Ideal Gas HW: Volume of Bubble Rising to Surface

In summary, a diver exhales a 1.00cm3 air bubble at a depth of 80.0m below the surface of the sea with a temperature of 5.0°C. The surface temperature of the sea is 20.0°C. Using the formula PV=nRT, the volume of the bubble just before it breaks the surface can be found by setting PiVi/Ti=PfVf/Tf and solving for Vf. However, this method has not been successful so far and the help offices and professor are unavailable. There is confusion over the units and conversions being used, and it is unclear why the volume of the bubble should decrease as it rises to the surface.
  • #1
ibnuts
15
0
Hey, my friend and I have worked on this problem for about 2 hours straight (from our online HW) and can't get the answer right. The help offices and the professor are closed/gone for the weekend. Can anyone please help?! :eek:

At 80.0m below the surface of the sea (density = 1.025g/cm3) where the temperature is 5.0°C, a diver exhales an air bubble having a volume of 1.00cm3.

If the surface temperature of the sea is 20.0°C, what is the volume of the bubble just before it breaks the surface? [in]cm3


We've tried using PV=nRT and setting PiVi/Ti=PfVf/Tf and solving for Vf (using pressure formula Pressure at sea level+(density*g*depth) to find the pressure at 80.0m and then using atmospheric pressure for Pf, but to no avail.

Perhaps someone can tell us what to do and/or give an answer to check against all that we got?

Thanks.
 
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  • #2
Did you convert to kelvin?
 
  • #3
whozum said:
Did you convert to kelvin?

Yup. I did the following:

PiVi/Ti = PfVf/Tf => Vf = (Tf/Ti)*(Pi/Pf)*(Vi)
=(Tf/Ti)*((Pi)/((Pi)+(density of sea water*gravity*initial depth of water)))*(initial volume of bubble)
= (293.15/278.15)*((1.013*10^5)/((1.013*10^5)+(1025*9.8*80.0)))*(1.00)

=[my answer was wrong]

Note: I just used 1.00 above as Vi because all the other units cancel out. I did also try using 0.001L and then converting to mL, but that's the same as just multiplying by 1

Please, someone help!
 
  • #4
Thats really confusing.

[tex] \frac{P_iV_i}{T_i} = \frac{P_fV_f}{T_f} [/tex]

[tex] V_f = \frac{P_iV_iT_f}{P_fT_i}[/tex]

[tex] T_i = 278.15, T_f = 293.15, V_i = 0.001L, P_f = 1atm, P_i = x [/tex]

Note that the pressure at the sufrace of the water is pretty much 1atm, there is no water above it to push down. The initialpresure, when it was 80m underwater however, can be found using the pressure eqaution. Also be careful to use SI units when finding this, g = 9.8m/s^2, and density = 1000kg/m^3.
 
  • #5
That's exactly what I did, and I got the answer wrong. By the way, how did you get the Pi etc to show like that?

Also, to whoever the ops of this site are, this is College HW NOT K-12

EDIT: Don't worry about explaining the Pi, I clicked on message and got the .pdf file explaining how-to.
 
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  • #6
whozum said:
Thats really confusing.

[tex] \frac{P_iV_i}{T_i} = \frac{P_fV_f}{T_f} [/tex]

[tex] V_f = \frac{P_iV_iT_f}{P_fT_i}[/tex]

[tex] T_i = 278.15, T_f = 293.15, V_i = 0.001L, P_f = 1atm, P_i = x [/tex]

I'm also wondering about this formula, which is in fact what I used before and got wrong.

Using the pressure formula, the [tex] P_f [/tex] will be far higher than the [tex] P_i [/tex], meaning the final answer for [tex] V_f [/tex] will be very low (dividing a small # by a very large #). This doesn't make sense though since the volume should increase as the bubble rises to the surface under decreasing pressure.
 
  • #7
You have [itex] P_f = 1.013x 10^5[/itex]. What units are these? Atmospheric pressure is just 1atm in SI.
 
  • #8
The initial pressure, 80m underwater, is higher than the pressure at the surface. This iswhere you are messing up.
 
  • #9
that's in Pa. I tried at one point converting to atm and it didn't work, but I'll try again just to make sure. I seem to be getting different answers using the computer calc than with my TI-89 anyway, so I'll just make sure.
 
