An air bubble of volume 1.0 ; cm^{3} rises fr | vedclass

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(B) Initial volume of the air bubble,$V_{1} = 1.0 \; cm^{3} = 1.0 	imes 10^{-6} \; m^{3}$.Depth of the lake,$d = 40 \; m$.Initial temperature at the

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$5.26$

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(B) Initial volume of the air bubble,$V_{1} = 1.0 \; cm^{3} = 1.0 	imes 10^{-6} \; m^{3}$.Depth of the lake,$d = 40 \; m$.Initial temperature at the bottom,$T_{1} = 12 \; ^{\circ}C = 285 \; K$.Final temperature at the surface,$T_{2} = 35 \; ^{\circ}C = 308 \; K$.Pressure at the surface,$P_{2} = 1 \; atm = 1.013 	imes 10^{5} \; Pa$.Pressure at the bottom,$P_{1} = P_{2} + d ho g$,where $ho = 10^{3} \; kg/m^{3}$ and $g = 9.8 \; m/s^{2}$.$P_{1} = 1.013 	imes 10^{5} + (40 	imes 10^{3} 	imes 9.8) = 1.013 	imes 10^{5} + 3.92 	imes 10^{5} = 4.933 	imes 10^{5} \; Pa$.Using the ideal gas law,$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$.$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}} = \frac{(4.933 	imes 10^{5}) 	imes (1.0 	imes 10^{-6}) 	imes 308}{285 	imes (1.013 	imes 10^{5})}$.$V_{2} \approx 5.26 	imes 10^{-6} \; m^{3} = 5.26 \; cm^{3}$.

An air bubble of volume $1.0 \; cm^{3}$ rises from the bottom of a lake $40 \; m$ deep at a temperature of $12 \; ^{\circ}C$. To what volume (in $cm^{3}$) does it grow when it reaches the surface,which is at a temperature of $35 \; ^{\circ}C$?

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