An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
This problem can by solved using the fact that ‘the number of moles/molecules of the air bubble remains unchanged as the bubbles rises from bottom to the surface’.
Given:
Volume of the air bubble (V1) = 1.0 cm3 = 1.0 × 10 –6 m3
Bubble rises to height (h) = 40 m
Temperature at a depth of 40 m (T1) = 12°C = 285 K
Temperature at the surface (T2) = 35°C = 308 K
The pressure on the surface (P2) = 1 atm = 1 × 1.01 × 105 Pa
The pressure at the depth of 40 m (P1) = 1 atm + ρgh
(Where, ρ is the density of water = 103 kg/m3
g is the acceleration due to gravity = 9.8 m/s2 )
∴ P1 = (1.013 × 10 5 pa)+ (40 m × 10 3 kg/m3 × 9.8 m/s2 )
⇒ P1 = 493300 Pa
We have 
= 
( = number of moles/molecules of the air bubble)
(Where, V2 is the volume of the air bubble when it reaches the surface )
∴ V2 = 
= 
⇒ V2 = 5.263 × 10 –6 m3
⇒ V2 = 5.263 cm3
When the air bubble reaches the surface, its volume is 5.263 cm3