The pressure at the surface of the sea is P0 ~ 105 Pa. You have just dived below

Question

The pressure at the surface of the sea is P0 ~ 105 Pa. You have just dived below the surface to a depth of d ~ 100 m and find that the pressure has increased to , where ? ~ 103 kg/m3 is the density of water. Then you fart, producing a single gas bubble with an initial volume of V1 = 10-4 m3! This bubble begins to rise. Just before reaching the surface, its volume is V0. Assuming that the temperature and number of gas molecules remains constant during its ascent, use the ideal gas law, , to find V0/V1, i.e. how many times has the bubble's initial volume increased?

Explanation / Answer

Increase of pressure at 100 m depth = 100*1000*9.81 Pa = 9.81*10^5 Pa. We have (9.81+1)*(10^5)*(10^-4) = V0*10^5 or V0 = 10.81*10^-4 m^3 = 1.081*10^-3. It has increased by 10.81 times

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