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

We need to find the volume of an air bubble when it reaches the surface from 40 m depth.

Apply Boyle's Law (constant temperature). Since the temperature is the same at the bottom and surface:

$$P_1 V_1 = P_2 V_2$$

Calculate pressure at the bottom: $$P_1 = P_{\text{atm}} + \rho g h = 1 \times 10^5 + 1000 \times 10 \times 40 = 1 \times 10^5 + 4 \times 10^5 = 5 \times 10^5 \text{ Pa}$$

Pressure at the surface: $$P_2 = P_{\text{atm}} = 1 \times 10^5 \text{ Pa}$$

Calculate the volume at the surface: $$V_2 = V_1 \times \frac{P_1}{P_2} = 1 \times \frac{5 \times 10^5}{1 \times 10^5} = 5 \text{ cm}^3$$

The correct answer is Option D: 5 cm$$^3$$.