Reef Chemistry Question of the Day #135 Speed of diffusion of molecules

Randy Holmes-Farley · Jun 27, 2015

Reef Chemistry Question of the Day [HASHTAG]#135[/HASHTAG]

So from the last question of the day we saw that a typical oxygen molecule in the air travels at about 1,000 miles an hour.

But from real life experience we know that gasses do not diffuse that fast in normal air. For example, open a can of paint or burn something on a stove and it takes a considerable time for someone across the room or in another room to smell it.

Why?

The answer is because while they are moving at a screaming fast pace, they bump into each other a lot, sending them off at angles to their initial direction.

How far do you think an oxygen molecule in normal air at room temperature travels before bumping into another molecule?

Let's think of the answer in terms of the molecular diameter of oxygen, so how many lengths of its own size can it travel before hitting something?

A. 18 diameters
B. 180 diameters
C. 1,800 diameters
D. 1,800,000 diameters
E. 1,800,000,000 diameters

As with last time, this isn't something you should know, but is fun to speculate about, and will help understand gas exchange.

Good luck!

.

JimWelsh · Jun 28, 2015

The distance you are asking about is the "Mean free path". In kinetic theory, the Mean free path l = 1/(sqrt(2) * n * σ), where n = the number of molecules per unit volume and σ = the effective cross-sectional area of the spherical particles, in other words, how close they have to be in order to bump into each other.

The value of n = p / (kB * T), where p is the pressure in Pascals, kB is the Boltzmann constant (which is just the R constant used in the previous question divided by Avogadro's number = 1.38E-23), and T is the temperature in degrees Kelvin. At sea level, p = 101325, so at room temperature n = 101325 / (1.38E-23 * 298.15) = 2.46E25 molecules / cubic meter.

The value of σ = pi * d squared, where d is the average diameter of the molecules. Assuming a value of 3.711 Angstroms for the value d for mean diameter of the molecules in dry air (The Civil Engineering Handbook, Second Edition), then we get σ = 3.1416 * (3.711E-10)^2 = 4.33E-19 square meters.

Plugging n and σ into the original equation gives a Mean free path value of l = 1/(1.414 * 2.46E25 * 4.33E-19) = 6.64E-08 meters = 66.4 nm.

The diameter of the O2 molecule is slightly smaller than the mean dry air molecule, at about 3.6 Angstroms, so expressed in terms of O2 diameters, the Mean free path = 66.4 / 0.36 = 184, so my answer is B: 180 diameters.

JimWelsh · Jun 29, 2015

Thinking about the math above, and trying to better understand exactly what it is saying for my own benefit, I have the following, hopefully clarifying, thoughts.

First, let's consider the case of a fast-moving molecule M whizzing along a straight line L through a bunch of other stationary molecules with random positions. If the molecules are all d = 3.711 angstroms in diameter, then any other molecule that is within d angstroms of line L will collide with molecule M. The question is how far, on average, does molecule M have to move along line L before such a collision can be expected to occur?

Note that molecule M will have moved down the center of a cylinder C with a radius of d (the radius of one molecule M = d/2, plus the radius of the other molecule that M collided with = d/2; d/2 + d/2 = d) a certain distance, on average, before colliding with another molecule. In other words, cylinder C contains the space within which another molecule may have existed that would have collided with molecule M; if the other molecule were outside cylinder C then a collision is impossible.

From the calculations I posted above, we see that there are 2.46E25 molecules per cubic meter. That means that there is, on average, one molecule every volume V = 1/2.46E25 = 4.063E-26 cubic meters. That means that molecule M will, on average, have to move along cylinder C with radius d the length necessary to give cylinder C a volume of V before a colliding with another molecule. Remember that the formula for the volume of a cylinder is V = pi * r^2 * l, where r is the radius and l is the length. From the previous post, we see that σ is the same as the "pi * r^2" part, so the length l of the cylinder with volume V is l = V / σ. In this case, V = 4.063E-26, and σ = 4.33E-19, so l = 4.063E-26 / 4.33E-19 = 9.38E-8 meters = 93.8 nm.

So, in this case, the Mean free path = 93.8 nm.

Remember that I said that this case was a fast-moving molecule and the other molecules were stationary and randomly-destributed. That case follows a Mean free path formula of l = 1/(n * σ). In the previous post, the formula is l = 1/(sqrt(2) * n * σ). That extra "sqrt(2)" factor basically accounts for the fact that our molecule M, and all the other molecules, are moving about in all different directions at the same average speed in a "Maxwell distribution". After skipping over a whole bunch more complicated math (which you can read about here: http://books.google.com/books?id=Cbp5JP2OTrwC&pg=PA88), the net effect of all those molecules moving about is to make a collision the square root of 2 = 1.414 times more likely, which makes the Mean free path 1/1.414 = 0.707 times as long. Therefore, the Mean free path of our Oxygen molecule is 93.8 * 0.707 = 66.3 nm, which is basically the same answer as the 66.4 derived above, after rounding errors.

JimWelsh · Jul 9, 2015

Do I know how to kill a thread, or what?!?!?

Randy Holmes-Farley · Jul 19, 2015

As Jim correctly pointed out, the mean distance that a molecule in air travels before hitting something is only 180 diameters.

So, imagine you are in your car and you are going 1000 miles an hour, but you only go a couple of hundred feet before hitting something and bouncing off it. That is what it is like to be a molecule in the gas phase!