Question

Derivation of kinetic gas theory in physical chemistry

Answer 1

The result you quoted is the average translational kinetic energy for an ideal gas.

First, let's sketch out a rough derivation for the average kinetic energy of a particles of an ideal gas using nothing more than high school physics. The gas molecules just undergo translations, and don't rotate/vibrate. Also, they don't interact with each other.

Imagine a rectangular box which contains an ideal gas; the gas molecules have a speed vxvx in the xx-direction. Since the box has a uniform cross-section of area AA, a volume element spanned by a gas molecule in the box is V′=|vxΔt|AV′=|vxΔt|A. On average half the molecules move to the right, and half the molecules move to the left, since there is no preferred direction of motion. Thus, the number of molecules colliding with the container walls is

no. of molecules =12.nNaV.|vxΔt|A=12.nNaV.|vxΔt|A

The momentum transferred per molecule is 2mvx2mvx, so the total momentum transferred is ΔP=12.nNaV.|vxΔt|2mvxΔP=12.nNaV.|vxΔt|2mvx

Since force is rate of change of momentum, F=limΔt→0ΔPΔt=nMAv2xVF=limΔt→0ΔPΔt=nMAvx2V

Pressure is force per unit area, so the average pressure is P=nM⟨v2x⟩VP=nM⟨vx2⟩V

Now, note that the total velocity is v2=v2x+v2y+v2zv2=vx2+vy2+vz2, and since there is no preferred direction of motion, one can assume all three components of velocity to be equal. v2rms=3⟨v2x⟩vrms2=3⟨vx2⟩

Thus, P=nMv2rms3VP=nMvrms23V also from the ideal gas equation P=NkTVP=NkTV we get mv2rms3=kTmvrms23=kT

Ekinetic=mv2rms2=3kT2Ekinetic=mvrms22=3kT2 Which is the result you wanted.