Derive an expression for kinetic energy of gas molecules.

Explanation

According to kinetic gas equation
`PV=1/3 mnc^2`
For mole is gas , number of molecules , n=N.
When N=Avogadro's Number
Then , `PV=1/3 mNC^2`, where mH=gram molecular mass 'M' of the gas , (mN=M)
`thereforePV=1/2MC^2=2/3(1/2MC^2)=2/3E_k.....(1)`
where `E_k` is K.E. of one mole of gas .
Ideal gas equation for 1 mole of a gas is
PV=RT......(2)
From (1) and (2) we get , `2/3 E_k=RT (or) E_k=3/2 RT`
Since'R' is a constant
`E_k prop T`
IT means that, at a given temperature, 1 mole of any gas will have the same kinetic energy.
Dividing throughout by 'N' (Avogadro's number)
`E_k/N=3/2(R/N)T=3/2kT`
`R/N=k`, where k is called, Boltzman constant.
It is the gas constant per molecule.
`therefore ` K.E. of 'n' moles of gas `=nE_k=3/2 nRT`