An air bubble of volume $1 \mathrm{~cm}^{3}$ rises from the bottom of a lake $40 \mathrm{~m}$ deep to the surface at a temperature of $12^{\circ} \mathrm{C}$. The atmospheric pressure is $1 \times 10^{5} \mathrm{~Pa}$ the density of water is $1000 \mathrm{~kg} / \mathrm{m}^{3}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$. There is no difference of the temperature of water at the depth of $40 \mathrm{~m}$ and on the surface. The volume of air bubble when it reaches the surface will be:

<p>The volume of the air bubble changes due to the change in pressure as it rises from the bottom of the lake to the surface. We can use Boyle&#39;s Law to calculate the change in volume, which states that the product of pressure and volume is constant for a given mass of confined gas held at a constant temperature:</p> <p>$P_1V_1 = P_2V_2$</p> <p>where $P_1$ and $V_1$ are the pressure and volume at the bottom of the lake and $P_2$ and $V_2$ are the pressure and volume at the surface of the lake.</p> <p>At the bottom of the lake, the pressure is the atmospheric pressure plus the pressure due to the water column above the bubble:</p> <p>$P_1 = P_{\text{atm}} + \rho gh$</p> <p>where $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $h$ is the height of the water column. Substituting the given values, we get:</p> <p>$P_1 = 1 \times 10^{5} \text{ Pa} + 1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 40 \text{ m} = 5 \times 10^{5} \text{ Pa}$</p> <p>At the surface of the lake, the pressure is the atmospheric pressure:</p> <p>$P_2 = P_{\text{atm}} = 1 \times 10^{5} \text{ Pa}$</p> <p>The initial volume of the bubble is:</p> <p>$V_1 = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3$</p> <p>Substituting these values into Boyle&#39;s Law and solving for $V_2$, we get:</p> <p>$V_2 = \frac{P_1V_1}{P_2} = \frac{5 \times 10^{5} \text{ Pa} \times 1 \times 10^{-6} \text{ m}^3}{1 \times 10^{5} \text{ Pa}} = 5 \times 10^{-6} \text{ m}^3 = 5 \text{ cm}^3$</p> <p>So, the volume of the air bubble when it reaches the surface is 5 cm³.</p>