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.

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