Van der Waals Equation: Real Gases Explained with Examples

Real gases do not obey the ideal gas law perfectly, because gas molecules attract each other and occupy space themselves — two facts the ideal law ignores. The van der Waals equation fixes both: it adds a pressure correction for attractions and subtracts an excluded volume for the molecules' own size.

This matters because every industrial use of compressed gases — from LPG cylinders to refrigeration — involves conditions where the ideal law is noticeably wrong. The equation also explains liquefaction: the same attractive forces it describes are what let gases condense into liquids.

Key Concepts

• van der Waals equation: (P + a·n²/V²)(V − nb) = nRT. It reduces to the ideal gas law when a = 0 and b = 0.
• Constant a: measures the strength of intermolecular attractions, in L²·bar/mol² (or Pa·m⁶/mol²). Larger a means stronger attraction and easier liquefaction.
• Constant b: the excluded volume per mole, in L/mol — roughly four times the actual volume of the molecules. Larger b means bigger molecules.
• Pressure correction (a·n²/V²): attractions pull molecules back from the walls, lowering the measured pressure, so this term is added to P to get the "ideal" pressure.
• Volume correction (nb): molecules cannot move through space already occupied by other molecules, so the free volume is V − nb.
• Typical values: for CO₂, a = 3.59 L²·bar/mol² and b = 0.0427 L/mol.

Formulas

• van der Waals equation: (P + a·n²/V²)(V − nb) = nRT, with P in bar (or Pa), V in litres (or m³), T in kelvin, R = 0.08314 L·bar/mol·K.
• Solving for pressure: P = nRT/(V − nb) − a·n²/V².
• Units check: a·n²/V² has pressure units; nb has volume units.

Assumptions: a and b are constant for a given gas (in reality they drift slightly with temperature); the equation is for gases, not liquids.

Worked Example

Calculate the pressure of 1.00 mol of CO₂ at 300 K in a 0.500 L vessel, using a = 3.59 L²·bar/mol² and b = 0.0427 L/mol. Compare with the ideal gas law.

1. Free volume: V − nb = 0.500 − 1.00 × 0.0427 = 0.4573 L.
2. Repulsion term: nRT/(V − nb) = 1.00 × 0.08314 × 300 / 0.4573 = 24.942 / 0.4573 = 54.55 bar.
3. Attraction term: a·n²/V² = 3.59 × 1.00 / (0.500)² = 3.59 / 0.25 = 14.36 bar.
4. van der Waals pressure: P = 54.55 − 14.36 = 40.19 bar.
5. Ideal gas pressure: P = nRT/V = 24.942 / 0.500 = 49.88 bar.

Real pressure ≈ 40.2 bar, about 20% below the ideal 49.9 bar — attractions dominate here.

Common Misconceptions

• Real gases always have lower pressure than ideal gases. Not always: at very high pressure the volume correction (molecular size) dominates and real pressure exceeds the ideal value.
• The constant a corrects for molecular size. It is the other way round: a corrects pressure for attractions, b corrects volume for molecular size.
• The van der Waals equation describes liquids too. It is a gas equation; near condensation it becomes unreliable and cubic equations of state take over.

Real-World Applications

• Designing LPG and CNG storage: real-gas corrections size cylinders and predict pressures safely.
• Refrigeration cycles (Joule–Thomson cooling) exploit the same intermolecular forces.
• Industrial gas liquefaction plants (oxygen, nitrogen, LNG) are engineered around these constants.
• High-pressure chemistry (e.g. Haber process) corrects equilibrium calculations for non-ideality.
• Comparing a-values predicts which gases liquefy most easily.

Add a·n²/V² to the pressure for attractions, subtract nb from the volume for molecular size, and the ideal gas law becomes a working model of real gases. The same two constants even predict each gas's critical point.