Thermodynamics Part 5

Concept:The air bubble follows the ideal gas law as it rises. Its volume changes due to decrease in pressure and increase in temperature.Explanation:At the bottom: $V_1 = 2.9\ \mathrm{cm}^3$, $T_1 = 17^\circ\mathrm{C} = 290\ \mathrm{K}$. Pressure $P_1 = P_0 + h\rho g = 10^5 + 5 \times 10^3 \times 10 = 1.5 \times 10^5\ \mathrm{Pa}$. At the surface: $T_2 = 27^\circ\mathrm{C} = 300\ \mathrm{K}$, $P_2 = P_0 = 10^5\ \mathrm{Pa}$. Using ideal gas law for constant moles: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$. Substitute values: $\frac{(1.5\times10^5)(2.9)}{290} = \frac{(10^5)V_2}{300}$. Cancel $10^5$: $\frac{1.5\times2.9}{290} = \frac{V_2}{300}$. Thus $V_2 = \frac{1.5\times2.9\times300}{290} = \frac{1305}{290} = 4.5\ \mathrm{cm}^3$. The decrease in pressure dominates over the slight temperature rise, causing volume expansion.Answer:$4.5\ \mathrm{cm}^3$; option D.