Thought experiment: Rising bubble in rigid container

[JimJim]

I came across a blog which posted a paradox and it really got me thinking. This is what it read:


You have a perfectly rigid jar, which is sealed and filled with a perfectly incompressible liquid. Within the liquid, at the bottom of the jar is a small bubble filled with an ideal gas. Since the gas is less dense than the liquid, it is buoyant and must rise.

However, since the liquid at the bottom of the jar must support the liquid above it, the pressure at the bottom is higher than the pressure at the top. So as the bubble rises, it moves from a region of high pressure to a region of low pressure.

But, the bubble is filled with an ideal gas, so as the pressure on it drops, it must expand.

But, the bubble is confined in an incompressible liquid within a rigid jar, so there is nowhere for it to expand into.

Therefore, the bubble can't expand.

Therefore, the bubble can't rise.

What's gone wrong?


At first I thought there would always be a buoyant force regardless if the bubble expands or not and hence it would simply rise. Any ideas?

 

[K^2]

If the fluid is incompressible, the bubble volume will remain constant. That means, the pressure within the bubble will remain constant. There is no contradiction between that and the bubble rising. The pressure within fluid will simply have to increase as the bubble goes up.

 

[JimJim]

The bubble would experience a larger pressure at the bottom than at its top so a larger force would be needed to keep the top from expanding than at the bottom. Wouldn't this create a net force downwards assuming the bubble is large enough that the difference in pressure is significant?

 

[K^2]

The forces aren't really acting on the bubble. The bubble, for all intents and purposes, is weightless. To really understand the effects of pressure differential, you would have to model the fluid flow due to pressure differential.

Without modeling fluid flow, we can only consider static cases, which are the bubble starting near the bottom, and bubble ending up near the top. It's easy to show that net energy has decreased, so the bubble will rise, and that the pressure in the fluid has increased, due to incompressibility of the fluid and ideal gas law.

 

[sophiecentaur]

If the fluid is incompressible, the bubble volume will remain constant. That means, the pressure within the bubble will remain constant. There is no contradiction between that and the bubble rising. The pressure within fluid will simply have to increase as the bubble goes up.

This just has to be the right answer. The fact that the bubble pressure must be the same throughout (because nothing is allowed to compress or expand) means that the pressure on the bottom just has to increase so that the hydrostatic pressure difference over the height of the column is maintained.
I just wish I could 'feel' that in a convincing way. I guess there is something involved with energy and gas laws. Potential has been reduced so that thermal energy has increased. This has increased the energy of the gas in the bubble which has maintained its volume despite the increase in pressure of the surrounding liquid.
Can someone put this in a better way so that I can feel I understand it better? This must be an isothermal process? Or must it?

 

[K^2]

Gas doesn't do any work, again, because volume doesn't change. So all of the energy must be absorbed by the fluid. Please, don't ask me to imagine what happens if the fluid also has zero viscosity, because my brain simply refuses to process that scenario.

 

[DrewD]

At the bottom there would be pressure [itex]P_0[/itex] and at the top the pressure would be [itex]P_1[/itex]. Approximately, the pressure in the bubble and outside of the bubble will be the same (apart from the minor difference that causes boyancy) or the net force on the bubble surface will not be zero which will cause it to change volume. Since [itex]P_0 > P_1[/itex]
[itex]P=\frac{\nu RT}{V}[/itex]
implies that the temperature will go down if the volume is kept constant. Am I making an error? The heat flow seems a bit strange, but it probably works out.

 

[256bits]

At the bottom there would be pressure [itex]P_0[/itex] and at the top the pressure would be [itex]P_1[/itex]. Approximately, the pressure in the bubble and outside of the bubble will be the same (apart from the minor difference that causes boyancy) or the net force on the bubble surface will not be zero which will cause it to change volume. Since [itex]P_0 > P_1[/itex]
[itex]P=\frac{\nu RT}{V}[/itex]
implies that the temperature will go down if the volume is kept constant. Am I making an error? The heat flow seems a bit strange, but it probably works out.