  • #10
whozum said:
You have [itex] P_f = 1.013x 10^5[/itex]. What units are these? Atmospheric pressure is just 1atm in SI.

I thought 1 Pa was 1 N/m^2, which is the product of density*height*gravity = pressure

EDIT: found http://www.unc.edu/~rowlett/units/dictP.html#pascal, Pa is in fact SI unit of pressure.

I don't think it would matter though, because if you divide both [tex]P_f[/tex] and [tex]P_i[/tex] by 101.325kPa (the conversion factor to atm), the units and conversion factors cancel each other out.

What I don't understand, perhaps you are getting a different answer, but according to the formula we have derived, doesn't it make the [tex]V_f[/tex] lower than the [tex]V_i[/tex] even though it's at higher depth (i.e. sea level)?
 
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  • #11
Thats correct, sorry I am starting to confuse my own units now.
 
  • #12
whozum said:
Thats correct, sorry I am starting to confuse my own units now.

That's okay. I edited too fast with link, obviously. Still, the rest of it stands. Why would the [tex]V_f[/tex] be lower than the [tex]V_i[/tex].

If you want, btw, I can direct you to the website that has this. My course website has guest log in.
 
  • #13
You should know that pressure under water is much higher than pressure above water. Also, note in the problem they tell you to convert back to cubic centimeters, did you remember to do that?
 
  • #14
Yeah, I converted back to cm^3. Also, I know the pressure under water is far higher, but shouldn't this mean the volume should increase as the bubble rises to the surface because it's under decreasing pressure? When I plug everything in, I get 0.152 mL as my [tex]V_f[/tex], which is far lower than the [tex]V_i[/tex] of 1.00 mL

(NOTE: the site changed the depth to 60.0 m for me now).
 
  • #15
ibnuts said:
Yup. I did the following:

PiVi/Ti = PfVf/Tf => Vf = (Tf/Ti)*(Pi/Pf)*(Vi)
=(Tf/Ti)*((Pi)/((Pi)+(density of sea water*gravity*initial depth of water)))*(initial volume of bubble)
= (293.15/278.15)*((1.013*10^5)/((1.013*10^5)+(1025*9.8*80.0)))*(1.00)

=[my answer was wrong]

Note: I just used 1.00 above as Vi because all the other units cancel out. I did also try using 0.001L and then converting to mL, but that's the same as just multiplying by 1

Please, someone help!

Don't get so involved in writing and subscripting an equation that you forget what quantities are represented by which subscript.
 
  • #16
volume and pressure are inversely proportional yes.


[tex] V_f = \frac{P_iV_iT_f}{P_fT_i}[/tex]

[tex] V_f = \frac{((1.013x10^5N/m^2)+(1000kg/m^3*9.8m/s^2*80m))(0.001L)(293.15K)}{(1.013x10^5N/m^2)(278.15K)}[/tex]

I believe you had your pressures backwards.
 
  • #17
Thanks very much both whozum and Bystander. My problem (and probably my friend's, too) was that I had [tex]P_i[/tex], which was the largest quantity, in the denominator rather than the numerator **stupid me.** That's what 6 hrs of physics HW does to you when you haven't taken your Adderall :biggrin:
 
  • #18
ibnuts said:
PiVi/Ti = PfVf/Tf => Vf = (Tf/Ti)*(Pi/Pf)*(Vi)
=(Tf/Ti)*((Pi)/((Pi) + (density of sea water*gravity*initial depth of water)[/B]))*(initial volume of bubble)
= (293.15/278.15)*((1.013*10^5)/((1.013*10^5)+(1025*9.8*80.0)))*(1.00)

=[my answer was wrong]

Why do you have a + there?
On the initial position you have all this water above the bubble pressing down that increases the pressure, but on the final position the water is below and does not press on it anymore.

Pf = Pi - pressure_exerted_by_water_between_initial_and_final_position
 
  • #19
gerben said:
Why do you have a + there?
On the initial position you have all this water above the bubble pressing down that increases the pressure, but on the final position the water is below and does not press on it anymore.

Pf = Pi - pressure_exerted_by_water_between_initial_and_final_position

Yeah, I swapped them by accident. I've finally gotten the right answer now with everyone's help though! Thanks!
 
