Question

Calculate the average molecular kinetic energy :
(a) per kilomole, (b) per kilogram, of oxygen at ${27}^{\circ }C$.
($R=8320J/KmoleK$, Avogadro's number $=6.03\times {10}^{26}molecules/Kmole$)

Answer 1

(a) $KE=\frac{3RT}{2N}$, where $R=8320J/Kmole$, $T=27+273=300K$ and $N=$ Avogadro number $=6.03\times {10}^{26}molecules/Kmole$

or, $KE=\frac{3\left(8320\right)300}{2(6.03\times {10}^{26})}$

$=\frac{{\left(4160\right)}^{}}{(0.67\times {10}^{24})}$

$=62\times {10}^{-22}J/molecule$
(b) $K.E.=\left(\frac{3}{2}\right)kT$

or, $K.E.=\left(\frac{3}{2}\right)(1.38\times {10}^{-23}J/K)(27+273)K=6.21\times {10}^{-21}J/molecule$.