Show that kinetic energy of molecule of gas is independent of the nature of gas but depends only on the temperature of gas molecule.

Solution

Consider 1 mole of an ideal gas at absolute temperature T, of volume V and molecular weight M.
Let N be the Avogadro's number and m be the mass of each molecule of the gas.
Then,
M = m X N
Let, C be the r.m.s velocity of the gas molecules, then pressure P exerted by ideal gas is,
P = $1 third M over V C squared$ or PV = $1 third M C squared$
Now, using the gas equation
PV = RT
Therefore,
$space space space space space space space space 1 third M C squared space equals space R T space rightwards double arrow space space space 1 half space M C squared space equals space 3 over 2 space R T$
Therefore, average K.E. of translation of one mole of the gas = $1 half space M C squared space equals space 3 over 2 space R T$
i.e., $1 half mNC squared space equals space 3 over 2 space straight R space straight T space space space space space left square bracket space because space straight M space equals space mN right square bracket space rightwards double arrow space 1 half mC squared space equals space 3 over 2 space open parentheses straight R over straight N close parentheses space straight T space equals space 3 over 2 kT space space space where comma space straight k space is space called space Boltzman space constant. space Therefore comma space Average space straight K. straight E space per space molecule space of space straight a space gas space equals space 1 half mC squared space equals space 3 over 2 straight K space straight T space$
That is, the average K.E of the gas is independent of the nature of the gas and is dependent only on the temperature of the gas.