Question

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 ?

Answer 1

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 = 10 3 kg/m3

g is the acceleration due to gravity = 9.8 m/s 2

∴P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa

We have: $\frac{P_1}{T}=\frac{P_2V_2}{T_2}$

Where, V2 is the volume of the air bubble when it reaches the surface

$V_2=\frac{P_1V_1T_2}{T_1P_2}$

$=\frac{(493300)(1.0x10^{-6})308}{285×1.013×10^5}$

= 5.263 × 10–6 m3 or 5.263 cm 3

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