BR question, Pressure inside balloon in water?

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justhanging

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The question stem basically says that a balloon goes from the surface into a tank of water at a uniform rate. Then asks how the volume of the balloon changes as you descend down the water?

A. The volume of the balloon decreases linearly
B. The volume of the balloon decreases a little at first than gradually more as it descends
C. The volume of the balloon decreases considerably at first than gradually less as it descends.

My understanding of the problem is as follows, and maybe someone can explain where my reasoning is wrong. The answer C btw.

First the pressure outside the balloon has to be the same as the pressure inside the balloon because the force outside and inside have to be the same. At the surface the balloon has same pressure as the atmosphere but it as descends down its pressure is equal to that of the fluid. Using bernoullis equation the pressure of the fluid changes as a function of height(depth) by:

Pressure of atm + DensityFluid*gravity*depth = Pressure of fluid
P(atm) + d*g*h = P(fluid)

if the pressure inside the balloon is the same as it is in the fluid than the pressure inside the balloon increases linearly as depth increases. So back to the original problem of how volume of the balloon changes, using
PV = nRt if pressure changes linearly than volume must do so as well, all else constant.

Where did I mess up in my understanding?
 
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what does the solution manual say about this question?

It is a CBT. It basically gives a bunch of numbers but doesn't say where it came from.
It saids:

As a balloon attached to the belt descends into the water, the relative change in gas pressure inside the balloon is greatest in the beginning and less as the balloon descends further into the water. For example, if the balloon has a 1.00 atmospheric pressure at the surface, then at a depth of 10 meters, the pressure will double to 2.00 atmospheres. It must reach a depth of 30 meters for the pressure to double again (to reach 4.00 atmospheres). This means that the volume does not decrease in a linear fashion, but instead shows the greatest change during the initial portion of the descent, and then less change as it continues to descend
 
My humble brain thinks that pressure does change linearly as you descend further due to

P(total)=P(atm)+pgh

However, due to pv=nrt, pressure and volume have an inverse relationship and not a linear one. You have to rephrase the equation as p=1/v.

or v=1/p, so if you graph this out, as pressure increase, volume decrease, non linearly.

I think this is correct unless somehow pressure of water increases exponentially as you go deeper, but my current understanding is that the pressure increases linearly as you go deeper.
 
My humble brain thinks that pressure does change linearly as you descend further due to

P(total)=P(atm)+pgh

However, due to pv=nrt, pressure and volume have an inverse relationship and not a linear one. You have to rephrase the equation as p=1/v.

or v=1/p, so if you graph this out, as pressure increase, volume decrease, non linearly.

I think this is correct unless somehow pressure of water increases exponentially as you go deeper, but my current understanding is that the pressure increases linearly as you go deeper.

Ahh I see, so pressure does change linearly but the volume doesn't. Good job zoner you expaination was much better than berkeley's.
 
Nice explanation.

You can also try using reasoning without formulas.

If you exert a crap load of pressure on something it will decrease the volume considerably. If you exert a second crap load of equal value pressure, do you expect to see twice the delta V as the original exertion?

You can apply this logic to squeezing a tennis ball in your hand.
 
Nice explanation.

You can also try using reasoning without formulas.

If you exert a crap load of pressure on something it will decrease the volume considerably. If you exert a second crap load of equal value pressure, do you expect to see twice the delta V as the original exertion?

You can apply this logic to squeezing a tennis ball in your hand.

True. You can also reason that if the volume decreases linearly than there will be a pressure where the volume is zero which is impossible. It should be as pressure approaches infinity, volume approaches zero. Which fits the equations.
 
This passage has some serious flaws.. First of all the whole set up is physically impossible, because there's no downward force acting on the system (only the upward buoyant force). The right arrow pointing downward in the diagram misleads readers to think that there is downward force on the right side (assuming a clockwise rotation) somehow exerted by the buoyant force on the left side, but it doesn't account for the buoyant force on the right side which is also present. In other words, even if some initial force is applied, the system will come to a stand-still (equilibrium) after a while because all of the forces (buoyant force) are pointing up.

