Funny balloon thought experiment

[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.

 

[Q_Goest]

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.

Do you think the pressure is the same wherever you go in the fluid? (it’s not) Is the pressure at the top of the container the same as the pressure at the bottom? (it’s not) Are you familiar at all with Bernoulli’s equation? (Pressure in a verticle column under gravity varies by rho*g*h)

So the column of water does not have a constant pressure throughout. The pressure is constant at any height, h, but increases the farther down we go.

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.

If you’re interested in learning about physics, please stop saying the vacuum pulls. A vacuum, regardless of how strong it is, does not pull. Are you familiar with absolute pressure? If the pressure is 1 psia (absolute pressure) it is producing a compressive force on the water equal to one pound of force over every square inch. If the pressure drops to zero (0 psia), there is no longer a compressive force caused by that pressure on the water. Force simply goes to zero when pressure equals zero absolute. At atmospheric pressure however, the force on the water is 14.7 pounds for every square inch of surface.

A layman however, may think of there being some kind of ‘pulling’ force caused by subatmospheric pressure, which is wrong. That’s just an incorrect mental representation of what’s happening, because what we often see is that a fluid such as air or water will flow into a vacuum. But the fluid isn’t being ‘pulled’ into the vacuum at all. It is being PUSHED into that space by atmospheric pressure acting at another location on the fluid and the pressure in the ‘vacuum’ space isn’t strong enough to keep it out!

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.

By “at first” I mean that assuming the balloon is at 14.7 psia at the top of the container, there is some depth to which the balloon can be taken during which the size of the balloon does not change significantly. The balloon pressure will actually increase very, very slightly, assuming no disolved air in the water, and the volume of the container is not extreamly large compared to the volume of the balloon. As the balloon goes down deeper into the water, the pressure of the water at every point also goes down. In so doing, the density of the water will decrease some very minute amount which is accommodated by the increase in the balloon’s air density.

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.

As the balloon sinks in the container, the air is compressed very, very slighty. This would cause it to warm very, very slightly. The process would follow a line of constant entropy.

In contrast, the water pressure at every point decays significantly, also following a line of constant entropy. Once the balloon reaches a level at which water at the upper surface can boil, a few things start to happen.
- The surface of the water begins to boil as the water passes through the saturation point. Imagine for example, taking the radiator cap off a car that has been running and is hot. The pressure inside the radiator drops suddenly and the water begins to boil.
- The boiling water actually gets cooler. The temperature where the water is boiling goes down and gets colder.

As the balloon goes deeper:
- Water vapor formed above the surface of the liquid water will begin to get colder faster and there would be a flow of heat from the warmer water to the colder water vapor and also to the colder water surface where the water cools from boiling.
- The balloon begins to get compressed as pressure in the water now is being pushed on by the water vapor at the top surface. Remember that although there is a partial vacuum at the top water surface, that vacuum pressure is PUSHING down on the water, not PULLING UP. As the balloon is compressed by the increased pressure, it gets warmer, again following a line of constant entropy. As it warms, heat transfer from the warmer balloon into the water would also start to occur.

 

[physical1]

As the balloon is compressed by the increased pressure

If you want to learn anything about physics, I suggest you stop using sentences that are doubling up on words. As the balloon is compressed already literally states that there is increased pressure - stating "compressed by pressure" is useless, like a double negative but instead a double positive. I forget the term for this error in English. As a human, I usually look over things like this - but since you nitpicked me, I am returning the favor.

I suggest if you want to have a conversation with a human being, you look in the mirror and nitpick your own pimples.

Are you familiar at all with Bernoulli's equation? (Pressure in a verticle column under gravity varies by rho*g*h)

I suggest if you want to learn physics you spell verticle using the "vertical" word from my dictionary.

As it warms, heat transfer from the warmer balloon into the water would also start to occur.

I mentioned insulation.

It is funny though, I wonder what the lurkers of this forum will think after reading these messages. Everything I said in my first few posts has been ridiculed, and yet everything is turning out to be quite close to what my hypothesis was. Sure I make a few statements in english that are not perfect, like every human. But to have some of my hypothesis(es) repeated back in my ear that were originally ridiculed.. is quite non-surprising actually. I am told to learn physics and what in fact has happened is many others on this forum, some with doctorate degrees, have learned more than I have. Where is that quotation about ridicule and self evidence. I have it stored somewhere. And one about self reflection.

