(a) Find the average translational kinetic energy of an oxygen molecule at 27 C (300 K). (b) At what temperature would the average KE be double? (c) Express the molar value. Solution: (a) Mean KE = (3/2) k T = 1.5 x (1.38 x 10^-23) x 300 = 6.21 x 10^-21 J per molecule. (b) KE proportional to T, so double KE needs 600 K. (c) Per mole: (3/2) R T = 1.5 x 8.31 x 300 = 3740 J/mol ~ 3.74 kJ/mol.