Having said that, assuming that the system somehow magically keeps moving , the answer for prob #41 still does not make sense. The answer says, 1. balloon will (try to) expand due to heat transfer, but 2. hot water will expand and 3. water pressure will keep it down so the balloon will not be able to expand and thus will be less buoyant because it's 2 reasons (less buoyant balloon) vs 1 reason (more buoyant balloon). This answer is simply wrong and unacceptable. (The whole 2 reasons vs 1 reason idea is just ridiculous without comparing the degree of effects of each. Yes, I know that the answer was trying to teach us to "pick and answer and move on", but that is not a valid reason to move on whatsoever) First of all, the hot water, being a liquid, will not expand much, resulting in a negligible change in water density -> negligible change in the buoyant force of the balloon. That leaves us with a question of whether the change in force (pressure) exerted by the air inside the balloon due to heat transfer from hot water to the balloon (which should be pretty quick because heat transfer in this case simply depends on the thermal conductivity of the very thin layer of rubber) is greater or less than the force exerted by the denser water on the exterior of the balloon (i.e. pressure). While it is true that the force exerted by the water will keep the volume of balloon down (hence less buoyant) at meaningful depths, near the surface, it will not be much. In other words, at or near the surface of water, the balloon could actually expand due to coming in contact with hot water, increasing its buoyancy, depending on how fast the belt is moving and whether the delta t (time it takes from balloon coming in contact with water until it is fully submerged) is long enough for the heat from the hot water to transfer to the air in the balloon and also depending on the size of temperature gradient. It would also depend on the thermal conductivity of the rubber layer (balloon), heat transfer coefficients, composition of air, elasticity of the balloon, atmospheric temperature and pressure, etc. but those would be beyond the scope of testing purported by MCAT anyway..

Having said all that, the most serious flaw of this answer comes from the fact that the answer does not answer the problem. The question asks for the effect of hot water on the balloon's buoyancy. Assuming the author did not know about the "water surface scenario" I presented above, the effect of hot water on the balloon's buoyancy should be negligible. First, the liquid water does not expand much, and even if it did so by a marginal amount and "tried" to make the balloon less buoyant, the pressure that it would exert on the balloon would decrease as well (term rho*g*h in Bernoulli equation) due to the same "expansion" of the water, counteracting the former effect.
 
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This passage has some serious flaws.. First of all the whole set up is physically impossible,

EXACTLY! That is the point of the passage. It is one big flaw. There is no such thing as perpetual motion, and you need to know that. It is physically impossible, and test takers are supposed to use that knowledge to see the flaws in the design. There are sentences like "One design that has been suggested for a perpetual motion machine" and "The inventor of the machine claims..." These should be your clue that it's a science passage with verbal reasoning qualities. Those type of passages happen, and if you get so upset at their infeasibility that you miss big picture, then you'll undermine your exam. It's like a question someone posted about a few years back where the author said that carbon formed five bonds and wanted to know about hybridization and bond angles. We all know it's false, but for the context of the passage you needed to (a) answer the questions based on this false assumption and (b) recognize which questions expected you to know that it was invalid.

because there's no downward force acting on the system (only the upward buoyant force). The right arrow pointing downward in the diagram misleads readers to think that there is downward force on the right side (assuming a clockwise rotation) somehow exerted by the buoyant force on the left side, but it doesn't account for the buoyant force on the right side which is also present. In other words, even if some initial force is applied, the system will come to a stand-still (equilibrium) after a while because all of the forces (buoyant force) are pointing up.

The system will come to rest, as will be true of all proposed perpetual motion machines, because of resistive forces (not because of the equilibrium you proposed). Energy will always be lost from these systems, so they'll come to rest. It has nothing to do with equal upward forces as you've proposed. You need to reconsider your perspective. You are correct that the buoyant force on the right side is equal in magnitude as the buoyant force on the left side, which means there is no net force (forces are in equilibrium). That does not mean it will come to a stand-still. Newton's First law tells us that an object in motion will remain in motion until a force acts on it. The buoyant forces are all upward (as you stated) and equally displaced on the right and left (as you pointed out), so the net buoyant force is zero and will neither speed up nor slow down the system. What will slow it down is the drag force, which opposes the direction of motion. The drag force will bring the system to a stand-still.

