Consider a situation where two gases occupying volumes $V_1$ and $V_2$ and having partial pressures $p_1$ and $p_2$, respectively, are kept separated in a rigid container. If the separator is removed, is it true that the first gas will occupy the whole volume and, similarly, the second gas will occupy the whole volume $(V_1 + V_2)$?

If yes, why?

If no, is the reason that the molecules of each gas needs some fixed space so in general they will occupy some quantity $V^ʹ$ and $V^{ʹʹ}$ in space such that $V^ʹ + V^{ʹʹ} = V_1 + V_2$?


1 Answer 1


One handy way of visualizing the behaviour of gasses in a container is to imagine them as ping-pong balls continually bouncing around in a vacuum sealed room, without gravity or friction. Pressure on the walls is caused by their bouncing off them, and room temperature is related to their average speed.

In the described situation, imagine that one gas is represented by red balls, and the other by blue balls.

At the instant the barrier is removed, all of one side of the room will be red and all of the other blue. Gradually, more of the balls of one colour will move into the area occupied by the other. Eventually, each colour will be uniformly distributed throughout the container.

In studying the behaviour or characteristics of one specific colour (type of gas), one can totally ignore the presence of any other colours.

The only effect of interaction between colours is mid-air collisions, which randomly transfer momentum and velocity between individual balls, but which make no over-all difference once a long-term stable state has been reached other than equalizing the temperatures of each of the colours.

And if different coloured balls happen to have different masses, again the only effect this has is in the transfer of momentum and velocity, and again it has no significance in the long-term.

The pressure caused by one colour is known as its partial pressure. The sum of the partial pressures of each of the colours always adds up to the total pressure (proof: turn out the light and all balls are the same colour).


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