physics
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?

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.