Why Kinetic energy of Gas molecules independent of Mass?

By Kinetic Theory of Gas,

$$K.E = \frac32 RT$$ (i.e it is independent of mass of the gas)

Its proof is as follows:

We know , $$P = \frac13 D\cdot v^2$$ (where $$D$$ is mass density and $$v$$ is average of the squared velocity of molecules)

Multiply both sides by $$V$$(Volume) $$PV = \frac13 Mv^2$$ Multiply and divide by $$2$$ in the rhs of the equation $$PV= \frac23 × \frac12 Mv²$$ $$PV= \frac23 × K.E ~~~~~~~~[1]$$ $$RT= \frac23 × K.E$$ $$K.E = \frac32 RT$$

But in [$$1$$] we used that $$\frac12 Mv^2= K.E$$ (i.e $$K.E$$ as a function of Mass). In the end we got $$K.E= \frac32 RT$$ (i.e $$K.E$$ is independent of Mass)

Please explain how $$K.E$$ of gases is independent of Mass

• Dupe-voters, here's your target: Kinetic Energy - dependence of mass Commented Mar 1, 2020 at 16:35
• @nitsua60 Unfortunately no one of the answers there could be considered an explanation. Commented Mar 1, 2020 at 20:44
• Doesn't R include molar mass, i.e., aren't the units proportional to kg/mole? Commented Mar 9, 2020 at 14:49

That is what thermal equilibrium is: when $$\frac{1}{\Omega} \frac{{\rm d}\Omega}{{\rm d}E}$$ is equal for two systems. Then their temperatures are the same.
So the number of microstates $$\Omega$$ att a certain energy depends on mass, but not the logarithmic derivative with respect to energy.