λ = 1/(√2 π d² n). Factors affecting mean free path.

Mean Free Path — What It Means and How to Calculate

Question

A gas molecule has diameter $d$ and the number density (molecules per unit volume) is $n$ . Derive the expression for the mean free path $\lambda$ , and state how it changes when:

(a) pressure is doubled at constant temperature
(b) temperature is doubled at constant pressure

Solution — Step by Step

Mean free path $\lambda$ is the average distance a molecule travels between two successive collisions. Think of a crowded corridor — the more people, and the fatter each person, the shorter the distance you cover before bumping into someone.

Imagine a molecule moving with average speed $\bar{v}$ . In time $t$ , it sweeps out a cylinder of radius $d$ (the molecular diameter) and length $\bar{v}t$ .

Any molecule whose centre lies within this cylinder gets hit. The volume swept = $\pi d^2 \bar{v} t$ .

Number of collisions = $\pi d^2 \bar{v} t \cdot n$ , where $n$ is number density.

The formula $\pi d^2 \bar{v} t \cdot n$ assumes all other molecules are stationary — they’re not. When we account for the relative speed between molecules (all moving with a Maxwell distribution), the average relative speed turns out to be $\sqrt{2}\,\bar{v}$ , not $\bar{v}$ .

So the corrected collision frequency = $\sqrt{2}\,\pi d^2 \bar{v} n$ .

\lambda = \frac{\text{distance per unit time}}{\text{collisions per unit time}} = \frac{\bar{v}}{\sqrt{2}\,\pi d^2 \bar{v}\, n}

The $\bar{v}$ cancels cleanly:

\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}

Case (a): Pressure doubled, temperature constant.
From the ideal gas law, $n = P/kT$ . Temperature is constant, so $n \propto P$ . Doubling $P$ doubles $n$ , and since $\lambda \propto 1/n$ , the mean free path halves.

Case (b): Temperature doubled, pressure constant.
At constant $P$ , doubling $T$ means $n = P/kT$ halves. So $\lambda$ doubles.

Why This Works

The key insight is that $\lambda$ depends only on number density $n$ and molecular size $d$ — not directly on speed. This is why $\bar{v}$ cancels out. A faster molecule sweeps more volume per second, but it also covers more distance per second, and the ratio stays the same.

The $\sqrt{2}$ correction is a statistical result from integrating over the Maxwell speed distribution for relative velocities. For board exams, just remember it’s there because both molecules are moving. For JEE, knowing why it’s $\sqrt{2}$ and not $2$ or $1$ is a common MCQ trick.

Number density $n$ connects to pressure via $P = nkT$ . This bridge between the microscopic ( $n$ , $d$ ) and macroscopic ( $P$ , $T$ ) is what makes this formula so useful in mixed problems.

Alternative Method

You can also write $\lambda$ in terms of pressure directly. Since $n = P/kT$ :

\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}

This form makes the pressure and temperature dependence immediately visible — $\lambda \propto T/P$ . Many JEE Main questions give you $P$ and $T$ directly, so having this form ready saves a step.

For numerical problems, $kT/P = V/N$ (volume per molecule). If the problem gives you molar volume, you can use $\lambda = RT/(\sqrt{2}\,\pi d^2 N_A P)$ — same thing, just with $R$ and $N_A$ instead of $k$ .

Common Mistake

Students often write $\lambda = 1/(\pi d^2 n)$ , dropping the $\sqrt{2}$ . In MCQs, one of the options is always this incorrect version. The $\sqrt{2}$ comes from the relative velocity correction and cannot be ignored — it changes the numerical answer by about 29%. In JEE Main 2024, a question directly tested whether students included this factor.

A second trap: confusing $d$ (molecular diameter) with $r$ (radius). The formula uses diameter because two molecules collide when their centres are within distance $d$ of each other — one radius from each side. Writing $r$ instead of $d$ gives a factor of 4 error in the denominator ( $\pi d^2$ vs $\pi r^2$ ).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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