The initial volume and depth of the gas bubble is V0 and H respectively. The bubble has a mass M and the gas can be assumed to be ideal. The gravitational field strength is a constant g. And the density of water is Dw.
What then, is the time taken for a bubble initially at rest at a depth H underwater rise to the surface? Assuming that there is no resistive force on the bubble and the temperature is constant during the rise. Try to avoid the presence of an atmosphere above the water surface, however if needed then take the atmospheric pressure at the water surface be Patm.
F = ma
Hydrostatic Pressure = height x density of liquid x g
Ideal gas law
The Attempt at a Solution
I tried finding the net force from the buoyant force and the weight of the bubble with the aid of Archimedes. But since when the bubble rises the pressure of its surroundings decreases and its volume should rise. So the buoyant force is not constant. I use the ideal gas law to find the volume in terms of H, V0 and h(t). Where h(t) is the depth at some time t. After some work I arrived at:
d2h/dt2 + [V0HDwg/M](1/h) -g = 0
The plan was to find h(t) then equate it to 0 and solve for t which will give the time for it to rise to the surface. But it is so complicated that I believe I must have made a mistake somewhere. So I am here to seek help.