Examples from the monatomic
ideal gas case
example 1
• In problem 3.2 you showed that
1
Computing Cv for Argon
including first 4 excited states
State
De, eV
wei
i=1
0
1
i=2
11.548
5
i=3
11.633
3
i=4
11.723
1
i=5
11.828
3
2
Example 2. Calculate the numerical value of the canonical partition
function for a single He atom (q) in a cubic box of edge 1 cm and the
probability of finding the He atom in a single energy level corresponding
to the mean kinetic energy of the molecule at 300K
a) The individual partition function for He is
 2mkT 
q

2
 h

3/ 2
Vw e1w n1
we can set wn1 = 1 and we1 = 1 since there is no degeneracy
for He (all the electrons fill the 1s orbital such that the spin
of the electrons are nondegenerate)
Let M = m ∙ 6.02x1023
3
b) the probability of finding the He atom in a single energy level
corresponding to the mean kinetic energy of the molecule at
300K
Pe 
e
e / kT
q
1.5
e
 26


2
.
84

10
0
24
7.82  10
c) The probability of finding He within ±1% of the mean energy:


e e / kT
P 0.99e  e  1.01e   Pe de  
de  9.76  10 3
0.99e
0.99e
q
1.01e
1.01e
4
Example 3. Consider the mixing of NA molecules of monatomic gas A and
NB molecules of monatomic gas B at constant volume V and temperature
T. Write the partition function for this system in terms of qA and qB.
Develop expressions for E, Cv, P, and the entropy S=S(xA, xB, P, T); xA and
xB are mole fractions. Also calculate the entropy of mixing.
q A (V , T ) N A q B (V , T ) N B
Q( N A , N B ,V , T ) 
N A!
N B!
A( N A , N B ,V , T )  kT ln Q  kTN A ln
with
qA  (
q Ae
q e
 kTN B ln B
NA
NB
2 πm A kT 3 / 2
) V
2
h
( A / T ) 3
E ( N A , N B ,V , T ) 
 kT ( N A  N B )
(1 / T ) 2
5
CV ( N A , N B ,V , T ) 
E 3
 k(N A  N B )
T 2
A kT ( N A  N B )
P

V
V
EA
q Ae5 / 2
qB e5 / 2
S ( N A , N B ,V , T ) 
 kN A ln
 kN B ln
T
NA
NB

 2 πm A kT 3 / 2 5 / 2 kT 
 2 πmB kT 3 / 2 5 / 2 kT 



S  k ( N A  N B )  x A ln (
) e
 xB ln (
) e
2
2

h
Px A 
h
PxB 



6
entropy of mixing
S ( N A , N B , P, T )  S ( N A , P, T )  S ( N B , P, T )

 2 πm A kT 3 / 2 5 / 2 kT 
 2 πmB kT 3 / 2 5 / 2 kT 
  xB ln (

 k ( N A  N B )  x A ln (
) e
) e
2
2
h
Px A 
h
PxB 



  2 πm A kT 3 / 2 5 / 2 kT 
 2 πmB kT 3 / 2 5 / 2 kT 
 kN A ln (
)
e

kN
ln
) e

(

B

2
2
h
P 
h
P 

 
 kN A x A ln x A  kxB N B ln xB
7
Example 4. An ideal gas of monatomic molecules on a line (for example in a
nanopore) where they cannot pass each other  the translational motion is
restricted to one dimension. The qm translational energy levels are:
h 2  l x2 
 2 
e (l x ) 
8m  L x 
Lx is the length of the line (pore). Find the translational contribution to the
partition function and the thermodynamic properties of this gas.
Replacing a summation with an
integral, we immediately obtain:

q  e
lx 0
 h2lx2 / 8 mkTL2x
2m kTL2x
2m kT


Lx
2
2
h
h
8
Example 5. A large polymer molecule is made up to N monomer
units, each of which can be either a helix (H) or a coiled (C) state
with energies eH and eC respectively. Assuming that the
conformation of a monomer unit is independent of all other
monomer units, determine the average fraction of monomers that
are in the H state as a function of a dimensionless T. How does the
statistical degeneracy comes into the result?
9
Example 5. A large polymer molecule is made up to N monomer
units, each of which can be either a helix (H) or a coiled (C) state
with energies eH and eC respectively. Assuming that the
conformation of a monomer unit is independent of all other
monomer units, determine the average fraction of monomers that
are in the H state as a function of a dimensionless T. How does the
statistical degeneracy come into the result?
The total energy of the polymer chain (N monomers) depends on the number of monomers in the helix state (n) and coiled state (N-n), i.e.,
E  ne h  ( N  n)e c
The number of combinations of n helix monomers and N-n coiled monomers is
N!
n!( N  n)!
ne h  ( N  n)e c
N!
Q
exp(
)
kT
n 1 n!( N  n)!
N
N
Therefore, the partition function of a polymer chain is

n 1
the last equality is obtain from binomial
expansion,
see eqn 5.2-7 in the textbook
e
e
N!
exp( h ) n exp( c ) N  n
n!( N  n)!
kT
kT
e
e 

 exp( h )  exp( c ) 
kT
kT 

N
10
The probability of finding the polymer having n
helix monomers is
ne h  ( N  n)e c
1
N!
p ( n) 
exp( 
)
Q n!( N  n)!
kT
The average
number
of helix
monomers is:
ne  ( N  n)e c
1 N
nN !
n   np(n)  
exp(  h
)
Q n 1 n!( N  n)!
kT
n 1
N
e
e
1 N
N ( N  1)!
 
exp(  h ) n exp(  c ) N n
Q n1 (n  1)!( N  n)!
kT
kT
e h N 1
e h n1
e c N n
N
( N  1)!
 exp(  ) 
exp(  ) exp(  )
Q
kT n 10 (n  1)!( N  n)!
kT
kT

N
eh
ec 

exp(

)

exp(

)

kT
kT 

exp( 
N
exp( 
eh
kT
eh
kT
N
exp( 
eh 
eh
ec 
) exp(  )  exp(  )
kT 
kT
kT 
)
)  exp( 
ec
kT
)
11
N 1
Similarly the average number of coiled
monomers is
N exp( 
N n 
exp( 
eh
kT
ec
kT
)
)  exp( 
ec
kT
)
The average number of
monomers that are in the helix
state can be rewritten as:
n
N exp( 
exp( 
eh
kT
eh
kT
)
)  exp( 
with
ec
kT
N

)
ec eh
kT
1  exp( 

ec eh
kT

)
N
1  exp( 
1
)
*
T
1
T*
12
Example 6. Compute the chemical potential of Avogadro’s
number of molecules of Ar at 298K and 1 bar pressure
First need to calculate the volume of 1 mole of an ideal gas at 298 K and 1 bar
3
bar
m
V  NRT / P  1 mol  8.314  105
 298 K/1 bar  0.02478 m3
mol K
q  1.88  10   39.945  298 
26
3/2
 0.02478  6.05  1030
q
q
per
molecule
or



RT
ln

  per mole
N
N
  3.995  104 joule/mol  39.95 kJ/mol
   kT ln 
13