 

[Q_Goest]

My post was not intended to ridicule you. In responding it seems clear you're not familiar with some basics, so showing you what misconceptions you have is just part of explaining the answer. Vacuum doesn't pull, and I'm not trying to poke fun of what you don't understanding.

 

[belliott4488]

physical1 - You've got a few misconceptions about this topic, which are causing the confusion that prompted your original question (as well as some follow-up questions you've posted). People are just trying to help you clear up these misconceptions so that you'll understand the answers, so trying to flame them the with insulting responses is not only rude, it defeats the presumed purpose of your post - to get a satisfactory answer to your question.

1. There is nothing wrong with referring to the creation of a vacuum in a thought experiment, since there is nothing physically wrong with a vacuum (speaking classically, at least, and I assume you're trying to invoke Quantum Mechanics to frame your problem). In practical terms, it is difficult to produce extremely low-pressure vacuums, but you can get arbitrarily close with the right equipment. As for answering your OP, however, a vacuum at the top of the container poses no physical problems; in fact it is what solves the problem.

2. Are you familiar with a water barometer? It's like a regular mercury barometer, but since water is so much less dense than mercury, you need a column of water some 10 m tall (at sea level). Nonetheless, in such a device you have a container with water and a vacuum above the surface of the water.

3. As others have mentioned, vacuums don't suck, and neither does your balloon as it contracts due to the increased pressure. Rather, greater pressure pushes from the other side. I used to pose the following problem in Intro. Physics, which I now pose to you: Suppose you are an astronaut on the surface of the Moon (where there is no atmosphere, and thus a vacuum), and you've brought along a soda in cup. Can you drink from this cup by sucking on a straw that goes from your mouth through a well-sealed port in your helmet and has its other end submerged in the soda?

 

[Doc Al]

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.

I think this "common sense" idea of a vacuum "pulling up" is at the root of the confusion in this thread. Q_Goest and belliott4488 have already expounded on that one.

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.

I'd love to know why you think this. (It's not clear to me at all.)

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.

Again: Why? Why does air need to get inside the container in order for the balloon to 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,

Right.

and the pressure in the balloon is equal to ambient pressure because the water is assumed to be incompressible.

Not sure what you mean by "ambient pressure".

(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.)

The sealed, rigid container is 1000 ft deep (say). Still think the air pressure within the balloon increases "very, very slightly" as the balloon is pulled to the bottom?

Perhaps you mean as the balloon just begins its descent, the first thing that happens is that the pressure between the water and the top of the container (presumed to equal 1 atm, but depends on how the tank was filled) quickly reduces as the water microscopically decompresses.

Then as the balloon continues to be pulled down, the pressure within it quickly increases as the balloon is compressed, which drops the water level. (As you describe, things are a bit more complicated due to the formation of water vapor--but I still don't see that as essential to the main point of this thought experiment. But well-described, by the way.)

Everything I said in my first few posts has been ridiculed, and yet everything is turning out to be quite close to what my hypothesis was.

Not quite, but I can see where you might have gotten that impression.

 

[LURCH]

A vacuum at the top of the container (localized) is hard to visualize for me also because according to Pascal...

This is where I found it helpfull to think about a barometer. It is a real-life example of a (very tall) tube of water with a vacuum at the top. This of course does not violate Pascal, as pressures remain the same across any cross-section of the container at a particular depth. I think that you would agree that Pascal is not saying pressure must be the same from top to bottom, right?

So, going back to the barometer, pressure is greatest at the bottom, and decreases by degrees as one travels farther up the pressure decreases untill, at the very top, there is vacuum. This is exactly the situation that exists within the imaginary container at the end of the thought experiment.

 

[Q_Goest]

I'd love to know why you think this. (It's not clear to me at all.)

Again: Why? Why does air need to get inside the container in order for the balloon to change volume?

In order for the balloon’s air pressure to change, there has to be some change of volume of the balloon. No volume change means no pressure change.

Imagine a tall, verticle, water filled pipe, 12” in diameter with a balloon just under the surface. Now imagine putting a cap on this so the water is up to the underside of the cap but instead of sealing the pipe, we put a hole in it. Now, as the balloon is pulled down the pipe, it shrinks due to the increased pressure and air is pulled in through the hole in the cap. The volume of air coming in through the hole is equal to the volume that the balloon is decreased by, and the balloon volume is decreased according to an isentropic compression of the air inside. The air warms up, the pressure increases and the volume decreases. For every dV of volume decrease in the balloon, there has to be an equal volume of air admitted through the hole in the cap.