  • #20
What's unclear to me about the problem is whether or not one should assume the gas has sufficient time to reach equilibrium temperature with the water.

I suspect that maybe the professor gave you the temperature at the surface of the water to make the question a "trick question", and that you are supposed to assume that the bubble rises so fast that very little heat is transferred between the air and the water - i.e. that the expansion is adiabatic.

So I'd suggest working out the problem using adiabatic expansion and seeing if that's the correct answer.

This would give you an expansion law of the form [tex]P V ^{\gamma}[/tex] = constant

http://farside.ph.utexas.edu/teaching/sm1/lectures/node52.html

You'd probably want to use the value of gamma for water vapor saturated air, but I don't know what that is offhand.
 
  • #21
pervect said:
What's unclear to me about the problem is whether or not one should assume the gas has sufficient time to reach equilibrium temperature with the water.

I suspect that maybe the professor gave you the temperature at the surface of the water to make the question a "trick question", and that you are supposed to assume that the bubble rises so fast that very little heat is transferred between the air and the water - i.e. that the expansion is adiabatic.

So I'd suggest working out the problem using adiabatic expansion and seeing if that's the correct answer.

This would give you an expansion law of the form [tex]P V ^{\gamma}[/tex] = constant

http://farside.ph.utexas.edu/teaching/sm1/lectures/node52.html

You'd probably want to use the value of gamma for water vapor saturated air, but I don't know what that is offhand.

Thanks for the reply, but he didn't make it a trick question. I just screwed up by swapping around my values.

BTW, should I post a question about another problem in this thread or in a new thread? Also, if I should post in a new thread, should I post in college or k-12, because someone moved this thread from college to k-12 for some odd reason.

EDIT: Don't need to any longer, figured it out.
 
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  • #22
This is college problem?i don't believe it...In India higher secondary school students have these kind of problems..infact these problems are considered easier...
 
  • #23
nomorevishnu said:
This is college problem?i don't believe it...In India higher secondary school students have these kind of problems..infact these problems are considered easier...


Yes, it is an easy problem which I made a mistake on, as I've said before, so stop posturing you a**. I wouldn't be bragging about living in a place that smells like crap where you'll make 1/30th of what I will haha.
 
  • #24
Easy there ibnuts.
 
  • #25
Stop it, both of you. If you can't behave yourself in your first few posts at PF, I'd advise you not to post anymore.

What did everyone get for their pressure at that depth? I am unfamiliar with the equation.
 
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  • #26
Normal pressure + Pressure of the water above, which is Force * Depth
 
  • #27
So where does density come in? It'd be easiest just to give me your number for pressure, I am trying to show another way of doing this.
 
  • #28
((1.013*10^5)+(1025*9.8*80.0)) is the density he reported in pascals
 
  • #29
ibnuts said:
Using the pressure formula, the [tex] P_f [/tex] will be far higher than the [tex] P_i [/tex], meaning the final answer for [tex] V_f [/tex] will be very low (dividing a small # by a very large #). This doesn't make sense though since the volume should increase as the bubble rises to the surface under decreasing pressure.

You first state that [tex] P_f >>P_i [/tex] and then you go on saying that the pressure is decreasing. Which statement is the true one?
 

1. What is an ideal gas?

An ideal gas is a theoretical gas that follows the gas laws perfectly, meaning that it has no intermolecular forces and its molecules have zero volume.

2. How is volume related to the behavior of an ideal gas?

According to the ideal gas law, the volume of an ideal gas is directly proportional to its temperature and pressure, and inversely proportional to its number of moles.

3. How does the volume of a bubble rising to the surface relate to ideal gas behavior?

The volume of a bubble rising to the surface is affected by the gas inside the bubble, which follows the ideal gas law. As the bubble rises, the pressure and temperature decrease, causing the volume of the gas to increase.

4. What factors affect the volume of a bubble rising to the surface?

The volume of a bubble rising to the surface is affected by the temperature, pressure, and number of moles of gas inside the bubble, as well as the external pressure and temperature of the surrounding environment.

5. How can the volume of a bubble rising to the surface be calculated?

The volume of a bubble rising to the surface can be calculated using the ideal gas law, PV=nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

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