Problem — Estimating collisions per second on a hand

Problem

Estimate the order of magnitude of the number of collisions per second that air molecules exert overall on the palm of your hand at room temperature and pressure. Use $P_\text{atm}\approx 10^5$ Pa, palm area $A\approx 10^{-2}$ m$^2$, average molar mass $29\cdot 10^{-3}$ kg/mol, $T = 300$ K. Hint: pressure is the flux of momentum.

Solution

Each collision transfers to the hand a momentum of order $\Delta p \sim m v$, where $m$ is the mass of a molecule and $v$ its typical speed. The force on the hand is $F = PA$, i.e. the momentum delivered per second. The number of collisions per second is therefore $N_\text{collisions} \sim F/\Delta p = PA/(m v)$.

Mass of one air molecule: $m = \frac{m_\text{mol}}{N_A} = \frac{29\cdot 10^{-3}}{6{,}02\cdot 10^{23}} \approx 4{,}8\cdot 10^{-26}\ \text{kg}$ Typical (root-mean-square) speed: $v \approx \sqrt{\frac{3RT}{m_\text{mol}}} = \sqrt{\frac{3\cdot 8{,}31\cdot 300}{29\cdot 10^{-3}}} \approx 5\cdot 10^{2}\ \text{m/s}$ Force exerted by the air on the palm: $F = PA = 10^{5}\cdot 10^{-2} = 10^{3}\ \text{N}$ Number of collisions per second: $N_\text{collisions} \sim \frac{F}{m v} = \frac{10^{3}}{4{,}8\cdot 10^{-26}\cdot 5\cdot 10^{2}} \approx 4\cdot 10^{25}$ As an order of magnitude: $\ev{N_\text{collisions} \sim 10^{25}\ \text{collisions/s}}$

Connections

Topics: Teoria cinetica dei gas Concepts: Pressione Skills: Stime di Fermi · Analisi dimensionale