Energy
Chapter 4 Calculation of Thermodynamic Properties
n n
n n
n n
a a
E
N N N
ε
ε
= =∑
=ε
∑
ε
nan
: the level energy
: the occupation number for the level An average energy of one molecule
1
2
ln
1 1
( )
N dq d q
E N N
q d d
d d
kT dT dT kT kT
ε β β
β β
−
= =− =−
= → = =−
The Boltzmann distribution for a series of non-degenerate energy levels
n
an e
N q
βε
= −
1 n; n
n
n n n n n
n n n n
a dq
P e e
N q d
βε βε
ε ε ε ε ε
β
− −
=
∑
=∑
=∑
− =∑
2
2
ln ln d q kT dT
d q E NkT
dT
ε
∴ =
=
1 dq d lnq
q d d
ε β β
∴ =− =−
Example:
Determine the total energy of an ensemble consisting of N particles that have only two energy levels separated by hv.
Figure 1. Depiction of the two-level system.
For the two-level system, the probability of occupying the excited energy level is
1
1 1
1 1
hv hv
hv
hv
a e e
P N q e
e
β β
β
β
− −
= = = −
+
= +
At T=0, P1=0
P1 increases until hv<<kT.
At high temperature, P1 → 0.5
Figure 2. Total energy as a function of temperature is presented for an ensemble consisting of units that have two energy
Figure 3. The probability of occupying the excited state in a two-level system is shown as
Canonical ensemble: N, V, T are held constant; No p-V work
o V
U U q
∴ − =
ln d Q
N U
d
ε
β
=− =−ln , d q
ε
d=−
β
Since.
E U= + pV At const V, E U= . Note:
ln ln ln ln !
! qN
Q N q N
N
= = −
( ) ( )
ln ln ln !
ln
d Q d d
N q N
d d d
d q N d
β β β
β
= −
=
ln
v
d Q
U dβ
= −
! qN
Q= N
Energy and the Canonical Partition Function
Example:
For an ensemble consisting of a mole of particles having two energy levels separated by hv=1x10−20 J, at what temperature will the internal energy of this system equal 1.00 kJ?
Energy and Molecular Energetic Degree of Freedom
tot t r v e
q = q q q q
( )
ln
ln ln
ln ln ln ln
ln ln ln ln
t r v e
V V V
t r v e
V
t r v e
V V V V
t r v e
d q q q q
d Q d q
U N N
d d d
N d q q q q
d
d q d q d q d q
N d d d d
U U U U
β β β
β
β β β β
= − = − = −
= − + + +
= − + + +
= + + +
Translations
t ,
t
t V
dq U N
q d
β
= −
3
2 2
3
where 3 , .
t 2
V h
q m
β π
= Λ Λ =
3
3 3
3 2
2
3 3
3 2 2
2
3 5
3 2 2
2
3
3 2 1
2
2
2
2 3
2
3 2
2 3
2
t t
t V
V
V
dq
N N d V
U q d V d
d m
N d h
m d
N h d
N m
h N m
h N
β β
π
β β
π β
β
π β
π β
β
−
−
−
Λ
∴ = − = − Λ
= − Λ
= − Λ
= − Λ −
= Λ
=
β
−13 3
Ut NkT = nRT
∴ =
3
2 2
3
2 h
m β π
Λ =
Rotations
For a diatomic molecule in the high-temperature limit within the rigid rotor approximation,
1 qr
σβ
hcB=
1
2 1
1
( )
r
r r
r V V V
N dq d d
U N hcB N
q
U NkT nRT
d d hcB d
N N
σβ β β
β β σβ β
β β β
−
− −
= − = − = −
= −
∴ = =
− =
The concept of equipartition 3
2
r
r
U nRT
U RT
=
=
(linear polyatomic)
(non-linear polyatomic)
Vibrations
1
v 1 hc
q = e−β ν
−
( )
( )( )( )
( )
2
1 1
1
1 1
1
v hc
v hc
v V V
hc hc hc
hc hc
hc
dq
N d
U N e
q d d e
N e hc e e
Nhc e e Nhc
β ν
β ν
β ν β ν β ν
β ν β ν
β ν
β β
ν ν
ν
−
−
− − − −
−
−
= − = − − −
= − − − −
= −
= −
Figure 4. The variation in average vibrational energy as a function of temperature where . ν =1000 cm−1
For the high-temperature limit, kT hc
ν
, i.e., T θ
v.2
1 1 1
2
,
x x
e x x x
x hc hc kT hc kT
β ν ν ν
= + + + ≅ +
= =
( )
1 1 1
v hc
Nhc Nhc Nhc N
e hc
U NkT n
c RT
β ν h
ν ν ν
β ν β ν β
= = = = =
− + =
−
v hc 1 U Nhc
eβ ν
= ν
−
v
hc k
θ
=ν
The high-temperature limit is applicable when T ≥10
θ
v.Review
3 0
2 1
5
tot t r v e
hc
hc
U U U U U
NkT NkT Nhc e NkT Nhc
β ν
β ν
ν ν
= + + +
= + + +
−
= +
−
For a diatomic molecule,
Heat Capacity
2 V
V V
U U
C k
T
β
β
∂ ∂
= ∂ = − ∂ Example:
Determine the heat capacity for an ensemble consisting of units that have only two energy levels separated by an arbitrary amount of energy hν.
