VanDerWaals Gas Law Calculator

Solve (P + a(n/V)²)(V - nb) = nRT for any variable with instant calculation and step-by-step reasoning.

L²·atm/mol²
L/mol
Result
21.7959 atm
Using: P = nRT/(V-nb) - a(n/V)²
1.van der Waals: (P + a(n/V)²)(V - nb) = nRT
2.P = nRT/(V - nb) - a(n/V)²
3.nRT/(V-nb) = 1.0000 × 0.08205746 × 273.15 / (1.0000 - 1.0000 × 0.03183) = 23.1509
4.a(n/V)² = 1.355 × (1.0000/1.0000)² = 1.35500
5.P = 23.1509 - 1.35500 = 21.7959 atm
6.Convert to atm: 21.7959 atm

Beyond ideal: the van der Waals equation

The van der Waals equation improves on the Ideal Gas Law by accounting for two effects that real gases exhibit: molecules have finite size, and molecules attract each other at close range. The equation: (P + a(n/V)²)(V - nb) = nRT. The parameter a corrects for intermolecular attraction; the parameter b corrects for the volume occupied by the molecules themselves.

Johannes Diderik van der Waals, a Dutch physicist, derived this equation in his 1873 doctoral thesis. It earned him the 1910 Nobel Prize in Physics. The equation was the first to describe both the gas and liquid phases of a substance with a single formula, and it successfully predicted the existence of the critical point — the temperature above which a gas cannot be liquefied by pressure alone.

The a and b parameters

The parameter a (units: L²·atm/mol²) represents the strength of attraction between molecules. Higher values mean stronger attraction. The parameter b (units: L/mol) represents the excluded volume — roughly four times the actual volume of the molecules. Some common values:

Gasa (L²·atm/mol²)b (L/mol)
Helium0.03410.0237
Nitrogen1.3900.0391
Oxygen1.3600.0318
CO₂3.5920.0427
Water vapor5.4640.0305

Notice helium has the smallest a value — it is the most ideal real gas because its weak interatomic forces make it nearly ideal even at relatively high pressure. Water vapor has the largest a because hydrogen bonding produces strong intermolecular attraction.

When Ideal fails: a quantitative comparison

Consider 1 mol of CO₂ in a 1.0 L container at 300 K. The Ideal Gas Law predicts P = nRT/V = 1 × 0.08205746 × 300 / 1.0 = 24.6 atm. The van der Waals equation with a = 3.592 and b = 0.0427 gives P = 1 × 0.08205746 × 300 / (1.0 - 1 × 0.0427) - 3.592 × 1² / 1.0² = 25.71 - 3.59 = 22.1 atm. The ideal prediction is 11 percent high — a significant error at this pressure. At 10 atm, the error drops below 1 percent.

Solving for volume: Newton's method

The van der Waals equation is cubic in V when solving for volume, meaning there is no simple algebraic solution. This calculator solves for pressure analytically and for volume using Newton's method, an iterative numerical technique. Starting from the ideal gas volume as the initial guess, the algorithm refines the estimate until it converges, typically within five to ten iterations. The step-by-step panel shows the initial guess and the converged result.

Why real gases deviate from ideal behavior

The Ideal Gas Law assumes two things that are not true for real gases: molecules have zero volume, and there are no forces between molecules. At low pressure, these assumptions are approximately correct because molecules are far apart. At high pressure, the finite volume of molecules becomes significant — the container is less empty than the Ideal Gas Law assumes, so the gas occupies less volume than predicted. This is the nb correction: V - nb in the van der Waals equation accounts for the space the molecules themselves take up.

At moderate pressure, intermolecular attraction reduces the pressure below the ideal prediction. Molecules near the container wall are pulled inward by neighboring molecules, reducing the force they exert on the wall. The a(n/V)² term accounts for this: it adds a correction to the measured pressure to get the effective pressure driving expansion. The correction is proportional to the square of the density because the probability of two molecules interacting depends on the square of the concentration.

The critical point: van der Waals's greatest prediction

The van der Waals equation predicts the existence of a critical temperature T_c, above which a gas cannot be liquefied by pressure alone. The critical point occurs where the P-V isotherm has an inflection point. For the van der Waals equation, the critical parameters are: T_c = 8a/(27Rb), P_c = a/(27b²), V_c = 3nb. For CO₂ (a = 3.592, b = 0.0427), T_c = 8 × 3.592 / (27 × 0.082057 × 0.0427) = 304.2 K (31.1 °C). The experimentally measured critical temperature of CO₂ is 304.1 K — within 0.03%. This remarkably accurate prediction, made in 1873 with only two parameters, is why van der Waals won the Nobel Prize.

The compressibility factor Z = PV/nRT measures deviation from ideal behavior. For an ideal gas, Z = 1 exactly. For real gases, Z varies with pressure and temperature. At the critical point, the van der Waals equation predicts Z_c = 3/8 = 0.375. Experimentally, most gases have Z_c between 0.25 and 0.31. The van der Waals equation overestimates the critical compressibility by about 20-50% for most substances, which is its primary quantitative weakness.

Beyond van der Waals: modern equations of state

The van der Waals equation was the first cubic equation of state, but it is not the last. The Redlich-Kwong equation (1949) improves the temperature dependence: P = RT/(V-b) - a/(T^0.5 V(V+b)). The Soave-Redlich-Kwong equation (1972) and Peng-Robinson equation (1976) further improve accuracy. The Peng-Robinson equation is the industry standard for natural gas processing and petrochemical design.

For the most accurate work, reference equations of state use dozens of parameters fitted to extensive experimental data. The GERG-2008 equation for natural gas uses 21 parameters and reproduces experimental measurements to within 0.1% across a wide range of conditions. But for educational purposes and quick estimates, the van der Waals equation remains the clearest — two parameters with straightforward physical meanings, capturing the essential physics of real gas behavior.

More worked examples

Example 2: Nitrogen at high pressure. 1 mol N₂ in a 0.200 L container at 300 K. Ideal: P = nRT/V = 1 × 0.08205746 × 300 / 0.200 = 123 atm. Van der Waals with a = 1.390, b = 0.0391: P = 0.08205746 × 300 / (0.200 - 0.0391) - 1.390/(0.200)² = 153.0 - 34.75 = 118 atm. The ideal prediction is 4% high.

Example 3: Compressibility comparison. For CO₂ at 10 atm and 300 K, the ideal molar volume is 2.46 L/mol. The van der Waals molar volume is 2.41 L/mol. The compressibility factor Z = PV/RT = 10 × 2.41 / (0.08205746 × 300) = 0.979. At this moderate pressure, the deviation is about 2%; at 100 atm the deviation exceeds 15%.

Frequently asked questions

What is the van der Waals equation? (P + a(n/V)²)(V - nb) = nRT. It corrects the Ideal Gas Law for molecular size (b) and attraction (a).

When should I use it instead of PV = nRT? At pressures above about 10 atm or near the condensation point, where real gas behavior deviates significantly from ideality.

What do a and b mean? a represents intermolecular attractive forces; b represents the volume occupied by one mole of molecules. Both are experimentally determined for each gas.

Why did van der Waals win the Nobel Prize? His equation was the first to describe both gas and liquid phases with a single formula, and it predicted the critical point. It launched the field of molecular physics.