an attraction that reduces the momentum transfer and hence the measured pressure below what it would be without attractions. Since this effect depends on how many molecules are pulling (proportional to density) and how many are being pulled (also proportional to density), the pressure correction scales as $(n/V)^2$ , written as $an^2/V^2$ . Both corrections are added back into the ideal gas relation: the observed pressure $p$ plus the "internal pressure" correction $an^2/V^2$ equals what an ideal gas would exert in the reduced free volume $V - nb$ .

Figure

Two molecular-scale effects that the van der Waals equation encodes

The consequence of both corrections together is dramatic: unlike the ideal gas law, the van der Waals equation is cubic in $V$ at fixed $p$ and $T$ . Below a critical temperature $T_c$ , this cubic has three real volume roots for certain pressures, producing the oscillating "van der Waals loop" that models the liquid, unstable, and gas branches of a real isotherm.

Figure

Van der Waals isotherms: loops below Tc, a smooth critical isotherm at Tc

The core relations

\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT

where: $p$ is pressure, $V$ total volume, $n$ moles, $T$ temperature, $a$ and $b$ are substance-specific constants.

p = \frac{nRT}{V-nb} - \frac{an^2}{V^2}

where: rearranged form giving $p$ explicitly as a function of $V$ and $T$ — the form used to plot isotherms.

p = \frac{RT}{V_m - b} - \frac{a}{V_m^2}

where: molar form, $V_m = V/n$ , useful for comparing substances per mole.

a = \frac{27R^2T_c^2}{64\,p_c}, \qquad b = \frac{RT_c}{8p_c}, \qquad p_c = \frac{a}{27b^2}, \qquad V_{m,c} = 3b, \qquad T_c = \frac{8a}{27Rb}

where: relations connecting $a$ , $b$ to the experimentally measurable critical constants $T_c$ , $p_c$ .

Z = \frac{pV_m}{RT} = \frac{V_m}{V_m - b} - \frac{a}{RTV_m}

where: $Z$ is the compressibility factor; this shows how $b$ pushes $Z$ above 1 (repulsion dominates) and $a$ pulls it below 1 (attraction dominates).

Derivation AFrom molecular corrections to the van der Waals equation

Goal. Starting from the ideal gas law, build in excluded volume and attractive pressure corrections to arrive at the van der Waals equation.

Step 1: Correct the volume. In the ideal gas law $pV = nRT$ , the volume $V$ is assumed fully accessible to every molecule as a point particle. Real molecules have finite diameter $d$ and cannot interpenetrate. For a pair of hard spheres of diameter $d$ , their centers cannot approach closer than $d$ , which excludes a sphere of volume $\frac{4}{3}\pi d^3$ around each molecule from its partner's center — shared between the pair, this gives an excluded volume per molecule of $4 \times \frac{4}{3}\pi (d/2)^3$ . Multiplying by Avogadro's number gives the molar excluded volume $b$ . The volume genuinely free for molecular motion becomes $V - nb$ , so the ideal gas law is rewritten as

p(V - nb) = nRT

Step 2: Correct the pressure. This corrected equation alone (sometimes called the "hard-sphere gas" or Clausius correction) already deviates from ideality, but it still predicts pressure diverging smoothly with no condensation. Real molecules also attract one another. A molecule striking the container wall feels a net inward pull from the bulk of the gas behind it, because it is surrounded by attracting neighbors on the bulk side only. This reduces its momentum transfer to the wall, so the measured pressure $p$ is lower than the pressure $p_{ideal}$ a non-attracting gas would show:

p = p_{ideal} - \frac{an^2}{V^2}

Step 3: Justify the density-squared scaling. The attractive force pulling a wall-bound molecule backward is proportional to the local number density of molecules doing the pulling, $n/V$ . But the frequency of molecules striking the wall is also proportional to $n/V$ . The net reduction in pressure is therefore proportional to the product of these two, $(n/V)^2$ , giving the $an^2/V^2$ term with $a$ as a proportionality constant reflecting the strength of intermolecular attraction.

Step 4: Combine. Solve Step 2 for $p_{ideal}$ :

p_{ideal} = p + \frac{an^2}{V^2}

Substitute into the Step 1 relation $p_{ideal}(V - nb) = nRT$ :

\boxed{\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT}

This is the van der Waals equation. Setting $a = b = 0$ recovers the ideal gas law exactly, confirming the corrections vanish for a hypothetical non-interacting, point-particle gas.

