funny balloon thought experiment

physical1

Thought experiment: a sealed container with water in it, along with a balloon. You are able to control the balloon with a string from the outside somehow, and this string however is still sealing perfectly. No leaks in the system.

The balloon is taken down to the very bottom of the container by you. You expended some energy to go against the buoyant force. When the balloon got to the bottom of the container, what happens?

Normally a balloon would shrink due to hydrostatic pressure. However in this case, the balloon cannot shrink because it would have to suck water into its place where it was taking up space before. It cannot suck water in to place because the container is sealed tight without any air in it. Would this put the whole container under vacuum? If it did, then the hydrostatic pressure would just press out that vacuum, no?

The balloon stays the same size? Can this be the case even though the balloon is very sensitive to pressure changes in typical settings? And if I had the ability to jump inside of the balloon, what would be inside? Just regular air pressure trapped inside not affected at all by hydrostatic pressure?

Cyrus

It's not sucking up water, the water level will rise as the balloon plunges into the water. Said differently, the air in the balloon will displace that same volume of water up, making the water level rise.

When the balloon is in the air portion of your tank, it's displacing a volume of air equal to the volume of the balloon.

When the balloon is submerged, it displaces a volume of water equal to the volume of the balloon.

There isn't a vacuum because that missing volume of air (case 1) got filled up with water (case 2) that rised as a result of putting the balloon into the water. But its the same volume of being replaced.

Unless the balloon goes *really* deep into this hypothetical tank, its volume will shrink due to pressure, so it will displace a smaller amount of water (thus make the water level rise) slightly less than the air it replaced. In that case, you wont create a "vacuum", but you will lower the pressure in the air portion.

You can solve this by thinking about the physics of what is happening - you don't need equations to arrive at the solution.

physical1

The balloon started out in a sealed container without any plunging or displacing. It was already there in the container ready to go at the top in the water. It is a special thought experiment where everything was placed into the sealed container already and we do not consider it plunging or being dropped into place. There is no open top on the container - it is sealed shut. No leaks and no room for the water to change volume.

It cannot possibly displace a smaller amount of water when it goes even hundreds of meters deep because the container is completely sealed and the water volume cannot change to a smaller amount. If the water volume could change then there would be a leak in the system or the container would be flexible (and if the container was flexible I wonder if it would deform - but if it did there would be a vacuum and it starts to become mind boggling since the vacuum would simply be battled back by the pressure).

I have to get all the little fine details of the thought experiment clearly laid out in order for you guys to see where I am going with this - basically the key is that the water is sealed and cannot change volume, unless it happened to implode the container somehow.

Doc Al

The volume of water in the container doesn't change. When the balloon is at the top, the water fills the container (water volume + balloon volume = container volume); when the balloon is at the bottom (and smaller) the water no longer fills the container (water volume + smaller balloon volume < container volume).

What's the problem?

Dadface

It's a nice thought experiment and reminiscent of the "cartesian diver".Try googling.

LURCH

So, are oyu saying a vacuum would form at the top of the container?

Doc Al

Sure, why not? (Ignoring the water vapor that would form.)

Cyrus

Yes, the tank is sealed - we all get that. Are you saying the balloon starts off inside water at the top position, or is it inside air at the top position? I thought it was sealed shut with air/water.

Second point, what do you mean "we cannot possibly displace a smaller amount of water as you go deeper" - yes, you absolutely can. Why would the volume of water have to change for this to happen - hint: it doesnt. It will create a vacuum equal in volume to the lost volume as the balloon contracts due to pressure from the water. I don't understand why you cant believe a vaccum would form. What 'pressure' is battling back the vaccum? The pressure is zero at the top of the water level. Google hydrostatics.

physical1

The water has no air above it, and no air dissolved in it.

My assumption, possibly wrong, is that water cannot be stretched out into a vacuum space since water is incompressible. There are no air bubbles in the water in this thought experiment. I thought fluids were incompressible - meaning they are also in-de-compressible.

The container is rigid solid and not flexible, so cannot collapse. However, I would also like to entertain the idea of a flexible container too.

Could water vapor form and we could have a special steam or mist? If so wouldn't it take a certain amount of sucking power to create vapor - kind of like a "breakdown point" (instead of boiling point) where it either does or does not form water vapor?

And if there is this vacuum, why doesn't the hydrostatic pressure just make contact with this vacuum and cancel it out, since pressure is throughout the whole system. I mean, we have a vacuum under pressure? Sounds contradictory :o)

Doc Al

That's a perfectly reasonable assumption. For all practical purposes, the volume of the water doesn't change. For some reason, you think this leads to some paradox.

