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What is the mean spacing between air molecules and what expresson relates the spacing to altitude?
Answer 1:

Interesting question! What you are talking about is also referred to as the "mean free path," which, as you've said, is the average distance a gas molecule can freely travel before hitting another one. I have detailed the equations below required for finding the mean free path as a function of altitude, however, I suggest you look at all of the provided links so that you know what every variable means and what assumptions these equations make.

mean_free_path = k_B * T / ( * d2 * sqrt(2) * pressure)

where k_B is Boltzmann's constant, T is temperature, d2 is the diameter of the gas molecule squared, sqrt(2) is the square root of 2, and pressure as a function of altitude is equal to:

pressure as a function of altitude = p_at_sealevel * (1 - (L * height_above_sealevel) / T_at_sea_level) ((g * MM)/(R * L))

Please note that ((g * MM)/(R * L)) is an exponent

where p_at_sealevel and T_at_sealevel are the pressure and temperature of air at sea level, g is the acceleration due to gravity, MM is the molar mass of the gas, R is the gas constant, and L is the temperature lapse rate.

Depending on what you're studying, you may want to plug in average values of the molar masses and molecular diameters where required.

For the molecular diameters, you can either look them up or estimate them in the following way:
Take the density of a gas and divide it by the molar mass to get the molar density or moles per volume of the gas. Take the inverse of the molar mass times Avogadro's number to get the volume of a single gas molecule. Then by reasonably assuming that the gas molecule is a sphere, you can solve for the gas molecule's radius with the following relationship:

radius = ((3/4pi) * MM*(Avogadro's # / density_of_gas)) (1/3)

There's a lot of tricky units in all of the above equations and dealing with Boltzmann's constant is always a pain, so please work carefully and patiently and you'll get it! Compare the answers you get with Wikipedia, which says that gas at room temperature and pressure will have a mean free path of about 70 nanometers. It took me about 5 or 6 tries to verify Wikipedia's answer, so please remember that science requires patience!

Answer 2:

Well, I can tell you how to calculate it.

The density of air varies with temperature as well as altitude, but (according to Wikipedia) it's about 1.3 kg/m3 at standard temperature and pressure, and it becomes less or more dense with temperature as an ideal gas.

The most common gas in the atmosphere is nitrogen, which has a mass of 0.028 grams per mole, where one mole is 6.022 x 1023 molecules. The atmosphere isn't entirely nitrogen, but the other common gasses in the atmosphere (oxygen, water vapor, and carbon dioxide, in that order) aren't so different as to make a whole lot of difference in the total density of molecules, so you can use that to get a rough estimate of the number of molecules per cubic meter.

Air pressure drops off at an approximately exponential rate as you go up from sea-level. Atmospheric pressure decreases by half with roughly 5.5 km of altitude.

Remember, however, that temperature will also go down as you go up in elevation. The rate at which this happens depends on the climate of the area you are in; in a desert, the atmospheric lapse rate is about 1 degree C per 100 m elevation. In a humid climate, it drops to 0.6 degrees/100 m.

Now, if the number of molecules that you calculated are evenly distributed in a cube, then the number of them that will be in any one direction in the cube will be equal to the cube-root of the total number of molecules. So, that's the number of molecules per meter of distance. One over that is the mean distance between molecules.

Answer 3:

The mean free path is not the same as the spacing between molecules. The spacing between molecules is just what it says: how close are the molecules from one another, and it is app.
--the cubic root of the inverse of the particle density.

The mean free path should be longer, in general, and it is related to the density as well as the collision cross section.

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