I thought of the same thing but it has the the implication that the bubble is cooler at the top than bottom. Would not temperature equalization after a period of time bring the pressure back up to Po?

 

[K^2]

Am I making an error?

Yes, you are. Temperature change requires either a heat flow or work done on gas. So temperature stays constant. Volume also stays constant. So the pressure on bubble's walls stays constant, and that means, pressure in the fluid increases. That's the only thing that can change without contradiction to something else.

 

[NascentOxygen]

::EDIT:: I missed the "rigidly sealed" significance. [strike]The bubble will expand as it rises, just as bubbles do in any liquid because of the decreasing pressure with decreasing depth. The gas in the bubble will cool as it expands, just as all volumes of gas do when they are allowed to expand without energy input.[/strike]

I'm not sure that the bubble would rise with much speed, though. Maybe it would rise with about the speed of a bubble in cold honey? Although the liquid may not be compressible, the gas in the bubble certainly is.

What would be interesting is to see a fish or a submarine trying to make its way through an incompressible liquid.

 

[K^2]

The bubble will expand as it rises, just as bubbles do in any liquid because of the decreasing pressure with decreasing depth.

Bubble can't expand. There is fluid in the way, which cannot be compressed. That's the whole point.

 

[NascentOxygen]

Bubble can't expand. There is fluid in the way, which cannot be compressed. That's the whole point.

I expect gravity will [appear to] move fluid out of the way to allow the bubble to rise.

Oh, I missed the significance of the container being so rigidly sealed; certainly in such a sealed container, the bubble won't expand. But it can still compress, if necessary, to allow the liquid to sink past it.

I'm thinking the bubbles may become elongated vertically, and may break up into many smaller ones, as they get forced towards the surface.

 

[K^2]

It doesn't matter, really. No matter how the bubble deforms, the total volume of bubble(s) will have to stay the same. The total amount of gas will have to stay the same. The temperature can only rise due to work on viscous fluid. So pressure in the bubble(s) cannot go down.

 

[mrspeedybob]

The bubble can be treated as a void. It could be vacuum for all it matters to this experiment. In any case, as long as the density of the bubble is less then the density of the liquid, the liquid will fall. The liquid in the top of the jar will fall into the void in the bottom of the jar. Since it is no longer in the top of the jar a void is left. Hence, the "bubble" moves from the bottom of the jar to the top.

 

[NascentOxygen]

The bubble can be treated as a void. It could be vacuum for all it matters to this experiment. In any case, as long as the density of the bubble is less then the density of the liquid, the liquid will fall.

In theory, sure. But if the bubble was also of an incompressible liquid, then I'm thinking drift upwards may be so slow as to be almost undetectable.

Just guessing though.

 

[sophiecentaur]

The upthrust would be the same and,with plenty of room around the sides, why shouldn't the bubble rise as expected (i.e. quite fast)? Its speed would be the same rather than accelerating as normal bubbles tend to do as they expand, of course.

 

[256bits]

I'm thinking the bubbles may become elongated vertically, and may break up into many smaller ones, as they get forced towards the surface.

You have hit upon a key aspect of this problem.
Release one big bubble and the fluid pressure increases only once.

Break up the larger bubble into smaller ones which release sequencially, only after the preceeding has moved to the top, and the fluid pressure will have an increase for each smaller bubble that has risen to the top.

While both scenarios result in the same amount of bubble moving from the bottom to the top, the second is not equal to the first - the second being a sort of pressure amplification depending upon the number of smaller bubbles, that being most likely a violation of some sort of law of physics akin to perpetual motion machines.

The question is: what will set the pressure within the system?
1. the initial pressure within the bubble that stays constant as the bubble moves up through the liquid
2. the liquid pressure represented by the height of the column of liquid

Item 1. is most likely true as the bubble cannot expand or contract within the incompressable liquid.

