Real gases have finite molecular volume and intermolecular attractions. How van der Waals constants a and b correct for this.

Why Does a Real Gas Deviate from Ideal Behaviour? — van der Waals Equation

Question

Real gases deviate from ideal behaviour. Explain why, and write the van der Waals equation with the significance of constants $a$ and $b$ .

(NCERT Class 11, Chapter 5 — also a favourite in JEE Main theory questions)

Solution — Step by Step

The ideal gas law $PV = nRT$ rests on two assumptions: molecules have zero volume, and there are no intermolecular forces between them. Real molecules are neither point masses nor completely non-interacting.

Inside a real gas, molecules attract each other. A molecule hitting the wall gets pulled back by its neighbours, so it hits with less force than expected. The observed pressure is lower than ideal pressure.

We correct this by adding back the “lost” pressure:

P_{\text{ideal}} = P_{\text{observed}} + \frac{an^2}{V^2}

The term $\frac{an^2}{V^2}$ is the pressure correction. The constant $a$ measures strength of intermolecular attraction — larger $a$ means stronger attractions (e.g., $\text{CO}_2$ has $a = 3.59$ vs $\text{H}_2$ at $0.24$ ).

Molecules occupy real space. The volume available for free movement is less than the container volume $V$ .

V_{\text{ideal}} = V_{\text{observed}} - nb

Here $b$ is the excluded volume per mole — roughly four times the actual volume of one mole of molecules. It’s purely a size correction with no relation to attraction.

Substituting both corrections into $PV = nRT$ :

\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT

This is the van der Waals equation. For 1 mole ( $n = 1$ ):

\left(P + \frac{a}{V^2}\right)(V - b) = RT

Getting units right is a common board exam ask.

$\frac{an^2}{V^2}$ must have units of pressure, so $a$ has units: $\text{atm·L}^2\text{·mol}^{-2}$ (or $\text{Pa·m}^6\text{·mol}^{-2}$ in SI)
$nb$ must have units of volume, so $b$ has units: $\text{L·mol}^{-1}$ (or $\text{m}^3\text{·mol}^{-1}$ )

Why This Works

The ideal gas law treats molecules as ghost particles — perfectly elastic billiard balls with no size and no feelings for each other. At low pressure and high temperature, real molecules are far apart and moving fast, so these assumptions nearly hold. That’s why real gases approach ideal behaviour at high $T$ and low $P$ .

At high pressure, molecules are cramped together — their actual volume becomes a significant fraction of $V$ , so the $b$ correction matters. At low temperature, molecules move slowly and attractive forces have time to act, so the $a$ correction becomes significant. This is why $\text{CO}_2$ liquefies easily (high $a$ ) while $\text{H}_2$ is notoriously hard to liquefy (tiny $a$ ).

A quick memory trick: $a$ is for Attraction (both start with ‘a’), $b$ is for Bulk (molecular size). In JEE, they sometimes give you $a$ and $b$ values and ask which gas is more easily liquefied — always pick the one with higher $a$ , since stronger attraction means easier condensation.

Alternative Method — Compressibility Factor Approach

Instead of correcting $P$ and $V$ separately, we can define the compressibility factor:

Z = \frac{PV}{nRT}

For an ideal gas, $Z = 1$ always. For real gases:

$Z < 1$ at moderate pressures → attractive forces dominate → gas is more compressible than ideal
$Z > 1$ at very high pressures → repulsive forces (size effect) dominate → gas resists compression

This approach is used in JEE Advanced-level problems where you’re given a $Z$ vs $P$ graph and asked to identify the gas or the dominant deviation. $\text{H}_2$ and $\text{He}$ show $Z > 1$ even at low pressures because their $a$ values are negligibly small — size effect wins from the start.

Common Mistake

Students often confuse what $a$ and $b$ correct for. $a$ corrects pressure (intermolecular attraction reduces pressure), but many write it as a volume correction. And $b$ corrects volume (molecular size reduces free volume), but students sometimes associate it with attraction because “bigger molecules attract more.” They don’t — attraction is entirely captured by $a$ . A molecule can have large $b$ (big size) but small $a$ (weak attraction), like noble gases.

Also watch the sign: the volume correction is $(V - nb)$ , not $(V + nb)$ . The free volume is less than $V$ , so we subtract. Writing $+nb$ is a guaranteed mark drop in board exams.

Question

Solution — Step by Step

Why This Works

Alternative Method — Compressibility Factor Approach

Common Mistake

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