Statistical Thermodynamics
Atkins, P. & de Paulo, J. (2009). Physical Chemistry. 9th edition.
Oxford: Oxford University Press. (Chapters 15, 16)
�The Distribution of Molecular States
�Configurations and Weights
�The Dominating Configuration
�Boltzmann Distribution
�Molecular Partition Function & its
� Partition Function for Independent Molecules
�Perfect Gas
� Translational, Electronic & Nuclear Partition Function
�Partition Function for Diatomic Gas
� Factorization
� Vibrational Partition Function
� Statistical Thermodynamics & Thermodynami Quantities
�The Internal Energy
�Heat Capacity
�Entropy
2
� By solving the Shrodinger equation (H =E ), one can calculate the energy of
molecules & then related the result to their structures.
� Thus, The next step is to see how this knowledge can be used to calculate the
properties of matter.
� Statistical thermodynamics provides the link between the individual molecular
properties and its bulk thermodynamic properties.
�It means that how the average behavior of large number of molecules will change.
� For example, the pressure of the gas depends on the average force exerted by its
molecules & there is no need to specify which molecules happen to be striking the
wall at any instant.
� Therefore, for a large number of molecules the change in any thermodynamic
properties can be negligible compared to the average values.
Introduction
3
The Distribution of Molecular States
� Consider a system composed of N molecules with a constant total energy, E.
� These molecules are independent, i.e. no interactions exist among the molecules.
� Countless collisions occur.
� It is hopeless to keep a track of positions, momenta, and internal energies of all
molecules.
� The Principle of equal a priori probabilities:
�All possibilities for the distribution of energy are equally probable.
� By this, the number of molecules and the total energy are kept the same.
�Here, “a Priori” means “as far as one knows”.
Democracy among microscopic states !!!
4
� For example, consider four molecules in a three-level system. The following
two arrangements have the same probability.
---------l-l-------- 2εεεε ---------l--------- 2εεεε---------l---------- εεεε ---------1-1-1---- εεεε---------l---------- 0 ------------------- 0
� Configurations and Weights
�Imagine that there are N molecules in the system.
� n0molecules with energy εεεε0; n1 molecules with energy εεεε1; n2 moleculeswith energy εεεε2, and so on.
� where εεεε0 < εεεε1 < εεεε2< ... are the energies of different states.
� The average number of molecules occupying a state is called population. So,
the energy of the state is denoted as εεεεi .
� The specific distribution of molecules is called configuration of the system,
denoted as { n0, n1, n2, ......}
5
� For example, in the system below, there is 17 molecules, and each molecule has
four states. The above configuration is thus: { 4, 6, 4, 3 }
1
2
3
4
5
6
7
89
10
1112
13
14
15
1617
1
2
3
4
STATE
6
� Now the question is how to determine the population ?
� For example,
�A configuration such as {N, 0, 0, ...} means that every molecule is in the
ground state, so there is only one way to achieve this configuration.
� But a configuration such as {N-2, 2, 0, ......} means that there are two
molecule is in the first excited state, and the rest in the ground state which can
be achieved in N(N-1)/2 ways.
�A configuration { n0, n1, n2, ......} can be achieved in W different ways.
�W is called the weight of the configuration and can be evaluated as follows,
W = N! / (n0! n1! n2! ...)
1. It means that there is N! different ways to arrange N molecules.
2. ni! arrangements of ni molecules with energy εεεεi correspond to the sameconfiguration.
7
� Question. Calculate the number of ways of distributing
20 objects into six boxes with the arrangement {1, 0, 3, 5,
10, 1}. Recall that; 0! = 1.
Answer:
20! / 1! 0! 3! 5! 10! 1! = 931,170,240
8
Question. Calculate the number of ways of distributing 3 objects a, b, and c
into two boxes with the arrangement {1, 2}.
Answer:
| a | b c |, | b | c a |, | c | a b |.
Therefore, there are three ways 3! / 1! 2!
| a | c b |, | b | a c |, | c | b a |
To eliminate over-counting of
these configurations
� But what if the number of objects is so large ?!!
