Calculate the depth to which a balloon full of Kr must be pushed underwater to make it sink to the bottom of the sea

  • #1
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Homework Statement:
Calculate the depth (in m) to which a balloon full of Kr must be pushed underwater to make it sink to the bottom of the sea. It takes place at 25 degrees Celsius, we know that p (atm) = 101 325 Pa, density (of sea H2O) = 1.04 g/cm3, g = 9.81 m/s2 and the M (Kr) = 83,8 g/mol, and that Kr acts as an ideal gas.
Relevant Equations:
p1*V1 = p2*V2
Hello.

Firstly, I've calculated the density of Kr ( = 3.74 g/dm3), and I know that the p (fluid) = ρ * h * g. And then I've used the following equation: p1*V1 = p2*V2, and therefore: p1*V1 = ρ * h * g * (m/ρ) => p1*V1 = h * g * m. (h = 3.0153 m) Is that correct? Please, how could I calculate the new density of Kr in the balloon? (To make sure my answer is correct.) Does anyone have any ideas?

Thank you!
 

Answers and Replies

  • #2
Lnewqban
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Welcome! :)

As the balloon is pushed down, the external pressure increases linearly, simultaneously reducing the volume and displaced water and bouyancy force resisting the pushing force.
All that happens at constant temperature and mass (and weight) of the gas.
I would find the point of balance between weight of the balloon plus gas and buoyancy force, then what volume and pressure correspond to that weight.
 
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  • #3
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Thank you! I've just realised what the problem was with my calculation and equation. Could you, please, illustrate how this idea of yours should be realised?
 
  • #4
kuruman
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Please, how could I calculate the new density of Kr in the balloon? (To make sure my answer is correct.) Does anyone have any ideas?
Since water just barely floats in water (:rolleyes:), Kr would have to have the density of water to float in it, no?
 
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  • #5
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Since water just barely floats in water (:rolleyes:), Kr would have to have the density of water to float in it, no?
Yes, but what do I make of that?
 
  • #6
kuruman
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How does the density of an ideal gas, which has constant mass, depend on pressure when the temperature is constant?
 
  • #7
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The higher the pressure, the higher the density of the gas. p1V1 = p2V2, p1V1 = p2(m/ρ). But we don't know p2. Is that the right direction?
 
  • #8
kuruman
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You don't know ##p_2##, but if you don't need to know it because you are not really interested in it. You are interested in the depth ##h## at which ##p_2## is such that the density of the gas matches the density of water. You can replace it with ##p_0+\rho_{\text{water}}~g~h## in any equation where it appears because that's what it is if the gas is in equilibrium. Here, ##p_0## is the pressure at the surface of the liquid, i.e. atmospheric pressure.
 
  • #9
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Well, then it may be: p1V1 = (p0 + ρgh)*(m(Kr)(Kr)). But I still can't calculate the depth or the density...
 
  • #10
kuruman
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Well, then it may be: p1V1 = (p0 + ρgh)*(m(Kr)(Kr)).
That's a good start but the equation is misleading. What does ##p_1## on the left stand for? In that expression, there are three variables that change as the balloon is pushed in deeper. One is the depth ##h## itself, and the other two are the gas pressure ##p## and the gas density ##\rho_{\text{Kr}}##. Solve your expression for ##\rho_{\text{Kr}}## in terms of ##h##. Don't forget the ideal gas law to replace ##pV##.
 
  • #11
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p1 is the pressure inside the balloon. Do you mean I have to replace pV with pV/nRT?
 
  • #12
kuruman
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p1 is the pressure inside the balloon.
How does that differ from the pressure outside the balloon when the gas is in equilibrium? You can only replace pV with pV/1.
 
  • #13
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Thanks for your effort, but I don't seem to understand this concept.
 
  • #14
kuruman
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Thanks for your effort, but I don't seem to understand this concept.
Precisely which concept is this? I will explain it to you if you let me know what bothers you.
 
  • #15
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I just can't get to any reasonable equation that would solve my problem. I know the density increases as the pressure does, but how does that help me? I've calculated the pressure in the balloon, and I know the mass of the gas inside (1 mole of Kr weighs 83,8 g, therefore, it's 22,41 l of Kr in the balloon. But that's all I know so far.
 
  • #16
Steve4Physics
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Thanks for your effort, but I don't seem to understand this concept.
Hope I’m not butting-in. Maybe this will help without giving away too much.

The information in the question is not entirely self-consistent so I'm guessing that you are meant to assume that the pressures inside and outside the balloon are equal, For example, this means the initial krypton pressure at sea-level is 1 atm.

For brevity, call the balloon+krypton ‘B’. And assume the mass of the balloon itself is negligible compared to the mass of the krypton.

Some questions…

Q1. You give the density of sea-water as 1.04g/cm³. What can you say about;

a) the density of an object that floats in sea-water?

b) the density of an object that sinks in sea-water?

c) the density of an object that’s on the borderline between floating and sinking in sea-water?
(This is what @kuruman nicely described as ‘barely floats’. It is called ‘neutral buoyancy’.)

