Question

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$^3$.

• A. 2.0
• B. 3.0
• C. 4.5
• D. 4.2

Hint:

Remember to convert Celsius to Kelvin. Pressure increases with depth: $P = P_{atm} + h\rho g$.

Answer 1

The Correct Option is C

To solve this problem, we will use the ideal gas law, assuming the process is isothermal and the pressure is the same at the bottom and top of the pool. The relationship between the initial and final states of the bubble is given by the formula for buoyancy due to change in temperature and pressure.

The relationship can be described by the equation:

\(P_1 V_1 / T_1 = P_2 V_2 / T_2\) 

Where:

• \(P_1\) and \(P_2\) are the pressures at the bottom and surface of the pool respectively.
• \(V_1\) is the initial volume of the bubble (2.9 cm³).
• \(V_2\) is the final volume of the bubble.
• \(T_1\) is the initial temperature (273 + 17 K).
• \(T_2\) is the final temperature (273 + 27 K).

Assuming the pressure at both levels is nearly equal due to low water pressure at 5 meters depth compared to atmospheric pressure effects, this equation can be simplified for temperature changes as \(P_1 ≈ P_2\):

\(V_1 / T_1 = V_2 / T_2\)

Plugging in the values:

\(2.9 / (273 + 17) = V_2 / (273 + 27)\)

Solving for \(V_2\):

\(V_2 = 2.9 \times \frac{(273 + 27)}{(273 + 17)}\)

Calculating numerically:

\(V_2 = 2.9 \times \frac{300}{290}\)

\(V_2 = 2.9 \times \frac{30}{29} ≈ 2.9 \times 1.0345 ≈ 2.9 \times 1.0345 ≈ 4.5 \, \text{cm}^3\)

Therefore, the volume of the bubble when it reaches the surface is approximately 4.5 cm³, which matches the option

4.5

.

 

Hence, the correct answer is 4.5 cm³.