What the mean free path is
In the ideal-gas picture, molecules are treated as point masses that move freely between collisions and the time spent colliding is negligible compared to the time between collisions. But molecules are not points — they are spheres of a small finite diameter $d$. That finite size guarantees collisions, and between collisions a molecule travels in a straight line at constant velocity. The path of a single molecule is therefore a sequence of straight segments joined by sharp deflections: a zig-zag.
The lengths of those straight segments vary from one collision to the next. The mean free path $\lambda$ is the average of these segment lengths — the average distance a molecule travels between two successive collisions. It is a single number that captures how far, on average, a molecule gets before its motion is interrupted. The larger $\lambda$ is, the more freely the gas spreads; for the rarefied gas in a highly evacuated tube, $\lambda$ can grow as large as the length of the tube itself.
The collision cylinder derivation
Follow one molecule of diameter $d$ moving with average speed $\langle v\rangle$, and freeze every other molecule for a moment. Our molecule will collide with any other molecule whose centre comes within a distance $d$ of its own centre. As it advances, it therefore sweeps out a cylinder of cross-sectional area $\pi d^2$ — the collision cross-section — and any molecule whose centre lies inside that cylinder is struck.
In a time $\Delta t$ the molecule travels a distance $\langle v\rangle\,\Delta t$, so the cylinder it sweeps has volume $\pi d^2\,\langle v\rangle\,\Delta t$. If $n$ is the number density (molecules per unit volume), the number of molecules inside this volume — and hence the number of collisions — is $n\,\pi d^2\,\langle v\rangle\,\Delta t$. The rate of collisions is therefore $n\,\pi d^2\,\langle v\rangle$, and the average time between two successive collisions is its reciprocal,
$$\tau = \frac{1}{n\,\pi\,\langle v\rangle\,d^2}.$$
The mean free path is the distance covered in this time, $\lambda = \langle v\rangle\,\tau$, which gives the first-pass result
$$\lambda = \langle v\rangle\,\tau = \frac{1}{n\,\pi d^2}.$$
The √2 from relative motion
The derivation above cheated once: it held all the other molecules still. In reality every molecule is moving, so a collision is governed not by our molecule's speed but by the average relative speed between molecules. For molecules with the same Maxwellian speed distribution, the average relative speed is $\langle v_r\rangle = \sqrt{2}\,\langle v\rangle$.
Replacing $\langle v\rangle$ by $\langle v_r\rangle = \sqrt 2\,\langle v\rangle$ in the collision rate raises the rate by a factor of $\sqrt 2$, and dividing the (still single-molecule) speed $\langle v\rangle$ by that larger rate shortens the free path by the same factor. NCERT's more exact treatment therefore gives
$$\boxed{\;\lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n}\;}$$
where $d$ is the molecular diameter and $n$ the number density. This is the formula NEET expects. The $\sqrt 2$ is not decorative — it is the signature that the other molecules are moving too.
Mean free path in terms of P and T
The number density $n$ is rarely handed to you directly; pressure and temperature are. Use the ideal-gas relation in its per-molecule form, $P = n k_B T$, so that $n = P/(k_B T)$. Substituting into the boxed formula,
$$\lambda = \frac{1}{\sqrt 2\,\pi d^2 n} = \frac{k_B T}{\sqrt 2\,\pi d^2 P}.$$
This is the working form for most problems: it lets you read off the temperature and pressure dependence at a glance, and it makes clear that $T$ and $P$ influence $\lambda$ only through $n$. Holding $T$ fixed, $\lambda$ falls as $P$ rises; holding $P$ fixed, $\lambda$ grows as $T$ rises. Both statements are just $n \propto P/T$ in disguise.
How λ depends on n, P, T and d
Four quantities can move the mean free path. Hold the others fixed and the dependence on each is simple — but the conditions matter, and that is where NEET sets its traps.
| Vary | Held fixed | Dependence | Reading |
|---|---|---|---|
| Number density $n$ | $d$ | $\lambda \propto 1/n$ | Denser gas → shorter free path. The most fundamental dependence. |
| Pressure $P$ | $T,\,d$ | $\lambda \propto 1/P$ | Compress at constant $T$ → $n$ rises → $\lambda$ falls. |
| Temperature $T$ | $P,\,d$ | $\lambda \propto T$ | Heat at constant $P$ → gas expands → $n$ falls → $\lambda$ rises. |
| Molecular diameter $d$ | $n$ | $\lambda \propto 1/d^2$ | Bigger molecule → larger cross-section $\pi d^2$ → shorter free path. |
| Molecular mass $m$ | — | $\lambda$ independent of $m$ | Mass never appears in the formula. Speed cancels out. |
The single safest statement is the one in terms of number density: $\lambda$ is inversely proportional to $n$, full stop. The pressure and temperature laws are corollaries that hold only when the other of the pair is fixed. The diameter enters as a square because the collision cross-section is an area, $\pi d^2$.
