My dad has a really annoying (in my opinion) habit of crushing bottles of carbonated drinks so that there’s as little air as possible in the bottle. He claims that this helps keep it carbonated, but I believe that the amount of air left over in a partly-drunk 2-liter bottle doesn’t have any major effect, if any effect at all, on its carbonation. Who’s right?
You are. He’s actually causing the carbonation to come out of solution faster, since the force of the bottle trying to push back out to its normal shape creates a partial vacuum over the soda. The same way that releasing a sqeezed eyedropper bulb will suck fluid up into the dropper tube.
Here’s a recent thread on the same topic.
A number of years ago, I ran across an item designed to presserve the carbonation in 2-liter soda bottles. It was basically a little air pump that screwed on in place of the lid. You just pumped this a couple of time to increase the pressure in the partially empty bottle, thus holding the CO2 in solution. I’ve only seen them in those cheesy mail-order catalogs, but Google should turn something up.
You would think so, but maybe it doesn’t work this way.
Think of the case where the bottle is squeezed so much it locks into a conformation that would take a greater amount of pressure to return to its original state. This is easy to do. The bottle bends over, with a deep crease on one side and only a bend opposite.
Perhaps in this case the amount of pressure required to overcome the deformation is greater than that provided by the carbonated lolly-water. Since the head space above the liquid is decreased, more CO2 remains dissolved in the liquid, ala Le Chatelliers principle.
So the effect may depend on the type of bottle deformation. One case would draw the CO2 out by reducing the pressure above the CO2 solution (as described by QED). The other would reduce the rate of dissolution of the gas by holding a significant pressure, with reduced headspace, above the solution.
Wouldn’t increasing the pressure of the air merely allow the CO[sub]2[/sub] to come out of solution into denser air? Seems like the only foolproof method would be to pump more CO[sub]2[/sub] in to replace the air.
Actually, you’re both wrong.
When you squeeze the bottle you provide a method for MORE gas to escape.
It is no longer held in by the bottle’s rigidity, but will push out the bottle until the bottle retains its shape and all carbonation is in the space above the liquid.
Cecil really needs to lay down the answer to this question. The two previous threads on this, that I know of, are surprisingly venom filled.
No that is not right.
Did you understand my post?
Facts not supplied, but critical to the correct answer:[ol]Is the crushing done on a full, partially full, or empty bottle?Is the cap on the bottle at the time?[/ol]
Facts not supplied, but critical to the correct answer:[ol][li]Is the crushing done on a full, partially full, or empty bottle?[/ol][/li][/QUOTE]
That is, what is the air/fluid ratio? (If there is a lot or air, the answer could be different from just a little.)
Yes and no. Pumping ordinary air into the gas-space of the bottle will not have any effect on the amount of CO[sub]2[/sub] in the liquid (assuming no CO[sub]2[/sub] at all in the air you pump in; probably a reasonable approximation). However, you don’t really care about the amount of carbon dioxide specifically in the liquid; you probably care primarily about the amount of fizziness, which will be the total amount of gas in the liquid. Whether that fizziness is carbon dioxide, or nitrogen, or hydrogen, or anything else, it’s still fizziness. And an air pump on the cap would, in principle, increase the total fizziness.
Not a physicist, but I do this myself, and, subjectively, it does seem to increase the residual fizziness in the remaining liquid.
I should do some proper testing, perhaps.
I dont have time to look up the relative solubilities of CO[sub]2[/sub], O[sub]2[/sub] and N[sub]2[/sub], but the point of the device is it should prevent the CO[sub]2[/sub] from escaping in the first place. Or more precisely it pushes the equilibrium CO[sub]2/sub <-> CO[sub]2[/sub] (aq) to the right. Of course, an equilibrium of all 3 gasses would exist between the two phases, but the CO[sub]2[/sub] is most soluble in water, so it will be more readily effected.
To give an idea of the low solubility of nitrogen - in nitrogenated beers, for example, its concentration is only of the order of mg/L, whereas carbon dioxide in beers is of the order of g/L
I just came across this thread.
Looks like this topic has been done to death.
Read Squinks post #3.