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?
A)31,800 meters
B)10,600 meters
C)21,100 meters
D)42,300 meters

Question

Quizplus · Accepted Answer

The average kinetic energy of an oxygen molecule at 800 K is:$$K = (3/2)kT = (3/2)(1.38 	imes 10^{-23} 	ext{ J/K})(800 	ext{ K}) = 1.66 	imes 10^{-20} 	ext{ J}$$The kinetic energy of the oxygen molecule at the bottom of the fall is:$$K = mgh$$where m is the mass of the oxygen molecule, g is the acceleration due to gravity, and h is the height of the fall.Setting these two equations equal to each other, we get:$$mgh = 1.66 	imes 10^{-20} 	ext{ J}$$Solving for h, we get:$$h = \frac{1.66 	imes 10^{-20} 	ext{ J}}{mg}$$The mass of an oxygen molecule is:$$m = 2(16 	ext{ amu}) = 32 	ext{ amu} = 5.31 	imes 10^{-26} 	ext{ kg}$$The acceleration due to gravity is:$$g = 9.81 	ext{ m/s}^2$$Substituting these values into the equation for h, we get:$$h = \frac{1.66 	imes 10^{-20} 	ext{ J}}{(5.31 	imes 10^{-26} 	ext{ kg})(9.81 	ext{ m/s}^2)} = 31,800 	ext{ meters}$$Therefore, the oxygen molecule must fall from a height of 31,800 meters so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at 800 K.

An Oxygen Molecule Falls in a Vacuum