An air bubble of volume `20cm^3` forms in a lake at a depth of 40 m below the water surface.What will be its volume when it rises just below the water surface?(Standard atmospheric pressure=76 cmHg)

Explanation

Volume of air bubble at a depth of 40 cm. `V_1=20 cm^3`.
Pressure on the bubble there
`p_1`= atmospheric pressure +pressure due to 4000 cm of water column
`=76 times 13.6 times 980 +4000 times 1 times 980`
`=(76 times 13.6 +4000) times 980 dyn.cm^-2`
Let the volume of the bubble just below the surface be `V_2`,pressure on the bubble just below the surface
`p_2=76 times 13.6 times 980 dyn.cm^2`
From Boyle.s law
`p_1V_1=p_2V_2`
or, `980times(76times13.6+4000)times20=V_2times76times13.6times980`
`thereforeV_2=((76 times13.6+4000))/(76 times 13.6)=97.4 cm^3`