Kvant Physics Problem 178

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 23m14s
Source on kvant.digital

Problem

In the journal, it was mistakenly published under number 180.

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

The vessel is at room temperature.

A. A. Borovoy

Setup and Assumptions

A cubic vessel of edge length $a = 1~\text{cm} = 10^{-2}~\text{m}$ contains $n$ gas molecules at room temperature. Each molecule is treated as a point particle moving randomly with thermal speed $v$ in a rarefied gas approximation, so intermolecular correlations are neglected except through equilibrium kinetic theory.

The molecules collide with the walls. Upon collision, a molecule remains attached to the wall for a fixed residence time $\tau = 10^{-2}~\text{s}$, after which it returns to the gas phase with unchanged statistical properties.

The gas is assumed to be in equilibrium, so the molecular velocity distribution is isotropic. The quantity to determine is the mean number of molecules simultaneously located on the walls.

Physical Principles

The kinetic theory flux of molecules to a surface is given by

$\Phi = \frac{1}{4} n_{\text{vol}} v,$

where $n_{\text{vol}} = \frac{n}{V}$ is the number density and $v$ is the mean molecular speed.

The total collision rate with all walls of a container of surface area $S$ is

$\dot{N}_{\text{coll}} = \Phi S.$

For a cube of edge $a$, the volume and surface area are

$V = a^3, \quad S = 6a^2.$

The mean number of particles residing on the walls in steady state equals the collision rate multiplied by the residence time,

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

The thermal speed at room temperature for a light gas is of order

$v \sim 5 \times 10^2~\text{m/s}.$

Derivation

The number density in the cube is

$n_{\text{vol}} = \frac{n}{a^3}.$

The flux to all walls is

$\dot{N}_{\text{coll}} = \frac{1}{4} \frac{n}{a^3} v \cdot 6a^2.$

Simplifying the geometric factor gives

$\dot{N}_{\text{coll}} = \frac{6}{4} \frac{n v}{a} = \frac{3}{2} \frac{n v}{a}.$

The mean number of molecules on the walls is then

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

Substituting $a = 10^{-2}\text{m}$, $\tau = 10^{-2}\text{s}$, and $v \approx 5 \times 10^2~\text{m/s}$ yields

$\frac{v \tau}{a} = \frac{(5 \times 10^2~\text{m/s})(10^{-2}\text{s})}{10^{-2}\text{m}} = 5 \times 10^2.$

Therefore,

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

Since a physical number of adsorbed molecules cannot exceed the total number present, the steady-state occupancy saturates at the upper bound set by particle conservation, leading to

$N_{\text{wall}} \sim n.$

Result

The kinetic-theory estimate gives

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

With $v = 5 \times 10^2~\text{m/s}$, $\tau = 10^{-2}\text{s}$, and $a = 10^{-2}\text{m}$,

$N_{\text{wall}} \approx 7.5 \times 10^2 n.$

Because the expression exceeds the total number of molecules, the physically consistent estimate is

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

Sanity Checks

Dimensional consistency follows from $v \tau / a$, where $(\text{m/s}) \cdot \text{s} / \text{m}$ is dimensionless, ensuring $N_{\text{wall}}$ has units of a pure count proportional to $n$.

The characteristic time for a molecule to traverse the cube is $t_{\text{cross}} \sim a/v \sim 10^{-2}\text{m} / (5 \times 10^2\text{m/s}) \approx 2 \times 10^{-5}\text{s}$. The residence time $\tau = 10^{-2}\text{s}$ exceeds this by a factor of about $500$, implying that once a molecule reaches a wall, it remains there long compared with the time needed to leave the bulk and return.

The naive flux-based estimate producing $7.5 \times 10^2 n$ fails because it implicitly assumes independent collision events without accounting for the finite population constraint. The breakdown occurs in interpreting the product of collision rate and residence time as a small perturbation; here it produces a probability greater than unity, indicating saturation of the wall population.