Now if you seal the hole and do the same thing, no air can get in. So as the balloon sinks, if it were to be compressed, there would need to be air or something getting into the pipe to take up the decrease in volume of the balloon. But since nothing can get in, the balloon's volume can’t decrease. If the balloon's volume can’t decrease, its pressure can’t increase. If the pressure can’t increase, the water pressure on the outside of the balloon must be in equilibrium with the pressure of the air inside. It’s this last sentence that’s hard to digest. It’s counterintuitive to think of a balloon going down in some water filled column and the pressure of the water filled column changing equally all along the length as it goes down. But we could do the same thing by removing the balloon, adding a piston to the top of the pipe and simply pulling up on the piston. In this case, the pressure at the top of the pipe drops according to how hard we pull on the piston. As the pressure at the top of the pipe changes, the pressure all along the length of the verticle pipe also changes the same amount.

In the case of the piston, there is a small amount of motion needed to change the pressure. It’s going to be extremely small since water is nearly incompressible, but it will be there. And the more water volume you have, the more this piston is going to move. The piston movement is dependant on the bulk modulus of the water.

Similarly, the balloon is decreasing in volume very slighty as it sinks in the pipe because all along the length of the verticle pipe, the pressure is dropping just as it would if you had a piston on top. The small increase in volume we saw with the piston movement will have to be the same as the small decrease in volume that occurs in the balloon. The water column pressure is dropping along the entire length in both cases, and because of the bulk modulus of the water, there is a very small but finite volume increase in the water which is compensated for by either the movement of the piston or the compression of the balloon.

Not sure what you mean by "ambient pressure".

The sealed, rigid container is 1000 ft deep (say). Still think the air pressure within the balloon increases "very, very slightly" as the balloon is pulled to the bottom?

Assuming the bulk modulus of water is always positive (pretty sure that’s a solid assumption), then regardless of what pressure the verticle pipe starts out at, if you increase the volume of the pipe, either using a piston or a balloon, the pressure along the length of the pipe will drop. Similarly, decreasing the volume of the pipe increases pressure, though the change in volume is small for such an incompressible media as water.

Note also that as soon as the pressure at the top of this verticle column drops to the boiling point of the water, it will be forced to boil. Regerdless of whether a piston or balloon is used to change the volume of the container, the water is going through an isentropic process as long as there is no heat transferred from the environment (ie: adiabatic container).

 

[Doc Al]

In order for the balloon’s air pressure to change, there has to be some change of volume of the balloon. No volume change means no pressure change.

Sounds good.

Imagine a tall, verticle, water filled pipe, 12” in diameter with a balloon just under the surface. Now imagine putting a cap on this so the water is up to the underside of the cap but instead of sealing the pipe, we put a hole in it. Now, as the balloon is pulled down the pipe, it shrinks due to the increased pressure and air is pulled in through the hole in the cap.

Sure. (Better to say that the air is pushed in through the hole by the atmosphere, not pulled in, of course.)

Now if you seal the hole and do the same thing, no air can get in.

This is true.

So as the balloon sinks, if it were to be compressed, there would need to be air or something getting into the pipe to take up the decrease in volume of the balloon.

Why is that? (This is the point that I'd like to see physically justified.)

But since nothing can get in, the balloon's volume can’t decrease.

Again, I'm not seeing this.

If the balloon's volume can’t decrease, its pressure can’t increase.

I agree with this.

If the pressure can’t increase, the water pressure on the outside of the balloon must be in equilibrium with the pressure of the air inside. It’s this last sentence that’s hard to digest. It’s counterintuitive to think of a balloon going down in some water filled column and the pressure of the water filled column changing equally all along the length as it goes down.

Quite counterintuitive, if I understand what you're saying. Are you saying that if I pull the balloon down 1000 meters (a difference in hydrostatic pressure of about 100 atmospheres) in that closed container, that somehow that huge water pressure at the bottom is negated and that the balloon is not compressed?

 

[Q_Goest]

Sure. (Better to say that the air is pushed in through the hole by the atmosphere, not pulled in, of course.)

<laughing> Touche, you got me there!

q_goest said: So as the balloon sinks, if it were to be compressed, there would need to be air or something getting into the pipe to take up the decrease in volume of the balloon.

Why is that? (This is the point that I'd like to see physically justified.)

