A bubble of air is rising up through the ocean. When it is at a depth of 20.0 m below the surface, where the temperature is 5.00°C, its volume is 0.80 cm3. What is the bubble's volume (in cm3) just before it hits the ocean surface, where the temperature is 20.0°C? Assume the average density of sea water is 1,025 kg/m3. Hint: Use Pascal's Principle (textbook Eq. 14.4) to determine the pressure at the depth of 20.0 m below the surface.

the volume is 0.253 cm³

Explanation:

The pressure underwater is related with the pressure in the surface through Pascal's law:

P (h) = Po + ρgh

where Po = pressure at a depth h under the surface (we assume = 1atm=101325 Pa), ρ = density of water, g = gravity, h = depth at h meters)

replacing values

P (h) = Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa

Also assuming that the bubble behaves as an ideal gas

PV=nRT

where

P = absolute pressure, V = gas volume, n = number of moles of gas, R = ideal gas constant, T = absolute temperature

therefore assuming that the mass of the bubble is the same (it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have

at the surface) PoVo=nRTo

at the depth h) PV=nRT

dividing both equations

(P/Po) (V/Vo) = (T/To)

or

V=Vo * (Po/P) (T/To) = 0.80 cm³ * (101325 Pa/302225 Pa) * (277K/293K) = 0.253 cm³

V = 0.253 cm³