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an air bubble of volume 1 0 cm3 rises from the bot...

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:

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 + dpg

Where, p 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: P₁V₁/ T₁ = P₂V₂/ T₂

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

V₂ = P₁V₁T₂/ T₁P₂

= 493300 x (1.0 x 10^-6) 308 / 285 x 1.013 x 10⁵

= 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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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?

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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?

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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?