When I pick up my scuba tanks from the dive shop, I measure the nitrogen and oxygen
levels in it. I check that they are close to what I specified and calculate the
maximum depth it's safe to go to to ensure acute oxygen toxicity is not an
issue. I also need to know the gas mix so that when I do the dive I
can accurately calculate my decompression obligations (or try to avoid
having any at all.) One is taught to not measure the mix too
soon after the tank is filled to ensure they have mixed thoroughly.
When I was a kid I was taught that the temperature of a gas was a measure of
the average speed of the molecules and that the speeds of the molecules in
the gas would be normally distributed about the mean speed, some moving very
slowly and some very fast. I had always
imagined the average speed to be quite zippy and that a mix of gasses
would be pretty evenly mixed quite quickly.
According to this answer How fast do molecules move in objects? the
molecules in a gas at 300 K, roughly the temperature on a hot day, the molecules
move at a speed of about 300 m/s.
So my question is, if one added to 8 litres of nitrogen 2 litres of oxygen, how quickly
would the mixture be thoroughly mixed? At least mixed enough that any
measurement would be the same to within .1 % throughout the cylinder. (That's
a tenth of one percent.)
So the cylinder starts like this (unmixed)
I have no idea how to calculate this.
Best Answer
This is all about diffusion. The speed of individual molecules is not relevant, because they collide with one another and change direction so frequently (at least at standard temperature and pressure) that this speed does not at all characterize the diffusion of one species into another.
Engineers have tabulated rate coefficients that describe the rate of diffusion of various gases through air, for example: https://www.engineeringtoolbox.com/air-diffusion-coefficient-gas-mixture-temperature-d_2010.html
This doesn't give the rate you'd want, but we can get the ballpark studying a similar rate, Argon diffusing through air.
Say you've got a can with Argon on the bottom, air on top, and a 1 cm mixed layer between otherwise pure gases.
J = (D = 0.189 cm^2/s) * (1.7 kg/m^3 Argon at STP)/(1cm) = 31.8 kg/cm^2/s
This is the mass flux of argon through the barrier. Multiply by some area A, Divide by argon density at STP (1.7 kg/m^3), and divide by A again to get argon flow per unit area, areas cancel and we have: = 1.89 cm/s. Note that the Argon mass actually canceled out here too, basically the mix rate just relates to how thick the boundary layer starts out. Initially, when it is infinitesimal, the rate is infinite, since the rate is just D=0.189 cm^2/s divided by the boundary layer thickness L, which I assumed to start at 1 cm.
This means that the pure argon below diffuses up into the pure air through the boundary at like 2 cm/s. Of course one second later the boundary layer is 3 cm thick instead of 1, so the rate slows 3x. Three seconds later it is five centimeters thick. You have to solve a differential equation to really get your answer of how long, and the notion of a firm boundary between pure and boundary layer is just an approximation. But roughly... continuing this pattern you hit 21 cm thick "boundary layer" after 100 seconds, which I'm guessing is close to your tank size. Double or triple that for the boundary layer to further mix up to your .1% requirement, and we're at 5 minutes.
Notably, given this surprisingly slow timescale, it probably does help to shake up the tank. I suspect that Argon is a slightly worse case than N2 and O2, but I don't really know. Comparing other gases on the engineering toolbox link, seems like D roughly goes with mass, lighter mass higher D, but Argon isn't very different from air anyway (40 v 29).