Having said that, assuming that the system somehow magically keeps moving , the answer for prob #41 still does not make sense. The answer says, 1. balloon will (try to) expand due to heat transfer, but 2. hot water will expand and 3. water pressure will keep it down so the balloon will not be able to expand and thus will be less buoyant because it's 2 reasons (less buoyant balloon) vs 1 reason (more buoyant balloon). This answer is simply wrong and unacceptable.

Huh? Let's slow down a bit and reason through the choices. A few things to consider. First, the machine is fixed in terms of its position. Second, water expands when heated, so the water level will rise as it is heated and the machine will be more submerged. Choice A states "Less of the apparatus will be in the water.", which is clearly false. So choice A is eliminated without any argument.

Let's skip to choice C, "Each balloon will deform into a shape that maximizes its surface area." This has nothing to do with water temperature and all to do with minimizing forces acting on the object. A balloon in motion is free to contort in a way to minimize the drag force acting on it. This will not lead to maximization of the surface area with greater water temperature. There is no reason to choice C.

Choice D states "The belt will no longer be in contact with the freely turning wheels." If anything, by heating up the water you in turn heat up the belt, which makes it expand and rub more against the belt. It's certainly not going to break free from the wheel. Choice D is out.

So this leads us to look at choice B, "The balloons will become less buoyant." The buoyant force on the balloon depends on three things, g (which did not change), density of the surrounding medium (which decreased with heating), and the volume of the submerged balloon (which is open to debate as we are not given enough details in the passage). You can argue the volume of the balloon as (a) increasing due to the air temp inside increasing and (b) decreasing, due to being more submerged in the water and thus subject to a greater pressure. The volume change is an unknown and we don't have enough information, but we definitely know that the density of the water decreased, so that would lead us to believe that the balloons would be less buoyant.

That is as good as we are going to do on this question, and that is EXACTLY what the explanation is telling you. This question serves two purposes: (1) to address some physics principles and (2) try to convince you to cut your losses and make a best choice on a couple questions. You have become angry at a question, and the impact it can have on you is devastating. That's EXACTLY the point of this question. Let go of your rage and see the bigger picture here. Choice B is the best answer, and as the answer explanation states, pick it and move on. I really want to quote the answer explanation, because it tells you what the point of this question was.

  • "You aren’t looking for an absolutely perfect answer. You are seeking the best answer, and given that choices A, C, and D are not it, choice B must be the best answer despite some uncertainty. The MCAT will have some questions with ambiguity, so it is important that you get to a point where you ignore the ambiguity and select the answer they want you to choose as quickly as possible, and then move on.

(The whole 2 reasons vs 1 reason idea is just ridiculous without comparing the degree of effects of each. Yes, I know that the answer was trying to teach us to "pick and answer and move on", but that is not a valid reason to move on whatsoever)

It's not ridiculous, and is actually a huge lesson test takers need to learn. Spending excess time on a question like this is the kiss of death on a timed standardized exam.

First of all, the hot water, being a liquid, will not expand much, resulting in a negligible change in water density -> negligible change in the buoyant force of the balloon.

How much did the water heat up? How much does water expand when heated by x degrees? You say the expnasion of water is insignificant, but without knowing the actual temperatures, how can you say this? The first question cannot be answered, because the info was not given. The second question is one you need to consider more carefully before deciding that the density of water does not change to any significant extent. Water at 100 degrees C has a density of 0.958 g/mL while water at 25 degrees C has a density of 0.997 g/mL. That's a roughly 4% change in density.

