Kinetic Theory : Calculation Of The Pressure Of A Gas

Let a mass M of gas be contained in a cube of side Z, and let G be the speed of the molecules, assumed provisionally to be the same for all. Let there be N particles in a unit cube. We may suppose that on the average one-third of the particles are moving (in both directions) perpendicular to each pair of faces. The number of impacts per second made by a particle on any one face is G/2L, since it traverses a distance 2L between each impact with that face. The total number of impacts per second on the face is G/2L*(NL³/3) = NGL²/6. The momentum of the molecule, of mass m, before impact is mG after impact it is -mG, hence the change of momentum is 2mG. The pressure per unit area due to all the particles is, therefore: (NGL²/6)*2mG/L², where L² is the area of the face. Thus:

p=⅓mNG².

But mNL³ = M, and L³ = V, the volume of the cube, hence, since M/V = D = density of the gas:

p = ⅓DG²

or pV = ⅓MG².

The mass of gas striking one cm² of the wall per second = ½mNū = ½Dū, where ū is the mean velocity normal to the wall, in the molecular shower. It can be shown, by the usual method of calculating average values in the integral calculus, that if the mean speed of the gas molecules is Ω, the average component in any direction is ½Ω, and that Ω = 4/√(6π)*G = 0.921G. G is called the mean square speed, and is such that the total kinetic energy of translational motion of the molecules is ½G² per gm. of gas.

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