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Pressure of Shaken Soda does change!

  1. Mar 11, 2014 #1
    Before submitting this thread for deletion (claiming it's been asked a jillion times before and answered definitively), please read my question thoroughly. I tried my best to find a proper answer and will cite what I found, but in the end, these answers were "lacking"...

    I know that the ACS claims that the pressure of a shaken soda can is unchanged, but I have a few problems with this claim:
    • Tests with larger bottles DO show palpable differences: I did a few simple double blind tests with 2-liter bottles of soda sitting for a day at room temperature and my blinded testers were able to identify the shaken soda more than 90% of the time, suggesting to me the effect is easier to detect in larger bottles than in small (rigid) cans.
    • Equilibrium seems to be assumed: And (ignoring the theoretical impracticality of true equilibrium), a shaken can is NOT in equilibrium for at least a few minutes after shaking. So I claim the argument that pressure doesn't change is based on shaking the can as vigorously as you want, and then letting it sit there and returning to equilibrium (an hour or so later at least), after which, the pressure would have reduced to its "pre-shaken" state in agreement with Henry's Law.

    Of many posts, I found only one on this site that might confirm my claim: here a user "bystander" explains that the pressure WILL CHANGE (he didn't specify whether higher or lower) shortly after shaking due to both low pressure effects (he says "cavitation")... his use of the scientific jargon is really confusing as it seems he is purposely using fancy language to sound erudite.
    • does he say pressure drops? it's possible he is claiming pressure actually drops (instead of my claim it rises). I can't tell too well, even after reading the entire thread. If cavitation creates pockets of low-pressure, this should lower the pressure in the can, and this seems not to happen, or I misinterpreted what he is saying (which is likely).

    so here is my suggestion: for the first 90 seconds after shaking a can, the pressure in the can definitely rises, but by a small amount (maybe 5-10 psi), but this is enough to be palpable in a larger soda bottle. This is the result of shaking, which, similar to higher pressure (and the use of sonication in degassing of liquids), reduces carbon dioxide solubility in the liquid, resulting in more gas coming out. Some of this is what forms the bubbles in the water, but more will simply add to the pressure of the gas as it escapes the liquid.

    This would be entirely consistent with the assertion the guys at the ACS makes that the pressure will not increase at equilibrium, since equilibrium will not be achieved for at least an hour after shaking... but my point is seconds after shaking, the pressure definitely increases.

    This seems plausible (and even testable, with the right equipment) to me... is it what happens shortly after shaking?:confused:
  2. jcsd
  3. Mar 11, 2014 #2

    Simon Bridge

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    The ACS article you cite does not claim that there is zero pressure difference, just that the difference is very small - too small to tell by a blind, randomized, squeeze-test, on a soda can.

    This seems to be what you are saying above too.

    I have an unopened PET bottle in the pantry - hang on.
    ... actually I can't tell the difference.
    And I used a guage to work out how much force had to be applied to get 1cm deformation before and after.
    I only have the one bottle so I can't repeat it yet.

    The heart of the idea is that shaking the soda bottle distributes nucleation sites (bits of dust/minerals etc) through the bottle - increasing the places that CO2 can come out of solution should the pressure drop. That's how the bubbles burst out of solution when you open the bottle after shaking.

    Shaking the PET bottle side-to-side, I see a lot of small bubbles close to the gas-pocket in the kneck but none at all at the bottom. If the pressure changed, gas bubbles would form everywhere.

    You may think that CO2 can build up more in the gas-pocket - so the higher pressure there results in higher overall pressure? But if the bottle is unopened, the pressure there should already be as high as it can go. You may get miniscule (<<5-10psi) random changes by this approach.

    If there were no gas pocket, thet it should be plain - for a bubble to form it has to displace something. If the bottle is totally full of liquid it can't - therefore no bubbles will form and the whole thing stays the same.

    Which leaves us to contemplate your test - it seems to be an anomaly: so you should look carefully at the setup: were the bottles initially in equilibrium? Were they new? Was there anything that could tip the subject off about which one was shaken? (foam at the top? slight wobble? you didn't leave the room before they entered to choose? poor randomization?)

    If you have repeatable results, you should document the entire process in video, careful to show every step.
    It will be interesting to see ... the theory is for specific conditions that may not exist for all real-life soda brands.
    If we find some it would mean "something else is going on" - which needs closer investigation.

    This is one of those things that goes on for ever with someone saying "but what about..."
    What you really want is some sort of measure of internal pressure without opening the bottle - the deformation force test is simple enough.
    Last edited: Mar 11, 2014
  4. Mar 12, 2014 #3
    It's all about equilibrium (or not)

    Thanks for the good ideas, Simon.

    The first few seconds after a bottle is vigorously shaken:
    • A state of NON-equilibrium: I posit that the ACS, and most other folks' argument for "no pressure change" rests on the system being in a state of equilibrium. The first lesson hammered into my head in Thermodynamics was "when a system is out of equilibrium, all assumptions that were valid when it was in equilibrium are in doubt, if not discarded."
    • Shaken liquid is (slightly) less soluble: This effect is used in ultrasonic degassers which use vibrations to reduce the need for heat, as both reduce gas solubility in a liquid. This makes sense as shaking increases the kinetic energy of the gas in the liquid, resulting in more gas escaping.
    • less gas in the liquid = more gas = pressure increase: The gas that would be released from the liquid is small, but enough to elevate the ullage by 5 mL (my estimate). This would increase the pressure by about 5psi.

    As time moves on, equilibrium will be re-established: The gas returns to the liquid: this is the reason behind the ACS claim that "pressure doesn't change". Once the gas returns, everything goes back to what it was.

    Testing my claims:
    I wish I wasn't so broke, so I could acquire a pressure gauge and test it more precisely. Nevertheless, I will videotape the blind testing again in a few days and post the results, as I believe there is value. I am also brainstorming how to make my own "pressure gauge" as I'm sure I can come up with something close, as long as I don't have to massively engineer a new contraption.
    Since I am biased to believe my proposal, my test will be aimed at falsifying my own claim and I will ask my assistant to carry out the interpretation.

    But, isn't it reasonable to conclude (based on the first three points) that pressure should be temporarily higher?
  5. Mar 12, 2014 #4


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    There is a huge difference between ultrasound and shaking a bottle. Shaking it could actually add new gas bubbles to the liquid instead of removing them.
    I don't see where a pressure increase should come from, but I cannot rule it out either.
  6. Mar 12, 2014 #5

    Simon Bridge

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    ... I think it is intuitively plausible.
    However, like mfb, I do not see a physically plausible mechanism for shaking the bottle changing the pressure under the conditions stated. The system need not be in equilibrium for the pressure to be constant for eg.

    My own quick test was unable to duplicate your results - and I didn't wait for a restored equilibrium.
    So I'd need to see more testing before going against the established wisdom on this.
    You do have to be very careful about how you do the test - the situation the claims refer to is very specific.
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