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At what depth will an open, upside down bottle full of air start to sink by itself?

  1. Oct 8, 2014 #1
    This was an extra question in an exam I took today. The temperature of the air in the bottle remains constant.

    I know this has something to do with buoyancy and ideal gases since temperature was mentioned in the question, and I think I could have easily solved the question, had the submerged object been a solid.

    The bottle starts filling up with water as the hydrostatic pressure increases as the bottle is pushed deeper and deeper into water, so the "average density" of the contents of the bottle start approaching the density of water, as the ratio of water to air increases inside the bottle. The formula for hydrostatic pressure ph=(ro)*g*h has the required depth of water in it. The pressure of air increases as the volume of the air pocket in the bottle decreases. This much I understand.

    However this is as far as I got. The mass of the bottle was something like 0,380kg; and it's volume was 0,55 liters (= 0,55dm3 = 0,00055m3).

    What information do I take advantage of to solve this type of question? My brain is jamming up right now.
     
  2. jcsd
  3. Oct 8, 2014 #2

    Borek

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    How does the volume of the gas inside change with depth? How is this volume related to the buoyancy?
     
  4. Oct 8, 2014 #3
    Well, at any given depth, h, the pressure on both sides of the water-air "interface" inside the bottle is equal, and should be normal air pressure + hydrostatic pressure at that depth: p0 + ph = p0 + roH2Ogh. Since we're dealing with an ideal gas, where T is constant, p0V0 = p1V1. Here V1 = (p0V0)/p1 = (p0V0)/(p0 + roH2Ogh)

    Buoyancy Nair = roH2O * g * Vair, right? So Vair = Nair/(roH2O*g).

    I'm starting to see that these two could be combined to produce V1 = Vair = N/(roH2O*g) = (p0V0)/(p0 + roH2Ogh).

    Solve for h, and we get: N(p0 + roH2Ogh) = (p0V0)(roH2O*g)
    => (Np0 + NroH2Ogh) = (p0V0)(roH2O*g)
    => h = ((p0V0)(roH2O*g) - (Np0) / (NroH2Og)

    With this I'm only taking the buoyancy of the "solid" water block inside the bottle into account, however. The bottle has a weight G = mg, but we can't calculate a buoyancy for it since we don't know enough about it. I now know we need to find a point at which Nair + Nbottle = G, but again, we don't know Nbottle. What to do...?
     
  5. Oct 8, 2014 #4

    Borek

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    Volume and the density of the glass are not given, which most likely means you are expected to ignore them.
     
  6. Oct 8, 2014 #5

    mfb

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    Is this the full, word-by-word problem statement? I think there is a serious problem with it.
     
  7. Oct 8, 2014 #6
    The title is what the problem statement was, although I was paraphrasing because I don't have the problem sheet with me. I only left out (in the title) the given values for the mass of the bottle (for calculating it's weight), the volume of the bottle (and therefore of the air at normal pressure) and the fact that the temperature of the air inside the bottle remains constant at 4 degrees C (another distraction, the only relevant part is the constant temperature, not the specific value), as the water pressure does work on it, as the bottle is pushed underwater.

    Regardless, I'm going to bed now. It's 1am over here and I have work tomorrow. :(

    I'll try working on this later tomorrow, after I've gotten the exam paper back with the original question. Just to make sure i didn't miss anything.
     
  8. Oct 8, 2014 #7
    You're very close to having it solved. You've already done the hard part. As Borek indicated in #4, the volume of glass is negligible, and you already know the weight of the glass. So all you need to do now is to determine the buoyant force, and find the depth at which the buoyant force is equal to the combined weight of the bottle plus water in the bottle.

    Chet
     
  9. Oct 9, 2014 #8

    NTW

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    • Physics Forums Rules and Guidelines prohibit giving the complete solution. We are to help the person work through the solution on his own by giving hints and asking leading questions.
    A rough-and-ready estimate can be made as follows:

    A rule of the thumb, well known to divers, is that pressure increases by 1 atm for every 33 feet (10m). Now, the volume decreases in inverse proportion to the pressure (depth). Hence, the buoyant force F, expressed in kg-f at any given depth in meters is F=55/depth*0,1 => for a buoyant force of 0,38 kg-f, the depth is 55/(0,38*0,1)=1447m
     
  10. Oct 9, 2014 #9

    Borek

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    Without any checks I can tell you this number is orders of magnitude wrong.
     
  11. Oct 9, 2014 #10

    NTW

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    Well, for round numbers, let's take -for example- 1000m, the pressure at that depth is 100 times greater that at the surface; hence, the volume will decrease to 1/100 of the original, so 55 liters become 0,55 liters, that have a buoyant force of 0,55 kg-f. So, at that depth, the bottle doesn't sink, but tends to float. And, at 2000m, the pressure at that depth is 200 times greater that at the surface; hence, the volume will decrease to 1/200 of the original, so 55 liters become 0,275 liters, that have a buoyant force of 0,275 kg-f. So, at that depth, the bottle doesn't float, but tends to sink.

    The equilibrium point is in-between... 1447m is -roughly- right, IMHO...
     
  12. Oct 9, 2014 #11

    Borek

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    Where did you got 55 liters from?
     
  13. Oct 9, 2014 #12
    The volume of the bottle was 0.55 liters, not 55 liters, and that volume doesn't change. Only the volume of air in the bottle and the volume of water in the bottle change.

