An air bubble of volume $2.9 \mathrm{~cm}^3$ rises from the bottom of a swimming pool of 5 m deep. A

Question

An air bubble of volume $2.9 \mathrm{~cm}^3$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is $17^{\circ} \mathrm{C}$. The volume of the bubble when it reaches the surface, where the water temperature is $27^{\circ} \mathrm{C}$, is $\_\_\_\_$ $\mathrm{cm}^3$. $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right.$, density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3$, and 1 atm pressure is $\left.10^5 \mathrm{~Pa}\right)$

Professor PC Thomas Classes · Answer

Applying the ideal gas equation $P_1 V_1 / T_1 = P_2 V_2 / T_2$. At the bottom, the pressure is $P_1 = P_{atm} + ho g h = 10^5 + (10^3)(10)(5) = 1.5 	imes 10^5 	ext{ Pa}$. The temperature is $T_1 = 17^\circ	ext{C} + 273 = 290 	ext{ K}$, and $V_1 = 2.9 	ext{ cm}^3$. At the surface, the pressure is $P_2 = 10^5 	ext{ Pa}$ and the temperature is $T_2 = 27^\circ	ext{C} + 273 = 300 	ext{ K}$. Solving for $V_2$: $V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{(1.5 	imes 10^5) 	imes 2.9 	imes 300}{10^5 	imes 290} = 1.5 	imes \frac{2.9}{290} 	imes 300 = 1.5 	imes 3 = 4.5 	ext{ cm}^3$.