Suppose you open a bottle of perfume at one end of a room. Very roughly, how much time would pass before a person at the other end of the room could smell the perfume, if diffusion were the only transport mechanism? Do you think diffusion is the dominant transport mechanism in this situation?

Assume a perfume bottle is opened at one end of a long $10m$room. Schroeder provides $D=2\times {10}^{-5}{\mathrm{m}}^2·{\mathrm{s}}^{-1}$for $\mathrm{CO}$At room temperature and atmospheric pressure, molecules in air Because a perfume molecule is likely larger than a CO molecule, its diffusion constant would be lower, say,localid="1650259068074" $D={10}^{-5}{\mathrm{m}}^2·{\mathrm{s}}^{-1}$.We can calculate how far a perfume molecule will diffuse by taking$\Delta x=10\mathrm{m}$to be the distance that diffusion has occurred over This region's volume is :

$V=A\mathrm{\Delta }x$

where A denotes the room's cross sectional area If N is the total number of molecules in this region, the particle density is as follows:

$n=\frac{N}{V}=\frac{N}{A\mathrm{\Delta }x}$

The flux can be calculated by dividing the time it takes$∆t$for the volume to acquire the N molecules by the volume $A∆t$, yielding:

$J_x=\frac{N}{A\mathrm{\Delta }t}$

$J_x=-D\frac{dn}{dx}$

substitute from equations$J_x=\frac{N}{A\mathrm{\Delta }t}$and $n=\frac{N}{V}=\frac{N}{A\mathrm{\Delta }x}$into equation$J_x=\frac{N}{A\mathrm{\Delta }t}$, so:

$\frac{N}{A\mathrm{\Delta }t}=-D\frac{d}{dx}\left[\frac{N}{A\mathrm{\Delta }x}\right]=D\left[\frac{N}{A(\mathrm{\Delta }x{)}^2}\right]$

$\to \mathrm{\Delta }t=\frac{(\mathrm{\Delta }x{)}^2}{D}$

substitute with,$\Delta x=10\mathrm{m}$and$D={10}^{-5}{\mathrm{m}}^2·{\mathrm{s}}^{-1}$, so:

which is approximately$116$days Certainly, most odors released at some point in a room (including those embarrassing to the emitter at times) travel much faster than that (often in less than a minute), implying that other processes (typically convection) are responsible for spreading them.