9
� Stirling’s Approximation:
When x is large: ln x! ≈≈≈≈ x ln x - x
x ln x! x ln x – x ln A
1 0.000 -1.000 0.081
2 0.693 -0.614 0.65
4 3.178 1.545 3.157
6 6.579 4.751 6.566
8 10.605 8.636 10.595
10 15.104 13.026 15.096
10
Note that : A = (2ππππ)1/2 (x+1/2)x e-x
Note that : A = (2ππππ)1/2 (x+1/2)x e-x
Therefore:
lnW ≈≈≈≈ ( N ln N - N ) - ∑∑∑∑ ( ni ln ni - ni ) = N ln N - ∑∑∑∑ ni ln ni
x ln x! x ln x - x ln A
1 19.987 17.819 19.980
16 30.672 28.361 30.666
20 42.336 39.915 42.332
30 74.658 72.036 74.656
11
The Dominating Configuration
� Imagine that there are N molecules distribute among two states. The
possible configurations are;
�{N, 0}
�{N-1, 1}, …
�{N-k, k}, …
�{1, N-1}
�{0, N}
�And their weights are 1; N; ... , N! / (N-k)! k!; ... ; N; 1, respectively.
� For example, N=8,
the weight distribution is:
0 1 2 3 4 5 6 7 8
0
20
40
60 N = 8
W
k12
0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6
0
2 0 0 0
4 0 0 0
6 0 0 0
8 0 0 0
1 0 0 0 0
1 2 0 0 0
N = 1 6
W
k
0 4 8 1 2 1 6 2 0 2 4 2 8 3 2
0 .0 0 E + 0 0 0
1 .0 0 E + 0 0 8
2 .0 0 E + 0 0 8
3 .0 0 E + 0 0 8
4 .0 0 E + 0 0 8
5 .0 0 E + 0 0 8
6 .0 0 E + 0 0 8
N = 3 2
k
13
�When N is even, the weight is maximum at k = N/2, therefore;
Wk=N/2 = N! / [(N/2)!]2
�When N is odd, the maximum is at k = N/2 ±±±± 1.
� Therefore, as N increases, the maximum becomes sharper!
� The weight for k = N/4 is Wk=N/4 = N! / [(N/4)! (3N/4)!]
� The ratio of the two weights is equal to:
R(N) ≡≡≡≡Wk=N/2 / Wk=N/4 =(N/4)! (3N/4)! / [(N/2)!]2
| N | 4 | 8 | 16 | 32 | 256 | 6.0 x 1023
|R(N) | 1.5 | 2.5 | 7.1 | 57.1 | 3.5 x 1014 | 2.6 x 103e+22
14
�Therefore, for a macroscopic molecular system ( N ~ 1023 ), there are
dominating configurations, so that the system is almost always found in
or near the dominating configurations, i.e. equilibrium.
�What if the system has more states?
� Constraints for the system
�There are two constraints for the system:
1. The total energy of a system is constant; ∑ni εεεεi = E = constant
2. The total number of molecules is conserved; ∑∑∑∑ ni = N = constant
� How to maximize W or lnW under these constraints?
15
� The Boltzmann distribution states;
ni / N = Pi = exp (αααα - βεβεβεβεi )
� Pi is the probability.
�β is defined as; 1/ββββ = kT ; Where k is the Boltzmann constant.
�αααα is a coefficient that makes the total probability equals ONE.
1 = ∑∑∑∑i ni / N = ∑∑∑∑i exp(αααα -βεβεβεβεi)
exp(αααα ) = 1 / ∑∑∑∑i exp(-βεβεβεβεi)
αααα = - ln [∑∑∑∑i exp(-βεβεβεβεi)]
The Boltzmann Distribution
16
� In summary, using the principle of equal a priori probabilities, the
probabilities of different configurations can be evaluated by simple counting.
� Then, the dominating configuration whose population follows the Boltzmann
distribution can be found.
� The Boltzmann distribution relates the macroscopic physical properties to the
microscopic molecular properties, and it is capable of explaining the equilibrium
properties of all materials.
� To find the most important configuration, one need to vary { ni } to seek the
maximum value ofW. But how?
� For example, consider a one-dimensional function such as F(x) = x2
dF/dx = 0
Is this enough?
X 17
�How about a two-dimensional case;
� For example, finding the minimum point of the surface of a half
watermelon which is F(x,y).
∂∂∂∂F/∂∂∂∂x = 0
∂∂∂∂F/∂∂∂∂y = 0
� If you have multi-dimensional function: F(x1, x2, …, xn);
∂∂∂∂F/∂∂∂∂xi = 0 ; i = 1,2,…,n
� To find the maximum value of W or lnW, we should have:
∂∂∂∂ lnW / ∂∂∂∂ni = 0 ; i=1,2,3,...
