When the air bubble is at depth of 40 m Volume of air bubble $V_{1}=1.0\ \mathrm{cm}^{3}=1.0 \times 10^{-6}\ \mathrm{m}^{3}$ Pressure of air bubble $P_1$ = 1 atm + 40 m depth of water = 1 atm + $h_1 \rho g$ Temperature of air bubble $T_{1}=12+273=285\ \mathrm{K}$When the air bubbles reaches at the surface of lake, pressure of air bubbles $P_{2}=1$ atm $=1.013 \times 10^{5}\ \mathrm{Pa}$Temperature of air bubble $T_2 = 35^{\circ}\mathrm{C} = 35 + 273 = 308\ \mathrm{K}$Volume of air bubbles = ?Now, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$$V_{2}= \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$ $= \frac{4.93 \times 10^{5} \times 1.0 \times 10^{-6} \times 308}{285 \times 1.01 \times 10^{5}}$ $=5.275 \times 10^{-6}\ \mathrm{m}^{3}$