Translational Heat Capacity
For an ideal gas,
,
3 2
3 2
t
t V t
V
U NkT
C dU Nk
dT
=
= =
Rotational Heat Capacity
3 2
(linear)
(non-linear)
r
r
U NkT
U NkT
=
=
,
,
3
(linear)
(non-linear)
V r
V r
C Nk
C Nk
=
=
Vibrational Heat Capacity
v hc 1 U Nhc
eβ ν
=
ν
−
( )
( )
( )
( )
2 ,
2
2
2
2 2
2
2 /
/ 2
1 1
1 1
1
v
v
v v
V v
V V
hc
V hc
hc
V hc
hc T v
T
dU dU
C k
dT d
Nk hc d
d e Nk hc hc e
e Nk hc e
e Nk e
T e
β ν
β ν β ν
β ν β ν
θ θ
β β
β ν β β ν ν
β ν
θ
∴ = = −
= − −
= − −
−
= −
=
−
For a polyatomic molecule, it has 3N-6 or 3N-5 vibrational dergee of freedom.
( )
, 3 6 or 3 5( )
,1
N N
V v tot V v m
m
C C
− −
=
=
∑
Figure 5. Evolution in the vibrational
contribution to CV as a function of temperature.
Calculations are for a molecule with three vibrational degrees of freedom as indicated in the top panel. Contribution for each
vibrational mode (top) and the total vibrational
Review
( )
( )
( )
( )
, , , ,
2 2
2
2 2
2
3
2 1
5
2 1
V tot V t V r V v
hc hc hc hc
C C C C
Nk Nk Nk hc e
e Nk Nk hc e
e
β ν β ν β ν β ν
β ν
β ν
= + +
= + +
−
= +
−
Figure 6. The constant volume heat capacity for gaseous HCl as a function of temperature.
The contributions of translational (yellow), rotational (orange), and vibrational (light blue) degrees of freedom to the heat capacity are shown.
For a diatomic molecule,
The Einstein Solid
The Einstein solid model was developed to describe the thermodynamic properties of atomic crystalline systems.
1. All of the harmonic oscillators are assumed to be separable. 2. The harmonic oscillators are assumed to be isoenergetic.
For a crystal containing N atoms, there are 3N vibrational degrees of freedom
( )
2 /
, 2
3 /
1
v
v
T v
V tot
T
C Nk e
T e
θ θ
θ
=
−
The total heat capacity is identical to that of a collection of 3N harmonic oscillators.
Figure 7. Comparison of CV for
diamond to the theoretical prediction of the Einstein solid model. The classical limit of 24.91 J/mol K is shown as the dashed line.
Note that the model predicts that the heat capacity should reach a limiting value of 24.91 J/mol K or 3R at high temperature.
This limiting law is known as the Dulong and Petit Law, and
represents the high-temperature or classical prediction for the heat capacity of such systems.
the Dulong and Petit Law
Figure 8. Comparison of CV versus for C, Cu, and Al. T Θ A plot of CV versus T/θv can describe the variation in heat capacity
for all solids.
(0) n n
n
U =U +
∑ ε
a(0) n n n n
n n
dU = dU +
∑
a dε
+∑ ε
dan n
n
dU =
∑ ε
da at constant volume (dεn=0) dU = dqrev =TdSn n
n
dS dU k da
T
β ε
∴ = =
∑
ln ln
n n 0
n n
W W
a a
βε
= ∂∂ +α
← ∂∂ + −α βε
= Entropy
←
entropy
a measure of the distribution of energy
partition function
⇒
⇓
( )
ln
ln ln
n n
n n n
n
n n
dS k W da k da
a
k W da k d W
a
∂
α
∴ = ∂ +
∂
= ∂ =
∑ ∑
∑
0, i.e., the number of molecules is constant
n n
da
← =
∑
ln S k W
∴ = the Boltzmann formula
( )
( )
ln ! ln ! ln !
