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 s⁻². 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:
Options

Question

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 s⁻². 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:

Options

G1 Physics · Accepted Answer

The temperature is the same at the bottom and the surface, so Boyle's law applies: P₁V₁ = P₂V₂ At depth h = 40 m: P₁ = P(atm) + ρgh = 1×10⁵ + (1000)(10)(40) = 5×10⁵ Pa At the surface: P₂ = 1×10⁵ Pa Given V₁ = 1 cm³: V₂ = P₁V₁/P₂ = (5×10⁵)(1)/(1×10⁵) = 5 cm³ Hence option D follows.