If two ideal gases with different molecular weights (i.e. nitrogen and argon) are injected into a closed container (of any given size, but assume ~1L), what will happen over time, assuming all external variables are held constant (no movement, no temperature change, no leaks, etc.)?
a) The two gases will mix evenly within the container; or,
b) the two gases will separate, so that the heavier gas is "on the bottom"?
Would the relative volume of the two gases have any effect one way or the other?
Further, assuming the two gases are initially in a "random/agitated" state (as a result of the physical forces of them being injected into the container), how long would it take for them to reach their end-state?
Would it make any difference if the two gases were injected together vs. one gas being present in the container and the second gas being injected afterward, displacing some of the "original" gas?
Best Answer
At any temperature $T$ the equipartition theorem tells us that a molecule in an ideal gas will have a kinetic energy of order $\tfrac{3}{2}kT$. Since kinetic energy is related to velocity by $E = \tfrac{1}{2}mv^2$ we can use this to estimate the velocity of gas molecules at room temperature, and for argon and nitrogen this works out to around 400 to 500 m/s.
So the gas molecules in your container are all whizzing around at hundreds of metres per second, and unsurprisingly this means they mix with each other. The speeds are so high that the different weights of the two types of molecule has essentially no effect in separating them.
Gravity does have an effect, and this is described by the barometric equation:
$$ \frac{\rho}{\rho_0} = \exp\left( \frac{-gM(h - h_0)}{RT} \right) $$
This relates the density $\rho$ to height $h$. $g$ is the gravitational acceleration and $M$ is the molar mass of the gas. Let's take your one litre container, in which case the height of the container is around 10cm, and the molar mass of nitrogen is 0.028kg. If we feed these values into the barometric equation we find the density ratio between the top and bottom of the container at room temperature is:
$$ \frac{\rho}{\rho_0} \approx 0.999989 $$
This is for nitrogen. The molar mass of argon is larger at 0.04kg, but this only reduces the ratio $\rho/\rho_0$ to 0.999984. So although in principle the argon to nitrogen ratio changes slightly from the bottom to the top of you container the change is vanishingly small. In practice the gas composition remains uniform.
If you make your container very large, e.g. about the thickness of Earth's atmosphere, then the change in composition does become measurably large, which is why the composition of Earth's atmosphere does change with height.
The second part of your question is a bit different since it starts with separated gases and asks how quickly they would mix. Given the gas molecules are travelling so fast you might think the mixing would be extremely rapid, but in fact this isn't the case. The trouble is that gas molecules collide with each other and ricochet back in random directions. The average distance a gas molecule travels before hitting another molecule is called the mean free path, and at room temperature and pressure the mean free path is around 0.1 microns.
So even though a nitrogen molecule is moving at about 500 m/s it's moving at random not in a straight line. If you start with two gases separated into different layers then the mixing is surprisingly slow. In fact as discussed in the question Why is it difficult to mix helium and nitrogen gases? it can be so slow that it takes days or weeks to happen.
Assuming you aren't stirring the mixture in any way the mixing takes place by diffusion, and will be described by the diffusion equation.