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.
No you cannot.
we can show that (via conservation of kinetic energy and momentum) the speed of the particle before and after collision is the same.
Only in the center of mass of two elastically colliding particles the momentum remains the same. Each pairwise collision has a different center of mass. In the laboratory frame, which is the frame one is trying to model the ideal gas, all the momentum might be taken by one of the particles, leaving the other motionless in the lab. This happens with billiard ball collisions all the time. See this analysis. So even if one made an experimental setup with all the particles of the ideal gas with the same speed, after the first scatter, speeds will change because they will not all be head on, there will be angles, and then the laboratory versus center of mass argument prevails.
The distribution functions for the ideal gas were given by Maxwell using simple and reasonable assumptions. Boltzman refined this.
It is a model, i.e. a theoretical formula, that has been validated by data over and over again.
Best Answer
Ok, I believe I figured this out. The pressure of each gas is proportional to the force with which the collide and the frequency of each collision. The force with which they collide is also proportional to the momentum of the particles. Therefore, the pressure is proportional to the collision frequency and momentum. The frequency of collision for the light particles is: $ {f_L} = \sqrt{x} {f_H} $, and the ratio of the pressures is $ {P_L} / {P_H} = (\sqrt{x}/x * \rho_H * \sqrt{x}{f_H}) / \rho_H * {f_H} = 1/1 = 1 $, therefore the pressures must be equal.