Question

Calculate the average molecular kinetic energy:
Per kilogram, of oxygen at ${27}^{o}C$
($R=8320J/KmolK$, Avogadro's number $=6.03\times {10}^{26}m$ molecules /$Kmole$)

Answer 1

$E=PV=nRT$

$=1\times 8320\times 6.03\times {10}^{26}\times 300$

$=1.505\times {10}^{33}J.$