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I am having an argument with my friend about how a nearly-full soda bottle should be stored in the fridge, with the goal of keeping the soda from going flat (i.e. keeping as much of the gas dissolved in the liquid as possible). Assume the bottle is a standard 2L bottle and its walls are impermeable so this is a purely static problem.

He thinks it doesn't make a difference whether the bottle is standing up vertically or lying flat horizontally. I think that standing up is better. My vague reasoning is that in either case, the plastic surface will feel the same amount of pressure $P$ from the contents; but when the bottle is lying horizontally, the shape of the air has a larger surface area and needs more stuff in it to attain the same pressure (since pressure if force over area).

My statement about the shape of the air is likely true by the isoperimetric inequality (or at least some heuristic which says that among shapes with the same volume, the ones "closer" to a sphere have less surface area), since the air-shape at the top of the bottle in the vertical case is closer to a sphere, than the shape when it is lying horizontally (in this case the air is like a long prism). However I'm not sure it's true that the pressure felt by the plastic is the same in both cases.

Who is right and why?

EDIT: thank you for your explanations. I understand how to answer this in terms of constant volumes now, although it isn't very intuitive to me. I wonder if there is an explanation involving the sum of the forces acting on the surface of the liquid being zero, for both the horizontal and vertical configurations.

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    $\begingroup$ His actual statement was that if there were a difference, it would be too small for us to tell without instruments (and I agree). I embellished the story to make a better problem. $\endgroup$
    – user25959
    Commented Dec 1, 2019 at 20:19
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    $\begingroup$ The real answer is you should squeeze the bottle while open and then close with as little air in it as possible. $\endgroup$
    – sirjonsnow
    Commented Dec 2, 2019 at 14:09
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    $\begingroup$ I'm voting to close this question as off-topic because it’s about food storage and not physics $\endgroup$
    – Kyle Kanos
    Commented Dec 2, 2019 at 15:17
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    $\begingroup$ @sirjonsnow, doesn't squeezing it just allow the CO2 more room to gas out before the vapor pressure is reached? $\endgroup$
    – JPhi1618
    Commented Dec 2, 2019 at 15:57
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    $\begingroup$ Actually, the best way is however it will fit in the refrigerator :-) $\endgroup$
    – jamesqf
    Commented Dec 2, 2019 at 17:27

4 Answers 4

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The volume of the bottle and the volume of the liquid are the same both ways. By subtraction, the means the gas volume is the same either way. The difference assumes in the question’s reasoning doesn’t exist.

There is more area in the sideways case. That’ll let equilibrium be reached more quickly. But it’ll reach the same equilibrium either way.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Commented Dec 4, 2019 at 10:07
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In an unopened 2L polyethylene terephathalate (PETE) bottle of soda the primary loss of CO2 is by diffusion of the CO2 through the walls of the plastic bottle. If you left a full bottle at room temperature, the half life of this process is about six months, meaning that a six-month old room temperature bottle of soda will be ~50% flat, even if you never open it. After a year, it will be 75% flat.

Keeping the bottle in the fridge naturally slows this process, so a chilled bottle of soda will retain its CO2 for much longer. And, in real life, you're not just watching a full bottle day after day - you're drinking out of it. The vast majority of the CO2 will be lost when you open the bottle. There is likely some differential between the diffusion rates at the liquid/PETE interface vs the gas/PETE interface, but in either case this is a process that takes months to years.

Each time you open the bottle, however, you lose a significant amount of CO2. The partition coefficient for CO2 in water is about 4, so the gas above the liquid will have about 1/5 of the CO2 concentration as compared to the liquid, which will have a concentration four times higher. So your bottle becomes about 20% flatter each time you open the bottle (this is a gross approximation! ***), assuming you leave enough time for it to return to equilibrium after closing the bottle.