That leaves Item 2. I would suggest that the fluid being incompressable exerts a pressure top, bottom and sides all equal and nowhere within the fluid is there a pressure differential, in relation to height of fluid as in real fluids.

So what is the final state of the system after the bubble has(or bubbles have) moved to the top - the same as it was initiallt except of course some heat generation which is accounted for by the increase in potential energy of the bubble and the decrease in potential energy of the fluid.

 

[Delta Kilo]

The bubble in the middle of the liquid is not in equilibrium. Pressures inside and outside the bubble are not equal. Air pressure inside is constant, liquid pressure outside is subject to hydrostatic gradient. The resulting net force pushes the bubble up.

 

[NascentOxygen]

The bubble in the middle of the liquid is not in equilibrium. Pressures inside and outside the bubble are not equal. Air pressure inside is [strike]constant[/strike] uniform, liquid pressure outside is subject to hydrostatic gradient. The resulting net force pushes the bubble up.

I think that's what you meant?

 

[sophiecentaur]

The bottom line of all this is that pressure Inside the container changes. But that, in itself, doesn't violate any conservation laws because there is no change in volume. So there is no paradox.

 

[russ_watters]

First the easy one: Buoyancy doesn't depend on compressibility of either the fluid or the object/bubble in it, so that is completely irrelevant to whether the bubble rises. It rises.

Now the other one is trickier and I was thinking of quibbling with the assumption of incompressibility, but I realize I shouldn't: the difference in compressibility between water and air is an issue whether we make that assumption or not.

So the issue with compressibility means that because we are not in a steady-state, the water pressure changes as the bubble rises...just not at the height of the bubble. The bubble has whatever pressure it needs to fill the volume that is void of water. And we'll have to assume that the pressure at the top of the container is high enough that the weight of the water doesn't pull a vacuum.

So let's say the bubble is 1psi and pressure due to depth at the bottom is 0.5psi. The total pressure is has to be 1psi, so the bubble is pressurizing the container, not just holding up the water. So the pressure at the top is 0.5 psi and at the bottom is 1psi. When the bubble rises to the top, its pressure is still 1psi, but now the pressure at the bottom is 0.5+1=1.5psi.

 

[DrewD]

Yes, you are. Temperature change requires either a heat flow or work done on gas.

yes. I was thinking that a decrease in potential would cause the temperature, of the liquid to go down first, but, even if that weren't completely wrong, I would have been backwards. Two strikes against me...

I was bothered by the thought that the fluid would have a greater capacity to do work since a small change in volume at a higher pressure would produce more work, but the fluid is incompressible and the gas has constant pressure... that's three strikes and I'm done with this one.

 

[sophiecentaur]

In the absence of viscosity, I guess the bubble would 'bounce' back down again as energy would have to be conserved.

 

[cragar]

so if it was in a super fluid it would bounce back down,
super-fluids have zero viscosity right?
actually now that I say it, it sounds bizarre

 

[chingel]

What makes the pressure increase in the container as the bubble rises?

 

[sophiecentaur]

Because the pressure on the bubble must be the same as it was at the bottom (or its size would change) and the hydrostatic pressure difference must be the same as before (the ρhg formula).

This will not lie down and it really is getting a bit silly because, if you take this simply and literally, the following situation must be the result.
So the pressure at the bottom, initially, would the original pressure on the bubble (when it was at the bottom), which was ρhg. For the bubble to be at the same pressure when it reaches the top, the new pressure at the top must be ρhg and that makes the pressure at the bottom 2ρhg. This is really counter intuitive because it would apply to ANY size of bubble. An interesting situation would arise for multiple bubbles because their individual sizes would all change during the process (I think) with the total volume staying the same.

I can't find anything wrong with the above argument except that it is based upon Incompressibility and a totally rigid container, so daft answers could well emerge - and seem to have, as with all immovable objects and irresistible force scenarios.

There is just one way out, as far as I can see and that could be the SHAPE of the bubble. The lowest potential shape at the top could well not be a sphere. But then, if a rigid sphere of light material were used, then the shape would have to stay the same.

