13.3 Mean free path and collision frequency

Mean Free Path and Gas Properties

Understanding Mean Free Path

The mean free path ($\lambda$) is the average distance a molecule travels between successive collisions with other molecules in a gas. It connects microscopic molecular behavior to macroscopic gas properties like viscosity and thermal conductivity.

Two molecular-level properties control $\lambda$:

• Number density ($n$): More molecules per unit volume means less room between them, so $\lambda$ shrinks.
• Molecular diameter ($d$): Bigger molecules present a larger cross-sectional target, making collisions more likely and shortening $\lambda$.

Because number density itself depends on the thermodynamic state, $\lambda$ also responds to pressure and temperature:

• Pressure is inversely proportional to $\lambda$. Doubling the pressure doubles $n$, which cuts $\lambda$ in half.
• Temperature is directly proportional to $\lambda$. At constant pressure, raising $T$ lowers $n$ (the gas expands), so molecules travel farther between hits. Quantitatively, $\lambda \propto T$ at fixed $P$, so doubling the temperature doubles the mean free path.

Calculating Mean Free Path

The standard kinetic-theory result for a single-component gas of hard-sphere molecules is:

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

The $\sqrt{2}$ factor accounts for the fact that other molecules are also moving; the relative speed between two molecules is $\sqrt{2}$ times the mean speed.

You can eliminate $n$ using the ideal gas law $n = P/(k_B T)$:

$\lambda = \frac{k_B\, T}{\sqrt{2}\,\pi\, d^2\, P}$

This form makes the proportionalities explicit: $\lambda \propto T/P$.

Worked example. Find $\lambda$ for $\text{N}_2$ at 300 K and $1.01 \times 10^5$ Pa, with $d = 3.7 \times 10^{-10}$ m.

1. Compute $d^2$: $(3.7 \times 10^{-10})^2 = 1.369 \times 10^{-19}\;\text{m}^2$
2. Compute the denominator: $\sqrt{2}\,\pi\,(1.369 \times 10^{-19})(1.01 \times 10^5) = 6.14 \times 10^{-14}$
3. Compute the numerator: $k_B T = (1.381 \times 10^{-23})(300) = 4.14 \times 10^{-21}\;\text{J}$
4. Divide: $\lambda = 4.14 \times 10^{-21} / 6.14 \times 10^{-14} \approx 6.7 \times 10^{-8}\;\text{m} \approx 67\;\text{nm}$

Unit consistency matters. Use pascals for $P$, kelvins for $T$, and meters for $d$ so that everything cancels cleanly to meters.

Knudsen Number and Flow Regimes

The Knudsen number compares the mean free path to a characteristic length scale $L$ of the system (pipe diameter, pore size, etc.):

$Kn = \frac{\lambda}{L}$

It tells you whether the gas "sees" the container walls or mostly just other molecules:

Collision Frequency in Gases

Understanding Collision Frequency

Collision frequency ($Z$) is the average number of collisions a single molecule experiences per unit time. While $\lambda$ tells you how far between collisions, $Z$ tells you how often they happen.

$Z = \sqrt{2}\,\pi\, d^2\, n\, \bar{v}$

where $\bar{v}$ is the mean molecular speed:

$\bar{v} = \sqrt{\frac{8\, k_B\, T}{\pi\, m}}$

Here $m$ is the mass of a single molecule. Notice that the combination $\sqrt{2}\,\pi\, d^2\, n$ is just $1/\lambda$, so there's an elegant connection:

$Z = \frac{\bar{v}}{\lambda}$

This makes intuitive sense: the collision rate equals how fast you're going divided by how far you travel between hits.

Factors Affecting Collision Frequency

• Temperature. Raising $T$ increases $\bar{v}$ (as $\sqrt{T}$) and, at constant pressure, decreases $n$ (as $1/T$). The net effect at constant $P$: $Z \propto 1/\sqrt{T}$, so collision frequency actually decreases with rising temperature at fixed pressure. At constant density (sealed rigid container), $n$ stays fixed and $Z \propto \sqrt{T}$, so collisions become more frequent.
• Pressure. At constant $T$, increasing $P$ raises $n$ proportionally, so $Z \propto P$. Double the pressure, double the collision frequency.
• Molecular size. Larger $d$ means a bigger collision cross-section ($\sigma = \pi d^2$). Under identical conditions, $\text{CO}_2$ molecules collide more often than $\text{H}_2$ molecules.
• Molecular mass. Heavier molecules move more slowly ($\bar{v} \propto 1/\sqrt{m}$), which reduces $Z$ compared to lighter molecules at the same $T$ and $n$.

Mean Free Path vs. Collision Frequency

How They Relate

Since $Z = \bar{v}/\lambda$, the two quantities are linked through the mean speed. But they don't always move in opposite directions, because $\bar{v}$ itself changes with conditions.

*At constant $n$, $\lambda = 1/(\sqrt{2}\,\pi d^2 n)$ has no $T$ dependence, so $\lambda$ stays the same while $\bar{v}$ (and therefore $Z$) rises.

Comparing Different Gases

When two gases are at the same $T$ and $P$, they share the same number density $n$ (ideal gas law). Differences in $\lambda$ and $Z$ then come entirely from $d$ and $m$:

• The gas with the larger diameter has a shorter $\lambda$ and (all else equal) a higher $Z$.
• The gas with the lighter molecules has a higher $\bar{v}$, which boosts $Z$ but doesn't affect $\lambda$ (since $\lambda$ doesn't depend on speed).

Connection to Transport Properties

Shorter mean free paths and higher collision frequencies mean molecules exchange momentum, energy, and mass more frequently over shorter distances. This directly feeds into the kinetic-theory expressions for:

• Viscosity ($\eta$): proportional to $\bar{v}\,\lambda$
• Thermal conductivity ($\kappa$): proportional to $\bar{v}\,\lambda$
• Diffusion coefficient ($D$): proportional to $\bar{v}\,\lambda$

All three transport coefficients scale as $\bar{v}\,\lambda$, which at constant pressure gives a $T^{3/2}$ dependence (since $\bar{v} \propto T^{1/2}$ and $\lambda \propto T$). That's why gas viscosity increases with temperature, unlike liquids.