Problem — Estimating collisions on a square centimetre

Problem

Estimate the number of collisions air molecules make each second on 1 cm$^2$ of wall at room temperature and pressure. (Hint: $n\bar{v}/4$ is the molecular flux.)

Solution

The molecular flux (number of collisions per unit area and time) is $\Phi = n\,\langle v\rangle/4$, where $n$ is the number density of the molecules and $\langle v\rangle$ their mean speed. Number density from $p = n\,k_B T$, with $p \approx 1{,}0\cdot 10^5$ Pa and $T \approx 300$ K: $n = \frac{p}{k_B T}=\frac{1{,}0\cdot 10^5}{1{,}38\cdot 10^{-23}\cdot 300}\approx 2{,}4\cdot 10^{25}\ \text{m}^{-3}$ Mean speed of air (molecules with $M\approx 29$ g/mol) $\langle v\rangle \approx 4{,}7\cdot 10^{2}$ m/s. The flux is then: $\Phi = \frac{n\,\langle v\rangle}{4}\approx \frac{2{,}4\cdot 10^{25}\cdot 4{,}7\cdot 10^{2}}{4}\approx 2{,}9\cdot 10^{27}\ \text{m}^{-2}\text{s}^{-1}$ On one square centimetre ($A = 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$): $N = \Phi\cdot A \approx 2{,}9\cdot 10^{27}\cdot 10^{-4}\approx 3\cdot 10^{23}\ \text{collisions/s}$ $\ev{N \sim 10^{23}\ \text{collisions/s on 1 cm}^2}$

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Topics: Kinetic theory of gases Concepts: Pressure Skills: Fermi estimates · Micro-macro interpretation