Worked examples

Example 1 — Foundational: pressure of CO₂ from the van der Waals equation. Find the pressure exerted by 1.00 mol of CO₂ confined to 0.500 L at 298 K, using van der Waals constants $a = 3.640\ \text{L}^2\text{atm mol}^{-2}$ and $b = 0.04267\ \text{L mol}^{-1}$ (standard tabulated values for CO₂). Step 1. Molar volume: since $n = 1.00$ mol, $V_m = V/n = 0.500$ L/mol. Step 2. Compute the repulsive (volume) term:

\frac{RT}{V_m - b} = \frac{(0.08206\ \text{L atm K}^{-1}\text{mol}^{-1})(298\ \text{K})}{0.500 - 0.04267\ \text{L/mol}}

= \frac{24.45\ \text{L atm/mol}}{0.4573\ \text{L/mol}} = 53.47\ \text{atm}

Step 3. Compute the attractive correction term:

\frac{a}{V_m^2} = \frac{3.640\ \text{L}^2\text{atm mol}^{-2}}{(0.500\ \text{L/mol})^2} = \frac{3.640}{0.250} = 14.56\ \text{atm}

Step 4. Combine:

p = 53.47\ \text{atm} - 14.56\ \text{atm} = 38.91\ \text{atm}

Step 5. Compare with the ideal gas prediction:

p_{ideal} = \frac{RT}{V_m} = \frac{24.45}{0.500} = 48.90\ \text{atm}

Sanity check: the van der Waals pressure (38.9 atm) is noticeably lower than the ideal value (48.9 atm), which makes sense — at this density (2 mol/L), attractive forces dominate over the modest volume correction, pulling the real pressure down by about 20%. This is physically reasonable for CO₂, a molecule with substantial dispersion interactions, at moderately high density.

Example 2 — Applied: extracting $a$ and $b$ from critical constants, and checking against tabulated values. Ammonia (NH₃) has experimentally measured critical constants $T_c = 405.5$ K and $p_c = 111.3$ atm. Estimate $a$ and $b$ , then estimate the critical molar volume, and compare against the accepted van der Waals value $a = 4.170\ \text{L}^2\text{atm mol}^{-2}$ , $b = 0.03707\ \text{L mol}^{-1}$ . Step 1. Compute $b$ from $b = RT_c/(8p_c)$ :

b = \frac{(0.08206\ \text{L atm K}^{-1}\text{mol}^{-1})(405.5\ \text{K})}{8 \times 111.3\ \text{atm}}

= \frac{33.28\ \text{L atm/mol}}{890.4\ \text{atm}} = 0.03738\ \text{L/mol}

This is within 0.8% of the tabulated 0.03707 L/mol — good agreement, since $T_c$ and $p_c$ are independently measured and only approximately consistent with a two-parameter model. Step 2. Compute $a$ from $a = 27R^2T_c^2/(64p_c)$ :

a = \frac{27 \times (0.08206)^2 \times (405.5)^2}{64 \times 111.3}

Numerator: $27 \times 0.006734 \times 164{,}430 = 27 \times 1107.5 = 29{,}903$ . Denominator: $64 \times 111.3 = 7123$ .

a = \frac{29{,}903}{7123} = 4.198\ \text{L}^2\text{atm mol}^{-2}

This is within 0.7% of the tabulated 4.170 L²atm/mol² — again excellent agreement. Step 3. Estimate the critical molar volume using $V_{m,c} = 3b$ :

V_{m,c} = 3 \times 0.03738\ \text{L/mol} = 0.1121\ \text{L/mol}

Sanity check: the experimental critical molar volume of ammonia is about 0.0724 L/mol. The van der Waals prediction (0.112 L/mol) overestimates it by roughly 55%, a well-known systematic failure of the model — the universal prediction $Z_c = p_cV_{m,c}/RT_c = 3/8 = 0.375$ is too high compared to real gases (typically 0.23–0.31). The $a$ and $b$ values themselves are trustworthy for pressure and energy estimates, but $V_{m,c}$ from $3b$ should be treated as qualitative only.

