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

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(B) According to Boyle's law,for a fixed mass of gas at constant temperature,pressure and volume are inversely proportional to each other,i.e.,$P \pro

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(B) According to Boyle's law,for a fixed mass of gas at constant temperature,pressure and volume are inversely proportional to each other,i.e.,$P \propto \frac{1}{V}$.Therefore,$P_1 V_1 = P_2 V_2$.At the bottom of the lake,the pressure $P_1$ is the sum of atmospheric pressure $P_0$ and the pressure due to the water column of height $h$: $P_1 = P_0 + h ho_w g$.At the surface,the pressure $P_2$ is equal to the atmospheric pressure $P_0$.Given: $h = 47.6 \ m = 4760 \ cm$,$V_1 = 50 \ cm^3$,$P_0 = 70 \ cm$ of $Hg = 70 	imes 13.6 	imes g \ 	ext{dynes/cm}^2$,$ho_w = 1 \ g/cm^3$.Substituting the values into the equation $P_0 V_1 + h ho_w g V_1 = P_0 V_2$:$V_2 = V_1 \left( 1 + \frac{h ho_w g}{P_0} ight) = V_1 \left( 1 + \frac{h ho_w g}{h_{Hg} ho_{Hg} g} ight) = V_1 \left( 1 + \frac{h ho_w}{h_{Hg} ho_{Hg}} ight)$.$V_2 = 50 \left( 1 + \frac{4760 	imes 1}{70 	imes 13.6} ight) = 50 \left( 1 + \frac{4760}{952} ight) = 50 (1 + 5) = 50 	imes 6 = 300 \ cm^3$.

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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