That leaves us with a question of whether the change in force (pressure) exerted by the air inside the balloon due to heat transfer from hot water to the balloon (which should be pretty quick because heat transfer in this case simply depends on the thermal conductivity of the very thin layer of rubber) is greater or less than the force exerted by the denser water on the exterior of the balloon (i.e. pressure). While it is true that the force exerted by the water will keep the volume of balloon down (hence less buoyant) at meaningful depths, near the surface, it will not be much. In other words, at or near the surface of water, the balloon could actually expand due to coming in contact with hot water, increasing its buoyancy, depending on how fast the belt is moving and whether the delta t (time it takes from balloon coming in contact with water until it is fully submerged) is long enough for the heat from the hot water to transfer to the air in the balloon and also depending on the size of temperature gradient. It would also depend on the thermal conductivity of the rubber layer (balloon), heat transfer coefficients, composition of air, elasticity of the balloon, atmospheric temperature and pressure, etc. but those would be beyond the scope of testing purported by MCAT anyway.

At least you are recognizing that you have extended beyond the scope. This question, to be fully answered, would take the analysis of several factors, and that's not what you should be doing on the MCAT. Apply basic POE and make a good choice. Your goal is to get into medical school, and getting as high of a score on the MCAT as possible helps your goal. This is achieved by doing enough on each question to arrive at a best answer quickly.

Having said all that, the most serious flaw of this answer comes from the fact that the answer does not answer the problem. The question asks for the effect of hot water on the balloon's buoyancy.

No, the question asks "As the temperature of the water in which the machine is suspended increases, what change will occur in the system?" You are given four choices, three of which are clearly wrong and one of which is ambiguous with the chance of being true. Your job is to choose the best answer to the question presented.

Assuming the author did not know about the "water surface scenario" I presented above, the effect of hot water on the balloon's buoyancy should be negligible. First, the liquid water does not expand much, and even if it did so by a marginal amount and "tried" to make the balloon less buoyant, the pressure that it would exert on the balloon would decrease as well (term rho*g*h in Bernoulli equation) due to the same "expansion" of the water, counteracting the former effect.

You are correct that there are factors going in opposite directions, which is exactly what the answer explanation was pointing out. You've decided the density change of water is neglible, when in fact it's not. But even if the temperature change is small and the density change is therefore small, there is a flaw in your logic. You've only considered the change in rho in the gauge pressure. Keep in mind that height has also changed, because the water level went up while the balloons remain mounted to the machine which hasn't moved. Perhaps if you consider what the gauge pressure is based on it will be more clear. It comes down to force per unit area. Because of expansion, the water level is rising, so the center of mass for the water has gone up. This means that every balloon in the water (near the surface and at maximum depth) will all be underneath a greater mass of water. This greater weight/area results in a greater external pressure that would reduce the volume of the balloon. This in turn would reduce the buoyancy.

The point is that the there are so many what ifs, and between the two of us and the author, there are no doubt a few we have missed. But it's beyond the scope of the question and purpose of the MCAT. No matter what we conclude, choice B (an ambiguous but likely valid statement) beats three invalid statements.
 
Thanks Berkeley Review for the prompt response. It's good to see that you guys stand behind your product. Continuing with your response though, there are some aspects that I disagree which I am not going to point out because there is really no point to do so anyway. However, I will just provide a crude example following your logic presented above:

I said that the balloon will expand near the surface of the water. Above the surface and at the surface (not fully submerged), it is intuitive that the balloon will enlarge if the heat gradient is large enough. Below the surface is a slightly trickier issue because of the denser water that is exerting pressure on the entire balloon. Or is it?

Following your example of where the water is heated from room temperature (298K) to a boiling temperature (373K), it is easy to do a crude calculation to find the volume of air inside the balloon.

Let's say we're trying to find the volume of the balloon at 1m depth from the surface. This is roughly equivalent of 0.1atm additional pressure due to water exerted on the balloon.

First, using P1V1 = P2V2, we can find the volume of the balloon at 1m depth at constant temperature. Assuming we start with 1L of air in atmosphere at room temperature, since we know that roughly 1m of water adds 0.1atm of pressure,
1atm*1L = 1.1atm*xL, where x = 0.91L of air inside balloon under 1m of water at 298K. This is what the volume of balloon would be at room temperature. Notice the volume decreased due to water pressure as it should have been.

