To determine in which pond the volume of the gas-filled balloon will be smaller at the bottom, we apply Boyle's Law, which states that for a fixed mass of gas at a constant temperature, the pressure $P$ and volume $V$ are inversely proportional ($PV = ext{constant}$). Let $V$ be the initial volume of the balloon at the surface of the ponds. For the first pond: Given, depth $h_1 = 20\ \text{m}$, atmospheric pressure $P_{atm1} = 1.5\ \text{atm} = 1.5 \times 1.013 \times 10^5\ \text{Pa} = 151950\ \text{Pa}$, water density $\rho_1 = 1050\ \text{kg m}^{-3}$, and $g = 9.8\ \text{ms}^{-2}$. The pressure at the bottom of the first pond is: $$ \begin{aligned} P_1' &= P_{atm1} + h_1 \rho_1 g \\ &= 151950 + (20 \times 1050 \times 9.8) \\ &= 151950 + 205800 \\ &= 357750\ \text{Pa} \end{aligned} $$ Using Boyle's Law, $P_{atm1} V = P_1' V_1'$, the volume at the bottom $V_1'$ is: $$ \begin{aligned} V_1' &= \frac{P_{atm1} V}{P_1'} \\ &= \frac{151950 V}{357750} \\ &\approx 0.4247 V \end{aligned} $$ For the second pond: Given, depth $h_2 = 25\ \text{m}$, atmospheric pressure $P_{atm2} = 1.2\ \text{atm} = 1.2 \times 1.013 \times 10^5\ \text{Pa} = 121560\ \text{Pa}$, and water density $\rho_2 = 1050\ \text{kg m}^{-3}$. The pressure at the bottom of the second pond is: $$ \begin{aligned} P_2' &= P_{atm2} + h_2 \rho_2 g \\ &= 121560 + (25 \times 1050 \times 9.8) \\ &= 121560 + 257250 \\ &= 378810\ \text{Pa} \end{aligned} $$ Using Boyle's Law, $P_{atm2} V = P_2' V_2'$, the volume at the bottom $V_2'$ is: $$ \begin{aligned} V_2' &= \frac{P_{atm2} V}{P_2'} \\ &= \frac{121560 V}{378810} \\ &\approx 0.3209 V \end{aligned} $$ Comparing the two volumes: Since $0.3209 V < 0.4247 V$, we have $V_2' < V_1'$. Therefore, the volume of the gas-filled balloon will be smaller at the bottom of the second pond.