Ok, good question. Note the assumptions we’ve made:
- The container is sealed so that nothing can get in or out.
- Container is infinitely rigid, therefore internal volume can’t change.
- Water is essentially incompressible. Let’s neglect water's bulk modulus for now and add that in later.
- The balloon’s pressure is in equilibrium with the water pressure. If the balloon’s pressure was more or less than the water, it wouldn’t be in static equilibrium and the balloon would have to expand or shrink respectively until it was in equilibrium. By equilibrium, I mean that the pressure on the inside of the balloon is essentially the same as the pressure on the outside.*

Now, let’s assume that as the balloon goes down, it shrinks in volume.
1. Since the container is sealed, neither air nor water can get into the container to make up for the volume left by the shrunken balloon.
2. Since the container is rigid, the container’s volume can’t decrease to make up for the volume.
3. Since the water is incompressible, the water can’t expand to make up for this decrease in volume.
4. All we’re left with is the knowledge that the water outside the balloon must be equal in pressure to the gas inside the balloon. So if nothing in the system can make up for the shrinking balloon volume, and if the pressure inside is the same outside the balloon, then we’re left with the assumption that the balloon pressure hasn’t changed but the water pressure did!

#4 leads us to the realization that the pressure all along the column must change as the balloon sinks. And we also know that when the local water pressure drops below the saturation point, it must boil, and the lowest pressure is at the upper most surface. So when the upper most surface reaches saturation, it must boil which creates water vapor above the column.

Quite counterintuitive, if I understand what you're saying. Are you saying that if I pull the balloon down 1000 meters (a difference in hydrostatic pressure of about 100 atmospheres) in that closed container, that somehow that huge water pressure at the bottom is negated and that the balloon is not compressed?

You’ve left out some assumptions. What should the pressure be at the top of the column when the balloon is at the top of the column?

For example, if the pressure at the top of the column was 15 psia, then the balloon would have to have 15 psia in it to be in equilibrium. Assuming roughly 2 feet per psi for water, then 30 feet down the pressure would be 30 psia. Now if the balloon sinks 30 feet, the pressure at that level has gone to 15 psia and the top of the column has dropped to 0 psia and begins to boil. Until it did that, the balloon couldn’t shrink. But once it passes that point, the water surface begins to boil producing water vapor at a much lower density than the liquid water. Now you have something that can expand and make up for the difference in the incompressibility of the water and rigidity of the container. Now as the balloon sinks, the decrease in balloon volume can be taken up by the increase in vapor space above the water. So in this case, when the balloon is 1000 feet down, it must have shrunken to the size of a pea.

The above is simplified by assuming the bulk modulus of water is zero. But since there is some finite bulk modulus, the water actually decreases in density slightly as pressure drops. Checking a database for water shows that the density decrease between 0.4 psia (aprox. boiling at 70 F) and 15 psia is only about 0.003%. If this is taken into account, we find that the balloon shrinks very slightly with a corresponding increase in pressure and the water expands very slightly with a corresponding decrease in pressure.

* Note that in real life, there is some minor difficulty with this assumption. Rho*g*h of air inside the balloon is not equal to rho*g*h of water outside. The difference has to be made up by internal stresses of the balloon, but that’s inconsequential. One could for example, instead of using a balloon, use a small cylinder containing air with a piston. In fact, it may be easier to visuallize using a cylinder and piston arrangement instead of a balloon.

 

[Doc Al]

Ok, good question. Note the assumptions we’ve made:
- The container is sealed so that nothing can get in or out.
- Container is infinitely rigid, therefore internal volume can’t change.
- Water is essentially incompressible. Let’s neglect water's bulk modulus for now and add that in later.
- The balloon’s pressure is in equilibrium with the water pressure. If the balloon’s pressure was more or less than the water, it wouldn’t be in static equilibrium and the balloon would have to expand or shrink respectively until it was in equilibrium. By equilibrium, I mean that the pressure on the inside of the balloon is essentially the same as the pressure on the outside.*

All good.

Now, let’s assume that as the balloon goes down, it shrinks in volume.
1. Since the container is sealed, neither air nor water can get into the container to make up for the volume left by the shrunken balloon.
2. Since the container is rigid, the container’s volume can’t decrease to make up for the volume.
3. Since the water is incompressible, the water can’t expand to make up for this decrease in volume.

All good.

4. All we’re left with is the knowledge that the water outside the balloon must be equal in pressure to the gas inside the balloon. So if nothing in the system can make up for the shrinking balloon volume, and if the pressure inside is the same outside the balloon, then we’re left with the assumption that the balloon pressure hasn’t changed but the water pressure did!

