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Asked by Manjulasuresh Last Modified
Answer 
Harshit Bhardwaj
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
∴P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa
We have: 
Where, V2 is the volume of the air bubble when it reaches the surface
= 5.263 × 10–6 m3 or 5.263 cm3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.
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