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