An air bubble of volume 2.9 cm^3 rises from | vedclass

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(D) For an air bubble rising in water, the number of moles remains constant. Using the ideal gas law, we have $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_

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$4.5$

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(D) For an air bubble rising in water, the number of moles remains constant. Using the ideal gas law, we have $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.At the bottom (state $1$): $P_1 = P_{atm} + ho gh = 10^5 + (10^3 	imes 10 	imes 5) = 1.5 	imes 10^5 \ Pa$.$V_1 = 2.9 \ cm^3$.$T_1 = 17 + 273 = 290 \ K$.At the surface (state $2$): $P_2 = P_{atm} = 10^5 \ Pa$.$T_2 = 27 + 273 = 300 \ K$.Substituting the values into the equation:$\frac{(1.5 	imes 10^5) 	imes 2.9}{290} = \frac{10^5 	imes V_2}{300}$.$V_2 = \frac{1.5 	imes 2.9 	imes 300}{290} = \frac{1.5 	imes 2.9 	imes 30}{29} = 1.5 	imes 0.1 	imes 30 = 4.5 \ cm^3$.

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