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Modern Engineering Thermodynamics

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18.5 Molecular Velocity Distributions 735<br />

3<br />

2<br />

Population growth<br />

(α > 0)<br />

N/N 0<br />

1<br />

Population decay<br />

(α < 0)<br />

FIGURE 18.4<br />

Population growth and decay as predicted by Eq. (18.17).<br />

The problem of determining the distribution of velocities among the molecules of a gas can be thought of as a<br />

population problem, except that we are no longer interested in how the population size N varies with time, but<br />

rather how it varies with molecular velocity. Using the general form of Eq. (18.16), we can postulate a velocity<br />

distribution population model as follows:<br />

dN V<br />

= f ðVÞN (18.18)<br />

dV<br />

where α has been replaced by a more general function f(V), called the velocity distribution function. The problem<br />

now is to find the mathematical form of f(V). For example, if we assume a Gaussian velocity distribution,<br />

we have<br />

<br />

Š<br />

f ðVÞ = pffiffiffiffiffi<br />

exp − V2<br />

2π δ 2δ 2<br />

where δ is the standard deviation.<br />

By utilizing the assumptions stated at the beginning of this section, Maxwell was able to show that f(V) isnot<br />

Gaussian, but instead has the following form:<br />

f ðVÞ =<br />

4 <br />

m 3/2V<br />

pffiffiffi<br />

2 exp − mV2<br />

(18.19)<br />

π 2kT<br />

2kT<br />

Substituting Eq. (18.19) into Eqs. (18.4) and (18.5), one finds that the average and root mean square velocities<br />

have the following simple formulae:<br />

rffiffiffiffiffiffiffiffi<br />

V avg =<br />

8kT<br />

πm<br />

(18.20)<br />

V rms =<br />

rffiffiffiffiffiffiffiffi<br />

3kT<br />

m<br />

(18.21)<br />

Figure 18.5 shows the shape of the distribution function f(V) for oxygen at 300 K as described by Eq. (18.19).<br />

We call the velocity at which f(V) has a maximum the most probable velocity V mp . It is determined by setting<br />

df ðVÞ/dV = 0 and solving for V = V mp , which, using Eq. (18.19), gives<br />

rffiffiffiffiffiffiffiffi<br />

2kT<br />

V mp =<br />

(18.22)<br />

m<br />

By comparing Eqs. (18.20), (18.21), and (18.22), it is clear that V mp < V avg < V rms , as shown in Figure 18.5.<br />

Z V2<br />

0<br />

0 1 2 3 4 5 6 7 8 9 10<br />

Let dN V = NðV 1 ! V 2 Þ be the number of molecules with velocities between V 1 and V 2 . Then, it follows from<br />

V 1<br />

Eq. (18.18) that<br />

NðV 1 ! V 2 Þ<br />

N<br />

Time<br />

Z V2<br />

= f ðVÞ dV<br />

V 1<br />

(18.23)

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