Given, Avogadro's number `N = 6.02 xx 10^23` and Boltzmann's constant `k = 1.38 xx 10^-23 J//K`.
(a) Calculate the average kinetic energy of translation of the molecules of an ideal gas at `0^@ C and at 100^@ C`.
(b) Also calculate the corresponding energies per mole of the gas.

Explanation

According to the kinetic theory, the average kinetic energy of translation per moecule of an ideal monoatomc gas at kelvin temperature T is `(3/2)` kT,
where k is Boltzmann's constant.
`At0^(@)C.` (T = 273 K ),
the kinetic energy of translation =`3/2`kt
`=3/2xx(1.38xx10^(-23))xx273`
`=5.65xx10^(-21)` J//"molecule"`
At `100^(@)C(T=373 K),` the energy is `3/2xx(1.38xx10^(-23))xx373=7.72xx10^(-21)J//"molecule"`
(ii) 1 mole of gas contains N `(=6.02xx10^(23))`mol
Therefore, at `0^(@)C,` the kinetic energy of translation of 1 mole of the gas id
`=(5.65xx10^(-21))(6.02xx10^(23))`
`3401J//mol` and at `100^(@)C.`
The kinetic energy of translation of 1 mol of gas is
`=(7.72xx10^(-21))(6.02xx10^(23))`
`4647J//mol`