Physical Chemistry Third Edition

408 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium

The result in Eq. (9.4-3) is a product of three factors. The factorg(v) is the probability

per unit volume in velocity space. The factor 4πv^2 is the area of one surface of the

spherical shell in Figure 9.11, and the factordvis the thickness of the shell. The product

4 πv^2 dvis the volume of the shell. The productg(v)4πv^2 dvequals the probability per

unit volume times the volume of the shell.

0 200

v/m s–1

vp

= 394 m s

–1

<v>

= 444 m s

–1

vrms

= 482 m s

–1

fv

(v

)

400 600 800 1000

Figure 9.12 The Probability Distri-
bution of Molecular Speeds for
Oxygen Molecules at 298 K.

Figure 9.12 shows this probability distribution of speeds for oxygen molecules at

298 K. The most probable speed, the mean speed, and the root-mean-square speed are

labeled on the speed axis. Compare this figure with Figure 9.7. The most probable

value of a velocity component is zero, while the most probable speed is nonzero and

the probability of zero speed is zero. This difference is due to the fact that the speed

probability density is equal to the area of the spherical shell in velocity space (equal to

4 πv^2 ) times the probability density of the velocities lying in the spherical shell. Zero

speed is improbable not because the velocity probability density is zero (it is at its

maximum value), but because the area of the spherical shell vanishes atv0.

EXAMPLE 9.8

Obtain a formula for themost probable speed,vp, by finding the value of the speed at which

the first derivative of the probability density in Eq. (9.4-4) vanishes.

dfv

dv



(

m

2 πkBT

) 3 / 2

4 π

[

2 ve−mv

(^2) / 2 kBT
−v^2 e−mv
(^2) / 2 kBT 2 mv
2 kBT
]
0 e−mv
(^2) / 2 kBT
(
2 v−
mv^3
kBT
)
e−mv
(^2) / 2 kBT
v
(
2 −
mv^2
kBT
)
This first factor of this expression vanishes whenv→∞, the second factor vanishes when
v0, and the third factor vanishes when
vp
√
2 kBT
m
(9.4-5a)
The extrema atv0 and atv→∞correspond to minimum values of the speed probability
distribution.
The most probable speed can also be written in terms of the molar mass,M, since
mM/NAvandkBR/NAv:
vp

√

2 RT

M

(9.4-5b)

EXAMPLE 9.9

Find the most probable speed of oxygen molecules at 298 K.

vp

√

2 RT

M



(

2(8.3145 J K−^1 mol−^1 )(298 K)

0 .031999 kg mol−^1

) 1 / 2

394 m s−^1