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

State 1 (Bottom): $P_1 = P_0 + \rho g h = 10^5 + 10^3(10)(5) = 1.5 \times 10^5$ Pa.
$T_1 = 17 + 273 = 290$ K. $V_1 = 2.9$ cm$^3$.
State 2 (Surface): $P_2 = 10^5$ Pa. $T_2 = 300$ K.
Ideal Gas Law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
$V_2 = V_1 \frac{P_1}{P_2} \frac{T_2}{T_1} = 2.9 \left( \frac{1.5}{1} \right) \left( \frac{300}{290} \right)$.
$V_2 = 2.9 \times 1.5 \times \frac{30}{29} = 1.5 \times 3 = 4.5$ cm$^3$.