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Question

A gas filled freely collapsible balloon is pushed fro the surface level of a lake to a depth of 50 m. Approximately what per cent of its original volume will be balloon finally have, assuming that the gas behaves ideally and temperature is same at the surface and at 50 m depth?

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Solution

P1= Pressure at the surface =1 atm
=760×13.6×981 dyne/cm2
P2= Pressure at a depth of 50 m
=76.0×13.6×981+(50×100)×1×981 dyne/cm2
981[76.0×13.6+5000]
=981×(6033.6)dyne/cm2
Now apply Boyle's law, P1V1=P2V2
or V2V1=P1P2=76.0×13.6×981981×6033.6=0.1713
% =0.1713×100=17.13.

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