An air bubble of volume 1 cm³ rises from the bottom of a lake 40 m deep to the surface at a temperature of 12° C. The atmospheric pressure is 1 × 10⁵ Pa the density of water is 1000 kg / m³ and g=10 m / s². There is no difference of the temperature of water at the depth of 40 m and on the surface. The volume of air bubble when it reaches the surface will be:

Solution

Physics sir · 3 steps

Same temperature means Boyle's law: find the pressure at 40 m and the volume scales up by that ratio.

1. Isothermal process. The water temperature is the same at the bottom and at the surface, so the air inside the bubble obeys Boyle's law, P₁V₁ = P₂V₂.
2. Pressure at the bottom. At depth h the bubble feels the atmosphere plus the water above it: P₁ = P₀ + ρ g h = 1 × 10⁵ + 1000 × 10 × 40 = 5 × 10⁵ Pa. At the surface, P₂ = P₀ = 1 × 10⁵ Pa.

   V₂ = P₁/P₂V₁ = (5 × 10⁵)/(1 × 10⁵) × 1 cm³ = 5 cm³

3. So. The bubble swells to 5 cm³ at the surface, which is option (d).

The idea to keep. Pressure under a liquid is P₀ + ρ g h, never just ρ g h. For a gas at constant temperature, the volume grows by exactly the factor the pressure falls.

Answer from the NTA final answer key. Solution by PrepWiser.