Modern Engineering Thermodynamics

U trans = 3 2 NkT (Continued )
18.3 Kinetic Theory of Gases 731
k = Boltzmann’s constant, 1.380 × 10 −23 J/(molecule·K)
m = M/N o , which is the mass of one molecule of the gas
Combining Eqs. (18.11) and (18.12), we find that pV = NkT, whereN ¼ m T =m and N is the total number of
molecules. Also, combining the second half of Eq. (18.11) and the first and last parts of Eq. (18.12), we get
V rms =
ffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi
p 3kT
3RT =
(18.13)
m
therefore, from Eq. (18.6), the kinetic theory interpretation of the translational total internal energy of an ideal gas is
U trans = 1 2 NmV rms 2 = 3 NkT (18.14)
2
We see from Eq. (18.14) that U trans depends only on temperature, which is in agreement with our definition of
an ideal gas given in Chapter 3.
EXAMPLE 18.1
The qualifying examination for a very exclusive preschool in the future to which you want to send your child requires that your
child answer questions on the kinetic theory of gases. For example, every five-year-old child must (Figure 18.2) be able to compute
a. The root mean square molecular velocity.
b. The total translational internal energy.
for 1.00 kg of nitrogen gas at 20.0°C. How is this done?
FIGURE 18.2
Example 18.1.
Solution
a. The kinetic theory root mean square molecular velocity is given by Eq. (18.13) as
p
V rms =
ffiffiffiffiffiffiffiffiffi
3RT
For nitrogen, R = 296 J/kg⋅K = 296 N ⋅ m/kg⋅K = 296 m 2 /s 2 ⋅K (from Table C.15b in Thermodynamic Tables to accompany
Modern Engineering Thermodynamics), then Eq. (18.13) gives
p
V rms =
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3½296m 2 /ðs 2 . KÞŠð20:0 + 273:15KÞ = 510:m/s
b. The kinetic theory total translational internal energy of a gas is given by Eq. (18.14) as