What is the average translational kinetic energy of an oxygen molecule at 600 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)
**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 600 = 1.242 × 10⁻²⁰ J. Substituting values gives 1.242 × 10⁻²⁰ J, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.
Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation