Question

An air bubble of volume 1 x 10^-6 m³ is at depth of 40 m from surface of a lake, where temperature is 283 K. When the bubble reaches at the surface of water, calculate its volume. Temperature at the surface of water is 27°C. (Atmospheric pressure = 1.01 x 10⁵ Pa)

Answer 1

Given: v₁ = 1 x 10^-6 m³, h = 40 m; P_a = 1.01 x 10⁵ Pa

∴ p = h.ρ_w.g = 40 x 10³ x 9.8

= 39.2 x 10⁴ = 3.92 x 10⁵ Pa

∴ P₁ = P_a + p = (1.01 + 3.92) x 10⁵

= 4.93 x 10⁵ Pa

T = 283 K; T2 = 27 + 273 = 300 K

P₂ = P_a = 1.01 x 10⁵ Pa; V₂ = ?

∵ From ideal gas equation 

= 5.1744 x 10^-6 m³

or V₂ = 5.17 x 10^-6 m³