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 into 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 won't 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]

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.

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]

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?

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

 

[Doc Al]

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

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

 

[Cyrus]

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.

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 can't believe a vacuum would form. What 'pressure' is battling back the vaccum? The pressure is zero at the top of the water level. Google hydrostatics.

 

[physical1]

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 can't believe a vacuum would form.

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 )

 

[Doc Al]

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.

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.

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

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]

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.

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 into 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]

The balloon tries to change the volume of the whole sealed system.

No it doesn't.

Inside the sealed system, if the balloon changes in volume (size) then something must fill that space that was taken away.

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.

For every action there is an opposite reaction (patronizing I know).

Newton's 3rd law is not relevant here.

With a sealed system, logically the water must expand if the balloon collapsed.

Not at all. What physical law suggests this?

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 into 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".

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]

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.

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 into 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]

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

Remove the air if you like--doesn't matter.

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

The water in your balloon experiment is not contracting either. Just like the ping pong balls, the water just shifts.

If you consider a diver who is jumping into 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.

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]

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.

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]

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.

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.

 

[Doc Al]

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).

Again, I have no idea where you are getting this.

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.

The bit of water vapor that would "fill" the space is irrelevant to the main point here. (If you fear the "vacuum", what do you think lies between those water vapor molecules?)

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

A diagram is not needed. The situation is clear; it's your reasoning that I'm trying understand.

How about we go back to your balloon experiment, only this time the water doesn't fill the entire container. Let's say there's an inch of "space" between the water level and the top of the container. Would that make a difference to you? When the balloon--which starts out just below the water surface--is pulled down to the bottom and compresses, the water level will go down. Just like with the diver and the pool. If you think there's a physically relevant difference, please point it out.

 

[physical1]

How about we go back to your balloon experiment, only this time the water doesn't fill the entire container. Let's say there's an inch of "space" between the water level and the top of the container. Would that make a difference to you?

You did not define what this inch of space was composed of.

If this inch of space was air, it would be physically different than liquid water being in that space. Why is it physically different? Because water is not compressible and maybe not decompressible either. In order for a vacuum to originate in that space, something has to "give way". Water does not give way. Air does.

If this inch of space was pure vacuum, I am afraid you have done something no man has done before, and I would like to ask you how you created that pure vacuum since they do not exist.

If you argue, that, well, no it is not a pure vacuum.. it is... composed of some water vapor.. then.. how did you first break down that water into vapor I will ask. If you argue that well, the space got full of a tiny bit of air, then I will just argue back that I said it was WATER not air.

Air can be changed easily (compressed and decompressed) in size. Water cannot. Air allows practical partial imperfect vacuums to exist. Water since it is not decompressible/compressible in liquid form, does not AFAIK, allow a vacuum to spontaneously exist out of no where. It seems that practical imperfect vacuums (ones we can actually create) need some other gas molecules to exist. Water is not a gas molecule, it is liquid.

If one has ever held a science syringe with thumb on end and pulled the syringe handle, one would know that it is very easy to change air volume in the syringe. However if one fills the plastic syringe with pure water and no air, the syringe gets locked when you try to pull it with thumb in place. Then, as you pull, the syringe busts and cracks or the handle rips into two pieces if you are strong enough. Often the syringe plastic will deform and collapse, imploding.

So, what component is the most likely to "give way" in the system I described? Probably the air inside the balloon. Instead of the air inside the balloon being compressed when under hydrostatic pressure, it will just.. I don't know, stretch out and have lower pressure. Still mind boggling me at this point though.

 

[Cyrus]

If the pressure at the top of the water is zero when normally full to the top... and then there is an empty space at the top with vacuum (after you reel the balloon down), why would the water 'go up'?

What has changed? - Nothing. Nothing has changed. The pressure at the top is zero. Has been, and always will be.

 

[physical1]

What has changed is that the balloon, a soft substance, has been placed under hydrostatic pressure. The air molecules in the balloon get affected by hydrostatic pressure, and there must be an equal and opposite reaction. Since the balloon rubber is sucked in and held in place by a water vacuum lock (forgive my pseudoscientific terms), I suspect the air will stretch out in the balloon and the water level will not go up (which it cannot since it is sealed shut). SOMETHING ELSE must give way since water cannot.

