# Thought experiment: Rising bubble in rigid container

1. Mar 27, 2012

### 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?

2. Mar 27, 2012

### 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.

3. Mar 27, 2012

### 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?

4. Mar 27, 2012

### 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.

5. Mar 27, 2012

### sophiecentaur

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?

6. Mar 27, 2012

### 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.

7. Mar 27, 2012

### DrewD

At the bottom there would be pressure $P_0$ and at the top the pressure would be $P_1$. 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 $P_0 > P_1$
$P=\frac{\nu RT}{V}$
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.

8. Mar 27, 2012

### 256bits

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?

9. Mar 27, 2012

### K^2

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.

10. Mar 27, 2012

### Staff: Mentor

::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.

Last edited by a moderator: Apr 17, 2017
11. Mar 27, 2012

### K^2

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

12. Mar 27, 2012

### Staff: Mentor

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.

13. Mar 27, 2012

### 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.

14. Mar 28, 2012

### 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.

15. Mar 28, 2012

### Staff: Mentor

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.

Last edited by a moderator: Apr 17, 2017
16. Mar 28, 2012

### 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.

17. Mar 28, 2012

### 256bits

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.

18. Mar 29, 2012

### 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.

19. Mar 29, 2012

### Staff: Mentor

I think that's what you meant?

20. Mar 29, 2012

### 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.