# Ideal Gas HW: Volume of Bubble Rising to Surface

1. Apr 22, 2005

### ibnuts

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?!

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.

2. Apr 22, 2005

### whozum

Did you convert to kelvin?

3. Apr 22, 2005

### ibnuts

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)

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

4. Apr 23, 2005

### whozum

Thats really confusing.

$$\frac{P_iV_i}{T_i} = \frac{P_fV_f}{T_f}$$

$$V_f = \frac{P_iV_iT_f}{P_fT_i}$$

$$T_i = 278.15, T_f = 293.15, V_i = 0.001L, P_f = 1atm, P_i = x$$

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. Apr 23, 2005

### ibnuts

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.

Last edited: Apr 23, 2005
6. Apr 23, 2005

### ibnuts

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

Using the pressure formula, the $$P_f$$ will be far higher than the $$P_i$$, meaning the final answer for $$V_f$$ 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. Apr 23, 2005

### whozum

You have $P_f = 1.013x 10^5$. What units are these? Atmospheric pressure is just 1atm in SI.

8. Apr 23, 2005

### whozum

The initial pressure, 80m underwater, is higher than the pressure at the surface. This iswhere you are messing up.

9. Apr 23, 2005

### ibnuts

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. Apr 23, 2005

### ibnuts

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

EDIT: found this link, Pa is in fact SI unit of pressure.

I don't think it would matter though, because if you divide both $$P_f$$ and $$P_i$$ 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 $$V_f$$ lower than the $$V_i$$ even though it's at higher depth (i.e. sea level)?

Last edited: Apr 23, 2005
11. Apr 23, 2005

### whozum

Thats correct, sorry Im starting to confuse my own units now.

12. Apr 23, 2005

### ibnuts

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

If you want, btw, I can direct you to the website that has this. My course website has guest log in.

13. Apr 23, 2005

### whozum

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. Apr 23, 2005

### ibnuts

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 $$V_f$$, which is far lower than the $$V_i$$ of 1.00 mL

(NOTE: the site changed the depth to 60.0 m for me now).

15. Apr 23, 2005

### Bystander

Don't get so involved in writing and subscripting an equation that you forget what quantities are represented by which subscript.

16. Apr 23, 2005

### whozum

volume and pressure are inversely proportional yes.

$$V_f = \frac{P_iV_iT_f}{P_fT_i}$$

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

17. Apr 23, 2005

### ibnuts

Thanks very much both whozum and Bystander. My problem (and probably my friend's, too) was that I had $$P_i$$, 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

18. Apr 23, 2005

### gerben

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. Apr 23, 2005

### ibnuts

Yeah, I swapped them by accident. I've finally gotten the right answer now with everyone's help though! Thanks!

20. Apr 23, 2005

### pervect

Staff Emeritus
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 transfered 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 $$P V ^{\gamma}$$ = 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.