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472 13 STATISTICAL THERMODYNAMICS
−10 −5 0 5 10
0.0
0.2
0.4
0.6
0.8
βε
S m
/R
Figure 13.19
(b) �e same calculation applies to a linear triatomic.
(c) For a non-linear triatomic there is one additional rotational mode giving
CV ,m = 3R. Hence γ = 1 + R/(3R) = 4/3 and cs = (3RT/4M)1/2 .
Taking the molar mass of air as 29.0 gmol−1 and using the value of γ for a
diatomic gives at 298 K
cs = (1.40RT
M
)
1/2
= (1.40 × (8.3145 JK−1mol−1) × (298 K)
29.0 × 10−3 kg mol−1
)
1/2
= 346 m s−1
P13E.18 It is convenient to rewrite the given expression for the energy by multiplying
the numerator and denominator by eε/kT to give
E = Nε
eε/kT + 1
(a) By de�nition CV ,m = (∂Um/∂T)V . Here Um is E with N = NA therefore
CV ,m = ( ∂
∂T
)
V
NAε
eε/kT + 1
= ε eε/kT
kT2
NAε
(eε/kT + 1)2
= NAk (
ε
kT
)
2 eε/kT
(eε/kT + 1)2
= R ( ε
kT
)
2 eε/kT
(eε/kT + 1)2
Multiplying the numerator and denominator of this expression by e−2ε/kT
gives the required expression.
(b) Figure 13.20 shows a plot of CV ,m as a function of the dimensionless pa-
rameter kT/ε.
(c) �ere is a maximum in the plot but it is not possible to �nd an analytic
expression for its position. Graphical work indicates that the maximum
is at kT/ε = 0.417 .