Problem — Diffusion of a scent

Problem

In a still room, perfume is released in one corner. Estimate the time it takes to reach the opposite corner at $L = 5$ m distance, based solely on molecular diffusion ($D_\text{air}\approx 0{,}2$ cm$^2$/s, diffusive path $\langle x^2\rangle\approx 6Dt$). Explain why in reality we smell perfumes much sooner.

Solution

In diffusion, the mean square displacement in three dimensions grows linearly with time: $\langle x^2\rangle\approx 6Dt$ I require the typical distance to reach $L$, i.e. $\langle x^2\rangle\approx L^2$, and solve for the time: $t=\frac{L^2}{6D}$ Converting the data to SI units: $L=5$ m, $D=0{,}2\ \text{cm}^2/\text{s}=0{,}2\cdot10^{-4}\ \text{m}^2/\text{s}=2\cdot10^{-5}\ \text{m}^2/\text{s}$. $t=\frac{(5)^2}{6\cdot 2\cdot10^{-5}}=\frac{25}{1{,}2\cdot10^{-4}}$ $t\approx 2{,}1\cdot10^5\ \text{s}$ In hours: $t=\frac{2{,}1\cdot10^5}{3600}\approx 58\ \text{hours}\ \sim\ \text{order of days}$ Molecular diffusion alone would therefore take hours (order of magnitude $\sim$ hours-days), because each molecule advances in a zig-zag, continually colliding with air molecules: even though it moves at hundreds of m/s, its net path is extremely slow.

Why in reality we smell it sooner. No room is ever truly still: temperature differences, people moving, draughts and convective motions create macroscopic flows that transport the scent by convection (bulk carrying of air), a mechanism orders of magnitude faster than molecular diffusion. This is why the smell arrives in seconds or minutes, not hours. $\ev{t\approx 2{,}1\cdot 10^5\ \text{s} \approx 58\ \text{hours (diffusion alone); in practice air currents carry it much sooner}}$

Connections

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