What is the final diameter of the bubble rising from the bottom of a lake?

[wolly6973]

Homework Statement

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.)

Homework Equations

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

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?

 

[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.

 

[wolly6973]

How do the atmospheric and hydrostatic pressures fit in?

 

[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 P_w+P_atm . On the Rhs you need take only atmospheric pressure, which you have done correctly. Do you follow ?

 

[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?

 

[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

 

[arunbg]

You are welcome :)