# I KTG general proof

1. Jul 19, 2016

### Shreyas Samudra

While deriving ideal gas equation - we take gas molecules to be contained in a cubical container (convinent shape) , but how do we derive it for a gas inside some arbitarily shaped container ?
i think this has 2 answers
1) Using maths - but it will be mostly impossible
2) or it will be a therotical proof (this is what i am intrested to know)

i could think of the ''therotical'' proof - Check it wheather its right

inside an arbitarily shaped container we can assume an imaginary cube shaped container (with imaginary boundaries , of course ! )
and i think that, the existence of those imaginary boundaries can be justified owing to large number of intermolecular collisions , in sense - the place where we have assumed the imaginary boundaries of the cube-shaped container , right there the randomly moving molecules are colliding so fast and so much that , resultantly we can assume a wall existing right there.

And hence again do the same cubical container proof

So is this right ?

And some other ideas of anybody ?

2. Jul 21, 2016

### Shreyas Samudra

??

3. Jul 24, 2016

### Andrew Mason

Interesting question, Shreyas.

Here are a couple of ways of looking at it:

1.We know from Newton that the centre of mass of the container and gas within it cannot change due to the thermal motion of the gas molecules. So for collisions with the walls of whatever shape, total components of momentum of all molecules before and after all impacts must average out to be zero, regardless of the container shape. Otherwise, the centre of mass would move. So the sum of all changes in momentum must be 0. And since there is not even a detectible jiggle of the container, all collisions must be completely random over very small periods of time.

So we assume that the direction of motion of gas molecules relative to the container walls is completely random. If it is always completely random, can you make it more random by arbitrarily changing the shape of the container walls?

I would say not. To find pressure, we are concerned only with the average change of momentum of a molecule on impact with the wall and that change is necessarily in the direction perpendicular to the plane that is tangent to the container wall at the point of impact. Since direction of a molecule before impact is completely random, if we make the angle of impact with the container wall at any point completely random, we will still end up with completely random collisions between the molecules and the container wall.

2. Once you prove that PV=nRT using the flat sided container one can see that it is the volume, not the shape of the container walls, that determines the equation of state.

AM

4. Jul 25, 2016

### Philip Wood

The approach I prefer is to consider the impact of molecules on a small patch (area A) of container wall. The container can be any shape you like. Let the outward normal to this patch be in the x direction. Then the molecules with velocity components u1, v1, w1 (say) that will hit A in time Delta t will be contained in an oblique cylinder of 'height' u1Delta t and 'base' area A, so if there are N1 molecules of this species in the container of volume V, there will be (N1/V)A u1Delta t molecules in the cylinder and the total x-momentum they bring up to the wall in time Delta t will be (N1/V)A mu12Delta t. It's not too difficult to proceed from there in the usual way. I found this derivation in a forgotten classic The Kinetic Theory of Gases by Sir James Jeans.

Last edited: Jul 25, 2016