14.4

*Diffusion

You can smell the fragrance of a perfume at some distance from an open bottle because perfume molecules evaporate from the liquid, where they are relatively concentrated, and spread out into the air, where they are less concentrated. During their journey, they collide with other molecules, so their paths resemble the zigzag paths characteristic of Brownian motion. The process in which molecules move from a region of higher concentration to one of lower concentration is called diffusion. Diffusion also occurs in liquids and solids, and Figure 14-13 illustrates ink diffusing through water. However, compared to the rate of diffusion in gases, the rate is generally smaller in liquids and even smaller in solids. The host medium, such as the air or water in the examples above, is referred to as the solvent, while the diffusing substance, like the perfume molecules or the ink in Figure 14-13, is known as the solute. Relatively speaking, diffusion is a slow process, even in a gas. Conceptual Example 7 illustrates why.

Figure 14-13 A drop of ink placed in water eventually becomes completely dispersed because of diffusion.

| Conceptual Example 7 | | | Why Diffusion Is Relatively Slow

In Example 6 we have seen that a gas molecule has a translational rms speed of hundreds of meters per second at room temperature. At such a speed, a molecule could travel across an ordinary room in just a fraction of a second. Yet, it often takes several seconds, and sometimes minutes, for the fragrance of a perfume to reach the other side of a room. Why does it take so long?
Reasoning and Solution When a perfume molecule diffuses through air, it makes millions of collisions each second with air molecules. As Figure 14-14 illustrates, the speed and direction of motion change abruptly as a result of each collision. Between collisions, the perfume molecule moves in a straight line at a constant speed. Although a perfume molecule does move very fast between collisions, it wanders only slowly away from the bottle because of the zigzag path resulting from the collisions. It would take a long time for a molecule to diffuse in this manner across a room. Usually, however, convection currents are present and carry the fragrance across the room in a matter of seconds or minutes.

Figure 14-14 A perfume molecule collides with millions of air molecules during its journey, so the path has a zigzag shape. Although the air molecules are shown as stationary, they are also moving.

Related Homework: Problems 40, 44 and 49

Diffusion is the basis for drug delivery systems that bypass the need to administer medication orally or via injections. Figure 14-15 shows one such system, the transdermal patch. The word “transdermal”means “across the skin.”Such patches, for example, are used to deliver nicotine in programs designed to help you stop smoking. The patch is attached to the skin using an adhesive, and the backing of the patch contains the drug within a reservoir. The concentration of the drug in the reservoir is relatively high, just like the concentration of perfume molecules above the liquid in a bottle. The drug diffuses slowly through a control membrane and directly into the skin, where its concentration is relatively low. Diffusion carries it into the blood vessels present in the skin. The purpose of the control membrane is to limit the rate of diffusion, which can also be adjusted in the reservoir by dissolving the drug in a neutral material to lower its initial concentration. Another diffusion-controlled drug delivery system utilizes capsules that are inserted surgically beneath the skin. Contraceptives are administered in this fashion, for instance. The drug in the capsule diffuses slowly into the bloodstream over extended periods that can be as long as a year.

Figure 14-15 Using diffusion, a transdermal patch delivers a drug directly into the skin, where it enters blood vessels. The backing contains the drug within the reservoir, and the control membrane limits the rate of diffusion into the skin.

The physics of drug delivery systems.

The diffusion process can be described in terms of the arrangement in Figure 14-16a. A hollow channel of length L and cross-sectional area A is filled with a fluid. The left end of the channel is connected to a container in which the solute concentration C2 is relatively high, while the right end is connected to a container in which the solute concentration C1 is lower. These concentrations are defined as the total mass of the solute molecules divided by the volume of the solution (e.g., 0.1 kg/m3). Because of the difference in concentration between the ends of the channel, , there is a net diffusion of the solute from the left end to the right end.

Figure 14-16 (a) Solute diffuses through the channel from the region of higher concentration to the region of lower concentration. (b) Heat is conducted along a bar whose ends are maintained at different temperatures.

Figure 14-16a is similar to Figure 13-8 for the conduction of heat along a bar, which, for convenience, is reproduced in Figure 14-16b. When the ends of the bar are maintained at different temperatures, T2 and T1, the heat Q conducted along the bar in a time t is

| (13.1)

where , and k is the thermal conductivity. Whereas conduction is the flow of heat from a region of higher temperature to a region of lower temperature, diffusion is the mass flow of solute from a region of higher concentration to a region of lower concentration. By analogy with Equation 13.1, it is possible to write an equation for diffusion: (1) replace Q by the mass m of solute that is diffusing through the channel, (2) replace by the difference in concentrations , and (3) replace k by a constant known as the diffusion constant D. The resulting equation, first formulated by the German physiologist Adolf Fick (1829–1901), is referred to as Fick’s law of diffusion.

| Fick’s Law of Diffusion
| The mass m of solute that diffuses in a time t through a solvent contained in a channel of length L and cross-sectional area A is*

| (14.8)

where is the concentration difference between the ends of the channel and D is the diffusion constant.
SI Unit for the Diffusion Constant: m2/s

It can be verified from Equation 14.8 that the diffusion constant has units of m2/s, the exact value depending on the nature of the solute and the solvent. For example, the diffusion Vconstant for ink in water is different from that for ink in benzene. Example 8 illustrates an important application of Fick’s law.

| Example 8 | | | Water Given Off by Plant Leaves

Large amounts of water can be given off by plants. It has been estimated, for instance, that a single sunflower plant can lose up to a pint of water a day during the growing season. Figure 14-17 shows a cross-sectional view of a leaf. Inside the leaf, water passes from the liquid phase to the vapor phase at the walls of the mesophyll cells. The water vapor then diffuses through the intercellular air spaces and eventually exits the leaf through small openings, called stomatal pores. The diffusion constant for water vapor in air is . A stomatal pore has a cross-sectional area of about and a length of about . The concentration of water vapor on the interior side of a pore is roughly , while that on the outside is approximately . Determine the mass of water vapor that passes through a stomatal pore in one hour.

Figure 14-17 A cross-sectional view of a leaf. Water vapor diffuses out of the leaf through a stomatal pore.

The physics of water loss from plant leaves.

Reasoning and Solution Fick’s law of diffusion shows that

| (14.8)

This amount of water may not seem significant. However, a single leaf may have as many as a million stomatal pores, so the water lost by an entire plant can be substantial.

Check Your Understanding 4

The same solute is diffusing through the same solvent in each case referred to in the table below, which gives the length and cross-sectional area of the diffusion channel. In each case, the concentration difference between the ends of the diffusion channel is the same. Rank the diffusion rates (in kg/s) in descending order (largest first). (The answer is given at the end of the book.)

| Length | Cross-Sectional Area
(a) |  | A
(b) | L |
(c) |  | 2A

Background:Fick’s law of diffusion is the key.
For similar questions (including calculational counterparts), consult Self-Assessment Test 14.1. The test is described next.

Test your understanding of the material in this chapter:

· | Molecular Mass, the Mole, and Avogadro’s Number
· | Kinetic Theory of Gases
· | Diffusion
· | The Ideal Gas Law

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.