calculate the kinetic energy of translation of molecules of nitrogen gas at STP moving with the most probable speed

1. An oxygen molecule has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at 27 ∘ C is given asE T = 23 kT= 23 ×1.38×10 −23 ×300=6.21×10 −21 J/mol2.  An oxygen molecule has total five degrees of freedom, hence its total kinetic energy is given asE T = 25 kT= 25 ×1.38×10 −23 ×300=10.35×10 −21 J/mol3. Total kinetic energy of one mole of oxygen is its internal energy, which can be given asU= 2f μRT= 25 ×1×8.314×300=6235.5J/mol1. An oxygen molecule has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at 27 ∘ C is given asE T = 23 kT= 23 ×1.38×10 −23 ×300=6.21×10 −21 J/mol2.  An oxygen molecule has total five degrees of freedom, hence its total kinetic energy is given asE T = 25 kT= 25 ×1.38×10 −23 ×300=10.35×10 −21 J/mol3. Total kinetic energy of one mole of oxygen is its internal energy, which can be given asU= 2f μRT= 25 ×1×8.314×300=6235.5J/mol

To use anything transparent