Question
An air bubble of volume $2.9{\text{ }cm}^{3}$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is $17^{\circ}C$. The volume of the bubble when it reaches the surface, where the water temperature is $27^{\circ}C$, is $\_\_\_\_$ ${cm}^{3}$. $\left( g = 10\text{ }m/s^{2} \right.\ $, density of water $= 10^{3}\text{ }kg/m^{3}$, and 1 atm pressure is $10^{5}\text{ }Pa$ )
Options
Solution
(4) 4.5
Sol. For an air bubble rising in water, the no. of moles remain constant
$${\frac{P_{1}{\text{ }V}_{1}}{{\text{ }T}_{1}} = \frac{P_{2}{\text{ }V}_{2}}{{\text{ }T}_{2}} }{\frac{\left( P_{atm} + \rho gh \right)2.9{\text{ }cm}^{3}}{290\text{ }K} = \frac{\left( P_{atm} \right)V_{2}}{300} }{V_{2} = 4.5{\text{ }cm}^{3}}$$
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