So, if you want to worry about keeping your soda from going flat, forget about which way you store it in the fridge - it doesn't matter. What you want to avoid is opening the bottle more frequently than necessary. How flat is your soda? How many times have you opened the bottle? Let's assume you remove 300mL from the bottle each time you open it :

  Openings   Fullness   Cumulative CO2 lost
  -----------------------------------------
  One          85%         4.2% 
  Two          70%         13.5% 
  Three        55%         28.2%
  Four         40%         47.8%
  Five         25%         70.2%
  Six          10%         90.8%

*** If we assume, at least, that a serving of soda is removed each time the bottle opens then there will be a ~15-85% fraction of gas/liquid at each opening. When the bottle is more full there will be a lower loss of CO2 as the volume required to be repressurized to equilibrium will be lower - as the bottle empties the CO2 loss per 'opening' likewise increases. A 20% loss of CO2 happens only when the volumes of liquid and gas are equal (ie: half empty bottle). A 75% full bottle would lose only 7.7% of its CO2, for example, but a 25% full bottle would lose a full 43% of its remaining CO2.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Commented Dec 4, 2019 at 18:04
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    $\begingroup$ Excellent answer. Much better than the accepted one because it actually answers the real life case. However, I have a nitpick: When you open a soda bottle, you do not remove the entire $CO_2$ from the gas fraction in the bottle. You immediately loose a certain amount of gas due to the pressure inside the bottle, but the bulk of the gas remains and is only diluted by the influx of air as you pour your drink. The remaining $CO_2$ will reduce the amount of $CO_2$ that needs to enter the gas phase to reestablish equilibrium. $\endgroup$
    – cmaster - reinstate monica
    Commented Dec 4, 2019 at 22:46
  • $\begingroup$ what if you open the bottle while holding it upside down (so water is near exit, not CO2)? Would you lose less CO2 then (or would bubbling action of incoming air make you lose more of the CO2)? $\endgroup$
    – Matija Nalis
    Commented Dec 4, 2019 at 23:46
  • $\begingroup$ @MatijaNalis I suggest you try it and find out. $\endgroup$
    – J...
    Commented Dec 5, 2019 at 0:00
  • $\begingroup$ For the losing 20% each time calculation, doesn't that require the volume of liquid and air to be the same, 50/50? I don't imagine a full bottle would lose 20% on the first open, but I'm not very familiar with the pressure dynamics involved. $\endgroup$
    – Mars
    Commented Dec 5, 2019 at 2:29
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According to Henry's law the mass of gas dissolved is proportional to the pressure on top of the liquid. Because the pressure is the same in both cases, it should not matter if the bottle is vertically or horizontally

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    $\begingroup$ Why is the pressure above the liquid the same in both cases? $\endgroup$
    – user25959
    Commented Dec 1, 2019 at 20:40
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    $\begingroup$ I get that the volumes don't change, but there could be some transfer of the gas from the liquid to the air depending on the position, so it's not clear to me the pressure will remain the same. (in above, all variables should have subscript $P_h$ or $P_v$ depending on horizontal or vertical, and not clear that the $n_h=n_v$) $\endgroup$
    – user25959
    Commented Dec 1, 2019 at 20:45
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    $\begingroup$ Unless you shake the bottle to mix the gas and the liquid out of equilibrium, there is no reason for gas to leave the liquid as you move the bottle because the pressure and Henry law will ensure this will not happen. If you shake it it will eventually reach a state of equilibrium that should be the same as the previous one. $\endgroup$
    – user65081
    Commented Dec 1, 2019 at 20:55
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    $\begingroup$ I am saying that it has the same pressure because n T and V are the same. The volume cannot change cause the liquid does not expand or shrink, the only that could change is n, the number of moles of gas outside the liquid, but this will not change unless the pressure changes first (because of Henry's law) $\endgroup$
    – user65081
    Commented Dec 1, 2019 at 21:26
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    $\begingroup$ @Taemyr Exactly. It's the difference in the pressure that drives the outgassing, and the equilibrium pressure is the same for both the vertical and horizontal positions (it doesn't depend on surface area, just the volume if all else is equal, which it is in the magical bottle in the question). The situation will tend towards the case where the dissolved gas pressure is the same as the non-dissolved gas pressure - the dissolved gas will lose pressure to the non-dissolved gas, until they're equal. $\endgroup$
    – Luaan
    Commented Dec 2, 2019 at 11:21
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I would like to comment on your argument that says that when it is upright, despite the same volume, the surface is smaller, hence the pressure.

This is where your reasonning is flawed. Indeed in your bottle, the pressure at the end is going to depend on the volume and temperature, not the surface area with the liquid.

Surface area with the liquid would have been interesting if the liquid was not, for all purpose here, incompressible, and thus the volume would have changed.

Here you are operating in a constant-volume situation.

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