If the air in the balloon gives way, then the balloon size stays the same. But that is why I came here - to find answers and confirmations (and back patting). In my original post, I hypothesize that the balloon will stay the same size.

 

[Cyrus]

What has changed is that the balloon, a soft substance, has been placed under hydrostatic pressure. The air molecules in the balloon get affected by hydrostatic pressure, and there must be an equal and opposite reaction. Since the balloon rubber is sucked in and held in place by a water vacuum lock (forgive my pseudoscientific terms), I suspect the air will stretch out in the balloon and the water level will not go up (which it cannot since it is sealed shut). SOMETHING ELSE must give way since water cannot.

If the air in the balloon gives way, then the balloon size stays the same. But that is why I came here - to find answers and confirmations (and back patting). In my original post, I hypothesize that the balloon will stay the same size.

Look, stop talking about the balloon. I don't care what's going on with the balloon. It's irrelevant. You are confusing too many things in your head. Think about ONE THING at a time. Are you listening to what I'm saying, or are you telling me?

You are wrong.

 

[Dadface]

How does boiling point of water vary with pressure?As the pressure decreases so does the boiling point so water will boil most readily into vacuum which will then cease to be a vacuum because it contains water vapour.Even if the water didn't boil it would evaporate and in the closed system being described here it will tend to saturation there being equilibrium being rate of evaporation and rate of condensation.The saturated vapour pressure is not necessarily negligible...as the critical point is approached it becomes more difficult to distinguish between the liquid and vapour phases their densities getting closer and closer.

 

[Cyrus]

how does boiling point of water vary with pressure

The boiling point will lower as the pressure decreases. Not relevant here, there is no water/air interface. The water can't boil.

 

[diazona]

If this inch of space was pure vacuum, I am afraid you have done something no man has done before, and I would like to ask you how you created that pure vacuum since they do not exist.

That is false. A pure vacuum certainly can exist. (well... in the real world, nobody has ever constructed a device that produces a perfect vacuum, but this is a thought experiment)

If you argue, that, well, no it is not a pure vacuum.. it is... composed of some water vapor.. then.. how did you first break down that water into vapor I will ask. If you argue that well, the space got full of a tiny bit of air, then I will just argue back that I said it was WATER not air.

Whenever you have water in a container with some free space above it, a small amount of the water will spontaneously vaporize. The "vapor pressure" of water is the equilibrium pressure of water vapor produced by this process. Under normal conditions, this pressure is pretty small so we usually pretend it doesn't exist.

 

[physical1]

You are wrong.

No, you are close minded and have no idea what the experiment is about, apparently.

 

[physical1]

That is false. A pure vacuum certainly can exist. (well... in the real world, nobody has ever constructed a device that produces a perfect vacuum, but this is a thought experiment)

A pure vacuum can certainly exist, yet you have no evidence it exists. This sounds like God believer to me.

 

[Cyrus]

No, you are close minded and have no idea what the experiment is about, apparently.

Don't ask any more questions if you are not willing to listen to the answer. I believe you will be banned from here shortly. Ciao.

This is why I hate "thought experiments" - typically very little thought goes into them, much like the airplane on a treadmill "thought experiment".

 

[m.e.t.a.]

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?

The answer to the first half of your question, "Can water be compressed?", is yes. All solids, liquids, gases and so on can be compressed. The reason you will see many PF users refer to water as "non-compressible" though is because it takes a very large increase in pressure (in the order of thousands of atmospheres) to compress a volume of water by just a few percent.

The graph on the following web page shows how the density of water varies with temperature at a variety of different pressures:

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

You can see that at, for example, room temperature, an increase in water pressure from 1 bar to 200 bar only results in about a -0.9 % change in volume. This is equivalent to the pressure a submarine feels in diving from sea level to a depth of 2 km. A reasonable depth for the water tank in this question might be 1 m. The density of the water at the bottom of the tank would only be ~0.0005 % greater than that of the water at the top. So you can see why people tend to disregard compression when dealing with water.

