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  • Define diffusion, osmosis, dialysis, and active transport.
  • Calculate diffusion rates.

Diffusion

There is something fishy about the ice cube from your freezer—how did it pick up those food odors? How does soaking a sprained ankle in Epsom salt reduce swelling? The answer to these questions are related to atomic and molecular transport phenomena—another mode of fluid motion. Atoms and molecules are in constant motion at any temperature. In fluids they move about randomly even in the absence of macroscopic flow. This motion is called a random walk and is illustrated in [link] . Diffusion is the movement of substances due to random thermal molecular motion. Fluids, like fish fumes or odors entering ice cubes, can even diffuse through solids.

Diffusion is a slow process over macroscopic distances. The densities of common materials are great enough that molecules cannot travel very far before having a collision that can scatter them in any direction, including straight backward. It can be shown that the average distance x rms size 12{x rSub { size 8{"rms"} } } {} that a molecule travels is proportional to the square root of time:

x rms = 2 Dt , size 12{x rSub { size 8{"rms"} } = sqrt {2 ital "Dt"} } {}

where x rms stands for the root-mean-square distance and is the statistical average for the process. The quantity D size 12{D} {} is the diffusion constant for the particular molecule in a specific medium. [link] lists representative values of D size 12{D} {} for various substances, in units of m 2 /s size 12{m rSup { size 8{2} } "/s"} {} .

The figure shows the path of a random walk. The random thermal motion of a molecule is shown to begin at a start point and then the particles move about zigzag in all directions and end up at the finish point. The distance between the start and finish point is shown as x. Continuous arrows show various directions of motion.
The random thermal motion of a molecule in a fluid in time t size 12{t} {} . This type of motion is called a random walk.
Diffusion constants for various molecules At 20°C and 1 atm
Diffusing molecule Medium D (m 2 /s)
Hydrogen ( H 2 ) Air 6.4 × 10 –5
Oxygen ( O 2 ) Air 1.8 × 10 –5
Oxygen ( O 2 ) Water 1.0 × 10 –9
Glucose ( C 6 H 12 O 6 ) Water 6.7 × 10 –10
Hemoglobin Water 6.9 × 10 –11
DNA Water 1.3 × 10 –12

Note that D size 12{D} {} gets progressively smaller for more massive molecules. This decrease is because the average molecular speed at a given temperature is inversely proportional to molecular mass. Thus the more massive molecules diffuse more slowly. Another interesting point is that D size 12{D} {} for oxygen in air is much greater than D size 12{D} {} for oxygen in water. In water, an oxygen molecule makes many more collisions in its random walk and is slowed considerably. In water, an oxygen molecule moves only about 40 μ m in 1 s. (Each molecule actually collides about 10 10 size 12{"10" rSup { size 8{"10"} } } {} times per second!). Finally, note that diffusion constants increase with temperature, because average molecular speed increases with temperature. This is because the average kinetic energy of molecules, 1 2 mv 2 size 12{ { { size 8{1} } over { size 8{2} } } ital "mv" rSup { size 8{2} } } {} , is proportional to absolute temperature.

Calculating diffusion: how long does glucose diffusion take?

Calculate the average time it takes a glucose molecule to move 1.0 cm in water.

Strategy

We can use x rms = 2 D t size 12{x rSub { size 8{"rms"} } = sqrt {2 ital "Dt"} } {} , the expression for the average distance moved in time t size 12{t} {} , and solve it for t size 12{t} {} . All other quantities are known.

Solution

Solving for t size 12{t} {} and substituting known values yields

t = x rms 2 2 D = ( 0.010 m ) 2 2 ( 6 . 7 × 10 10 m 2 /s ) = 7 . 5 × 10 4 s = 21 h .

Discussion

This is a remarkably long time for glucose to move a mere centimeter! For this reason, we stir sugar into water rather than waiting for it to diffuse.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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