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

a. An oxygen molecule has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at `27^(@) C` given is
`E_(T) = (3)/(2) kT = (3)/(2) xx 1.38 xx 10^(-23) xx 300`
`= 6.21 xx 10^(-21) J//"molecule"`
b. An oxygen molecule has total five degrees of freedom, hence its total kinetic energy is givne as
`E_(T) = (5)/(2) kT = (5)/(2) xx 1.38 xx 10^(-23) xx 300`
`= 10.35 xx 10^(-21) J//"molecule"`
c. Total kinetic energy of one mole of oxygen is its internal energy, which can be given as
`U = (f)/(2) mu RT = (5)/(2) xx 1 xx 8.314 xx 300 = 6235.5 J//"mole"`