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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)

enter image description here

I have no idea how to calculate this.

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    $\begingroup$ If there is no gravity effects (Archimedes force or so), then the mixing is determined with mutual diffusion. Look at the definition of the diffusion coefficient $D$. It is determined with the time between two successive collisions. And the relaxation time is determined with the solution of the diffusion equation. $\endgroup$ Jul 7, 2016 at 20:06
  • $\begingroup$ I suppose you should be able to get a ball park figure by doing some experiments with your scuba tank: as soon as the tank is filled measure the pressure, it will vary with time, but give you an average mixing time when the pressure reaches a steady state $\endgroup$
    – nluigi
    Jul 9, 2016 at 12:22

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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).

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  • $\begingroup$ @ Dave R thanks for this wonderful answer. The five minutes estimate is reassuring. The tanks have usually sat for a while, sometimes hours, before I measure them. Sometimes we're measuring them just after they've been filled, but having read this, I can easily arrange to wait fifteen minutes or so, and shake them up a bit, so the measurement should be accurate enough. $\endgroup$
    – Flynn
    Jan 26, 2021 at 18:15
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I did an experiment today. Approximately 12 minutes before the readings were steady.

Bottles: 2 x 12 liter doubles, standing upright Start pressure: 200 Atmospheres Start blend: Nitrox 33,4

I made a top off with air (nitrox 21)

Every minute I measured the percentage of oxygen After 10 minutes it was close. After 12 minutes it was steady. If the bottles lie down, they are likely to mix faster

Here you see the time of day and the reading

Time Oxygen percentage

11.13 22.0

11.14 26.9

11.15 27.3

11.16 28.1

11.17 29.0

11.19 30.5

11.21 31.0

11.22 31.1

11.23 31.3

11.24 31.4

11.25 31.5

11.26 31.5

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    $\begingroup$ Welcome! This experimental result is helpful. Please see this guidance about formatting tables, then edit your answer. $\endgroup$
    – rob
    Jan 14, 2023 at 16:38
  • $\begingroup$ @ Jesper Thanks for this. It is a very useful result. Usually, I'd be picking up the tanks hours after they were filled. So your experiment is very reassuring. Even if I were there as they were being filled, it's no problem to wait ten or fifteen minutes. $\endgroup$
    – Flynn
    Jan 18, 2023 at 18:45

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