Question

An air bubble of 21 cm3 volume is at the bottom of a lake 32 m deep where the temperature is 4.0°C. The bubble rises to the surface, which is at a temperature of 24°C. Take the temperature of the bubble's air to be the same as that of the surrounding water. Just as the bubble reaches the surface, what is its volume?

Answer 1

Depth of the lake, h = 32 m

As the bubble rises through the water, the water pressure decreases, causing the bubble to expand.
Water pressure = Density of water * g * depth
Density = 1000 kg/m^2
Water pressure at depth of 32 m = 1000 * 9.8 * 32 = 3.136 * 10^5 N/m^2

Standard atmospheric pressure = 1.0 * 10^5 N/m^2

So, total pressure on the balloon at depth of 32 m = 3.136 * 10^5 N/m^2 + 1.0 * 10^5 N/m^2

= 4.136 * 10^5 N/m^2

Pressure just below the surface of the lake = atmospheric pressure = 1.0 * 10^5 N/m^2
The ratio of expansion of the volume is dependent on the ratio of the pressures in N/m^2
Ratio of expansion of the volume = (4.136 * 10^5) /(1.0 * 10^5)


Now, as the bubble rises through the water, the temperature increases from 4.0°C to 24°C, causing the bubble to expand.
The ratio of expansion of the volume is dependent on the ratio of the temperatures in °K.
4.0°C + 273 = 277.0°K
24°C + 273 = 297°K

Ratio of expansion of the volume = 297 / 277.0

New volume = original volume * (4.136 * 10^5 /1.0 * 10^5) * (297 / 277.0)

Given that the original volume of the air bubble = 21 cm^3

New volume = 21 cm^3 *(4.136 * 10^5 /1.0 * 10^5) * (297 / 277.0) = 93.13 cm^3

Therefore, when the air bubble reaches the surface, its new volume = 93.13 cm^3 (Answer)