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147
CHAPTER 16
16.1 W =
j
g
g N N
j
j j j
!
( )! !−
n
W =
j
[
n
gj! -
n
(gj – Nj)! -
n
Nj!]
Employ Sterling’s formula.
( )
N
g N nN
g N
j
j
j j
j j
−
+−
2
n W
Nj
=
g
g N
j
j j
−
-
g
g N
j
j j
−
+
n
N
g N
j
j j
−
n W
Nj
= -
n
g
N
j
j
−
1
For W to be a maximum value,
d(
n
W) =
j
( )n W
Nj
dNj =
j
n
g
N
j
j
−
1
dNj = 0
Also:
148
j
dNj = 0
and
e e
ej
'+1
16.2 We recognize that both entropy, S, and thermodynamic probability, W, are measures of the
disorder of a system. Hence, there is a relation between S and W which was first established by
Boltzmann as follows:
S = k
n
W (P1)
g
j
N
j
150
Q
Tv
Hence, Eq. (P7) becomes
Q
n Q
VT
From Eq. (P7)
S
n Q
1
E
16.3 p = N k T
n Q
TT
n
Q =
n
Qtrans +
n
Qrot +
n
Qvib +
n
Qe
The only partition function that depends on V is Qtrans.
2
3 2
mkT
/
2
3 2
mkT
/
152
Divide by the mass, M.
pv = n K T
16.4 Qtrans =
j
kT
ejtrans
−
/
Where the sum is over all states rather than over all levels multiplied by the statistical weights.
Both are equivalent.
h2
n
n
n
1
2
2
2
3
2
153
In the limit, as n → , the area under the curve and the area of the sum of rectangles approach
each other. Hence, the above summations can be replaced by integrals.
From a table of standard integrals:
2
h
16.5 Qrot =
jikT
g e rot
−
/
h
2
154
h
IkTJ J− +
2
2
8
(!)
h J
IkT
−2 2
2
8
2
h
16.6 (a) First, determine the total number of molecules in one kilogram of N2. One kilogram-
mole of N2 has a mass of 28 kilograms (by definition). Hence, the number of moles in one
155
e
jkT−
/
1
7
2
hkT
ehkT
/
/−
1
R = R/M =
8314
28
joule kg mole K
kg kg mole
/
/
−
−
= 297 joule/kg K
h
k
=
3389K
157
T
7
2RT
3389
1
3389
/
/
T
eRT
T−
h
(K) (joule/kg) (joule/kg) (joule/kg)
300 3.12 x 105 1.25 x 101 3.12 x 105
158
7
h
kT e
h kT
2
/
159
16.7 If h/k vib, then Eqs. (16.49) and (16.51) become
3
vib
T
/
2
( )
T
e
vib
−
2
1
/
T
16.8 The thermodynamic probability for each species in the Boltzmann limit, where Nj << gj,
is given by Eq. (P3) in the solution for problem 13.2 .
g
j
A
160
g
j
B
j
N
j
AB
The total thermodynamic probability for the mixture of A, B and AB is
j
Nj
A
j
Nj
B
j
Nj
AB
where from Eq. (Q1) – (Q3),
n W
n W
A
g
j
A
Nj
B
N
j
B
Nj
AB
N
j
AB
Substituting into Eq. (Q4)
g
j
A
g
j
B
g
j
AB
j
j
161
N
j
AB
From Eqs. (Q9) – (Q11),
− +( )
'
162
g e
j
A
jAoA
− +
( )
Q
A
which is Eq. (13.54a)
From Eq. (Q13)
−
−
jBBo
−− −
or
