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?

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Solution

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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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?

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?