Real gases and the van der Waals equation

Real gases deviate from ideal behavior at high pressure and low temperature. The van der Waals equation corrects for particle volume and attractions.

Quick answer

Real gases deviate from PV = nRT at high pressure and low temperature because particles occupy real volume and attract each other. The van der Waals equation corrects for both: (P + a(n/V)^2)(V − nb) = nRT.

(P+a(nV)2)(V−nb)=nRT\left(P + a\left(\frac{n}{V}\right)^2\right)(V - nb) = nRT

Example: One mole of CO2 in a 0.500 L tank at 300 K would exert 49.2 atm by the ideal law but only about 39.5 atm by the van der Waals equation, because attractions between CO2 molecules soften the wall impacts.

What you will learn

  • Identify the conditions where real gases deviate from ideal behavior.
  • Explain deviations using particle volume and intermolecular attractions.
  • Apply the van der Waals equation to compute real-gas pressure.
  • Interpret the compressibility factor Z as a deviation measure.

Where the ideal law breaks down

The ideal gas law is a model, and models have jurisdictions. At ordinary pressures it predicts beautifully, but squeeze a gas into a steel cylinder at 200 atm or chill it toward liquefaction and the predictions drift. Measured pressures come out lower than nRT/VnRT/V at moderate pressure, then higher at very high pressure.

Kinetic molecular theory names the two broken assumptions. Postulate 1 claims particle volume is negligible, but at high pressure the particles themselves occupy a real share of the container. Postulate 4 claims no attractions, but slow, crowded particles feel each other's pull.

A worked example of the scale: one mole of gas at STP fills 22.4 L while its molecules occupy roughly 0.02 to 0.05 L of actual volume, a tenth of a percent, utterly negligible. Compress that mole to 0.200 L at high pressure and the molecules' own volume is now 10 to 25 percent of the container. The 'empty space' the ideal law assumes is no longer empty, and the law's answers stop matching reality.

The compressibility factor: Z tells the story

The compressibility factor compresses the whole story into one number:

Z=PVnRTZ = \frac{PV}{nRT}

For an ideal gas Z=1Z = 1 exactly. For real gases, ZZ dips below 1 at moderate pressure, where attractions let particles pull back from the walls and soften their impacts, then climbs above 1 at high pressure, where the particles' own bulk leaves less free room than the ideal law assumes.

Different gases trace different curves. Helium, with almost no attractions, barely dips below 1. Carbon dioxide, large and polarizable, plunges well below 1 before climbing. The Boyle temperature, where the dip and the climb momentarily cancel and Z=1Z = 1 over a range, is a characteristic fingerprint of each gas.

A worked example: nitrogen at 100 atm and 300 K has Z≈0.98Z \approx 0.98, so its real pressure-volume product is 2 percent below ideal. At 1000 atm, Z≈1.9Z \approx 1.9, nearly double. Pipeline engineers who sized a 1000-atm vessel with Z=1Z = 1 would be off by almost a factor of two, which is why compressibility charts hang on every gas-plant wall.

The van der Waals equation: patching both flaws

In 1873 van der Waals patched the ideal law with two measured constants per gas. Attractions reduce the wall impacts, so the observed pressure PP is lower than the ideal kinetic pressure by a(n/V)2a(n/V)^2, where a measures attraction strength. The particles' bulk removes free volume, so the free space is V−nbV - nb, where b measures particle size:

(P+a(nV)2)(V−nb)=nRT\left(P + a\left(\frac{n}{V}\right)^2\right)(V - nb) = nRT

For carbon dioxide, a=3.59 L2⋅atm/mol2a = 3.59\ \text{L}^2\text{·atm/mol}^2 and b=0.0427 L/molb = 0.0427\ \text{L/mol}.

A worked example: 1.00 mol CO2 in a 0.500 L vessel at 300 K. Ideal law: P=nRT/V=(1.00×0.08206×300)/0.500=49.2P = nRT/V = (1.00 \times 0.08206 \times 300)/0.500 = 49.2 atm. Van der Waals: first compute n/V=2.00n/V = 2.00 mol/L. The attraction term is a(n/V)2=3.59×4.00=14.36a(n/V)^2 = 3.59 \times 4.00 = 14.36 atm, and the free volume is 0.500−0.0427=0.45730.500 - 0.0427 = 0.4573 L. Then P+14.36=(0.08206×300)/0.4573=53.84P + 14.36 = (0.08206 \times 300)/0.4573 = 53.84, so P=39.5P = 39.5 atm. Attractions win this round: the real pressure is nearly 10 atm below ideal, a 20 percent correction no engineer can ignore.

