Lesson 5 · 25 min

Real gases

At high pressure and low temperature, molecular attraction and molecular size can no longer be ignored. The compressibility factor measures how far a gas is from ideal; the van der Waals equation explains why and predicts the critical point.

In this lesson

  • Measure deviations from ideality with the compressibility factor Z=PVm/RTZ = PV_\mathrm{m}/RT
  • Attribute Z<1Z < 1 to attraction and Z>1Z > 1 to molecular volume
  • Interpret aa and bb in the van der Waals equation and use it to predict the critical point
  • Judge when the ideal-gas approximation is good enough

A 40 L carbon dioxide cylinder is usually filled with about 25 kg of COX2\ce{CO2}. At 20 °C the pressure inside is only about 57 bar — most of the contents are liquid, with saturated vapour above. By the ideal-gas equation a 40 L cylinder at 57 bar would hold only about 4 kg, and even at the cylinder’s rated 150 bar only about 11 kg. The ideal-gas equation fails completely here: it assumes that molecules have no volume and do not attract one another, and it is exactly the attraction between molecules that lets COX2\ce{CO2} be squeezed into a liquid at room temperature.

Predict first

In the molecular-dynamics box, switch on attraction and lower the temperature. What happens to the measured PA/NkTPA/NkT?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

The compressibility factor

Z=PVmRT{Z=1idealZ<1attraction dominatesZ>1repulsion (molecular size) dominatesZ = \frac{PV_\mathrm{m}}{RT}\qquad\begin{cases} Z = 1 & \text{ideal} \\ Z < 1 & \text{attraction dominates} \\ Z > 1 & \text{repulsion (molecular size) dominates} \end{cases}
VmV_\mathrm{m}
molar volume V/nV/n
ZZ
compressibility factor; exactly 1 for an ideal gas
Compressibility factor and isotherms

Real gases: the van der Waals equation

(P+aVm2)(Vm−b)=RT\left(P + \frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m} - b) = RT

Ideal gas 0100200300400500600Pressure / bar0.511.5Z = PVm / RT
  • He (a = 0.0346, b = 0.0238)
  • N2 (a = 1.37, b = 0.0387)
  • CO2 (a = 3.64, b = 0.04267)
  • NH3 (a = 4.225, b = 0.0371)
Curves are computed from the van der Waals equation with a and b from the CRC Handbook. It captures both deviations qualitatively — attraction (Z < 1) and molecular volume (Z > 1) — but not quantitatively: it predicts a critical compressibility of 0.375, whereas real gases have about 0.23–0.30.
Compare gases: strongly attracting NHX3\ce{NH3} and COX2\ce{CO2} dip far below 1; helium, with almost no attraction, rises above 1 from the start. Raise the temperature and watch attraction fade.

The van der Waals equation

In 1873 van der Waals made two corrections to the ideal gas law:

  • molecules occupy space, so the volume available for motion is reduced by bb (the “excluded volume” per mole);
  • molecules attract one another, lowering the pressure by an amount proportional to the density squared, a/Vm2a/V_\mathrm{m}^2.

aa measures the strength of attraction and bb the size of the molecules; both are fitted to experiment.

(P+aVm2)(Vm−b)=RT\left(P + \frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m} - b) = RT
aa
attraction parameter, bar L²/mol² (He\ce{He} 0.035, COX2\ce{CO2} 3.64)
bb
excluded volume, L/mol (He\ce{He} 0.024, COX2\ce{CO2} 0.043)
Compressibility factor and isotherms

Real gases: the van der Waals equation

(P+aVm2)(Vm−b)=RT\left(P + \frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m} - b) = RT

00.10.20.30.40.5Molar volume Vm / (L mol⁻¹)050100150Pressure / bar
  • 260 K
  • 280 K
  • 304 K
  • 320 K
  • 350 K
  • vdW critical point 304 K, 74 bar

Below the critical temperature a van der Waals isotherm develops a loop; a real system instead boils at constant pressure there (Maxwell’s equal-area construction).

Curves are computed from the van der Waals equation with a and b from the CRC Handbook. It captures both deviations qualitatively — attraction (Z < 1) and molecular volume (Z > 1) — but not quantitatively: it predicts a critical compressibility of 0.375, whereas real gases have about 0.23–0.30.
Van der Waals isotherms of COX2\ce{CO2}. Above the critical temperature (about 304 K) they fall monotonically; below it a loop appears — where a real system splits into liquid and vapour.

Worked example

The critical point from van der Waals

From a=3.64a = 3.64 bar L²/mol² and b=0.04267b = 0.04267 L/mol, find the critical temperature and pressure of COX2\ce{CO2} and compare with measurement (304.1 K, 73.8 bar).

Concept check

Under which conditions does nitrogen behave most nearly ideally?

Try the calculation

With the van der Waals equation (a=1.37a = 1.37, b=0.0387b = 0.0387), find the pressure (bar) of 1.00 mol NX2\ce{N2} in 0.500 L at 300 K.

bar

Key takeaways

  • The compressibility factor Z=PVm/RTZ = PV_\mathrm{m}/RT measures non-ideality: Z<1Z < 1 attraction dominates, Z>1Z > 1 molecular volume dominates.
  • The van der Waals equation corrects the ideal gas law with aa (attraction) and bb (excluded volume) and predicts a critical point.
  • Gases are most ideal at high temperature and low pressure; deviations grow at low temperature and high pressure, up to liquefaction.
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Data and references

  1. W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016), “Van der Waals constants for gases” and “Critical constants”.
  2. NIST Chemistry WebBook, SRD 69: carbon dioxide critical point (304.13 K, 7.377 MPa).
  3. P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 1C (real gases).