How Does the Volume of a Diver's Exhaled Bubble Change from Deep Sea to Surface?

[latitude]

Homework Statement

Well, I did get an answer to this, but I don't think I did it right. It kinda seems wrong to me, so I wanted a second opinion :)

At 25 m below the surface of the sea (Density = 1025 kg/m^3), where the temperature is 5 degrees Celsius, a diver exhales an air bubble having a volume of 1 cm^3. If the surface temp of the sea is 20 degrees Celsius, what is the volume of the bubble just before it breaks the surface?

Homework Equations

Ideal gas: PV = nRT
P = Po + (DENSITY)gh ??Not sure if I should use this??

The Attempt at a Solution

P = Po + Density(g)(h) = 1.013 x 10^5 Pa + (1025)(9.81)(25)
P = 3.5 x 10^5 Pa (This seems fairly logical to me... kinda? But I'm not very logical :P)

Under the sea
(3.5 x 10^5)(0.01 m^3) = n(3.814)(278 degrees K)
n = 3.3
At surface
(1.013 x 10^5)(V) = (3.3)(3.814)(293)
V = 0.036 m^3, or 3.6 cm^3... that's fairly reasonable, isn't it?? But I'm not sure if using the P = Po + density(g)(h) was all right when the height is below the sea like that ??

 

[mgb_phys]

You can use pv=nRT remember that 'nR' isconstant.
Then you just need to find the pressure at 25m underwater, p = density * g * h

 

[latitude]

Forgive me if I sound dense (I may very well be!) but isn't that kind of what I did? If it's been smart enough to note they were constant, I could have done it with less work

 

[Borek]

But I'm not sure if using the P = Po + density(g)(h) was all right when the height is below the sea like that ??

Looks OK to me.