Question

An air bubble released by a deep-water diver, 120 m below the surface of a lake, has a volume of 1.65 cm3. The surface of the lake is at sea level, and the density of the lake water can be approximated as that of pure water. As the bubble rises to the surface, the temperature of the water and the number of air molecules in the bubble can each be approximated as constant. Find the volume (in cm) of the bubble just before it pops at the surface of the lake. 1 cm³

Answer 1

Application of Ideal Gas equation

② Po = Atmospheric pressure = lolxlos Pa th=120m Pressure at depth h , k = P + 5gh. where Igh is the gange pressure due to the lelguid column Pressure at Hust below the surface of lake I p. - lo Density of water f = 1000 kg/m3 Volume at depth h, v = 1,65 cm3 =1165x100 m3 I since Jemberature & no. of molecules I semain Constant, we can use ideal gas equahon Pivo = Pz Vz (P + 8gh) V = lov 11001x105°+ 1000 *9.81 X/20)x1, 65 x 10-6 - stofxios V V2 = 20.9 x 10-6 m3 TV = 20.9 cm