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

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

1 Answer 1


Because relative change in the number of microstates with energy does not depend on mass.

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.


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