What is mean free path and how does it depend on pressure and temperature

Question

What is mean free path? How does it depend on (a) pressure at constant temperature, and (b) temperature at constant pressure?

Solution — Step by Step

Mean free path ( $\lambda$ ) is the average distance a gas molecule travels between two successive collisions with other molecules.

Gas molecules are in constant, rapid, random motion — the Maxwell-Boltzmann distribution. They collide with each other frequently. After each collision, they change direction. The mean free path is the statistical average of all the individual path lengths between collisions.

For an ideal gas, the mean free path is:

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

where $d$ = diameter of the molecule, $n$ = number density (molecules per unit volume)

Here, $n = N/V$ (number of molecules per unit volume).

Using the ideal gas equation: $PV = NkT$ (where $k$ = Boltzmann constant), we get:

n = \frac{N}{V} = \frac{P}{kT}

Substituting into the mean free path formula:

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

At constant temperature ( $T$ = constant):

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

So $\lambda \propto \frac{1}{P}$ .

Interpretation: When pressure increases at constant temperature, more molecules are packed into the same volume (higher number density). More molecules in the same space means more frequent collisions → shorter mean free path.

Example: At sea level (atmospheric pressure), the mean free path of air molecules is ~70 nm. At 10 km altitude where pressure is about ¼ of sea level, the mean free path is ~280 nm (4 times larger).

At constant pressure ( $P$ = constant):

\lambda = \frac{kT}{\sqrt{2}\pi d^2 P} = \frac{k}{\sqrt{2}\pi d^2 P} \times T

So $\lambda \propto T$ .

Interpretation: At constant pressure, when temperature increases, the gas expands — molecules are farther apart (lower number density $n = P/kT$ decreases as $T$ increases). Fewer molecules per unit volume means fewer collisions → longer mean free path.

Wait, let’s check: if $P$ is constant and $T$ increases, from $n = P/(kT)$ , as $T \uparrow$ , $n \downarrow$ . With lower number density, molecules collide less often → $\lambda$ increases. This is consistent with $\lambda \propto T$ at constant $P$ .

Condition	Relationship	Physical Reason
$T$ constant; $P$ increases	$\lambda \propto \frac{1}{P}$	More molecules/volume → more collisions
$P$ constant; $T$ increases	$\lambda \propto T$	Molecules spread out → fewer collisions
$T$ constant; $n$ increases	$\lambda \propto \frac{1}{n}$	More molecules → more collisions (fundamental)

Why This Works

The formula $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$ comes from a simple model: imagine one molecule as a sphere of diameter $d$ moving through a sea of point particles. As it moves, it sweeps out a cylinder of radius $d$ — any molecule with its centre within this cylinder will be hit. The cross-sectional area of this cylinder is $\pi d^2$ . In distance $\ell$ , the volume swept is $\pi d^2 \ell$ . The mean number of collisions is $\pi d^2 \ell \times n$ . Setting this equal to 1 collision gives the mean free path. The $\sqrt{2}$ factor corrects for the fact that all molecules are moving (not just the test molecule).

Alternative Method — Collision Frequency Approach

Mean free path = speed / collision frequency

Collision frequency: $Z = \sqrt{2}\pi d^2 n \bar{v}$

where $\bar{v} = \sqrt{\frac{8kT}{\pi m}}$ is the mean speed.

\lambda = \frac{\bar{v}}{Z} = \frac{\bar{v}}{\sqrt{2}\pi d^2 n \bar{v}} = \frac{1}{\sqrt{2}\pi d^2 n}

Same result — consistent.

JEE Main and NEET both ask about the dependence of mean free path on P and T. The clean way to remember: mean free path depends on $\frac{T}{P}$ . If both T and P double, $\lambda$ stays the same (the effects cancel). This proportionality is directly tested in MCQ format.

Common Mistake

Students sometimes say “mean free path decreases with increasing temperature” — this is only true if the volume is constant (i.e., pressure increases). At constant pressure, increased temperature expands the gas and the mean free path INCREASES. Always specify what is being held constant when discussing how $\lambda$ varies.

Question

Solution — Step by Step

Why This Works

Alternative Method — Collision Frequency Approach

Common Mistake

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