Question

Interactive Solution 14.41 provides a model for problems of this type. The temperature near the surface of the earth is 304 K. A xenon atom (atomic mass = 131.29 u) has a kinetic energy equal to the average translational kinetic energy and is moving straight up. If the atom does not collide with any other atoms or molecules, then how high up would it go before coming to rest? Assume that the acceleration due to gravity is constant during the ascent. Number Units the tolerance is +/-2%

Answer 1

Given data.

Mass Of Xenon Gas Atom m =131.29 u

= 131.29e-3/6.022*10²³ kg

Assume Xenon is a ideal monatomic gas:
Total Kinetic Energy E_kinetic(T) = (3/2)*R*T

Since 1 Mole of any gas contains (6.022*10²³) atoms, each atomin 1 Mole of Xenon
gas at (T = 304 K) will have kinetic energy:
Calculate Kinetic Energy Per Xenon Atom

E_{kinetic_per_atom}=  (3/2)*R*T/(6.022*10²³) =
           = (3/2)*(8.314)*(304)/(6.022*10²³)
           = 6.3 e-21 Joules

By conservation of energy, single Xenon atom will travel distance"h" vertically .

Decrease In Kinetic Energy= Increase In Gravitational potential Energy

E_ki= m*g*h

h = E_ki/{m*g}
= (6.3e-21)/{(131.29e-3/6.022*10²³)*(9.8)}
= 2946.57 m