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:$$K = \frac{3}{2}kT = \frac{3}{2}(1.38 	imes 10^{-23} 	ext{ J/K})(800 	ext{ K}) = 1.66 	imes 10^{-20} 	ext{ J}$$The kinetic energy of an oxygen molecule falling from a height h is:$$K = mgh$$where m is the mass of the oxygen molecule, g is the acceleration due to gravity, and h is the height.The mass of an oxygen molecule is:$$m = \frac{32.0 	ext{ g/mol}}{6.022 	imes 10^{23} 	ext{ molecules/mol}} = 5.31 	imes 10^{-23} 	ext{ g}$$The acceleration due to gravity is:$$g = 9.81 	ext{ m/s}^2$$Setting the two expressions for kinetic energy equal to each other, we get:$$mgh = 1.66 	imes 10^{-20} 	ext{ J}$$Solving for h, we get:$$h = \frac{1.66 	imes 10^{-20} 	ext{ J}}{5.31 	imes 10^{-23} 	ext{ g} 	imes 9.81 	ext{ m/s}^2} = 31.8 	ext{ km}$$Therefore, the oxygen molecule must fall from a height of 31.8 km to have a kinetic energy equal to the average energy of an oxygen molecule at 800 K.

An Oxygen Molecule Falls in a Vacuum