Physical Chemistry Third Edition

9.3 The Velocity Probability Distribution 401

Solution

<κx>

1

2

kBT

〈

κ^2 x

〉



(m

2

) 2 〈

v^4 x

〉



(m

2

) 2 ( m

2 πkBT

) 1 / 2 ∫∞

−∞

v^4 xe−mv

(^2) x/ 2 kBT
dvx
 2
(m
2
) 2 ( m
2 πkBT
) 1 / 2 ∫∞
0
v^4 xe−mv
x^2 /^2 kBT
dvx
 2
(m
2
) 2 ( m
2 πkBT
) 1 / 2
3
8
√
π
(
2 kBT
m
) 5 / 2

3
16
( 2 kBT)^2 
3
4
k^2 BT^2
σκx
(
3
4
k^2 BT^2 −
1
4
k^2 BT^2
) 1 / 2

kBT
√
2
Exercise 9.8
a.Evaluateσκxfor oxygen molecules at 298 K.
b.Express this standard deviation as a fraction of thevxcontribution to the kinetic energy.
The probability thatulies between two valuesaandbcan be obtained by an
integration:
(
probability that
a<u<b

)



∫b

a

f(u)du (normalized distribution) (9.3-37)

The Gaussian probability distribution is an integrand function for which no antideriva-

tive has been found, so that a probability integral such as in Eq. (9.3-37) for a Gaussian

distribution must be carried out numerically or looked up in a table of values.

EXAMPLE 9.5

Find the probability thatvxlies between 0 and 500 m s−^1 for neon atoms at 300 K.

Solution

(probability)

(

m

2 pkBT

) 1 / 2 ∫500 m/s

0

e−mv

x^2 /^2 kBT

dvx

We change variables, letting

t(mv^2 x/ 2 kBT)^1 /^2 ; dt(m/ 2 kBT)^1 /^2 dvx

At the upper limit, we have

t

√

(3. 35 × 10 −^26 kg)(500 ms−^1 )^2

(2)(1. 3807 × 10 −^23 JK−^1 )(300 K)

 1. 0055

so that

(probability)

1

√

π

∫ 1. 0055

0

e−t

2

dt