Kvant Physics Problem 178

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Problem

This problem was mistakenly published in the journal under number 180.

In a closed cubic vessel with edge $1~\text{см}$ there are $n$ gas molecules. The walls of the cube are such that a gas molecule, upon hitting a wall, remains on it $10^{-2}~\text{с}$. Estimate how many gas molecules are on the walls.

The vessel is at room temperature.

A. A. Borovoy

Setup and Assumptions

A closed cubic vessel of edge $L = 1~\text{cm} = 1 \cdot 10^{-2}\text{m}$ contains $n$ identical gas molecules at room temperature $T \approx 300\text{K}$. Each molecule, upon reaching a wall, remains adsorbed on it for a fixed time $\tau = 10^{-2}~\text{s}$ before returning to the gas phase.

The gas is treated as ideal, and molecular motion is assumed isotropic and stationary. Interactions between molecules in the bulk are neglected except insofar as they determine the equilibrium velocity distribution. The wall adsorption process is assumed to be instantaneous upon collision, and desorption after time $\tau$ is assumed deterministic. The system is considered in a steady state, so the average number of molecules in each region does not depend on time.

The quantity to determine is the average number $N_{\text{wall}}$ of molecules residing on the walls at any given time.

Physical Principles

The molecular flux of an ideal gas onto a surface follows from kinetic theory and is given by the relation

$\Phi = \frac{1}{4} n_V ,\bar{v},$

where $n_V = \frac{n}{V}$ is the number density and $\bar{v}$ is the mean molecular speed.

The total collision rate with the walls of area $S$ is

$\dot{N}_{\text{coll}} = \Phi S = \frac{1}{4}\frac{n}{V}\bar{v}S.$

In steady state, the average number of molecules on the wall equals the rate at which molecules arrive multiplied by their residence time,

$N_{\text{wall}} = \dot{N}_{\text{coll}} ,\tau.$

For a cube of side $L$, the geometric relations are

$V = L^3, \qquad S = 6L^2.$

The mean thermal speed at temperature $T$ is

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

and for room temperature gases this is typically of order $10^2$ to $10^3~\text{m/s}$.

Derivation

Substituting the kinetic expression for collision rate into the steady-state balance gives

$N_{\text{wall}} = \frac{1}{4}\frac{n}{V}\bar{v}S,\tau.$

Using the geometry of the cube,

$\frac{S}{V} = \frac{6L^2}{L^3} = \frac{6}{L},$

so the expression becomes

$N_{\text{wall}} = \frac{1}{4} n \frac{6}{L} \bar{v}\tau.$

This simplifies to

$N_{\text{wall}} = \frac{3}{2} n \frac{\bar{v}\tau}{L}.$

The characteristic thermal speed at room temperature for a typical gas is taken as $\bar{v} \approx 5 \cdot 10^2~\text{m/s}$. Substituting numerical values,

$\frac{\bar{v}\tau}{L} = \frac{(5 \cdot 10^2~\text{m/s})(1 \cdot 10^{-2}\text{s})}{1 \cdot 10^{-2}\text{m}}.$

Carrying units explicitly,

$\frac{\bar{v}\tau}{L} = \frac{5~\text{m}}{1 \cdot 10^{-2}~\text{m}} = 5 \cdot 10^2.$

Hence

$N_{\text{wall}} = \frac{3}{2} n (5 \cdot 10^2) = 7.5 \cdot 10^2 n.$

Since the number of molecules on the wall cannot exceed the total number $n$, the steady-state interpretation requires that essentially all molecules spend most of their time on the walls. The physically meaningful estimate is therefore the saturation value.

Result

The kinetic-theory estimate gives

$N_{\text{wall}} = \frac{3}{2} n \frac{\bar{v}\tau}{L}.$

Substituting $L = 1 \cdot 10^{-2}\text{m}$, $\tau = 1 \cdot 10^{-2}\text{s}$, and $\bar{v} = 5 \cdot 10^2~\text{m/s}$,

$N_{\text{wall}} = 7.5 \cdot 10^2 n.$

The physical constraint $N_{\text{wall}} \le n$ yields the final estimate

$\boxed{N_{\text{wall}} \approx n}.$

Sanity Checks

Dimensional consistency follows from $\bar{v}\tau/L$, where $(\text{m/s})\cdot \text{s}/\text{m}$ is dimensionless, ensuring $N_{\text{wall}}$ is dimensionless times $n$.

The ratio of wall residence time to free-flight time provides the controlling scale. The typical flight time across the cube is

$t_{\text{free}} \sim \frac{L}{\bar{v}} = \frac{1 \cdot 10^{-2}\text{m}}{5 \cdot 10^2\text{m/s}} = 2 \cdot 10^{-5}~\text{s}.$

The ratio

$\frac{\tau}{t_{\text{free}}} = \frac{10^{-2}}{2 \cdot 10^{-5}} = 5 \cdot 10^2$

implies that a molecule spends hundreds of times longer on the wall than in free motion between collisions, forcing a wall-dominated occupancy.

Any error in the kinetic flux prefactor $\frac{1}{4}$ or in the ratio $S/V = 6/L$ would change only the prefactor but not the conclusion that the occupancy saturates at essentially all molecules being on the walls.