Hello everyone. Here is my problem:(adsbygoogle = window.adsbygoogle || []).push({});

At 25.1 meters below the surface of the sea (density is 1025 (kg)/m^3),

where the temp is 3.00 C, a diver exhales an air bubble that has a volume of 1.00 cm^3. If the surface temp of the sea is 15.00 C, what is the volume of the bubble just before it breaks the surface.

Here is what I got: (d=density)

pv=nrt under water P+dgh=p

PV=nrT surface so, (P+dgh)v=nrt and PV=nrT

The nr's cancel, so (PV/((P+dgh)(v)))=T/t

Simplifying, I get that V=((P+dgh)vT)/(Pt)

Only thing is, I don't have pressure at surface. I tried using 1.013X10^5 N/m^2 (1 atm) and got .174 m^3. And, I did remember to change to .01 meters from 1.00 cm^3. Answer seems reasonable but wrong. Any help would be appreciated.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Varying depths of water

**Physics Forums | Science Articles, Homework Help, Discussion**