Now, using P1V1/T1 = P2V2/T2, let's find the volume of the balloon at 373K 1m deep in the water. 1.1atm*0.91L/298K = 1.1atm*xL/373K, where x = 1.14L. That is roughly 25% increase in volume under water, or 1/1.25 = 80% of the original density -> 100%-80%= 20% reduction in density of air in the balloon, leading to increased buoyancy of the balloon. This is significantly larger than the decrease in density of the water which you suggested (4%). That is what I meant by negligible increase in water volume (relatively speaking).

Of course, we can again go into details like water height increasing and etc., but what I showed you was just a big picture.

You can see that the "effect of hot water on balloon's buoyancy" is in fact always "increased balloon buoyancy", not just at the surface like I previously mentioned but everywhere in the water column, because the water pressure (which by the way is same as the balloon pressure) will stay constant throughout the temperature change, if not decrease (due to the density decrease as you mentioned).

In other words, the density decrease of water can never exceed the density decrease of air inside balloon to promote "decreased buoyancy". And the answer choice B is not only partially incorrect, but entirely incorrect.


The ONLY way for decreased balloon buoyancy would be if the air inside balloon were unable to expand due to water pressure. But keeping everything else constant (such was water depth), which we definitely should, and changing only the temperature of water, the balloon will ALWAYS increase in volume, a lot more so than liquid water will increase in volume. Hence, the buoyancy of the balloon will ALWAYS increase.
 
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Thanks for taking the time to respond Kupa. Exhanges like these are always a good thing and I genuinely want to say thank you. You bring up good points and I have a feeing in the end that neither of us will budge on our view, because we are locked into seeing the ambiguities in the question in our own way. There is definitely ambiguity in this question, and I would ventrure to guess that if someone were to take the time to analyze each one, you'd find that it would all come down to the type of rubber the balloon was made out of.

And the truth of the matter is that in the last few years that I've tutored, no one has called out the flaws in this question so eloquently. I sincerely want to thank you and confess that after our exchange, I'll strongly suggest to my bosses that they add a few more terms into this question to make it work more clearly. To be perfectly honest, I had issue with the fact that water doesn't expand from 0 to 4 degreesC (it contracts), but I figured they'd shoot it down given that the purpose of their question was to emphasize POE and speed rather than physical science concepts.

I want to adjust for a few factors that were ignored in your Charles' law approach to the question, which will once again introduce ambiguity to the point where an ideal version of that law is not enough. I'm going to apply a middle value to my factors, which will hopefully give us a centered value to argue from.

For simplicity, I've chosen 27 degreesC and 57 degreesC, so the temperature change is 10%. This temperature change would result in a 1.8% change in the water density, and thus a rise in the water height of 1.8%. I'm going to consider a balloon exactly in the middle of the system, which will feel a 0.9% increase in the mass of water above it (assuming that if the top of the water increases by 1.8% and the bottom stays at the same point, then the middle goes up by 0.9%). I'm going to modify this number later to account for the rise in the water level caused by the balloon expansion.

Your assumption is that V1/T1 = V2/T2 applies for this system. In an ideal piston that exhibits no resistance to expansion or contraction, filled with an ideal gas, will follow that law. However, in a balloon that exhibits a restoring force when deformed, that law no longer applies. To what extend it deviates depends on the thickness of the balloon and the material of which it's made. As I'm sure you have personally experienced, it gets harder and harder to blow up a balloon as you continue to fill it. To look at the impact of the rubber and material thickness, consider blowing up a basketball or bike tire. They fill easily at first, but get much, much harder to expand as you pump more and more air inside, eventually reaching a point where it just won't expand anymore. In this question we are not told the degree to which the restoring force of the balloon increases, but at the surface of the water the restoring force will be its greatest (as the balloon is closest to maximum volume). The point is that your assumption of a 25% change in volume is for an ideal system and is over estimated in this case (to what extent would depend on the material and gas). Again, though, without those details in the question, we can't find an exact answer. If the balloon didn't expand, then it have a delta V of 0%. If it expands to the maximum (with a heating from 27 to 57 in degreesC), it would be a deltaV of 10%. It obviously expands because of the passage, so a 5% expansion for an ideal gas is a middle ground number we can use.