The answer to the second half of your question, "Can water be compressed into a solid?", is no* for water but yes for virtually all other substances. At standard atmospheric pressure, water reaches its highest density at about 4 ºC, as can be seen from the graph.

- m.e.t.a.


-------

* From Wikipedia: "Not all forms of ice are less dense than liquid water however, high density amorphous ice (HDA) and very high density amorphous ice (VHDA), for example, are both denser than pure liquid water."

(It seems that both HDA and VHDA can only be formed from ultra-pure water, which is then supercooled. As for whether compressing, instead of cooling, the ultra-purified water would result in some form of higher-than-water-density ice, I couldn't possibly imagine; the physics of water is extremely complex.)

 

[Dadface]

The boiling point will lower as the pressure decreases. Not relevant here, there is no water/air interface. The water can't boil.

I googled and watched a short film of water boiling at room temperature as the pressure was reduced.

That is false.


Whenever you have water in a container with some free space above it, a small amount of the water will spontaneously vaporize. The "vapor pressure" of water is the equilibrium pressure of water vapor produced by this process. Under normal conditions, this pressure is pretty small so we usually pretend it doesn't exist.

But it does exist and under some conditions it is not negligible

 

[Cyrus]

I googled and watched a short film of water boiling at room temperature as the pressure was reduced.

Did it boil in a sealed off container with no air in it?

 

[Dadface]

Did it boil in a sealed off container with no air in it?

Well the air was pumped out.I googled "how does the boiling point of water vary with pressure?"I remember trying it years ago with a syringe partly filled with water.I simply blocked the open end and pulled up the plunger but I wasn't sure whether any air bubbles were being sucked in through the seal.

 

[Cyrus]

Well the air was pumped out.I googled "how does the boiling point of water vary with pressure?"I remember trying it years ago with a syringe partly filled with water.I simply blocked the open end and pulled up the plunger but I wasn't sure whether any air bubbles were being sucked in through the seal.

That means it had air in it. The OP here is talking about a tank filled to the brim with water. No air is inside the tank anywhere. Not even a little.

 

[Dadface]

Water will boil more readily if there is no air.Try googling"boiling water under a vacuum.There's some nice films there

 

[Cyrus]

http://video.google.com/videoplay?docid=-5733115458174025512&hl=en

 

[Mapes]

Water is not a gas molecule, it is liquid.

Someone who would assert this is unlikely to be convinced no matter how many members and moderators try to explain the physics of the thought experiment. physical1, you're arguing from incredulity ("If this is the case then it means that water is in fact "decompressible" and can be ripped and pulled apart."), which is a pointless position to take in physics. Plenty of things occur in nature that are quite surprising when one first learns about them. But I'm not sure you're here to learn, or you would have examined your assumption about "decompressibility" and found it to be incorrect. (When one increases the volume of a container entirely filled with water, a vapor phase forms in the additional volume. At low pressure, it's the gas phase, not the liquid phase, of water that's most stable.) Instead, you seem to want only support for your misguided assumptions and hypotheses, and you're not going to get that support from the knowledgeable members of this forum.

 

[Dadface]

Physical,what actually happens during the descent and its aftermath is very complicated and depends,amongst other things, on the speed with which the baloon is dragged down.Eventually ,however, a state of dynamic equilibrium will be reached and the density of the vapour/ liquid will increase from the top of the container to the bottom there being a large increase at the vapour /liquid interface.

 

[Doc Al]

All this talk about the water vapor that would form above the water surface is largely irrelevant to the main point of this "thought experiment". The question is: What happens to the water level when the balloon is pulled to the bottom?

physical1 is under the impression that something prevents the water level from lowering. All this talk of the details of vapor formation is not helping to correct that misunderstanding.

 

[russ_watters]

Water will boil more readily if there is no air.Try googling"boiling water under a vacuum.There's some nice films there

No air does not automatically imply a vacuum. We should presume that the experiment is set up at atmospheric pressure, with the balloon providing the pressure inside the vessel. When the balloon is pulled under, the pressure will decrease, but it will not pull a vacuum until it gets to a certain depth.