18
� The method of Lagrange Multipliers
� It is used to minimize or maximize a function F(x1, x2, …, xn) under
constraints:
C1(x1, x2, …, xn) = Constant1
C2(x1, x2, …, xn) = Constant2.
.
.
Cm(x1, x2, …, xn) = Constantn
L = F(x1, x2, …, xn) - ∑∑∑∑i λλλλiCi(x1, x2, …, xn)
∂∂∂∂L/∂∂∂∂xi = 0 ; i=1,2, ...,
19
� Let’s see the water melon’s surface case. How to find the minimum
or maximum of F(x, y) under a constraint x = a ?
L = F(x, y) - λλλλ x
∂∂∂∂ L / ∂∂∂∂ x = 0
∂∂∂∂ L / ∂∂∂∂ y = 0
� Therefore;
dL = dF (x,y) - ∑∑∑∑i λλλλi dCi
� Under constraint of x=a, dCi= 0, thus dF = 0.
�It means that F is at its maximum or minimum.
20
�The method of Lagrangian Multiplier based on the Boltzmann
distribution function;
� First, construct a new function, L.
L = lnW + αααα ∑∑∑∑i ni - ββββ ∑∑∑∑i ni εεεεi
� Next, we need to find the maximum of L by changing ni .
� Finding αααα and ββββ is equivalent to finding the maximum of W under the
two constraints defined before, i.e.;
∂∂∂∂L/∂∂∂∂ni = ∂∂∂∂lnW/∂∂∂∂ni + αααα -βεβεβεβεi = 0
�Assignment 1. Prove this equation is correct.
21
� Since, ln W ≈≈≈≈ ( N ln N - N ) - ∑∑∑∑i ( ni ln ni - ni )
= N ln N - ∑∑∑∑i ni ln ni
∂∂∂∂lnW/∂∂∂∂ni = ∂∂∂∂(N ln N)/∂∂∂∂ni - ∂∂∂∂(ni ln ni)/ ∂∂∂∂ni
= - ln (ni/N)
� Therefore,
ln (ni / N) + αααα -βεβεβεβεi = 0
ni / N = exp(αααα -βεβεβεβεi)
� By this, E = N < εεεε > = 3N/2ββββ
- where < εεεε > is the average kinetic energy of a molecule.
Therefore, < εεεε > = <mv2/2> = 3/2ββββ
22
�On the other hand, according to the Maxwell distribution of speed,
the average kinetic energy of a molecule at an equilibrium is:
<mv2>/2 = 3kT/2
� This is actually the definition of the temperature where k is the
Boltzmann constant.
� Thus, 1 / ββββ = kT. This is the physical meaning of ββββ, the reciprocaltemperature.
23
Q. Consider a molecule whose ground state energy is -10.0 eV, the first excited
state energy -9.5 eV, the second excited state energy -1.0 eV. Calculate the
probability of finding the molecule in its first excited state at T = 300, 1000,
and 5000 K.
24
Hint;
The energies of second, third and higher excited states are much higher than those of
ground and first excited states, for instance, at T = 5,000 K, the probability of finding
the first excited state is,
p2 = exp(αααα-βεβεβεβε2) with εεεε2= -9.5 eV
And, the probability of finding the second excited state is,
p3 = exp(αααα-βεβεβεβε3) with εεεε3= -1.0 eV
If you calculate the ratio of probabilities between second & first excited states at this
temperature and then compared that to the first excited state, you will see the chance
that the molecule is in second excited state is exceedingly slim. Thus, we consider
only the ground and first excited states, a two-state problem.
Note that: [ 1 eV = 1.60 x 10-19 J; k = 1.38 x 10-23 J.K-1]
Therefore, the probability of finding the molecule in the first excited state is:
p = exp(-βεβεβεβε2) / [exp(-βεβεβεβε1) + exp(-βεβεβεβε2)]; where, εεεε1= -10.0 eV, and εεεε2 = -9.5 eV
(1) T = 300 K
(2) T = 1000 K
(3) T= 5000 K
×−
+
×−
=
300
116000.5exp1
300
116000.5exp
p
×−
+
×−
=
1000
116000.5exp1
1000
116000.5exp
p
×−
+
×−
=
5000
116000.5exp1
5000
116000.5exp
p
25
The Molecular Partition Function
� The Boltzmann distribution function can be written as;
pi = exp (-βεβεβεβεi) / q
� where pi is the probability of a molecule being found in a state i with
energy εεεεi.
