Posted by james on .
at 25.0m below the surface of the sea (density=1.025kg/m^3), where the temperature is 5.00C, a diver exhales an air bubble having a volume of 1.00cm^3. If the surface temperature of the sea is 20.0C, what is the volume of the bubble just before it breaks the surface?

physics 
drwls,
Your sea density is incorrect. It is 1.025*10^3 kg/m^3
First calculate the pressure at 25 m depth, making sure that you add atmospheric pressure Po to (density)*g*(depth).
P(25 m) = 1.01*10^5 + (1.025*10^3)(9.81)*25.0 = 1.01*10^5 + 2.51*10^5 = 3.52*10^5
Prssure at 25 m depth is about 3.5 times larger than at the surface.
The ratio of bubble volume at the surface to the volume at 25 m is
[T(surface)/T(25m)]*[P(25m)/P(surface)]
=(293/278)*(3.5)
Check my numbers.