!
ln ! ln ! ( ln ) ln
( ln ln )
ln ln
n n n
n
n n n n
n n
n n
n
n n n
n
S k N k N k a
a
k N k a k N N N k a a a
a N
k N N a a
k a N a a
= = −
= − = − − −
= −
= −
←
=
∑
∏ ∏
∑ ∑
∑
∑ ∑
ln ln
n n
n
n n
n
k a a
N k a P
= −
= −
∑
∑
where Pn is the probability of occupying energy level n.
ln ln ln
n
n n
P e q
q
βε
βε
−
= = − −
( )
ln
ln ln ln
n n
n
n n
n
n n n
n n
S k a P
k a q
k a k a q
k E kN q
βε
β ε
β
∴ = −
= − − −
= +
= +
∑
∑
∑ ∑
ln ln ln E N
k q T
E k Q T
U k Q T
= +
= +
= +
ln 2 ln
V V
Q d Q
U kT
β
dT ∂
= − ∂ =
( )
n ln
l
l n
n l
V
S U k Q T
d kT
d Q
kT k Q
d Q
T dT
∴ = = +
=
+
( )
( )
3
3
2 3/2
ln ln
1 3 ln
2 !
3 ln ln
2
5 ln ln
2
5 ln ln
2
5 ln ln ln
2
5 ln ln ln
2 2
V
N trans
trans
trans
d U
S kT Q k Q
dT T
NkT k q
T N
Nk Nk q k N N N Nk Nk q Nk N
Nk Nk V Nk N
Nk Nk V Nk Nk N
Nk Nk V Nk h Nk N
π
mkT
= = +
= +
= + − −
= + −
= + −
Λ
= + − Λ −
= + − −
Entropy of an Ideal Monatomic Gas
3
(← =U 2 NkT for monoatomic gas)
2/3 2 3/2
2/3 2/3 2 3/2
5 3
ln ln ln
2 2 2
5 3
ln ln ln
2 2 2
A
N h Nk Nk V Nk T Nk
mk n N h nR nR V nR T nR
mk
π
π
= + + −
= + + −
The Sackur-Tetrode Equation
5 5
2 2
3 3
3
2 2
where 3
ln ln 2
A A
Ve RTe
S nR nR
nN N p
h
π
mkT
= Λ = Λ
∴ Λ =
V1→V2 for an ideal monatomic gas
2 1 final initial ln
S S S nR V
∆ = − = V
∵ The second term of the Sackur-Tetrode equation depends only on volume while the other terms are unchanged.
T1→T2 ΔV=0
2 2
3 ln ln
final initial 2 V
T T
S S S nR nC
T T
∆ = − = =
Example:
Determine the standard molar entropy of Ne and Kr under standard thermodynamic conditions (298 K, Vm= 24.4 l = 0.0244 m3).
Residual Entropy
For CO, at thermodynamic standard temperature and pressure
cal exp
197.9 193.3
S J mol K
S J mol K
= ⋅
= ⋅
cal exp 4.6 residual entropy
S − S = J mol K⋅ =
Figure 9. The origin of residual entropy for CO. Each CO molecule in the
solid can have one of two possible orientations as illustrated by the central CO.
Each CO will have two possible directions such that the total number of
Because each CO molecule can assume one of two possible orientations, the entropy associated with this orientational order is
ln ln 2N ln 2 ln 2
S = k W = k = Nk = nR For 1 mol, S = Rln 2 = 5.76 J mol K⋅
Note: The third law of thermodynamics states that the entropy of a pure and crystalline substance is zero at 0 K.
Other Thermodynamic Functions
H U pV A U TS G H TS
= +
= −
= −
Helmholtz Energy
ln
ln
U ln
S k
A U TS U T U k Q
T T
T Q
Q k
= − = − +
= −
← = +
T
p A
V
∂
= −∂
(
ln)
ln
T
T
kT Q V
kT Q
V
∂ −
= − ∂
∂
= ∂
!
ln !
