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An air bubble of volume V rises from the bottom of a lake 40m deep to the surface at a temperature of 12°C. Calculate the volume of the air bubble when it reaches the surface. Given atmospheric pressure, density of water, and gravitational acceleration.

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The volume of the air bubble when it reaches the surface of the lake is 5.26 x 10⁻⁶ m³.

How to calculate the volume of the air bubble at the surface?

The volume of the air bubble at the surface is calculated by applying general gas equation.

P₁V₁/T₁ = P₂V₂/T₂

where;

  • P₁ is the pressure at the bottom of the lake
  • P₂ is the pressure at the surface of the lake
  • T₁ is the temperature at the bottom of the lake
  • T₂ is the temperature at the surface of the lake.
  • V₁ is the volume at the bottom of the lake
  • V₂ is the volume at the surface of the lake

The pressure at the bottom of the lake is calculated as;

P₁ = 1 atm + ρgh

P₁ = 1.013 x 10⁵ Pa + (1000 x 9.8 x 40)

P₁ = 493,300 Pa

The pressure at the surface of the lake:

P₂ = 1 atm

P₂ = 1.013 x 10⁵ Pa

The temperature at the bottom of the lake:

T₁ = 12 ⁰C + 273

T₁ = 285 K

The temperature at the surface of the lake:

T₂ = 35 ⁰C + 273

T₂ = 308 K

The volume at the bottom of the lake:

V₁ = 1 cm³

V₁ = 1 x 10⁻⁶ m³

The volume of the air bubble at the surface is calculated as;

V₂ = (P₁V₁T₂) / (P₂T₁)

V₂ = (493,300 x 1 x 10⁻⁶ x 308) / (1.013 x 10⁵ x 285)

V₂ = 5.26 x 10⁻⁶ m³

The complete question is below:

An air bubble of volume 1 cm³ rises from the bottom of a lake 40m deep to the surface at a temperature of 12°C. Calculate the volume of the air bubble when it reaches the surface, which is at a temperature of 35 ⁰C. (Given atmospheric pressure, 1 atm = 1.013 x 10⁵ Pa , density of water = 1000 kg/m³, and gravitational acceleration = 9.8 m/s²).

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