Limiting cases & common traps

Situation	What happens	Why
$V_m \gg b$ , low density	Reduces to ideal gas law	Excluded volume and attraction corrections both vanish relative to $V_m$
$a = 0$	Hard-sphere-like gas, $Z > 1$ always	Only repulsion (finite size) remains; pressure exceeds ideal
$b = 0$	$Z < 1$ possible at all densities	Only attraction remains, always lowering pressure below ideal
$T < T_c$	Cubic equation gives 3 real $V_m$ roots at some $p$	Loop region models metastable/unstable states; the physical isotherm uses a flat tie line (Maxwell construction) instead
$T = T_c$	Inflection point, single root of multiplicity 3	Defines the critical point exactly
High $p$ , $V_m \to b$	$p \to \infty$	Molecules cannot be compressed past their excluded volume

The chemist behind it

Johannes Diderik van der Waals was a Dutch physicist who, largely self-taught in advanced mathematics as a young man, earned his doctorate in 1873 at Leiden with a thesis titled "On the Continuity of the Gaseous and Liquid States." He was inspired by Andrews' experimental isotherms of carbon dioxide, which showed continuous transformation between gas and liquid near a critical point. Van der Waals's equation was the first to explain this continuity theoretically, uniting gas and liquid as limits of one equation of state rather than two separate substances. His work directly guided the liquefaction of hydrogen and helium by Kamerlingh Onnes, who considered van der Waals's theory essential to predicting the conditions needed. Van der Waals received the Nobel Prize in Physics in 1910.

Origins · 1 minJohannes Diderik van der Waals and how this was discovered

Johannes Diderik van der Waals

1837–1923 · Dutch physicist whose 1873 thesis unified gas and liquid behavior in one equation, earning the 1910 Nobel Prize in Physics

Ask the tutor: Want to see how the Maxwell equal-area construction converts a van der Waals loop into a real coexistence pressure, step by step?

Why does the van der Waals equation predict liquid-gas coexistence but the ideal gas law cannot?

The ideal gas law is linear in $1/V$ at fixed $T$ , so pressure decreases monotonically as volume increases — there is only ever one root, one phase. The van der Waals equation is cubic in $V_m$ , because both the $1/(V_m-b)$ term and the $1/V_m^2$ term appear together. Below $T_c$ , this cubic can have three real roots at a given pressure. The two outer roots correspond to physically stable liquid and gas molar volumes; the middle root is unstable and is replaced by a flat coexistence line in the real isotherm.

How are a and b actually measured for a real substance?

Two common routes exist. First, fit measured $p$ – $V_m$ – $T$ data directly to the van der Waals equation by regression. Second, use the critical constants $T_c$ and $p_c$ , which are relatively easy to measure precisely (the point where liquid and gas become indistinguishable), and apply $a = 27R^2T_c^2/(64p_c)$ and $b = RT_c/(8p_c)$ . Both routes give similar but not identical values, since real gases only approximately follow the two-parameter van der Waals form.

Does a larger a value always mean a substance liquefies more easily?

Generally yes — larger $a$ corresponds to stronger intermolecular attractions, which raises $T_c$ (since $T_c = 8a/27Rb$ ) and makes condensation occur at higher, more easily reached temperatures. Water has an unusually large $a$ (5.536 L²atm/mol²) due to hydrogen bonding, consistent with its high boiling point. However, $b$ matters too — comparisons across substances should track both parameters together, or better, the combination that gives $T_c$ directly.

Why is Zc = 3/8 for the van der Waals equation, and why don't real gases match it?

Setting $V_{m,c} = 3b$ , $T_c = 8a/(27Rb)$ , and $p_c = a/(27b^2)$ into $Z_c = p_cV_{m,c}/RT_c$ gives exactly $3/8 = 0.375$ for every substance, regardless of $a$ and $b$ — a universal prediction of the model. Real gases have $Z_c$ typically between 0.23 and 0.31, because the true intermolecular potential is more complex than a hard sphere plus mean-field attraction; more sophisticated equations of state (Redlich–Kwong, Peng–Robinson) improve on this.

How does the van der Waals equation connect to the compressibility factor Z?

Writing $Z = pV_m/(RT)$ and substituting the van der Waals pressure gives $Z = V_m/(V_m-b) - a/(RTV_m)$ . At low density ( $V_m$ large), $Z \to 1$ , recovering ideal behavior. At high temperature, the $b$ -term dominates and $Z>1$ ; at lower temperature and moderate density, the $a$ -term dominates and $Z<1$ . This directly explains the shape of experimental $Z$ versus $p$ curves covered in compressibility factor discussions.