b
Volume of the air bubble, \(V_{1}=1.0 cm ^{3}=1.0 \times 10^{-6} m ^{3}\)

Bubble rises to height, \(d=40 m\)

Temperature at a depth of \(40 m , T_{1}=12^{\circ} C =285 K\)

Temperature at the surface of the lake, \(T_{2}=35^{\circ} C =308 K\)

The pressure on the surface of the lake:

\(P_{2}=1 atm =1 \times 1.013 \times 10^{5} Pa\)

The pressure at the depth of \(40 m\)

\(P_{1}=1 atm +d\rho g\) Where \(, \rho\) is the density of

water \(=10^{3} kg / m ^{3} g\) is the acceleration due

to gravity \(=9.8 m / s ^{2}\)

\(\therefore P_{1}=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8=493300 Pa\)

We have: \(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

Where, \(V_{2}\) is the volume of the air bubble when it reaches the surface \(V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}\)

\(=\frac{(493300)\left(1.0 \times 10^{-6}\right) 308}{285 \times 1.013 \times 10^{5}}\)

\(=5.263 \times 10^{-6}\; m ^{3}\) or \(5.263 \;cm ^{3}\)

Therefore, when the air bubble reaches the surface, its volume becomes \(5.263 \;cm ^{3} .\)