Diffusion: a couple of examples

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.67 – 1.68.

The process of diffusion obeys Fick’s law

$\displaystyle J_{x}=-D\frac{dn}{dx} \ \ \ \ \ (1)$

where ${J_{x}}$ is the number of molecules per unit area diffusing in the ${x}$ direction per unit time, ${D}$ is the diffusion constant and ${n=N/V}$ is the particle density.

We can use Fick’s law to do a few estimates of how long various types of diffusion processes take.

Example 1 Schroeder gives the diffusion constant ${D}$ for sucrose (table sugar) in water at room temperature as ${D=5\times10^{-10}\mbox{m}^{2}\mbox{s}^{-1}}$. If we take ${\Delta t=60\mbox{ s}}$, how far can we expect a molecule of sucrose to diffuse? We can estimate this by taking ${\Delta x}$ to be the distance over which diffusion has taken place. The volume of this region is ${V=A\Delta x}$ where ${A}$ is the cross sectional area of the container. If ${N}$ is the total number of molecules within this region, then, approximately,

$\displaystyle n\approx\frac{N}{A\Delta x} \ \ \ \ \ (2)$

The flux can be calculated by taking ${\Delta t}$ to be the time taken for the volume ${A\Delta x}$ to acquire the ${N}$ molecules, so we get, again approximately

$\displaystyle J_{x}\approx\frac{N}{A\Delta t} \ \ \ \ \ (3)$

From Fick’s law we get

 $\displaystyle \frac{N}{A\Delta t}$ $\displaystyle \approx$ $\displaystyle D\frac{N}{A\left(\Delta x\right)^{2}}\ \ \ \ \ (4)$ $\displaystyle \Delta x$ $\displaystyle \approx$ $\displaystyle \sqrt{D\Delta t} \ \ \ \ \ (5)$

With the values given, we have

$\displaystyle \Delta x\approx1.7\times10^{-4}\mbox{ m} \ \ \ \ \ (6)$

or about a fifth of a millimetre.

Example 2 As an example of diffusion in air, suppose a bottle of perfume is opened at one end of a room 10 m long. Schroeder gives ${D=2\times10^{-5}\mbox{ m}^{2}\mbox{s}^{-1}}$ for CO molecules in air at room temperature and atmospheric pressure. A perfume molecule is probably larger than a CO molecule, so its diffusion constant would be smaller, say, ${D=10^{-5}}$. Using the same approximate formula as in the previous example, the time it would take for the perfume to diffuse to the other end of the room is

$\displaystyle \Delta t\approx\frac{\left(\Delta x\right)^{2}}{D}=10^{7}\mbox{ s} \ \ \ \ \ (7)$

which is about a third of a year. Certainly most odours released at some point in a room (including those embarrassing to the emitter, at times) travel the length of the room considerably faster than that (often less than a minute), so other processes (typically convection) are responsible for spreading them.