Collision frequency
The companion quantity to the mean free path is the collision frequency $\nu$ — the number of collisions a molecule undergoes per second. Since a molecule covers one free path $\lambda$ in the average time $\tau$ between collisions, and moves at speed $v$,
$$\nu = \frac{1}{\tau} = \frac{v}{\lambda} = \sqrt 2\,\pi d^2 n\, v.$$
Either the mean speed $\langle v\rangle$ or the rms speed $v_{\text{rms}}$ may be used depending on the level of the problem; for NEET estimates the distinction rarely matters and $\nu = v_{\text{rms}}/\lambda$ is the standard working relation. Note the inverse link: a long free path means infrequent collisions, and a short free path means a high collision frequency. Both follow from the same cylinder geometry.
The $v$ in $\nu = v/\lambda$ comes straight from the speed distribution — revise $v_{\text{rms}}$, $\bar v$ and $v_{\text{mp}}$ in molecular speeds.
Typical values for air
Numbers anchor the idea. For air molecules at STP, NCERT takes an average speed $\langle v\rangle \approx 485\ \text{m s}^{-1}$ and a number density
$$n = \frac{6.02\times10^{23}}{22.4\times10^{-3}} \approx 2.7\times10^{25}\ \text{m}^{-3}.$$
With a molecular diameter $d = 2\times10^{-10}\ \text{m}$, the formulae give a time between collisions $\tau \approx 6.1\times10^{-10}\ \text{s}$ and a mean free path
$$\lambda \approx 2.9\times10^{-7}\ \text{m} \approx 1500\,d.$$
So a typical air molecule travels roughly 1500 times its own diameter — about a third of a micrometre — between collisions. That the free path so dwarfs the molecular size is precisely what gives gases their free-flowing, compressible behaviour and explains why a gas cannot be confined without a container. The corresponding collision frequency is of order $\nu \sim 10^9\!-\!10^{10}$ collisions per second.
Worked example — nitrogen (NCERT Exercise 12.10)
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0$ atm and $17^{\circ}\text{C}$. Take the radius of a nitrogen molecule to be roughly $1.0$ Å, so $d = 2.0\times10^{-10}\ \text{m}$. Molecular mass of $\text{N}_2 = 28.0\ \text{u}$. ($k_B = 1.38\times10^{-23}\ \text{J K}^{-1}$, $1\ \text{atm} = 1.013\times10^{5}\ \text{Pa}$.)
Step 1 — number density. $T = 17 + 273 = 290\ \text{K}$ and $P = 2.0\times1.013\times10^{5} = 2.026\times10^{5}\ \text{Pa}$. From $n = P/(k_B T)$, $$n = \frac{2.026\times10^{5}}{(1.38\times10^{-23})(290)} \approx 5.06\times10^{25}\ \text{m}^{-3}.$$
Step 2 — mean free path. With $\pi d^2 = \pi(2.0\times10^{-10})^2 = 1.26\times10^{-19}\ \text{m}^2$, $$\lambda = \frac{1}{\sqrt 2\,\pi d^2 n} = \frac{1}{(1.414)(1.26\times10^{-19})(5.06\times10^{25})} \approx 1.11\times10^{-7}\ \text{m}.$$
Step 3 — molecular speed. Using the rms speed of $\text{N}_2$ ($M = 28\times10^{-3}\ \text{kg mol}^{-1}$, $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$), $$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3(8.31)(290)}{28\times10^{-3}}} \approx 508\ \text{m s}^{-1}.$$
Step 4 — collision frequency. $\displaystyle \nu = \frac{v_{\text{rms}}}{\lambda} = \frac{508}{1.11\times10^{-7}} \approx 4.6\times10^{9}\ \text{s}^{-1}.$
Answer. $\lambda \approx 1.1\times10^{-7}\ \text{m}$ and $\nu \approx 4.6\times10^{9}\ \text{collisions s}^{-1}$. The time between collisions is $\tau = 1/\nu \approx 2.2\times10^{-10}\ \text{s}$; the time of a collision itself ($\sim d/v_{\text{rms}} \sim 4\times10^{-13}\ \text{s}$) is some 500 times shorter, which justifies the kinetic-theory assumption that collision time is negligible.
The example rewards the working form $\lambda = k_B T/(\sqrt 2\,\pi d^2 P)$: doubling the pressure to $2.0$ atm halves the free path relative to $1$ atm, which is why the nitrogen result ($1.1\times10^{-7}\ \text{m}$) sits below the STP air value ($2.9\times10^{-7}\ \text{m}$) even before the difference in temperature and diameter is counted.
Mean free path in one breath
- $\lambda$ is the average distance a molecule travels between two successive collisions.
- Formula: $\lambda = \dfrac{1}{\sqrt 2\,\pi d^2 n} = \dfrac{k_B T}{\sqrt 2\,\pi d^2 P}$. The $\sqrt 2$ is the average relative speed factor.
- Collision cross-section $= \pi d^2$; that is why $\lambda \propto 1/d^2$.
- Dependences: $\lambda \propto 1/n$ always; $\lambda \propto 1/P$ at constant $T$; $\lambda \propto T$ at constant $P$. Independent of molecular mass.
- Collision frequency $\nu = v/\lambda = \sqrt 2\,\pi d^2 n\, v$.
- For air at STP $\lambda \approx 2.9\times10^{-7}\ \text{m} \approx 1500\,d$ — far larger than the molecular size.