Reading a gas's personality in a and b

The constants a and b are a personality profile for each gas. Large a means strong attractions: big, polarizable molecules like xenon (a=4.19a = 4.19) or butane cling together and deviate early. Large b means bulky particles that steal container space sooner.

Helium sits at the other extreme: a=0.034a = 0.034 and b=0.0237b = 0.0237, tiny on both counts, which is why helium stays near-ideal down to temperatures that liquefy everything else. This is also why helium is the gas of choice for deep-diving mixtures and cryogenics.

A worked example comparing personalities: 1.00 mol of helium in the same 0.500 L vessel at 300 K. Ideal: 49.2 atm as before. Van der Waals: a(n/V)2=0.034×4.00=0.136a(n/V)^2 = 0.034 \times 4.00 = 0.136 atm, free volume 0.500−0.0237=0.47630.500 - 0.0237 = 0.4763 L, so P+0.136=24.618/0.4763=51.68P + 0.136 = 24.618/0.4763 = 51.68 and P=51.5P = 51.5 atm. Helium reads above ideal because its volume correction outweighs its feeble attractions, while CO2 read far below. Same conditions, opposite deviations: the constants decide.

Liquefaction and why engineers care

The deepest payoff of the corrections is liquefaction. Cool a real gas enough and attractions overwhelm thermal motion: particles clump, ZZ collapses, and the gas condenses. The ideal law, blind to attractions, can never predict this; van der Waals can, approximately, including the critical point where liquid and gas merge.

This is industrial reality. Natural gas pipelines, CO2 fire extinguishers, and liquid-nitrogen dewars all operate where Z≠1Z \neq 1, and their designs use equations of state descended from van der Waals with more fitted constants.

A worked example of the stakes: a cylinder rated for 150 atm is filled with CO2 to an intended 140 atm at 300 K using the ideal law, requiring n=PV/RT=(140×10.0)/(0.08206×300)=56.9n = PV/RT = (140 \times 10.0)/(0.08206 \times 300) = 56.9 mol in a 10.0 L tank. But at that density attractions pull the real pressure well below 140 atm, tempting an operator to add more gas to 'reach' 140. Warm the tank on a hot day and the attractions weaken while thermal pressure surges, potentially overshooting the rating. Real-gas literacy is safety equipment.

Common mistakes

✗ The ideal gas law is exact for all gases at STP.

It is an excellent approximation at STP, but every real gas deviates slightly. Precision work, like calibrating instruments, must correct for it.

✗ Deviations mean the kinetic theory is wrong.

The theory is right about motion and collisions; only two of its five postulates, zero particle volume and zero attractions, are idealizations. The van der Waals equation keeps the theory and patches those two.

✗ Attractions always lower the measured pressure.

At moderate pressure attractions dominate and pressure reads low, but at very high pressure the particles' own volume dominates and pressure reads high. The deviation changes sign.

Real-world applications

  • Designing high-pressure natural gas pipelines

  • Predicting when industrial gases will liquefy

  • Modeling planetary atmospheres under extreme pressure

  • Sizing CO2 systems for carbonation and fire suppression

Frequently asked questions

When do gases deviate most from ideal behavior?

High pressure. Squeezing a gas to hundreds of atmospheres makes the particles' own volume a real fraction of the container and forces attractions into play. Low temperature causes deviations too, near condensation.

Which gases have the largest van der Waals constants?

Large, easily polarized molecules like xenon or butane have strong attractions (large a), while small molecules like helium have tiny ones. The constants are measured per gas and tabulated.

Is the van der Waals equation exact?

It is better but still approximate. It misses details like non-spherical molecules and temperature-dependent attractions, so engineers use more elaborate equations of state for critical designs.

Why does CO2 deviate strongly in a fire extinguisher?

Because the pressure is high enough that molecular volume matters and the temperature low enough that attractions matter. Both corrections in the van der Waals equation become significant there.