# Change in bubble size

1. Dec 3, 2006

### wolly6973

1. The problem statement, all variables and given/known data

A bubble rises from the bottom of a lake of depth 90 m, where the temperature is 4°C. The water temperature at the surface is 19°C. If the bubble's initial diameter is 1.00 mm, what is its diameter when it reaches the surface? (Ignore the surface tension of water. Assume the bubble warms as it rises to the same temperature as the water and retains a spherical shape. Assume Patm = 1.0 atm.)

2. Relevant equations

P1V1/T2=P2V2/V2
Vsphere = 4/3 pi r^3
P=pgd

3. The attempt at a solution

(1000*9.8*90/1000)*4/3*pi*.5^3/4=101.3*4/3*r^3/19

I get 3.458, which isn't right. Am I missing something?

2. Dec 3, 2006

### arunbg

You are missing pi on the rhs and also you have not taken into consideration the overlying atmospheric pressure, along with the hydrostatic pressure.

3. Dec 3, 2006

### wolly6973

How do the atmospheric and hydrostatic pressures fit in?

4. Dec 3, 2006

### arunbg

On the Lhs you simply took water pressure as the overall pressure, and did not include the pressure due to air above water, so overall pressure would be Pw+Patm . On the Rhs you need take only atmospheric pressure, which you have done correctly. Do you follow ?

5. Dec 3, 2006

### wolly6973

So my equation should be
(882+101.3)*4/3*pi*.5^3/4=101.3*4/3*pi*r^3/19
Which gives r=2.846
So the final diameter should be 5.692 mm
This however is incorrect.
Am I still missing something?

6. Dec 3, 2006

### wolly6973

I got it figured out. I had to convert the pressures from kPa to Pa, and the Temperature into Kelvins. Thanks for your help

7. Dec 4, 2006

### arunbg

You are welcome :)