Why Real Gases Break the Ideal Gas Law: Z, a and b Explained
Your textbook says PV = nRT, yet a gas cylinder at 100 atm refuses to obey it. If the law is a "law", why does it fail, and how do we know by how much? That question is exactly what this concept answers, and it matters for both NEET and JEE (Class 11, States of Matter).
Quick Summary
An ideal gas obeys PV = nRT at all pressures and temperatures; a real gas only approximately.
The ideal model assumes (1) molecules have negligible volume and (2) no attractions between them.
Real gases deviate most at high pressure and low temperature, where both assumptions break.
The compressibility factor Z = PV/nRT measures deviation: Z = 1 is ideal, Z < 1 means attraction dominates, Z > 1 means molecular size dominates.
The van der Waals equation corrects PV = nRT with two constants, a (attraction) and b (size).
Where the ideal gas law comes from
You already know Boyle's law (P ∝ 1/V at fixed T), Charles's law (V ∝ T at fixed P) and Avogadro's law (V ∝ n at fixed T and P). Combine the three and you get PV = nRT, where P is pressure, V is volume, n is the number of moles, T is the absolute temperature in kelvin and R is the gas constant (0.0821 L atm mol⁻¹ K⁻¹, or 8.314 J mol⁻¹ K⁻¹).
Behind this sits a model: molecules are point-like dots that never attract or repel each other, and they collide perfectly elastically with the walls. Those two assumptions are the whole story of deviation. When they are almost true, the law works. When they are false, the law fails.
Why real gases deviate
Molecules do occupy space. At low pressure the gas is mostly empty space, so the molecular volume is negligible. Squeeze the gas to high pressure and the molecules themselves take up a noticeable share of the container. The free space available for movement is then smaller than V, so the real pressure is higher than the ideal prediction. This pushes Z above 1.
Molecules do attract each other. A molecule about to hit the wall is pulled back by its neighbours, so it hits the wall more softly. The measured pressure is therefore lower than the ideal value. This pushes Z below 1. Attraction matters most at moderate pressure and low temperature, when molecules move slowly enough for the pull to matter.
So there is a contest. At low pressure, attraction wins slightly (Z dips below 1). At very high pressure, size wins (Z rises above 1). At high temperature, molecules move so fast that attraction is irrelevant, so gases behave ideally.
The van der Waals equation
Van der Waals fixed both assumptions in one equation, written for n moles as: (P + an²/V²)(V − nb) = nRT.
nb is subtracted from V because the molecules' own volume is not available for motion. b is the excluded volume per mole, in L mol⁻¹.
an²/V² is added to P because attraction lowers the measured pressure; we add it back to get the pressure the gas would have had without attraction. a measures the strength of attraction, in L² atm mol⁻².
A larger a means the gas liquefies more easily. That is why CO₂ and NH₃ (large a) are easier to liquefy than H₂ and He (tiny a).
Worked example
Question. One mole of a gas occupies 0.200 L at 300 K and exerts a measured pressure of 100 atm. Find Z and say what dominates. Use R = 0.0821 L atm mol⁻¹ K⁻¹.
Ideal value of nRT: 1 × 0.0821 × 300 = 24.63 L atm.
Actual PV = 100 × 0.200 = 20.0 L atm.
Z = PV/nRT = 20.0 ÷ 24.63 ≈ 0.81.
Since Z < 1, the gas is more compressible than ideal, so intermolecular attraction dominates under these conditions.
For comparison, the ideal pressure would be nRT/V = 24.63 ÷ 0.200 ≈ 123 atm. The real gas shows 100 atm, about 19% lower, because attraction is pulling molecules back from the walls.
Mistakes students actually make
Using temperature in °C in PV = nRT. Always convert to kelvin: T(K) = t(°C) + 273.
Mixing units of R with units of P and V. If P is in atm and V in litres, use 0.0821.
Reversing the corrections: writing (P − an²/V²) or (V + nb). Remember, attraction makes P too low so you add; volume is reduced so you subtract.
Saying "Z > 1 means attraction". It is the opposite: Z > 1 means repulsion or size effects dominate.
Believing the ideal gas law fails at high temperature. It works better there.
How it is asked in NEET and JEE
Conceptual: "At what conditions do real gases show maximum deviation?" (high P, low T).
Interpreting Z versus P graphs, including why H₂ and He show Z > 1 right from low pressure (very small a).
Comparing the a or b values of two gases to predict which liquefies more easily or which is bigger.
Units of a and b, or calculating Z from given data as in the example above.
Pro Tip: Remember "a for attraction, b for bulk". The b term matters when volume is tiny (high pressure), and the a term matters when molecules are slow (low temperature). If a question gives you both a high temperature and a low pressure, the answer is almost always "behaves ideally", whatever the gas is. Connecting a and b to the two broken assumptions lets you rebuild the whole van der Waals equation in seconds, even if you forget it, much like the way we reason out concepts step by step at Sukriti NeetJee instead of memorising them.
Frequently Asked Questions
Does any real gas behave perfectly ideally?
No gas is perfectly ideal, but gases like He and H₂ come very close at ordinary conditions because their molecules are small and attractions are extremely weak. Every gas approaches ideal behaviour as pressure tends to zero.
What is the Boyle temperature?
It is the temperature at which a real gas obeys Boyle's law over a wide range of low pressures, because the effects of attraction and molecular size cancel. At that temperature Z stays close to 1 as pressure rises from zero.
Why does a larger value of a mean easier liquefaction?
A larger a means stronger intermolecular attraction. Stronger attraction lets molecules hold together once they are cooled or compressed, so the gas condenses at a higher temperature.
Why is Z greater than 1 for hydrogen even at low pressure?
Hydrogen's attraction constant a is very small, so the attraction effect is negligible. The molecular-size effect (the b term) then dominates, and Z is above 1 across the pressure range.
Is the van der Waals equation exact?
No, it is a better approximation than PV = nRT but still an approximation. It fits well for many gases away from the critical point but fails to describe all behaviour near liquefaction.
Stuck on a Z versus P graph or a van der Waals problem? Drop your doubt in the comments, and share this with a fellow aspirant who is revising States of Matter.
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