 

[russ_watters]

All this talk about the water vapor that would form above the water surface is largely irrelevant to the main point of this "thought experiment". The question is: What happens to the water level when the balloon is pulled to the bottom?

physical1 is under the impression that something prevents the water level from lowering. All this talk of the details of vapor formation is not helping to correct that misunderstanding.

That "something" is the air pressure inside the balloon.

 

[Doc Al]

That "something" is the air pressure inside the balloon.

Say what? As the balloon is lowered, the air pressure increases as its volume decreases. The air pressure doesn't prevent the water level from lowering.

 

[Dadface]

All this talk about the water vapor that would form above the water surface is largely irrelevant to the main point of this "thought experiment". The question is: What happens to the water level when the balloon is pulled to the bottom?

physical1 is under the impression that something prevents the water level from lowering. All this talk of the details of vapor formation is not helping to correct that misunderstanding.

Okay physical1 The main point is that the balloon will shrink and the water level will fall.Once that sinks in you may want to consider the other details.

 

[LURCH]

Wow, this thread has gotten huge since I was last here!

Doc; when I asked, "So, are you saying a vacuum would form at the top o fthe container?" you replied

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

That was the conclusion I had reached, but I wanted to check and see if others had come to the same answer. It is, after all, a bit counterintuitive to think of a balloon being compressed within a sealed container that has a vacuum at the top. On intuition, one could be forgiven for thinking that the vacuum at the top would pull the water up, causing the water level to rise and fill the container, and the balloon to keep its original volume.

It does help if one pictures the experiment beginning with a slight vacuum at the top of the container (like a water barometer) as you suggested. Although this differs from the original thought experiment, it makes a good preliminary exercise. If one is willing to picture the experiment this way, then it is not a difficult step to progress from there to the conditions described in the OP.

 

[Mapes]

Equilibrium state 1: a (relatively small) air-filled balloon with volume V_{B1} is at the top of a sealed rigid container . The volume of the water is V_W. The volume of the container is V_{B1}+V_W. The pressure outside the balloon at the top of the container is P_{top}, which is the (saturated, equilibrium) vapor pressure of water at that temperature. The pressure inside the balloon is P_{top}+P_B, where the excess pressure is related to the strain energy of the stretched balloon material. The pressure at the bottom of the container is P_{top}+\rho g h.

Equilibrium state 2: the balloon is at the bottom of the container (having been pulled down by some arbitrary mechanism). The vapor pressure of the water is unchanged, being a function of temperature only, so the pressure is still P_{top} at the top and P_{top}+\rho g h outside the balloon at the bottom of the container. So the balloon pressure is larger after reaching equilibrium, its volume V_{B2} smaller. There is vapor at the top of the container with volume V_{B2}-V_{B1} and pressure P_{top}. The volume of the water is still approximately V_W; some amount of water evaporated to form the vapor, but since the density of the vapor is orders of magnitude larger than the liquid, let's assume the amount was negligible.

For simplicity, we might assume P_{top} is negligible (as some have in this thread), but it's important to understand that vapor, not a vacuum, lies above the water in the second case.

Any problems with this?

EDIT: This is just an attempt to assign variables to the scenario explained first by Doc Al, then by Cyrus, Dadface, and diazone. It may be of use to physical1, it that person is still reading the thread.

 

[Q_Goest]

Hi physical1. Interesting thought experiment! I think people are getting confused about the experimental set up. Here's what I understand you mean:
- There is a perfectly sealed container with water and a balloon inside.
- The starting point is such that the balloon is at the top of the container and at ambient pressure.
- The balloon is moved from the top of the container to a point near the bottom of this sealed container. How it is moved is irrelevant.

The question then is, what happens to the pressure inside the container.

Assuming the container with the balloon at the top is at atmospheric pressure, the pressure at the bottom is simply described by rho*g*h, just as it would under normal atmospheric conditions.

But because there can’t be any air getting into nor out of the container, and because water is essentially incompressible, then as the balloon is moved down toward the bottom of the container, the pressure inside the balloon can’t change. If pressure in the balloon increased, the volume would decrease, but since water is incompressible and there is no way for air to get inside the container, the balloon can’t (at first) change volume.

