How Does Temperature and Depth Affect the Volume of an Air Bubble in Water?

[hils0005]

[SOLVED] Ideal gases

an air bubble of volume 20cm^3 is at the bottom of a lake 40m deep where the temp is
4C, the bubble rises to the surface which is at temp 20C, take the temp of the buble to match that of the surrounding water, just as the bubble reaches the surface, what is it's volume?

Homework Equations

pV=nRT
p(f)V(f)/p(i)V(i)=nRT(f)/nRT(i)
final pressure= 1.01*10^5Pa
V(i)=20x10^-6m^3
T(i)=277K
T(f)=293K

The Attempt at a Solution

would need to determine the pressure on the bubble at the bottom of the lake, which I can't remember how to determine
p=hp(density)g?
p=40m(1000kg/m^3)9.8m/s^2
p=39.2x10^4

Vf=(T(f)*(p(i)V(i))/(T(i)p(f))
Vf=(293 * 39.2x10^4 * .00002)/(277*101000)
Vf=2297.12/2.7977x10^7=8.2x10^-5m^3

This shows the volume decreased, wouldn't the volume increase?

 

[mgb_phys]

You don't actaully need the value of nR, just set PV/T equal for the top and bottom.
The extra pressure of the water is just the depth * density * g, so for fresh water each 9.8m is 1 extra atmosphere, remember to add on the atmospheric pressure.

As you said the bubble should expand - It's always worth thinking about the answer before you start punching numbers.
The pressure goes down by a factor of around 5 and the absolute temperate goes up by less than 10% so you are looking for an expansion of around 5.

 

[hils0005]

ok
so p=1.01 x 10^5Pa + 39.2x10^4
p=4.93 x 10^6
(293*4.93x10^6*.00002)/(277*101000)
V(f)=28889.8/27977000
V(f)=103cm^3
I think that is the correct answer-Thanks