Second, the expansion of the submerged balloon you speak of will cause an additional increase in the water level (again we don't know to what extent), so the balloons end up more submerged not only because of the reduction in water density, but also because of the submerged balloons expanding and pushing the water up. There is no way to know the exact amount without knowing the dimensions of the tank and volumes of the balloon, but it's a factor to some extent. We'll make it a small one.

These two factors are significant enough to support a realistic scenario where the balloon becomes less buoyant. If we consider the 27 to 57 change I proposed, then the following are realistic numbers: 10% max expansion with no restoring force issue with the balloon will be reduced based on the material's natural restoring constant and the magnitude to which is extends. Let's use F = kx to determine our restoring term. If the volume were to increase by 10%, then x would increase by 2.1% in each direction. Now this again is ideal and we don't know k, but let's just say it reduces the expansion to 1.1% in all directions (about half of the maximum). 1.011^3 = 1.033. So that gives us an expansion of the balloon of roughly 3.3%. If the gas behaves ideally, we'll get 3.3%, and if it doesn't it would be more in the area of 3.0%. Again, this is an estimate that can't be amde without knowing something about the gas. Deviations from ideality can be as large as 20% and as small as 0.000001%, so I went with a 10% decrease. This gives us a 3.0% increase in volume that could be larger or smaller, depending on the exact system.

So let's say the water height increases by the 1.8% due to the change in water density and for simplicity add 0.2% for the impact of the expanding balloons. This gives us a 2% increase in the water level, which will amount to a 1% increase in external pressure on the average balloon (in the middle), and thus a drop of 1% in the volume (using 1.0 for the density of water to make the gauge pressure calculation easier).

So now we can use Archimedes' Principle to see how the buoyancy changed. The volume of the balloon goes up by about 2.0% (again, this assumes an average restoring term, a 10% maximum increase in volume due to temp when we ignore other factors, a 1% decrease in volume due to pressure when we ignore other factors, and an average deviation from ideal gas behavior).

B = rhowater x Vdisplaced x g

rho goes down by 1.8% and Vdisplaced goes up by 2%.​

I think I have been fair in taking the middle value for every assumpiton, and in the end the two factors are so close that no definite conclusion can be drawn. I am certainly not disputing that there is ambiguity in this question, and neither is BR if you read through their explanation. You have convinced me that they need to put a restoring term for the balloon and the exact temperature change in their passage or question.

Hopefully you are convinced that there is enough ambiguity that it could go either way with choice B.

So let's just say this was an exact MCAT question on your test. Having completely over-analyzed this question as much as we have, what do you think the best answer is? Not a correct answer, because I think we can all agree (including BR if you read their explanation) that there is no correct answer. But what is the best answer?
 
oh man!

So is my reasoning incorrect?

Your reasoning is fine and exactly what you should be thinking.

Kupa and I are arguing over a different question. If you read the explanation to #41 it should explain why there can be an argument about what is correct.
 
Your reasoning is fine and exactly what you should be thinking.

Kupa and I are arguing over a different question. If you read the explanation to #41 it should explain why there can be an argument about what is correct.

Thanks for acknowledging the need for modification. Your responses really show that you guys try to stand behind your product. Aside from the elasticity of the balloon which is a counteracting factor to increased balloon volume (I stated this in my first post), significant depth could also prevent the gas from behaving ideally (say, 10000m below surface, at 1000atm), so there are certainly some instances where the buoyancy of the balloon would decrease, especially when the temperature gradient is not large enough.

The purpose of my post was to show that without going beyond the scope of knowledge required by MCAT, the balloon would intuitively gain more buoyancy through increased temperature.

This problem did teach me one thing that i should not overthink in the actual MCAT, since the actual MCAT would not be 100% invulnerable to such errors either. In that aspect, it's been a valuable lesson. Keep up the good work Berkeley Review.
 
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