The pressure of the water therefore, is equal to the pressure of the balloon at the depth of the balloon. You can think of the balloon as a pressure gage at this point. It is measuring the pressure of the water at the given depth. Whatever the location of the balloon, the pressure of the water at that location must equal the pressure of the balloon, and the pressure in the balloon is equal to ambient pressure because the water is assumed to be incompressible. (in reality, there's some bulk modulus for water which will tend to decrease the water density slightly so that the air pressure increases very, very slightly as the balloon sinks in the container.)

Now you can determine the pressure of the water at any other location knowing the pressure changes according to rho*g*h. That means the pressure ABOVE the balloon is LESS than the pressure of the balloon and the pressure BELOW the balloon is GREATER than the pressure of the balloon.

When the pressure at the top of the container drops to the boiling point of that water (pressure and temperature hits the saturation point on a TS diagram), the water will begin boiling at that point and water vapor will begin to appear at the top of the container. At this point, the balloon will have to shrink.

To go much further than this, we’d have to start using thermodynamics to determine the equilibrium state of the water and air in balloon, but pressure at any level is determined using thermo & Bernoulli's equation when the balloon is below this point.

 

[Mapes]

Q_goest's very nice treatment is the most general I've seen, since it can accommodate initial container pressures higher than the saturated vapor pressure (my comment at #44 assumes an initial pressure less than or equal the vapor pressure, which is around 3 kPa at room temperature).

So Q_goest, if we start with a container with an initial pressure of 1 atm at the top, it would seem that the balloon could sink a certain distance with a relatively minuscule (but finite) decrease in volume. Meanwhile, the pressure at the top is decreasing from 1 atm to 3 kPa. When the pressure at the top reaches 3 kPa, the water begins evaporating; if the balloon continues to sink, its rate of volume decrease is much larger, being a function now of vapor compressibility rather than liquid compressibility. Does this track with your thinking?

 

[Q_Goest]

Hi Mapes,

So Q_goest, if we start with a container with an initial pressure of 1 atm at the top, it would seem that the balloon could sink a certain distance with a relatively minuscule (but finite) decrease in volume. Meanwhile, the pressure at the top is decreasing from 1 atm to 3 kPa. When the pressure at the top reaches 3 kPa, the water begins evaporating; if the balloon continues to sink, its rate of volume decrease is much larger, being a function now of vapor compressibility rather than liquid compressibility. Does this track with your thinking?

Yes, you've got it.

 

[Cyrus]

The airplane takes off. Q.E.D.

 

[physical1]

A vacuum at the top of the container (localized) is hard to visualize for me also because according to Pascal, pressure applied at the top is transmitted undiminished and does not "localize" in little areas. If this were true a hydraulic jack would not work. Unless, this is a different situation... let me know.

I visualize that vacuum pulling on the water from the top, also pulling on the balloon.. and hence, the "shrinking" of the balloon is defeated by balloon expansion - and we are back at zero - canceled out.

I still recommend everyone play around with a water filled plastic syringe, and you will understand where I come from when i say a vacuum is extremely hard to form in a sealed system with water lock in place. When you play with the syringe try to imagine what a little shrinking balloon will do. Think if there was a balloon in that syringe too, as soon as you pull on the syringe, it will transmit to the water, which will then pull on the balloon outer edges. This brings the balloon into expansion when gravity was trying to contract it. Game canceled.

the balloon can’t (at first) change volume.

Sad to say that this is what I said all along.

By "at first" do you mean that even under tiny amounts of vacuum not significant enough (which is what a balloon might cause), water will find a way to vaporize? I think it will not be the case and only super duper thick strong potentially powerful balloons at super depths could work for that.

 

[physical1]

Can anything in the system heat up or cool down by the way - i.e. some pressure or vacuum causes something else to change temperature. like the air in the balloon, or the water itself, etc. That would be interesting if a refridgerator, freezer, or heater was formed out of this.

If a volume stays constant but pressure increases, a temperature change could occur. i.e. if the balloon is locked into position, but the pressure increases inside - the (specially insulated) balloon becomes hot inside, yet the water starts to freeze or cool down a human. Please disprove this, it can't be true.