� q is called the molecular partition function;
q = ∑∑∑∑i exp (-βεβεβεβεi)
� The summation is over all possible states (not just the energy levels).
� For example, if the energy level is gi-fold degenerate, then the molecular
partition function can be re- written as;
q = ∑∑∑∑i gi exp(-βεβεβεβεi)26
� It means that:
�As T→→→→ 0, q→→→→ g0
Meaning at T = 0, the partition function is equal to the degeneracy of
the ground state.
�As T→→→→ ∞∞∞∞, q→→→→ the total number of states.
� Therefore, the molecular partition function gives an indication of the
average number of states that are thermally accessible to a molecule at
the temperature of the system.
� The larger the value of the partition function is, the more the number
of thermally accessible states is.
� Thus, The relationship between q and αααα: exp(αααα) = q-1
27
e-ββββεεεε q = e
-ββββεεεε + e
-2ββββεεεε + e
-3ββββεεεε + ......
= q - 1
Therefore,
q = 1 / ( 1 - e-ββββεεεε )
29
Partition Function for Independent Molecules
� Consider a system which is composed of N identical & indistinguishable
molecules with no interaction between them.
�We may generalize the molecular partition function q to the partition
function of the system Q (β=1/kT);
Q = ∑∑∑∑i exp(-ββββEi)
� where Ei is the energy of a state i of the system, and summation is over
all the states.
� Since there is no interaction among molecules, Ei can be expressed as;
Ei = εεεεi(1) + εεεεi(2) +εεεεi(3) + … + εεεεi(N)
� where εεεεi( j) is the energy of molecule j in its molecular state.
30
The partition function Q:
Q = ∑∑∑∑i exp[-βεβεβεβεi(1) - βεβεβεβεi(2) - βεβεβεβεi(3) - … -βεβεβεβεi(N)]
= {∑∑∑∑i exp[-βεβεβεβεi(1)]}{∑∑∑∑i exp[-βεβεβεβεi(2)]} … {∑∑∑∑i exp[-βεβεβεβεi(N)]}
= {∑∑∑∑i exp(-βεβεβεβεi)}N
= qN
� where q ≡≡≡≡ ∑∑∑∑i exp(-βεβεβεβεi) is the molecular partition function.
� This equality (Q= qN) is satisfied because the molecules are
independent of each other.
31
Perfect Gas
� Perfect gas is an idealized gas where an individual molecule is treated as a point
mass which randomly moving and no interaction exists among molecules (PV=nRT).
�Molecular energy; the total energy of each molecule is defined as the sum of the
translational (kinetic) and internal energies;
� The internal energies include rotational, vibrational, electronic, and nuclear
energies
trans intε ε ε= +
int rot vib elect nuclε ε ε ε ε= + + +
electronic vibrational rotational translational32
Translational Partition Function (qT) of a monoatomic gas
�A molecule moves in a three-dimensional space, but first consider one-
dimensional case.
� Imagine a molecule of mass m moves along the x direction between x= 0 &
x= X, but it has been confined in the y- and z- direction. What is its partition
function qx?
� From particle in a box calculation, the energy levels are;
En = n2h2 / (8mX2) n = 1, 2, 3, …
� If the lowest energy equals to zero point energy, the relative energies can
then be expressed as;
εεεεn = (n2-1)εεεε with εεεε = h2 / (8mX2)
qx = ∑∑∑∑n exp [ -(n2-1)βεβεβεβε ]
33
� βεβεβεβε is very small, thus;
qx = ∫∫∫∫ dn exp [ -(n2-1)βεβεβεβε ]
= ∫∫∫∫ dn exp [ -n2βεβεβεβε ] = (2ππππm/h2ββββ2)1/2 X
� Now consider a molecule of mass m, free to move in a container of volume,
V=XYZ.
� Its partition function qT can be expressed as;
qT = qx qy qz
= (2ππππm/h2ββββ2)1/2 X (2ππππm/h2ββββ2)1/2 Y (2ππππm/h2ββββ2)1/2 Z
= (2ππππm/h2ββββ2)3/2 XYZ = (2ππππm/h2ββββ2)3/2 V
= V/ΛΛΛΛ3
� where, ΛΛΛΛ = h(ββββ/2ππππm)1/2 or (h2/2ππππmkT)1/2, the thermal wavelength. The thermalwavelength is small compared with the linear dimension of the container.
