1 (463)

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 455
0 5 10 15 20 25
0
5
10
15
20
T/θ
(⟨
ε⟩
−
E 1
,0
)/
kθ
Figure 13.11
P13C.8 (a) �e mean molecular energy is given by [13C.4a–549]
⟨ε⟩ = −(1/q)(∂q/∂β)V
where β = 1/kT and q is the partition function given by [13A.11–535],
q = ∑i e−βε i . For the uniformly spaced set of energy levels, ε j = jε, the
partition function is given by [13B.2a–539], q = ∑ j e−β jε = 1/(1 − e−βε).
With these results the mean energy is calculated as
⟨ε⟩ = −1
q
dq
dβ
= −(1 − e−βε) −εe−βε
(1 − e−βε)2
= εe−βε
(1 − e−βε)
= ε
(eβε − 1)
Setting ⟨ε⟩ = aε
aε = ε/(eβε − 1)
eβε = 1 + 1/a
thus β = 1
ε
ln(1 + 1
a
)
Given ⟨ε⟩ = ε, it follows that a = 1. Because T = 1/kβ and ε = hcν̃
T = ε
k ln(2)
= hcν̃
k ln(2)
= (6.6261 × 10−34 J s) × (2.9979 × 1010 cms−1) × (50 cm−1)
(1.3806 × 10−23 JK−1) × ln(2)
= 1.0 × 102 K
(b) If ⟨ε⟩ = aε it follows from the above discussion that βε = ln(1 + 1/a),
therefore
q = 1
1 − e−βε =
1
1 − e− ln(1+1/a)
= 1
1 − a/(a + 1)
= a + 1