Calculate (a) the average kinetic energy of translation of an oxygen molecule at `27^(@) C` (b) the total kinetic energy of an oxygen molecule at `27^(@)C` (c ) the total kinetic energy in joule of one mole of oxygen at `27^(@)`. Given Avogadro's number `= 6.02 xx 10^(23)` and Boltzmann's constant `= 1.38 xx 10^(-23) J//(mol-K)`.

Explanation

An oxygen molecuel has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at `27^(@)C` is given as
`E_(T)=(3)/(2)kT=(3)/(2)xx1.38xx10^(-23)xx300`
`=6.21xx10^(-21)` J/mole