Problem — Fermi estimate: collisions on the skin

Problem

Estimate the number of collisions per second that air molecules make on the palm of a hand ($A\approx 100$ cm$^2$), under ambient conditions. Hint: the molecular flux per unit area is $\tfrac14\,n\,\bar v$.

Solution

The number of collisions per second is the molecular flux $\Phi=\tfrac14 n\bar v$ multiplied by the area: $\dot N=\frac14\,n\,\bar v\,A$ Number density $n=N/V$ at $p=1{,}013\cdot10^5$ Pa and $T=293$ K: $n=\frac{p}{k_B T}=\frac{1{,}013\cdot10^5}{1{,}38\cdot10^{-23}\cdot 293}=2{,}5\cdot10^{25}\ \text{m}^{-3}$ Average speed of air (mass of one molecule $m\approx 4{,}8\cdot10^{-26}$ kg, given $M=29$ g/mol): $\bar v=\sqrt{\frac{8k_B T}{\pi m}}=\sqrt{\frac{8\cdot1{,}38\cdot10^{-23}\cdot293}{\pi\cdot4{,}8\cdot10^{-26}}}\approx 4{,}6\cdot10^2\ \text{m/s}$ Area $A=100\ \text{cm}^2=1{,}0\cdot10^{-2}\ \text{m}^2$. Putting it all together: $\dot N=\frac14\cdot 2{,}5\cdot10^{25}\cdot 4{,}6\cdot10^{2}\cdot 1{,}0\cdot10^{-2}$ $\dot N\approx 2{,}9\cdot10^{25}\ \text{collisions/s}\approx 3\cdot10^{25}\ \text{collisions/s}$ Roughly $10^{25}$ collisions per second on one hand: it is this continuous rain of collisions that produces, averaged over time, the constant atmospheric pressure we feel. $\ev{\dot N\approx 3\cdot10^{25}\ \text{collisions/s on } 100\ \text{cm}^2}$

Connections

Topics: Teoria cinetica dei gas Concepts: Pressione Skills: Stime di Fermi · Interpretazione micro-macro