(1) You assume that something must "make up" for the shrinking balloon volume. Why is that? This is the key assumption that requires justification--at least to me.
(2) As the balloon is pulled down, both water pressure and balloon pressure increase as the balloon is compressed. No mystery there.

(I understand your point about water vapor forming at the surface, but to me that is a side issue.)

 

[Q_Goest]

(1) You assume that something must "make up" for the shrinking balloon volume. Why is that? This is the key assumption that requires justification--at least to me.

V_a + V_w = V_c
where
V_a = Volume of the air (balloon)
V_w = Volume of the water
V_c = Volume of the container

V_c doesn't change.
Water is incompressible, so density remains constant and V_w is a constant
V_a can't change unless there's a change in either the container or water.
Since the container can't change volume, V_a can only change if V_w changes.

 

[Doc Al]

V_a can't change unless there's a change in either the container or water.

I don't see a justification for that statement.

The only volumes that are fixed are V_w & V_c; V_a can certainly change. It happens to be true that V_a + V_w = V_c at the beginning, when the balloon is at the top of the container, but it's not true in general. Generally, V_a + V_w ≤ V_c.

 

[Q_Goest]

V_a + V_w ≤ V_c.

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

 

[Doc Al]

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

Simple: V_a + V_w + Remaining "Empty" Space = V_c.

 

[Dadface]

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

Because Va +Vw+Vs=Vc

Here Vs=volume above the water surface caused by the balloon shrinking

 

[Dadface]

Stretching the thought experiment further if the water was replaced by a perfectly non volatile liquid the space above the water surface would be vacuum.

 

[belliott4488]

V_a + V_w = V_c
where
V_a = Volume of the air (balloon)
V_w = Volume of the water
V_c = Volume of the container

V_c doesn't change.
Water is incompressible, so density remains constant and V_w is a constant
V_a can't change unless there's a change in either the container or water.
Since the container can't change volume, V_a can only change if V_w changes.

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

I'm not sure if Doc Al is really not understanding your objection, of if he's just trying to draw you out. I think I see what you're objecting to, and since I think it's the same thing that was bothering the OP, I'm going to go ahead and address it.

My impression is that you believe that since V_a + V_w = V_c at the start that this must still be true after the balloon has been pulled to the bottom of the container. Moreover, I get the impression you object to the notion that there is a vacuum at the top of the container, of volume equal to change in V_a as the balloon is compressed. Is this correct?

The point I and others have been trying to make is that there is nothing wrong with a vacuum being created as the balloon is compressed by the greater hydrostatic pressure at the bottom of the container - in fact, it must be created in order for the balloon to compress, as you and physical1 have pointed out. This means that when the balloon is at the bottom of the container you have,

V'_a + V_v + V_w = V_c

where V'_a is the (new) smaller volume of the compressed balloon
and V_v is the volume of the vacuum at the top of the container.

I think there's an intuitive resistance to the creation of a vacuum since our experience in life is that it takes a lot of work to lower the air pressure in a container - by pulling on a piston, for example. That's only true if there is atmospheric pressure on the other side of the piston, however. The force you feel is not due to the low pressure "pulling", but rather on the higher atmospheric pressure on the other side "pushing". Air pressure only ever pushes; it cannot pull.

[Note that I am ignoring the presence of some water vapor in the new volume at the top of the container that has been mentioned in this thread. There will undoubtedly be some vapor, so of course it will not be a perfect volume. This has no important effect on the explanation of the outcome of the thought experiment, other than to replace the word "vacuum" with "very low pressure volume of water vapor." Whatever the pressure of this volume - approximately zero, or simply very low - that pressure is added to the pressure of the column of water above the balloon's new depth when calculating the pressure in the balloon.]

 

[Doc Al]

I'm not sure if Doc Al is really not understanding your objection, of if he's just trying to draw you out. I think I see what you're objecting to, and since I think it's the same thing that was bothering the OP, I'm going to go ahead and address it.

I admit that I was trying to draw out the physical reasoning (or lack thereof, as I see it), as I have been throughout this thread. The rest of your post expresses my point exactly. Well done.

 

[Q_Goest]

Simple: V_a + V_w + Remaining "Empty" Space = V_c.

Empty space? If it's empty, it's vacuum. But a vacuum can't be in equilibrium with water, it will boil. For the case of a liquid that can't boil, that's a slightly different issue. It doesn't really matter as long as the surface of the water doesn't boil.

