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Modern Engineering Thermodynamics

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732 CHAPTER 18: Introduction to Statistical <strong>Thermodynamics</strong><br />

EXAMPLE 18.1 (Continued )<br />

Since the total mass of gas present is m total = m molecule N, where N is the number of molecules present, and m molecule =<br />

molecular mass/Avogadro’s number = M/N o ; for nitrogen,<br />

then,<br />

m molecule = M N o<br />

=<br />

N =<br />

28:0 kg/kgmole<br />

6:022 × 10 26 molecules/kgmole = 4:65 × 10−26 kg/molecule<br />

m<br />

1:00 kg<br />

=<br />

m molecule 4:65 × 10 −26 kg/molecule = 2:15 × 1025 molecules<br />

The total translational internal energy in 1.00 kg of nitrogen gas is<br />

U trans = 3 2 NkT = 3 h<br />

<br />

<br />

i<br />

2 ð2:15 × 1025 moleculesÞ 1:380 × 10 −23 J<br />

ð20:0 + 273:15 KÞ<br />

molecule.K<br />

= 131,000 J = 131 kJ<br />

Note that the kinetic theory internal energy is independent of gas pressure and depends only on gas temperature, which<br />

is a characteristic of ideal gas behavior.<br />

Exercises<br />

1. Determine the root mean square velocity of the nitrogen gas in Example 18.1 when the temperature is increased from<br />

20.0°C to 200.°C. Answer: V rms = 648 m/s.<br />

2. Suppose the nitrogen gas in Example 18.1 is replaced with water vapor. Determine the root mean square velocity and<br />

mass of a water vapor molecule at 20.0°C. Answer: (m molecule ) water = 2.99 × 10 −26 kg and (V rms ) water = 637 m/s.<br />

3. Determine the number of water molecules in 1.00 m 3 of saturated water vapor at 100.°C. Answer: N water = 2.00 × 10 25<br />

water molecules.<br />

18.4 INTERMOLECULAR COLLISIONS<br />

To better understand molecule-molecule collisions, imagine that all the molecules except one are frozen in<br />

space. Then, the moving molecule travels through this stationary forest of molecules colliding with them at<br />

random. This model can be further simplified if the moving molecule is enlarged to twice its normal diameter<br />

while all the stationary molecules are reduced to points of zero diameter (see Figure 18.3). The area swept out<br />

by the motion of the enlarged molecule is called its molecular collision cross-section, given by<br />

σ = πð2rÞ 2 = 4πr 2 :<br />

Moving molecule<br />

enlarged to twice<br />

its normal diameter<br />

Stationary molecules whose<br />

diameters have been shrunk<br />

to a point<br />

The moving molecule’s<br />

collision cross-section<br />

FIGURE 18.3<br />

A simplified model illustrating the molecular collision cross-section.

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