University Physics I - Classical Mechanics, 2019

322 CHAPTER 13. THERMODYNAMICS
So, 1 kg of aluminum gives off 44.4 kJ of thermal energy and its temperature drops almost 50 ◦ C,
from 80 ◦ Cto30.6 ◦ C, whereas 1 kg of water takes in the same amount of thermal energy and its
temperature only rises about 10.6 ◦ C.
13.6.2 Equipartition of energy
Estimate the speed of an oxygen molecule in air at room temperature (about 300 K).
Solution
Recall that in Section 13.2.2 I mentioned that the average translational kinetic energy of a molecule
in a system at a temperature T is 3 2 k BT (Eq. (13.7), where k B , Boltzmann’s constant, is equal to
1.38 × 10 −23 J/K. So, at T = 300 K, a molecule of oxygen (or of anything else, for that matter)
should have, on average, a kinetic energy of
〈K trans 〉 = 3 2 k BT = 3 2 × 1.38 × 10−23 × 300 J = 6.21 × 10 −21 J (13.24)
Since K = 1 2 mv2 , we can figure out the average value of v 2 if we know the mass of an oxygen
molecule. This is something you can look up, or derive like this: One mole of oxygen atoms has
a mass of 16 grams (16 is the atomic mass number of oxygen) and contains Avogadro’s number
of atoms, 6.02 × 10 23 . So a single atom has a mass of 0.016 kg/6.02 × 10 23 =2.66 × 10 −26 kg. A
molecule of oxygen contains two atoms, so it has twice the mass, m =5.32 × 10 −26 kg. Then,
〈v 2 〉 = 2〈K trans〉
m
= 2 × 6.21 × 10−21 J
5.32 × 10 −26 kg
=2.33 × 105
m2
s 2 (13.25)
The square root of this will give us what is called the “root mean square” velocity, or v rms :
√
v rms = 2.33 × 10 5 m2
s 2 = 483 m (13.26)
s
This is of the same order of magnitude as (but larger than) the speed of sound in air at room
temperature (about 340 m/s, as you may recall from Chapter 12).