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452 13 STATISTICAL THERMODYNAMICS
�e equipartition value is ⟨εV⟩ = 9kT , because there are two quadratic terms
for a harmonic oscillator, and nine modes in total. �ese two expressions for
the energy are plotted as a function ofT in Fig. 13.9.�e value from the equipar-
tition theorem comes within 5% of the exact value at 3.92 × 103 K .
0 1 000 2 000 3 000 4 000
0.0
20.0
40.0
T/K
⟨ε
V
⟩×
10
20
/J
exact
equipartition
Figure 13.9
E13C.7(b) �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 [13B.1b–538], q =
∑i g ie−βε i , where g i is degeneracy and the corresponding energy is given as
ε i = hcν̃ i . At T = 2000 K
βhc = (6.6261 × 10−34 J s) × (2.9979 × 1010 cms−1)
(1.3806 × 10−23 JK−1) × (1900 K)
= 7.19... × 10−4 cm
�erefore the electronic partition function is
qE = g0 + g1e−βhc ν̃1 + g2e−βhc ν̃2 = 5.80...
�erefore the mean energy is
⟨εE⟩ = − 1
qE
(
∂qE
∂β
)
V
= hc
qE
(g1 ν̃1e−βhc ν̃1 + g2 ν̃2e−βhc ν̃2)
= (6.6261 × 10−34 J s) × (2.9979 × 1010 cms−1)
5.80...
× [1 × (850 cm−1) × e−(7.19...×10
−4 cm)×(850 cm−1)
+5 × (1100 cm−1) × e−(7.19...×10
−4 cm)×(1100 cm−1)]
= 1.01 × 10−20 J
Solutions to problems
P13C.2 �e mean molecular energy is given by [13C.2–549], ⟨ε⟩ = (1/q)∑i ε ie−βε i ,
where β = 1/kT , and q is the partition function given by [13A.11–535], q =