An air bubble of volume 2.9 cm³ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 °C. The volume of the bubble when it reaches the surface, where the water temperature is 27 °C, i $s____$ cm³.
(g = 10 m/s², density of water = 10³ kg/m³, and 1 atm pressure is 10⁵ Pa)

Question

An air bubble of volume 2.9 cm³ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 °C. The volume of the bubble when it reaches the surface, where the water temperature is 27 °C, i $s____$ cm³.

(g = 10 m/s², density of water = 10³ kg/m³, and 1 atm pressure is 10⁵ Pa)

G1 Physics · Accepted Answer

The number of moles of gas in the bubble remains constant. Hence, P₁V₁/T₁ = P₂V₂/T₂. At the bottom, T₁ = 17+273 = 290 K. At the surface, T₂ = 27+273 = 300 K. Pressure at the bottom is P₁ = P(atm)+ρgh = 10⁵ + 10³×10×5 = 1.5×10⁵ Pa. At the surface, P₂ = 10⁵ Pa. Given, V₁ = 2.9 cm³. Therefore, V₂ = P₁V₁T₂/(P₂T₁) = (1.5×10⁵)(2.9)(300)/ [(10⁵)(290)] = 4.5 cm³. Correct option: D.