14.2) Show that for an assembly of N particles that obeys Maxwell-Boltzmann statistics,
the occupation numbers for the most probable distribution are given by:
 ∂ ln Z 

N J = − NkT 

 ∂ε j T
re − organized the above equation
 ∂ ln Z 

N J = − NkT 

 ∂ε j 
Z = ∑ gJ ⋅ e
−ε J
kT
 ∂Z

 ∂ε J
−ε

1 
 = g j ⋅ e kT ×  −

 kT 
T
−ε
NkT
 1 
NJ = −
× g j ⋅ e kT ×  −

Z
 kT 
−ε
N
= ⋅ g j ⋅ e kT (this is the MB distribution. The most probable one )
Z
J
J
J
14.3a) Show that for an ideal gas of N molecules,
g j Z εkT
Z (kT ) 2  2πm 
= e where =


Nj N
N
P  h2 
5
J
Sol:
3
2
PV = nRT = NkT
V kT
=
N
P
The partition function (14.12 )
3
Z V  2πmkT 
 2πmkT  2
= 
Z =V


2
h
N
N  h2 


From MB distribution
3
2
N J N −kTε
g J Z +kTε
= ⋅e →
= ⋅e
gJ Z
NJ N
J
J
g J kT  2πmkT  2 εkT (kT ) 2  2πmkT 
=

 ⋅e =


2
NJ
P  h
P  h2 

3
5
J
14.3b) for ε J
3
2
( )
= 3 kT , T = 300k , P = 103 Pa, and m = 10 −26 kg , calculate
2
gJ
NJ
Sol:
(kT )
P
5
2
3
 2πmkT  2 εkT
×
 ⋅e
2
h


J
(1.38 × 300 × 10 )
=
5
3
− 26
 2π × 10  2 3 2
×
 ⋅e
2
− 68
103
 6.62 × 10 
= 1.1048 × 10 −54 × 0.0542 × 1063 × 4.538
− 23
2
= 2.67 × 108
14.4) calculate the entropy of a kilomole of neon gas at standard temperature and
−26
pressure. The mass of a neon atom is 3.35 × 10 kg
Sol:
Use equation 14.14
3
 5
V (2πmkT ) 2  
S = NK  + ln 

3
Nh
 2

 
3
− 26
− 23
2 
 5
 22.4
×
×
×
×
×
(
)
π
2
3
.
35
10
1
.
381
10
273


= nR  + ln
×
26
− 34 3

(6.62 ×10 )
 2
 6.02 ×10

(
 5
 22.4 × (2π × 3.35 ×1.381× 27.3)3 2
= nR  + ln 
6.02 × 6.62 3 ×10 −76
 2

) ×10


− 72
 

 
5

= 8.314 J ⋅ kilomole + ln (1.608 ×1013 )
2

= 273.6 J ⋅ kilomole
14.5) calculate the chemical potential in electron volts for a kilomole of helium gas at
−27
standard temperature and pressure. The mass of a helium atom is 6.65 × 10 kg
Sol:
µ = − KT ln
Z
N
(14.3)
3
 2πmkT  2
Z =V

2
 h

3
 V  2πmkT  2 

µ = −kT ln 
 N  h 2  


at STP 1 kilomole gas has 22.4m 3 volume N = 6.02 × 10 26
 22.4
 2π × 6.65 × 10 −27 × 1.381× 10 −23 × 273 
− 23
×
µ = −1.381 × 10 × 273 × ln 

26
− 34 2
×
6
.
02
10
(
)
×
6
.
62
10



3
2



 22.4 × (2π × 6.65 × 10 −27 × 1.381× 10 −23 × 273) 2 × 10 −7
= −8.617 × 10 eVk × 273 × ln 
6.02 × 6.62 × 6.62 × 10 −42

3
−5
−1
14.6) A tank contains one kilomole of argon gas at 1 atm and 300 k. The mass of an
−26
argon is 6.63 × 10 kg
a) what is the internal energy of the gas in joules? What is the average energy of a
molecule in eV?
Sol:
U=
3
3
3
NkT = nRT = × 8.314 × 103 × 300 = 3.74 × 106 J
2
2
2
b) What is the partition function Z?
sol:
need to calculate Z
3
 2πmkT  2
Z =V

2
 h

nRT 3.413 × 300k × 103
=
= 24.9m 3
V=
5
P
10
3
 2 × 3.14 × 6.63 × 10 −26 × 1.381× 10 −23 × 300  2
Z = 24.9
 = 6.15 × 1033
− 34 2
(6.62 × 10 )


c) What is the chemical potential µ in eV?
Sol:
33
×
6
.
15
10
µ = −kT ln Z N = −8.617 × 10 × 300eV × ln
6.02 × 10 26
= −0.4168eV
−5
d) What is
Sol:
NJ
gJ
?
N J N −ε
= ⋅e
gJ Z
J
kT
need to assume e
−ε J
kT
is of the order of unity
N J 6.02 × 10 26
=
= 1 × 10 −7
33
g J 6.15 × 10
1
or use ε J = kT
2
−1
NJ
= 1 × 10 −7 ⋅ e 2 = 6 × 10 −8
gJ
14.7)
The partition function of a system that obeys Maxwell-Boltzmann statistics is
given by Ζ = aVT 4 , where a is a constant. Calculate U, P, and S.
Solution
According to Maxwell-Boltzmann’s Statistics
z = avT 4
where v = Volume
T = Temperature
a = Cons tan t
Calculate
Energy
= U =?
Pressure
= P =?
Entropy
= S =?
Calculating of U
We know that
 ∂ ln z 
U = ΝκΤ 2 

 ∂Τ  v
z = avT 4
(
from (book )
)
ln Z = ln avT 4
ln Z = ln av + 4 ln T
∂ ln Z
4
=0+
∂T
T
4
 ∂ ln Z 

 =
 ∂T  v T
put this value in eg 14.7
U = NkT 2 ⋅
Ans:
4
T
U = 4 NkT
Calculating of P
We know that
 ∂ ln Z 
P = NkT 

 ∂v  Τ
We know that
14.10
Z = avT 4
ln Z = ln a + ln v + 4 ln T
ln z = ln a + ln v + 4 ln T
1
 ∂ ln z 

 =
 ∂v  Τ v
Put this in equ 14.10
1
P = NkT ⋅
v
NkT
P=
v
Calculating of S
We know that
U
14.1
S = + ΝΚ (ln z − ln N + 1)
T
put the values of U and ln z
4ΝκΤ
+ ΝΚ ln avΤ 4 − ln Ν + 1
S=
Τ
S = 4ΝΚ + ΝΚ ln avΤ 4 − ΝΚ ln Ν + ΝΚ
P = 5ΝΚ + ΝΚ ln avΤ 4 − ln Ν
(
(
avΤ 4
Ν

avΤ 4 
P = ΝΚ 5 + ln

Ν 

= 5ΝΚ + ΝΚ ln
)
)
14.8) An ideal monatomic gas consists of N atoms in a volume V. The gas is allowed to
expand isothermally to fill a volume 2V. show that the entropy change of Nk ln2
Sol:
Use equation 14.14
3
 5
V (2πmkT ) 2  
S = NK  + ln 

Nh 3
 2
 

3
 5
 Vi (2πmkT ) 2  
S x = Nk  + ln  ×

h3
 2
 
N
3
 5
 2Vi (2πmkT ) 2  
S f = Nk  + ln 
⋅

3
2
N
h

 

  2V 
∆S = S f − S i = Nk ln i  = Nk ln 2
  Vi 
or use equation (14.15)