CK-12-Physics - Intermediate

13.2. Kinetic Theory of Temperature http://www.ck12.org

Illustrative Example 1

Determine the average velocity of random motion of an oxygen molecule at 22. 0 ◦C. The mass of an oxygen molecule

is 5. 36 × 10 −^26 kg, and the constantk= 1. 38 × 10 −^23 kgJ.

Answer:Recall that the temperature must be expressed in degrees Kelvin.

TK=TC+ 273 → 22 + 273 = 295 K.

KE=^12 m(vt)^2 =^32 kT→m(vt)^2 = 3 kT→(vt)^2 =^3 mkT→

v=

√

3 kT

m =

√

(^3) ( 1. 38 × 10 −^23 KJ)( 295 K)
( 5. 36 × 10 −^26 kg) =^477.^34 →^477
m
s
Internal Energy
The temperature of a gas is directly proportional to the average kinetic energy of the particles of the gas. But the
total kinetic energy of the molecules of a gas is a measure of the internal energy, or thermal energy of the gas.
The internal energyUof a gas is, therefore, the product of the number of moleculesNof the gas and the average
kinetic energy of each molecule:
U=N

( 1

2 mv

2 )

But according to the kinetic theory,

1

2 mv

(^2) = 3
2 kT
Therefore, the internal energy can be directly related to the temperature of a gas as
U=N

( 3

2 kT

)

→U=^32 NkT

The internal energy is, therefore, dependent upon the temperature and the number of particles within a substance.

This means that a small quantity of substance at a high temperature may have less internal energy than a large

quantity of substance at a low temperature. For example, a boiling pot of water has a higher temperature than a large

cool mountain lake. But the internal energy of the lake is considerably larger than the internal energy of the boiling

water. Even though the average kinetic energy of the molecules of the lake water is smaller than that of the boiling

water, there’s a good deal more molecules in the lake.

http://demonstrations.wolfram.com/ThermalMotionInASolid/