91影视

Assume a perfume bottle is opened at one end of a long $10m$room. Schroeder provides $D=2脳{10}^{-5}{\mathrm{m}}^2路{\mathrm{s}}^{-1}$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" $D={10}^{-5}{\mathrm{m}}^2路{\mathrm{s}}^{-1}$.We can calculate how far a perfume molecule will diffuse by taking$螖x=10\mathrm{m}$to be the distance that diffusion has occurred over This region's volume is :

$V=A\mathrm{螖}x$

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:

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

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

$J_x=\frac{N}{A\mathrm{螖}t}$

$J_x=鈭�D\frac{dn}{dx}$

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

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

$鈫�\mathrm{螖}t=\frac{(\mathrm{螖}x{)}^2}{D}$

substitute with,$螖x=10\mathrm{m}$and$D={10}^{-5}{\mathrm{m}}^2路{\mathrm{s}}^{-1}$, 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.