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

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 4.5

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):

P_1 = P_{atm} + \rho gh = 10^5 + 10^3(10)(5) = 1.5 \times 10^5

Pa
At surface:

P_2 = 10^5

Pa
Initial temperature:

T_1 = 17°C = 290

K, Final temperature:

T_2 = 27°C = 300

K
Using ideal gas law:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 2.9 \times \frac{1.5 \times 10^5}{10^5} \times \frac{300}{290}

= 2.9 \times 1.5 \times \frac{30}{29} = 2.9 \times 1.5 \times 1.034 \approx 4.5

cm³

An air bubble of volume 2.9 ~cm^ 3 rises from the bottom of a...

Options

Correct answer

Step-by-step solution

More from Mechanical Properties of Fluids