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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 221
(a)
∆E(298 K) = 5.06... × 10−6 Jm−3
e(1.44...×103 K)/(298 K) − 1
= 3.96 × 10−8 Jm−3
(b)
∆E(4 K) = 5.06... × 10−6 Jm−3
e(1.44...×103 K)/(4 K) − 1
= 5.12 × 10−163 Jm−3
P7A.4 �e Planck distribution is [7A.6a–239],
ρ(λ, T) = 8πhc
λ5 (ehc/λkT − 1)
�e value of λ at which ρ is at a maximum is found by solving dρ/dλ = 0.
dρ
dλ
= 8πhc d
dλ
(λ−5(ehc/λkT − 1)−1)
= 8πhc [(ehc/λkT − 1)−1 dλ−5
dλ
+ λ−5
d(ehc/λkT − 1)−1
dλ
]
= 8πhc [−5λ−6(ehc/λkT − 1)−1 + λ−5(ehc/λkT − 1)−2 hc
λ2kT
ehc/λkT]
= 8πhc
λ7(ehc/λkT − 1)
[5λ + hcehc/λkT
kT(ehc/λkT − 1)
]
�us, at λ = λmax,
xmaxexmax − 5(exmax − 1) = 0
where xmax = hc/λmaxkT .�is equation is solved numerically by xmax = 4.965,
giving
λmaxT = hc
4.965k
= (6.6261 × 10−34 J s) × (2.9979 × 108ms−1)
4.965 × (1.3806 × 10−23 JK−1)
= 2.9 × 10−3 K m
as in the text.
P7A.6 �e total energy absorbed by the atmosphere is 70% of the total energy incident
on the top of the atmosphere
343 Wm2 × 0.7 = 240.1 Wm2
At equilibrium, the total energy absorbed is equal to the total energy emitted,
which is determined by the Stefan–Boltzmann law
Eabs = Eem = 5.672 × 10−8(T/K)4Wm2
�is is rearranged to
T = (240.1 K4Wm2/5.672 × 10−8Wm2)1/4 = 255 K
Wien’s law [7A.1–238], λmaxT = 2.9 × 10−3 mK, is rearranged to give
λmax = (2.9 × 10−3 mK)/(255... K) = 1.14 × 10−5 m