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, is ____ cm³. ( g = 10 m/s², density of water = 10³ kg/m³, and 1 atm pressure is 10⁵ Pa )

An air bubble of volume 2.9 cm³ rises from the bottom of a swimming pool of 5 m deep. Find the volume of the bubble at the surface.

Q. 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 $17^\circ C$. The volume of the bubble when it reaches the surface, where the water temperature is $27^\circ C$, is ____ $cm^3$.

$(g=10\,m/s^2,\ \rho_{water}=10^3\,kg/m^3,\ 1\,atm=10^5\,Pa)$

A. $2.0$

B. $4.2$

C. $3.0$

D. $4.5$

Correct Answer: 4.5

Explanation

For the air bubble, the combined gas law applies:

$$ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} $$

Pressure at the bottom of the pool:

$$ P_1 = P_{atm} + \rho g h $$

$$ P_1 = 10^5 + (10^3)(10)(5) = 1.5\times10^5\,Pa $$

Pressure at the surface:

$$ P_2 = 10^5\,Pa $$

Temperatures in Kelvin:

$$ T_1 = 17 + 273 = 290\,K $$

$$ T_2 = 27 + 273 = 300\,K $$

Substituting values:

$$ \frac{1.5\times10^5 \times 2.9}{290} = \frac{10^5 \times V_2}{300} $$

Solving,

$$ V_2 = \frac{1.5\times2.9\times300}{290} $$

$$ V_2 = 4.5\,cm^3 $$

Final Answer: $$ \boxed{4.5\ cm^3} $$

Explanation

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