Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T1 = 12°C = 285 K

Temperature at the surface of the lake, T2 = 35°C = 308 K

The pressure on the surface of the lake,

P2 = 1 atm = 1 ×1.013 × 105 Pa

The pressure at the depth of 40 m,

P1 = 1 atm + dρg

where,

ρ is the density of water = 103 kg/m3

g is the acceleration due to gravity = 9.8 m/s2

Therefore,

P1 = 1.013 × 105 + 40 × 103 × 9.8

= 493300 Pa

We have,

P1V1 / T1 = P2V2 / T2

where,

V2 is the volume of the air bubble when it reaches the surface.

V2 = P1V1T2 / T1P2

= 493300 × 1 × 10-6 × 308 / (285 × 1.013 × 105)

= 5.263 × 10–6 m3

= 5.263 cm3

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.