How long does it take to equalize pressure of two gas containers? I'm looking for an equation that will tell me how long it takes for two gases at two different pressures to equalize their pressures.
As a more concrete question, Container 1 has gas at pressure P1. Container 2 has the same gas at P2. There's a hole of area A connecting the two containers. How long does it take before both containers are equalized in pressure?
If we need to know the volumes and/or temperatures as well, call them V1, V2, T1, and T2. However, I'm hoping we won't need those because I'd also like to use the equation for finding the time to zero pressure, such as a decompressing spacecraft. But perhaps in that case we can use V2 = infinity?
Does it matter what the contents of the gas are? My intuition says no, as long both containers hold gases of the same molecular composition.
Sorry I can't show my work. The only thermodynamics equation I have memorized is the ideal gas law, which isn't enough here. In addition, I know virtually no fluid-dynamics equations. A google attempt brought up not much more than how to pop your ears.
 A: You can get nothing out of equilibrium thermodynamic considerations for the rate at which pressure will equalize. What will matter is the speed of sound in the gas, as that is the rate at which density fluctuations travel in a fluid and assuming an equation of state, say $p(\rho)=\rho^{\gamma}$, the pressure is then enslaved to the density. So the sound waves then are basically pressure waves and the velocity of sound will determine the time scale of relaxation or equalization of pressure in the two containers. If the fluids are assumed to be incompressible, then the pressure equalizes instantly as the speed of sound is effectively infinite. In case of dense fluids (which are almost always assumed to be incompressible), Pascal's law dictates equalization of pressure upon contact of fluids, though diffusion will moderate the mass efflux, independent of pressure.
A: Unless the two containers are separate, i.e. have a wall sealing them off completely, the right set of tools for this question is fluid-dynamics rather than thermodynamics.
for the sealed off problem, assuming ideal gasses, the end state  for the coupled baths will be that of equal temperature. in that case it is essential you have the right number of particles and right volume for them to have equal pressure like so:
$$
T_1 =T_2 \Rightarrow \frac{P_1 V_1}{N_1 R}=\frac{P_2 V_2}{N_2 R}
$$
If you then require equal pressures you are left with the relations:
$$
\frac{V_1}{N_1}=\frac{V_2}{N_2}
$$
Which is a "fine-tuning problem".
However, if you have a wall not completely sealing off the two baths, you are better off using Navier-Stokes equations for fluid dynamics, and require a stable end-state.
NS equations are essentially force equations, and as such are time dependant.
Another way you MIGHT want to look at is using the grand canonical ensemble, which is well suited to deal with particles that are freely moving between one bath to the other. 
A: It is a case of flow through an orifice.
It depends on the shape and area of the orifice, and on the viscocity of the fluid.
At a low ratio of pressure to viscocity, flow rate is proportional to pressure.
At a high ratio of pressure to viscocity, flow rate is proportional to square root of pressure.
You're going to have to write a differential equation, and probably solve it numerically.
The pressure difference will approach, but never actually reach, zero, so you need a threshold that you label "zero".
A: I once had the equation but lost it. However, all the answers given seem to be avoiding the question. Since the flow slows as the pressure of the two vessels approach one another we are dealing with an equation of the form
$$t = C e^x$$
Sorry that's the best I can do.
