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I'm basing this off of a home experiment I did to demonstrate Torricelli's law.

If I take a regular plastic water bottle and poke a small hole at the bottom, water will not flow out. This was because the bottle cap was still on and thus any small changes will cause the pressure inside to not be equal to the pressure outside and suction will occur. If I take off the cap, water starts flowing out with a speed $\sqrt{2gh}$. If I put on the cap again, water flow stops. This is all fairly standard.

I then tried to conduct this experiment but by pouring soda into a plastic bottle. I recently read that shaking a soda bottle does not increase the air pressure inside. However, when I shook the bottle and then poked a hole (with the cap still on), soda started to flow out! When I took off the cap, the motion of the soda surprisingly stopped! (I'm guessing this second part was due to surface tension taking effect due to a low bond number). When I put the lid back on, nothing happened.

The logical conclusion I made from this was that shaking a bottle does indeed change the air pressure inside (perhaps by a tiny amount). When the cap was taken off, the pressure inside was equal to the atmospheric pressure which was why nothing changed if when I put the cap back on.

Is this correct? Or is there something else I'm missing here?

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Shaking the soda bottle makes the dissolved gas escape the liquid. As long as the lid is on, this can't happen (or happens for a little bit and then stops): the air in the bottle is already saturated with gas.

Then you poke a hole in the bottle, below the surface. The liquid may come out, reducing its volume in the bottle, thus expanding the volume of the air inside the bottle. In a non-fizzy beverage, the expanding air reduces its pressure and "sucks the liquids back in", as equilibrium is established. No liquid flows out of the hole.

In a fizzy beverage, the expansion of the air volume makes it so that it isn't saturated anymore; the gas solved in the liquid can move out of the solution; the air in the bottle thus is both expanding and gaining particles. As long as these two factors balance each other, the air can expand without dropping in pressure, allowing the liquid to flow out. When the separation of the gas from the solution can't supply a number of molecules large enough to keep the pressure constant, we would fall back into the "non-fizzy beverage" case, stopping the flow.

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I recently read that shaking a soda bottle does not increase the air pressure inside.

You're correct that shaking it doesn't further increase the pressure. However, the pressure is (even before shaking) already well above atmospheric. When you poke your hole, substances near the hole will be under unequal pressures and will be forced outside.

The logical conclusion I made from this was that shaking a bottle does indeed change the air pressure inside (perhaps by a tiny amount).

I think you'd need to poke the hole without shaking to test this. Otherwise, what are you comparing it to?

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There is a YouTube video from Veritasium with a nice explanation. Amusingly, it came out 4 days before your post.

Veritasium on soda

  • The pressure is indeed unchanged. He measures it!
  • When opening after shaking, however, the $CO_2$ quickly comes out of solution.
  • This effect is due to lots of small bubbles in the liquid. Opening means dropping the pressure, so the gas now 'has the tendency' to come out. This process can take quite a while to reach equilibrium. The much larger total surface area of all the bubbles leads to this process being much faster.
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Shaking the bottle does not increase air pressure but increases the pressure of the carbon dioxide gas above the liquid. Some of the co2 dissolved in the liquid is "undissolved" by shaking the bottle.

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  • $\begingroup$ I'm not sure why partial pressures matter. Bernoulli's equation depends on the "air pressure", not the pressure due to a specific gas. $\endgroup$ – QiLin Xue Nov 24 '19 at 19:07
  • $\begingroup$ What If the fluid is not surounded by air but by another gas? Where do you get the idea that Bernoulli is restricted to air pressure? How about a soda dispenser(siphon)? Air pressure is same inside and outside but the liquid is still coming out of the bottle. $\endgroup$ – nasu Nov 24 '19 at 19:15
  • $\begingroup$ What's a low bond number? $\endgroup$ – Gert Nov 24 '19 at 19:17
  • $\begingroup$ @nasu Bernoulli's Principle states that $P+0.5\rho v^2+\rho gh=\text{constant}$ where $P$ is just the air pressure. You said that for siphons: "Air pressure is same inside and outside but the liquid is still coming out of the bottle." I don't really think this is where my problem is. As I stated, when I tried the experiment with a regular water bottle, water wasn't coming out when the lid was closed but did when the lid was opened. However, the opposite occurred when I did it with soda. $\endgroup$ – QiLin Xue Nov 24 '19 at 19:26
  • $\begingroup$ @Gert See en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_number - I don't think it's too important for this question; but if it's below a certain threshold, the fluid just wouldn't come out of the bottle if the hole was small enough. $\endgroup$ – QiLin Xue Nov 24 '19 at 19:27

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