An air bubble of volume 1 cm ^ 3 rises from the bottom of a lake...

The pressure at the bottom of the lake can be calculated as follows:

$\begin{cases}{P}_{bottom} & = & {P}_{0} + ρgh \\ & = & \left({10}^{5} + 1000 \times 10 \times 40\right)Pa \\ & = & 5 \times {10}^{5}Pa\end{cases}$

As the temperature of water remains constant, it can be written that

${P}_{bottom}{V}_{bottom} = {P}_{top}{V}_{top} . . . ( 1 )$

Substitute the values of the known parameters into equation (1) to calculate the required volume of the bubble when it reaches the surface.

$\begin{cases}5 \times {10}^{5} \times 1 & = & 1 \times {10}^{5} \times {V}_{top} \\ & ⇒ & {V}_{top} = \dfrac{5 \times {10}^{5} \times 1}{1 \times {10}^{5}}{cm}^{3} \\ & = & 5{cm}^{3}\end{cases}$