Real Gases and the Compressibility Factor
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Direct answer
Divide pV by nRT for a real gas and you rarely get 1. The ratio is the compressibility factor Z, the single best index of non-ideality: Z = 1 ideal; Z below 1 means attractive forces dominate and the gas is more compressible than ideal (methane, carbon dioxide at moderate pressure); Z above 1 means the molecules' own volume and repulsions dominate at high pressure. Hydrogen and helium show Z above 1 at ordinary temperatures outright — their attractions are so weak that their critical temperatures (about 33 K and 5 K) sit hopelessly below room temperature. The van der Waals equation, (p + an^2/V^2)(V − nb) = nRT, corrects the ideal gas law for exactly these two failings, with a measuring attraction and b the excluded volume.
What you must remember
- Z and its reading: Z = pV/nRT; a dip below 1 grows as temperature falls toward the critical temperature — the Z-p curves for CO2 dip far deeper than nitrogen's shallow dip.
- van der Waals pair: a (units bar L^2 mol^-2) scales attraction and raises the effective pressure; b (L mol^-1) is the excluded volume, roughly four times the molecules' own volume.
- Critical constants from a and b: Tc = 8a/27Rb, pc = a/27b^2, Vc = 3b — three derived results JEE Advanced recycles.
- Boyle temperature: the temperature where Z stays near 1 over a range of pressures, attraction balancing repulsion; above it, Z exceeds 1 from the start.
- Liquefaction rule: no pressure liquefies a gas above its Tc — the critical point caps compression alone.
- Syllabus position: the 2023-24 rationalisation removed States of Matter from the JEE Main listing, but JEE Advanced retains gaseous and liquid-state behaviour including van der Waals — check the current bulletin and prepare accordingly.
Running the van der Waals numbers
Hold one mole of CO2 in 5 L at 300 K, with a = 3.59 atm L^2 mol^-2 and b = 0.0427 L mol^-1. Ideal gas: p = RT/V = 0.0821 × 300/5 = 4.93 atm. van der Waals: p = RT/(V − b) − a/V^2 = 24.63/4.957 − 3.59/25 = 4.97 − 0.14 = 4.83 atm. The real gas exerts less pressure because the attraction term subtracts, and Z = 4.83/4.93 = 0.98 — mildly attractive behaviour, exactly what a moderate-pressure CO2 curve predicts.
Now squeeze the same gas. As V shrinks toward a few b, the V − b denominator inflates the pressure faster than a/V^2 can subtract, p overshoots the ideal value, and Z climbs through 1 — the high-pressure side of every Z-p graph. One equation, both regimes: the a term owns the dip, the b term owns the rise, and their competition is the whole figure.
How the exam frames it
The recurring misread is direction: Z below 1 means the gas is easier to compress than ideal (attractions help the squeeze), not harder — inverted readings eliminate half the class on a single MCQ. The second staple is graph interpretation: given Z-p curves at several temperatures for N2 or CH4, identify which isotherm sits highest (highest temperature), which dips deepest (lowest), and where each crosses Z = 1. Third comes the a-and-b reasoning: CO2 has larger a than N2 (stronger attractions, more liquefiable, higher Tc at 304 K against nitrogen's 77 K), while Ne sits lowest on both counts — ranking questions built entirely from these constants. For JEE Advanced, the critical-constant derivations and the b-equals-four-times-molecular-volume argument are permanent residents.
Frequently asked questions
What does a compressibility factor below 1 signify?
Attractive intermolecular forces dominate, so the real gas exerts lower pressure than ideal and is more compressible than the ideal gas law predicts.
Why do H2 and He show Z above 1 at ordinary temperatures?
Their weak attractions and very low critical temperatures mean repulsion and excluded volume win at every ordinary pressure, keeping Z above 1.
What do the van der Waals constants a and b physically represent?
a measures the strength of intermolecular attraction (pressure correction), b the excluded volume per mole — roughly four times the volume of the molecules themselves.
What is the Boyle temperature of a gas?
The temperature at which Z remains approximately 1 over a range of pressures, because attraction and repulsion effects cancel — above it the gas shows positive deviation throughout.
Why can't a gas be liquefied above its critical temperature?
No pressure can overcome thermal motion enough to condense it; compression alone only works at or below Tc, above which the fluid becomes supercritical.