N
N
Q q
N p kT q
V N
=
∂
= ∂
ln ln ! ln
ln
N
T N
T
T
kT q N
V V
kT q
V
NkT q
V
∂ ∂
= ∂ − ∂
∂
= ∂
∂
= ∂ For a monatomic gas,
3
q = V Λ
ln 3
T
p NkT V
V
∂
∴ = ∂ Λ
ln ln 3
ln
( ideal gas law)
T
T
NkT V
V V
NkT V
V NkT nRT
V V
∂ ∂
= ∂ − ∂ Λ
∂
= ∂
= = ←
Enthalpy
( )
2
ln ln ln l
l n
ln n n
l
V T
V T
V T
V T
Q V A
V
kT Q
Q V
V
Q Q
kT VkT
T
Q Q
T
V
kT Vk
T V
β β
∂ ∂
= −∂ + −∂
∂ ∂ −
= −∂ + − ∂
∂ ∂
= ∂ + ∂
∂ ∂
+
∂ ∂
=
H = +U pV
Gibbs Energy
= +
T
p A
V
← = − ∂
∂
(
← = −A kT lnQ)
ln ln ln ln
T T
kT Q V Q kT Q
V Q VkT
V
∂
∂
− − ∂
= − +
∂
=
For an ideal gas,
ln
ln !
ln ln !
ln ( ln )
ln ln
ln ln
N
N
kT Q NkT kT q NkT
N
kT q kT N NkT
NkT q kT N N N NkT NkT q NkT N
N q q
kT N nRT
N
= − +
= − +
= − + +
= − + − +
= − +
= − = −
G = +A pV
Example:
Calculate the Gibbs energy for 1 mol of Ar at 298.15 K and standard pressure (105 Pa), assuming that the gas demonstrates ideal behavior.
For the reaction aA bB cC dD+ = + ,
0 0 0 0
0
0 0 0 0
0 0
0
ln ln ln ln
ln ln ln ln
ln
C D A B
c d a b
C D A B
c d
C D
A
q q q q
G c RT d RT a RT b RT
N N N N
q q q q
RT N N N N
q q
N N
RT
q N
∆ = − + − − − − −
=− + − −
=−
a 0 b
qB
N
0 0 0 0 0
0 ln
C D A B
G cG dG aG bG
G RT K
∆ = + − −
∆ =−
Chemical Equilibrium
0 0
0 0
c d
C D
p a b
A B
q q
N N
K
q q
N N
∴ =
The vibrational and electronic ground state are not equivalent for all species, since the presence of a bond between the two atoms in the molecule lowers the energy of the molecule relative to the separated atomic fragments.
Figure 10. The ground- state potential energy curve for a diatomic
( ) ( ) ( 2 )
(1 2 )
n D D D
D
D
hcv hcv
v
n
hcv hcv
v
q e e e e
e e e
e q
βε β ε β ε β ε
βε β β
βε
− − − − − + − − +
− −
′ = = + + +
= + + +
=
∑
Because the energy of the atomic fragments is defined as zero, the ground vibrational state is lower than zero by an amount equal to the dissociation energy of the molecule, .
ε
D0 0
( )
0 0
0 0
0 0
C D A B
c d
C D
c d a b
p a b
A B
c d
C D
a b
A B
q q
N N
K e
q q
N N
q q
N N
e
q q
N N
β ε ε ε ε
β ε
+ − −
− ∆
∴ =
=
Figure 11. The statistical interpretation of equilibrium. (a) Reactant and product species having equal ground-state energies are
depicted. However, the energy spacings of the product are less than the reactant such that more product states are available at a given temperature. Therefore, equilibrium will lie with the product. (b) Reactant and product species having equal state spacings are
depicted. In this case, the product states are higher in energy than those of the reactant such that equilibrium lies with the reactant.
Example:
What is the general form of the equilibrium constant for the dissociation of a diatomic molecule?
X ( )2 g 2X( )g
Summary
( ) ( )
2 1/2 3
2 1 2
2
1 2
1/2 1/2 1/2
3 5
1
2
1 ; ;
8
1 1 1
1 1
total t r v e
t
r
r
A B C
v hc
N
v total v i
i
q q q q q
V h
q m
h m m
q B I r
hcB cI m m
q hcB hcB hcB
q e
q q
β ν
β π
µ µ
σβ π
π
σ β β β
−
−
=
=
= Λ Λ =
= = = =
+
=
= −
=
or
where
where for diatomic
for non-linear polyatomic
3N 6
N N
Q q Q q
−
=
=
∑
for distinguishable for indistinguishable
Canonical ensemble
,
ln
( ln )
ln
ln
ln ln
ln
V
V
T
T
V T
d Q
U d
S d kT Q dT
A kT Q p kT Q
V
G kT Q V Q
V kT Q
N
β
µ
= −
=
= −
= ∂
∂
= − − ∂
∂
= − ∂
∂