Q2. Complete this sentence:
For B to sink in sea-water, B’s density must be greater than __________.

Q3. What volume must B have, in order to have that density?

Can you then take it from there?
 
  • #17
Steve4Physics
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I just can't get to any reasonable equation that would solve my problem. I know the density increases as the pressure does, but how does that help me? I've calculated the pressure in the balloon, and I know the mass of the gas inside (1 mole of Kr weighs 83,8 g, therefore, it's 22,41 l of Kr in the balloon. But that's all I know so far.
In your orginal (Post #1) question, there isn't enough information to find the amount of krypton. You will need to post the complete question.
 
  • #18
hutchphd
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In your orginal (Post #1) question, there isn't enough information to find the amount of krypton. You will need to post the complete question.
The amount doesn't matter!
 
  • #19
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Hope I’m not butting-in. Maybe this will help without giving away too much.

The information in the question is not entirely self-consistent so I'm guessing that you are meant to assume that the pressures inside and outside the balloon are equal, For example, this means the initial krypton pressure at sea-level is 1 atm.

For brevity, call the balloon+krypton ‘B’. And assume the mass of the balloon itself is negligible compared to the mass of the krypton.

Some questions…

Q1. You give the density of sea-water as 1.04g/cm³. What can you say about;

a) the density of an object that floats in sea-water?

b) the density of an object that sinks in sea-water?

c) the density of an object that’s on the borderline between floating and sinking in sea-water?
(This is what @kuruman nicely described as ‘barely floats’. It is called ‘neutral buoyancy’.)

Q2. Complete this sentence:
For B to sink in sea-water, B’s density must be greater than __________.

Q3. What volume must B have, in order to have that density?

Can you then take it from there?
The density of Kr must be higher than the density of seawater. And it's about 3,74 kg/m3 compared to that of seawater 1040 kg/m3. To question 3 - the original volume is molar volume, so 22,41 dm3. But I truly don't know how I should calculate the new density or the corresponding depth. I know it has to do with the ideal gas law, buoyant force, hydrostatic force...
 
  • #20
Steve4Physics
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The amount doesn't matter!
In Post #15, @morechem28 stated “I've calculated the pressure in the balloon, and I know the mass of the gas inside”. This can’t be correct if the original homework statement was correct. I think @morechem28 has made a mistake such as wrongly assuming the ballo0n contains 1 mole of krypton.

That’s what I was trying to point out.

But note, the amount might matter. For example (using a really strong balloon!) if we were given the amount and initial volume, we might find that the initial density (hence presure) was extremely high - the krypton might even already be dense enough to sink if released at the surface!

But I agree, the amount probably doesn't matter - we are probably meant to assume that the initial krypton pressure is atmospheric.

EDIT. I was on the wrong track. We simply need to assume the initial krypton density is less than that of sea-water (common sense assumption), then amount of gas doesn't matter.
 
Last edited:
  • #21
Steve4Physics
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The density of Kr must be higher than the density of seawater. And it's about 3,74 kg/m3 compared to that of seawater 1040 kg/m3. To question 3 - the original volume is molar volume, so 22,41 dm3. But I truly don't know how I should calculate the new density or the corresponding depth. I know it has to do with the ideal gas law, buoyant force, hydrostatic force...
But you haven't answered the questions I asked in Post #16!

Here's another question:
A liquid has a density of 3g/cm³. An object sinks when placed in the liquid. What can you say about the density of the obect?

You may want to revisit Post #16.
 
  • #22
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Of course, I know that the balloon won't sink till its density isn't higher than water's. Kr is less dense than seawater, so we have to consider the pressure of water that will press on the balloon to make its density higher at a particular depth.
 
  • #23
kuruman
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To @morechem28:
It seems that you are looking for an equation when what you need is a strategy to build one. Here is a strategy that worked for me.

1. Start with the ideal gas law either ##pV=nRT## or ##pV=Nk_BT##.
2. Write the total mass ##m## of the gas in terms of the number of moles ##n## or the number of atoms ##N## and the appropriate constants. You don't need to know its numerical value, just use ##m##.
3. Arrange your expression to have ##\dfrac{m}{V}## on the left side and the rest of the stuff on the right side of the equation.
4. Note that the left side is ##\rho_{\text{Kr}}## (that is why you don't need a value for ##m##) and recognize that this equation gives you the density of the gas at any pressure and temperature. Here the temperature is given.
5. Use that equation to find the pressure (and hence the depth) at which the gas density matches the density of water.

Let's see some work.
 
  • #24
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Thank you. So, I've got m(Kr)/V(Kr) = p*M(Kr)/(R*T). And then I'd replace p with ρ*g*h. But then I get 908,54 = h*m(Ar). Am I right?
 
  • #25
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Oh, yes. I am sorry, I didn't realise m(Kr) cancel each other on each side.
 

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