Question

Calculate

1. the average kinetic energy of translation of an oxygen molecule at ${27}^{\circ }C$,
2. the total kinetic energy of an oxygen molecule at ${27}^{\circ }C$,
3. the total kinetic energy in joule of one mole of oxygen at ${27}^{\circ }C$.

Given Avogadro's number$=6.02\times {10}^{23}$ and Boltzmann's constant$=1.38\times {10}^{-23}J/(mol-K)$.

Answer 1

1. An oxygen molecule has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at ${27}^{\circ }C$ is given as
$E_T=\frac{3}{2}kT=\frac{3}{2}\times 1.38\times {10}^{-23}\times 300$
$=6.21\times {10}^{-21}J/mol$
2. An oxygen molecule has total five degrees of freedom, hence its total kinetic energy is given as
$E_T=\frac{5}{2}kT=\frac{5}{2}\times 1.38\times {10}^{-23}\times 300$
$=10.35\times {10}^{-21}J/mol$
3. Total kinetic energy of one mole of oxygen is its internal energy, which can be given as
$U=\frac{f}{2}\mu RT=\frac{5}{2}\times 1\times 8.314\times 300=6235.5J/mol$