Why does the speed of an air bubble rising in a liquid increase? An air bubble rises from the bottom of a tank containing a viscous fluid. 
I think it speed should continuously decrease, because while moving upward, the bubble has to go against the gravity too. So, it will decrease continuously. But it’s said that the speed increases up to a certain limit and then becomes constant.
 A: Essentially, there are two forces acting on your bubble:


*

*Buoyancy, which pushes the bubble upwards. This is an effect of gravity pushing the fluid downwards more strongly than the bubble.

*Forces through viscosity, which is caused by movement, acts in the opposite direction, and increases with speed.
So, unless the bubble moves, there is no force pushing it downwards. Once the bubble is accelerated upwards by buoyancy, it experiences forces through viscosity that decrease the acceleration until both forces are in equilibrium, at which point the bubble acquires its maximum speed. So, the speed does not decrease, only the acceleration.
A: The speed will increase in 2 steps.
First and shorter step:
 is the initial acceleration as most answers here refers to until it reach terminal velocity which might again increase under certain circumstances (step 2)
Second step:
If the bubble was first released with a certain size (kinda small) and under great depth. The bubble will increase in size as it ascends due to the fact that it was containing compressed gas equal to the fluid pressure when it was first released. The bigger size will be faster to ascend (Google the reason for that)
A: A bubble in a viscous liquid experiences two forces:


*

*BUOYANCY. This is approximately constant - equal to the weight of the displaced liquid. In fact, the bubble will expand (as the hydrostatic pressure decreases as the bubble gets closer to the surface) so this force will increase as the bubble approaches the surface

*DRAG. Depending on the Reynolds number of the flow (laminar or turbulent) this force will either scale linearly, or quadratically, with velocity. Either way - the faster the bubble goes, the greater the force of drag.


When the two forces are in equilibrium, the velocity of the bubble will not change. When buoyancy is larger than drag (which will be the case for sufficiently small velocity - i.e. definitely at the start of the bubble's rise) the bubble will accelerate; as the speed increases, the drag increases also.
Incidentally, for laminar flow it is possible to calculate the "inertia" of the spherical bubble. Even though the bubble is filled with gas, and you would expect it to have very low inertia (mass), it will not accelerate very rapidly. The reason is that as the bubble moves up, liquid has to flow down and around the bubble. If the bubble accelerates, the liquid around it must also accelerate. It turns out that there is an exact mathematical solution for the case of laminar flow around a sphere, and that the inertia of the sphere is equal to half the mass of the displaced liquid. And since the weight of the liquid is equal to the force of buoyancy, it follows that the initial acceleration of the bubble is $2g$ - it "falls up" at twice the acceleration of gravity. But it will reach a steady state velocity pretty quickly. And then, as the bubble gets closer to the surface and expands slightly, there will be a further acceleration (which is normally ignored).
Mathematically, Stokes drag for a sphere is given by 
$$F = 6\pi~\mu~ R ~v$$
Where $\mu$ is the dynamic viscosity, $R$ the radius of the sphere, and $v$ the velocity. This is valid for laminar flow - that is, when the viscous forces are greater than the inertial forces in the liquid. This is usually described by the Reynolds number,
$$Re = \frac{\rho v L}{\mu}$$
Where $L$ is the "characteristic length scale" (which you can consider equal to the diameter of the sphere, $2R$). "Purely laminar flow" is considered to exist when the Reynolds number is less than 10; at much higher values, flow will be turbulent and drag will be given by
$$F_d = \frac12 \rho v^2 A C_D$$
Where $A$ is the projected area ($\pi R^2$ for a sphere) and $C_D$ is the drag coefficient (approximately 0.5 for a sphere, but that depends again on the Reynolds number).
Combining this with the buoyancy equation:
$$F_b = \frac43 \pi R^3 \rho g$$
Where $\rho$ is the density of the liquid (assuming we can ignore the mass of the air in the bubble) and $g$ is the gravitational acceleration. We can then solve for the terminal velocity of the bubble:
$$6\pi \mu R v = \frac43 \pi R^3 \rho g\\
v= \frac29 \frac{\rho g R^2}{\mu}$$
You can google "terminal velocity" or "Stokes drag" to learn much more about these things. See for example Stokes' Law and Reynolds number
