# funny balloon thought experiment

by physical1
Tags: balloon, experiment, funny
 PF Patron P: 1,887 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.
 Mentor P: 40,240 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.
Mentor
P: 21,652
 Quote by Dadface 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.
Mentor
P: 21,652
 Quote by 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.
That "something" is the air pressure inside the balloon.
Mentor
P: 40,240
 Quote by russ_watters 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.
PF Patron
P: 1,887
 Quote by 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.
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.
P: 2,501
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
 Quote by Doc Al 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.
 PF Patron HW Helper Sci Advisor P: 2,532 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.
 PF Patron HW Helper Sci Advisor P: 2,791 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.
 PF Patron HW Helper Sci Advisor P: 2,532 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?
PF Patron
HW Helper
P: 2,791
Hi Mapes,
 Quote by 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.
 P: 4,778 The airplane takes off. Q.E.D.
P: 41
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.

 Quote by Q_Goest 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.
 P: 41 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.
PF Patron
HW Helper
P: 2,791
 Quote by 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.
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.

 Quote by physical1 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!

 Quote by physical1 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 accomodated by the increase in the balloon’s air density.

 Quote by 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.
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.
P: 41
 Quote by Q_Goest 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.

 Quote by Q_Goest 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.

 Quote by Q_Goest 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.
 PF Patron HW Helper Sci Advisor P: 2,791 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.