An air bubble of volume 2.9 ~cm ³ 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 , is _ _ _ _ cm ³ . ( g =10 ~m / s ² . , density of water =10³ ~kg / m ³ , and 1 atm pressure is .10⁵ ~Pa )

Step-by-step solution

An air bubble rises through water with initial volume 2.9 cm³ at the bottom. Calculating pressures at both locations: At bottom (5 m depth): P₁ = P_ atm + gh = 10^5 + 10^3(10)(5) = 1.5 10^5 Pa At surface: P₂ = 10^5 Pa Initial temperature: T₁ = 17°C = 290 K, Final temperature: T₂ = 27°C = 300 K Using ideal gas law: P₁V₁ T₁ = P₂V₂ T₂ V₂ = V₁ P₁ P₂ T₂ T₁ = 2.9 1.5 10^5 10^5 300 290 = 2.9 1.5 30 29 = 2.9 1.5 1.034 4.5 cm³