There's no pressure in the vacuum. Don't forget that there's a rigid container surrounding the water, balloon, and any extra space created by the shrinking of the balloon.

physical1

The balloon tries to change the volume of the whole sealed system. Inside the sealed system, if the balloon changes in volume (size) then something must fill that space that was taken away. For every action there is an opposite reaction (patronizing I know).

With a sealed system, logically the water must expand if the balloon collapsed. The container could try to collapse and the system container shape could change (contract) in response to the balloon changing size. If the container is rigid then that could not happen. (A bit of an off topic aside, but if the container contracted would not the atmosphere have to expand, since the atmosphere is just a container itself that must have an equal and opposite reaction. If this atmosphere is expanding then is the universe contracting?).

Possible outcomes:
1, Nothing collapses at all, including the balloon, and the balloon could simply stay there under hydrostatic pressure at a certain depth without being affected at all in size. This is a paradox to me because we are applying pressure to compressible air, and yet nothing is happening in response.

2. balloon may never have been able to be placed there in the first place and while pulling the balloon down in the initial setup stages of the experiment, it gets stuck in some sort of "lock" as soon as you try to move it - and your brain catches on fire.

3. the air inside the balloon becomes a vacuum and nothing happens with the balloon size. In this case hydrostatic pressure is creating a vacuum? Huhhhh?

4. "water vapor" spontaneously forms out of the "water liquid", but I did not think that water was de-compressible into vapor. (my spell checker thinks "decompressible" is not a word :-)

5. some other behavior occurs, or I am mis-visualizing something

6. I try the experiment in my home, and get sucked into a black hole and die. I would prefer to have it done in the mind ahead of time.

7. there is a "nothing" space which spontaneously is created (a pure vacuum) at the top of the container composed of pure "nothing" which fills up the space lost by the compressed air. Sorry, doesn't make sense how "nothing" can exist and still be "something". There would need to be some substance in there mixed in to hold that nothing there (water vapor, or oxygen gas that was originally dissolved in the water, or some other substance that escaped the water and turned into a gas). This also begs the question about vacuums - how can they take up space if they are "nothing". Completely illogical, since nothing cannot take up space if it is really "nothing". If "nothing" is some space, then "nothing" is not really "nothing", it is indeed "something".

For every milli-litre of air that compresses, something must in return fill that space that was lost.

Doc Al

No it doesn't.
In this particular case, the water just shifts to take up the change in volume. And that means that the water level drops; it used to be filled to the top with water, but now it's not.
Newton's 3rd law is not relevant here.

Not at all. What physical law suggests this?

Overly philosophical and melodramatic fear of "nothing".

Try this example on for size: You and an empty box are in a sealed container jam packed with ping pong balls. The ping pong balls go right to the top of the container, no room to spare. You manage to open the box and ping pong balls fall into it, filling it up. So what happens to the level of the ping pong balls in the container? Assuming you have no issue with this example, point out what you think the difference is between it and your balloon experiment.

physical1

The human and the air in the box exchanges positions with the ping pong balls.

Nothing is contracting. That is a different thought experiment all together.

If you consider a diver who is jumping in to a pool, the pool is open system. When the diver goes deep his body starts to contract. But in a sealed system if his body started to contract then he could not suck anything into his old bigger space he took up since the seal would "blow", if the seal wasn't strong enough. But I prefer to not use a diver because humans are more complex than a simple balloon.

Dadface

There's quite a bit more to this problem and let me throw in a thought that may be relevant but which I need to think about in greater detail.It does seem ,at first sight, that a vacuum will be formed but then the water will evaporate into that vacuum and possibly even start to boil.This means that the space will not remain a true vacuum but will become a saturated vapour.Taking it further if the conditions are such that the liquid vapour interface is at its critical point then the two states will be indistiguishable.

Doc Al

Remove the air if you like--doesn't matter.
The water in your balloon experiment is not contracting either. Just like the ping pong balls, the water just shifts.
Huh? Please explain the physical reasoning behing your thinking. Note that in a rigid, sealed system you are isolated from the outside, so there's no atmospheric pressure to contend with.

physical1

If this is the case then it means that water is in fact "decompressible" and can be ripped and pulled apart. But, water is not "compressible" the other way, meaning it cannot form a solid by being compressed?

physical1

Atmospheric pressure was never a consideration. If a diver goes down and contracts in size due to maybe his lungs getting compressed with the air inside them, then the water level goes down in the pool. But I should not even bring this diver up. Humans are much more complex to involve in an experiment as they have both air, solids, and liquids in them.

If the system is sealed off then as air in the the divers lungs contract he has to pull something into that place. Nothing is there to pull if the system is sealed off and rigid - water is not decompressible (well maybe it is). I can see maybe the dissolved oxygen in the water being ripped out, or maybe some loose water vapor molecules. In this case the water volume would change - and take up the space which was lost.

Possibly I am not very good with words - experiments are hard to explain. I could draw some diagrams but it will take time.