A bubble having a diameter of 1.00 cm is released from the bottom of a swimming pool where the depth is 5.00 m. What will the diameter of the bubble be when it reaches the surface? The temperature of the water at the surface is 20.0°C, whereas it is 15.0°C at the bottom. (The density of water is 1.00 × 10³ kg/m³.)

Question

Quizplus · Accepted Answer

Using the ideal gas law, we have: 
P1V1/T1 = P2V2/T2
where P1 = P2 = atmospheric pressure (since the bubble reaches the surface)
V1 = (1/2)^3 π = 0.524 cm^3 (since the diameter is 1 cm)
T1 = 15°C + 273.15 = 288.15 K
T2 = 20°C + 273.15 = 293.15 K
Now, we need to find V2, the volume of the bubble at the surface. To find the pressure at the bottom of the pool, we use P = ρgh, where ρ is the density of water, g is the acceleration due to gravity, and h is the depth of the pool. Thus, 
P1 = ρgh = (1.00 × 10^3 kg/m^3)(9.81 m/s^2)(5.00 m) = 4.91×10^4 Pa
Plugging in all the values, we get:
(4.91×10^4 Pa)(0.524 cm^3)/(288.15 K) = (1 atm)(V2)/(293.15 K)
Solving for V2, we get V2 = 0.570 cm^3. 
Thus, the new diameter is given by
V2 = (1/2)^3 π = 0.570 cm^3 
= (d/2)^3 π 
Solving for d, we get d = 1.15 cm.

A Bubble Having a Diameter of 1