    Chet
     
  14. Oct 9, 2014 #13

    NTW

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    That's terrible... No wonder you stated that the solution was orders of magnitude wrong...

    However, the approach is right. Repeating it with the right volume:
    (...) the buoyant force F, expressed in kg-f at any given depth in meters is F=0,55/depth*0,1 => for a buoyant force of 0,38 kg-f, the depth is 0,55/(0,38*0,1)=14,47m
     
  15. Oct 9, 2014 #14

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    0,55 liters has to be the volume of the air in the bottle. For the bottle material itself, it would mean a density of 380g/550cm3 = 0,69... Unreal...
     
  16. Oct 9, 2014 #15
    I stand by what I said. The 0.55 liters is the initial volume of air in the bottle. The problem statement doesn't say anything about the volume of glass comprising the bottle, but implies that it is negligible.

    Chet
     
  17. Oct 9, 2014 #16

    mfb

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    Oh sorry, I missed the given mass of the bottle in post #1. If we would neglect this the bottle would never sink, as compressed air does not exceed the density of water at the same pressure.
     
  18. Oct 10, 2014 #17
    Ok, I'm going to continue from where i left off two days ago (#3):

    Set the buoyancy equal to the weight of the bottle: G = Nair => roH2OgVair = mg

    => VAir = mg/roH2Og = m/roH20 = 0,380kg/1000kg/m3 = 0,000380m3

    By substituting this to the original solution for N, N = roH2O * g * Vair = roH2Og(m/roH20) = 1000kg/m3*9,81m/s2*0,000380m3 = 3,7278N (= G = mg)

    Then, by substituting this to the equation solved for height (or depth in this case), we get:

    h = ((p0V0)(roH2O*g) - (Np0)) / (NroH2Og)
    = ((101325Pa*0,00055m3*1000kg/m3*9,81m/s2)-(3,7278N*101325Pa))/(3,7278N*1000kg/m3*9,81m/s2)
    = 4,62075...m ~ 4,6m

    And there you have it. This turned out to be the correct answer as well. I would have written all of this in LATEX, but I figured it would have been too time consuming from my perspective, having to learn it again after a long break. I hope this is readable to you guys.
     
  19. Oct 10, 2014 #18
    I should add, that only the volume of the air inside the bottle matters, since it's buoyancy is doing the lifting, and Nair decreases as Vair decreases, as the theoretical pressure difference between the top and the bottom of the "solid" air block decreases. That was the thing I was supposed to figure out in the beginning.
     
  20. Oct 10, 2014 #19

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    Let's see... Verifying the solution by a different approach... At 4,6m, and if we assume that the pressure rises linearly with the depth, at a rate of 1 atm for every 10 meters, the increase of pressure for 4,6 meters will be 0,46 atm. At the surface, we have P*V=k; i.e. 1*0,55 = k = 0,55. Solving for V at 4,6 meters mean V=0,55/(1+0,46) => V=0,38liters. As that volume of water weighs precisely 0,38kg-f, the bottle neither sinks nor floats...
     
  21. Oct 10, 2014 #20

    Borek

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    Somewhere between 0 and 10 meters is what I expected just by the logic (at 10 meters there will be only a half volume left, and it will be already less than 0.38 L which is more or less volume needed for 0.38 kg).
     
  22. Oct 10, 2014 #21

    Orodruin

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    Just to note, even if it is assumed negligible in the given problem, it would not be negligible in a real world setting. Typical glass has a specific density in the region of 2.5, which would mean that the buoyant force on the glass itself would cancel around 2/5 of the weight of the bottle and only 3/5 of the weight would have to be provided by the buoyant force of the air inside.
     
  23. Oct 10, 2014 #22
    The point of the question was to find the point at which the bottle stays still in the water. Pushing it any further than that is going to cause it to sink, as the volume of the air is no longer sufficient enough to cause enough buoyancy.

    2,5 is the density for (window) glass given in the book of formulas and tables we were allowed to take with us into the exam. However, it is not stated, that the bottle is made of window glass, so the density could be different for the bottle material. There's also the thing about this being a high school (=Finnish lukio) level physics problem, so it can be expected that some things are to be ignored. The answer sheet handed to us yesterday states as much.

    EDIT: Just to clarify, I finished this problem without looking at the answer sheet. I'm no cheater. :)
     
    Last edited: Oct 10, 2014
  24. Oct 10, 2014 #23

    NTW

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    Of course... But that point is one of unstable equilibrium. A little less deep, and the bottle will float away... A little deeper, and the bottle will sink to the bottom...
     
  25. Oct 12, 2014 #24

    Orodruin

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    Yes, this is why I referred to a real world setting - just in case someone decides to try this out. (It could be quite instructional in scuba diving training to emphasize the importance of neutral buoyancy with an empty BCD.) Glass seems to be a reasonable assumption given the weight and size of the bottle.

    Regardless, I would like to personally lecture any teacher marking such a consideration wrong in the importance of actually thinking than just applying what the teacher expects.
     
  26. Oct 13, 2014 #25
    The thing is, the bottle is what really threw me off with this question. I knew it must have buoyancy, but the required information to calculate it wasn't given, so I became totally confused and thought there must be something else to the question, and got stuck. Then again, I'm not sure I'd know how to calculate buoyancy for open, hollow objects at this stage anyways.
     
    Last edited: Oct 13, 2014
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