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Mon 12 Dec, 2005 12:07 am
A diver releases an air bubble of volume 2.0 cm cubed from a depth of 15 m below the surface of a lake, where the temperature is 7.0 degrees Celsius. What is the volume of the bubble when it reaches just below the surface of the lake, where the temperature is 20 degrees Celsius?
First, let's decide on what gas law you want to use. I propose using the ideal gas law (PV=nRT) but your instructor may have issued you guidance on this already. Since R never changes and n is constant for this problem, you can write this as
(P1)(V1)/(T1) = (P2)(V2)/(T2)
You are given V1, T1 and T2 (convert deg C into deg K by adding 273). P2 is atmospheric pressure, so you're given that also. All you need to find V2 is to find P1. The pressure due to a column of water is (density x gravitational constant x height), so you have to look up the density of water at 7 degC (easy on the internet) and you probably need to convert some units so that P1 and P2 are in the same units.
That's everything you need. Hope that helps. Post back if you need more help.
I think thats probably a perfect answer engineer, and far too generous if you ask me.
But being the pedant I am, the questioner wants the volume of the bubble just below the surface. It will be a bit more than atmospheric pressure there and hence the volume will be less than it would be if it were for instance a soap bubble actually floating above the water surface. If we take the bubble to be at a depth equivalent of one radius, then P2 will be a function of V2 and things get a bit more complicated.