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So, soda is under pressure and has gas dissolved in it. But, when you open it, the gas is still dissolved in it. But, if we wait a few hours, the gas has escaped into the atmosphere.

What factors determine the rate at which gas escapes the soda-gas solution?

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    $\begingroup$ Don't cross post engineering.stackexchange.com/users/6488/mohammad-athar $\endgroup$ – paparazzo Jan 11 '18 at 0:21
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    $\begingroup$ There is an ongoing close vote on the engineering SE, it will nearly surely close the question there, I think there is no reason to close the post here (considering its score). $\endgroup$ – user259412 Jan 11 '18 at 5:55
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    $\begingroup$ Isn't an engineering answer fundamentally different from a physics answer? $\endgroup$ – Mohammad Athar Jan 11 '18 at 12:37
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When you open the bottle and reduce the pressure you now have a supersaturated solution of carbon dioxide in water so it is energetically favourable for the gas to come out of solution.

However for the gas to come out of solution you have to form a bubble and the mechanism by which this happens is called nucleation. But there is an energy barrier that prevents tiny bubbles from forming, and as a result bubbles will only form when there is something to help them nucleate. For more on this see:

Anyhow, if you look carefully at the bubbles coming out of an opened bottle of soda you'll see they aren't forming randomly. Typically you'll see streams of bubbles coming from an area where there is some aid to nucleation e.g. a defect on the glass wall of the bottle.

This means the escape of the gas involves two steps:

  1. formation of bubbles at a nucleus

  2. diffusion of the carbon dioxide through the water to the nucleus

And it's step 2 that makes the escape of the carbon dioxide take a while. Where an aid to nucleation exists, e.g. at a defect on the glass wall, formation of a bubble will be very quick. However formation of the bubble will quickly use up the dissolved carbon dioxide in its vicinity and formation of further bubbles has to wait for more carbon dioxide to diffuse to the nucleation site. Diffusion of gases dissolved in water is surprisingly slow, and even though the gases typically only have to diffuse a few millimetres they still take a while to do it.

If you want to get the carbon dioxide out quickly the best way to do it is to increase the number of nucleation sites so you get many more nuclei and hence reduce the distance the dissolved gas has to diffuse. The traditional way to do this is to drop a mint into the bottle of soda, but stand well back when you do it.

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    $\begingroup$ How does the mint increase the nucleation rate? It sounds like the diffusion of CO2 through the water is the rate-limiting step. Unless the mint dissolves to create more evenly-spread nucleation points, it's not clear to me how the mint produces such effervescence. $\endgroup$ – Nuclear Wang Jan 10 '18 at 18:00
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    $\begingroup$ Interesting. If you prevented nucleation what is the upper bound on diffusion at STP? I wonder how slow diffusion could be if we used some sort of highly polished, non-reactive vessel? Would be a great product. The soda cup! LOL -> physics.stackexchange.com/q/229684/96218 $\endgroup$ – James Jan 10 '18 at 19:35
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    $\begingroup$ @R.. Offhand, I doubt it — all tapping would do is create more potential nucleation sites from turbulence and deformation of the container wall (and that's if you tap hard). But ask away! $\endgroup$ – detly Jan 10 '18 at 21:51
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    $\begingroup$ Turbulence is exactly what causes that. Not because it creates more nucleation sites, but because it creates more fluid flow over the nucleation sites which are already present. Since the fluid is flowing, the slowness of diffusion is less of a barrier to nucleation. $\endgroup$ – Beanluc Jan 10 '18 at 22:08
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    $\begingroup$ @R.. When you open the can, all the tiny bubbles on the nucleation sites expand rapidly and float to the surface - pushing any liquid above them as well. Tapping the can is an attempt to knock the bubbles off the sides of the can so they rise to the top, thus when you open the can, less liquid is pushed out. $\endgroup$ – Bilkokuya Jan 11 '18 at 10:56

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