An inverted bell lying at the bottom of a lak | vedclass

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(B) According to Boyle's law,for a constant temperature,the pressure and volume of a gas are inversely proportional: $P_1V_1 = P_2V_2$.At the bottom o

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$300$

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(B) According to Boyle's law,for a constant temperature,the pressure and volume of a gas are inversely proportional: $P_1V_1 = P_2V_2$.At the bottom of the lake,the total pressure $P_1$ is the sum of atmospheric pressure and the pressure due to the water column: $P_1 = P_{atm} + h ho_w g$.Given: $P_{atm} = 70 \, cm$ of $Hg = 70 	imes 13.6 \, g/cm^2 	imes g$,$h = 47.6 \, m = 4760 \, cm$,$ho_w = 1 \, g/cm^3$,$V_1 = 50 \, cm^3$.At the surface,$P_2 = P_{atm} = 70 	imes 13.6 \, g/cm^2 	imes g$.Substituting the values into $V_2 = V_1 	imes (P_1 / P_2)$:$V_2 = 50 	imes \frac{(70 	imes 13.6 	imes g) + (4760 	imes 1 	imes g)}{70 	imes 13.6 	imes g}$$V_2 = 50 	imes \frac{952 + 4760}{952} = 50 	imes \frac{5712}{952} = 50 	imes 6 = 300 \, cm^3$.

An inverted bell lying at the bottom of a lake $47.6 \, m$ deep has $50 \, cm^3$ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be ......... $cm^3$ (atmospheric pressure $= 70 \, cm$ of $Hg$ and density of $Hg = 13.6 \, g/cm^3$).

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