Regardless; let's change the liquid for the sake of discussion. I think this should help shed some light on things.

Thus far we’ve discussed water which has some vapor pressure at ambient temperature that’s well above a perfect vacuum. It’s around 0.3 to 0.4 psia at 70 F.

What if we used a liquid that could not convert to a gas? Vacuum pump oils for example are highly refined, long chain hydrocarbons with a vapor pressure in the micron range. For the sake of argument, let’s just consider a hypothetical liquid such as hypothetical vacuum pump oil that does not boil/does not vaporize at any low pressure. We'll aslo hypothesize that this liquid is completely incompressible so it rules out any bulk modulus affects. I think these affects are probably getting in the way of understanding what's going on anyway, so let's just ignore them.

For our experiment, we use the balloon and this non-boiling liquid. Let’s start at atmospheric pressure at the upper surface, say 15 psia for the sake of argument, and the balloon is at the top of the container. Let’s also assume the liquid is roughly the same as water, so for every 2 feet of head we get 1 psi. So 30 feet down in this liquid, the pressure is 30 psia (ie: rho*g*h +P_s where P_s is the pressure at the surface = 15 psia at the surface, plus 15 psi head).

As we pull the balloon down 2 feet, we might think the balloon shrinks due to increased head. But if it did, we’d need to come up with that volume somewhere else in our system (ie: We may think there's an additional volume = V_v (vacuum volume)).

-> If a vacuum space formed above the liquid, and assuming Bernoulli’s still holds (which it does) then the surface pressure would be 0 psia, so the pressure 2 feet down (per rho*g*h + P_s where P_s =0 for vacuum) would be only 1 psia. But the balloon had 15 psia in it to begin with, hence it won’t be at 1 psia if it shrinks. Pressure has to be greater than 15 psia and the conclusion is it hasn’t shrunk.

In other words, if the surface pressure drops instantly from 15 psia to 0 psia, because there's a vacuum there, we have a sudden change in pressure ALL THE WAY DOWN to the bottom of the container.

So we are left with the balloon not shrinking and not expanding. We are left with the surface pressure decreasing by 1 psi, not suddenly going into vacuum. There is no sudden step change in pressure at the upper surface.

When the balloon gets down to 30 feet, the pressure at the top of the column finally reaches 0 psia (rho*g*h = 15 psi). At this point, a vacuum just starts to form, and the pressure 30 feet down is only 15 psia instead of the 30 psia originally. Everything is still in equilibrium, and the balloon hasn't changed volume yet.

As the balloon goes deeper, a vacuum space finally begins to form above the liquid as the depth of the balloon goes down more than 30 feet and pressure in the balloon exceeds 15 psia. Now, as the balloon goes deeper and contracts, the volume that it decreases by can be made up by the volume of the vacuum at the top of the container.

What everyone is missing is the Bernoulli equation for a static column of liquid, P(at some point below the surface) = P_s + rho*g*h

 

[Doc Al]

In other words, if the surface pressure drops instantly from 15 psia to 0 psia, because there's a vacuum there, we have a sudden change in pressure ALL THE WAY DOWN to the bottom of the container.

So we are left with the balloon not shrinking and not expanding. We are left with the surface pressure decreasing by 1 psi, not suddenly going into vacuum. There is no sudden step change in pressure at the upper surface.

When the balloon gets down to 30 feet, the pressure at the top of the column finally reaches 0 psia (rho*g*h = 15 psi). At this point, a vacuum just starts to form, and the pressure 30 feet down is only 15 psia instead of the 30 psia originally. Everything is still in equilibrium, and the balloon hasn't changed volume yet.

As the balloon goes deeper, a vacuum space finally begins to form above the liquid as the depth of the balloon goes down more than 30 feet and pressure in the balloon exceeds 15 psia. Now, as the balloon goes deeper and contracts, the volume that it decreases by can be made up by the volume of the vacuum at the top of the container.

I agree 100%! In fact I started to make the same point--somewhat obliquely--in post #55. (All the talk about vapor formation must have obscured the issue.) The first thing that happens as the balloon lowers is that the overpressure (the 1 atmosphere at the top of the water) must be reduced throughout the fluid. As I was imagining the balloon being lowered 1000 m, that bit was minimized in my mind as it was resolved in the first 10 m.

So I think we agree after all. And thanks for your patience and clarification. (Whew!)