Perhaps the whole thing is basically a nonsense and the thought experiment would have to involve something that was not irresistible - like the liquid, and that would allow a real result.

 

[russ_watters]

It makes more sense if you select 0 for the bubble's pressure...

 

[sophiecentaur]

Zero Volume?

 

[chingel]

Maybe the pressure must increase, because if that is the only explanation that's it, but what exactly makes it increase? What will apply the force and why will it do so?

 

[256bits]

Real fluids compress a bit any the stress between atoms or molecules gives the pressure (I think, correct me if I am wrong )

I just don't see how any ideal incompressible fluid can have a pressure anywhere between 0 or infinity. Any attempt to change the volume, would mean the pressure skyrockets.

So conversely, what happens when the pressure changes as is proposed here by the bubble rising. C= V (dV/dP) , or dP = (VdV)/ C , but C=0 ( and dV=0 ) so mathematically it doesn't make sense.

 

[K^2]

256bits, it doesn't have to be perfectly incompressible. Compared to a gas, just about any real fluid will be near enough incompressible to get the same behavior as stated in the problem. So yes, real fluid will give a little, and the pressure in the bubble will drop slightly, but the change will be negligible, and you can easily have a setup where that pressure drop is much smaller than pressure differential between bottom and top of the container, resulting in an overall pressure increase in fluid as the bubble rises.

 

[sophiecentaur]

@k^2
OK, allowing for some compressibility in the liquid takes us away from a total nonsense scenario.
But the requirement is that the bubble volume doesn't change either so the pressure on it must remain the same. So the increase in pressure at the bottom of the container must be a factor of nearly two, as I described earlier.
Perhaps the trap door into this problem lies in asking how the bubble got to the bottom in the first place. It would need to have been introduced at that level as some liquid was drawn off through a hole in the top. Would that have involved Work Done? and a consequent change in Potential? I think so. It would correspond to an amount of liquid, the volume of the bubble being lifted up to the top(??) of the container. (Or possibly to just above the bubble). This volume of liquid would then fall to the bottom of the container, displacing the bubble and reducing the potential again.

 

[K^2]

sophiecentaur, you are making it way too complicated. In this entire problem, the only work being done is the work done by rising bubble against viscosity of the fluid. The exact nature of that process is complex, but we know that fluid will reach a steady state, and absorb all that work as heat. (And yes, real fluid will expand slightly due to the added heat, slightly reducing volume of the bubble, but these are tiny, tiny corrections.)

The only problems the initial statement suffers are problems of idealization, which are typical for this type of problem. If you consider constraint forces during collision of two rigid bodies, you run into the same kind of problem. What do you do? You consider energy and momentum. Same thing here. Except what we get to work with are energy and ideal gas law.

How the bubble actually gets to the bottom is irrelevant. We have an initial state. Question is how that state evolves.

 

[sophiecentaur]

Yes, you are right but I am thinking of the non ideal case. The relative nature of a gas and a liquid (and steel) allows this, I think. So, although it may be an irrelevant conclusion, it seems that the pressure must nearly double in the non ideal case.

 

[K^2]

Only if at the beginning of experiment the top of the fluid was not under pressure, and all of the pressure was accounted for by the gradient. That's not necessarily the case. Say, I start with a 1m tall cylinder filled with air at 1atm and sealed, with exception of a small inlet valve. Through that valve, I start forcing in liquid until the cylinder is 90% filled with liquid. I then allow the temperature in the cylinder to equalize with room temperature. The pressure of the gas is now 10atm. The pressure gradient in fluid is just under 0.1atm. If I now quickly invert the cylinder, perhaps by the top end so that centrifugal effect helps me to end up with bubble almost perfectly at the bottom, I have the setup for our problem. Bottom of the cylinder is at 10atm, while top is at 9.9atm. Once the bubble rises, the top is at 10atm, and the bottom is at 10.1atm. Increase of only about 1%.