Molecular speeds: An oxygen molecule falls in a vacuum. From what height must it fall so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at 800 K? (The Boltzmann constant is 1.38 × 10^-23 J/K, the molecular weight of oxygen is 32.0 g/mol, and Avogadro's number is 6.022 × 10²³ molecules/mol.) A) 31.8 km B) 10.6 km C) 21.1 km D) 42.3 km

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The average kinetic energy of an oxygen molecule at 800 K is:$$KE = \frac{3}{2}kT = \frac{3}{2}(1.38 	imes 10^{-23} 	ext{ J/K})(800 	ext{ K}) = 1.66 	imes 10^{-20} 	ext{ J}$$The potential energy of the oxygen molecule at a height h is:$$PE = mgh = (32.0 	ext{ g/mol})(9.81 	ext{ m/s}^2)h$$Setting the kinetic energy equal to the potential energy, we get:$$1.66 	imes 10^{-20} 	ext{ J} = (32.0 	ext{ g/mol})(9.81 	ext{ m/s}^2)h$$Solving for h, we get:$$h = \frac{1.66 	imes 10^{-20} 	ext{ J}}{(32.0 	ext{ g/mol})(9.81 	ext{ m/s}^2)} = 31.8 	ext{ km}$$

Molecular Speeds: an Oxygen Molecule Falls in a Vacuum