� Partition function of a perfect gas (N molecules); Q = (qT) N= V N / ΛΛΛΛ3N34
� For example, calculate the molecular partition function q for He in a cubical
box with sides 10 cm at 298 K.
� Answer;
� The volume of the box is V=0.001m3. The molar mass of the He molecule is
0.004/(6.022×1023)=6.6466×10-27kg. Substituting these numbers and the proper
natural constants, we have;
�
3/ 227 23
27
34 2
2 6.6466 10 1.38 10 2980.001 7.820 10
(6.626 10 )q
π − −
−
⋅ × × × ×= × = × ×
p
NkT
p
nRTV ==
( )
⋅×
=−
Pa
Kmolkg102052.8 2
523
17
tp
TMN
q
A
Nn
N=
35
� Electronic Partition Function
� The electronic partition function is usually written as a sum over
energy levels rather than a sum over states (Where gi is the
degeneracy of the ith electronic level);
�Recall that the first electronic level is the ground state & the
remaining energy levels are expressed in terms of their spacing
relative to that.
�At ordinary temperatures, usually only the ground state and perhaps
the first excited state need to be considered.
electi
iq g eβε−= ∑
12
elect 1 2q g g eβ ε− ∆= + +L
36
� Nuclear Partition Function
�The nuclear partition function has a form similar to that of the
electronic partition function.
� Nuclear energy levels are separated by millions of electron volts.
� It means that temperatures on the order of 1010 K need to be
reached before the first excited state becomes populated to any
significant degree.
�Therefore, at ordinary temperatures we can simply express the
nuclear partition function in terms of the degeneracy of the ground
state.
electi
iq g eβε−= ∑
12
elect 1 2q g g eβ ε− ∆= + +L
37
!/ NqQ N=
Diatomic Gas
� Consider a diatomic gas made of two atoms A and B with N identical
molecules. A and B may be the same (homonuclear) or different
(heteronuclear).
� The mass of a diatomic molecule is M. These molecules are
indistinguishable. Thus, the partition function Q of the gas can be
expressed in terms of the molecular partition function q;
� The molecular partition q;
�where, εεεεi is the energy of a molecular state i, β=1/kT, and ΣΣΣΣì is thesummation over all the molecular states.
exp( )ii
q βε= −∑
39
)()()()()( jjjjj EVRT εεεεε +++=
Molecular Partition Function; Factorization
� The energy of a molecule j is the sum of contributions from its different
modes of motion; T (translation), R (rotation), V (vibration), E, (electronic)
� The separation of the electronic and vibrational motions is justified by
different time scales of electronic and atomic dynamics.
� Rotational motion is approximated based on rigid rotor in QM. It
means rotary motion is about the center of mass.
� Vibrational motion is approximated based on harmonic oscillator. It
means relative vibratory motion of the two atoms.
exp( ) exp[ ( )]
[ exp( )][ exp( )][ exp( )][ exp( )]
T R V E
i i i i ii i
T R V E
i i i ii i i i
T R V E
q
q q q q
βε β ε ε ε ε
βε βε βε βε
= − = − + + +
= − − − −
=
∑ ∑∑ ∑ ∑ ∑
41
The translational partition function of a molecule
ΣΣΣΣì sums over all the translational states of a molecule.
The electronic partition function of a molecule
ΣΣΣΣ ì sums over all the electronic states of a molecule.
∑ −≡i
T
i
Tq )exp( βε
The vibrational partition function of a molecule
ΣΣΣΣ ì sums over all the vibrational states of a molecule.
The rotational partition function of a molecule
ΣΣΣΣ ì sums over all the rotational states of a molecule.
∑ −≡i
R
i
Rq )exp( βε
∑ −≡i
V
i
Vq )exp( βε
∑ −≡i
E
i
Eq )exp( βε
42
�A polyatomic molecule containing N atoms has 3N degrees of freedom
totally.
� Three of these degrees of freedom can be assigned to translational
motion.
� Two (linear molecules) or three (for nonlinear molecules) are for
rotational motion.
� Thus, there are (3N-5) (linear molecule) and 3N-6 (nonlinear molecule)
for vibrational motion.43
� For example, CO2 has 3×3-5 = 4 degrees of freedom of vibration;
nonlinear molecule of H2O has 3×3-6 = 3 degrees of freedom of
vibration.
44
� Recall rigid rotor & harmonic oscillator energy levels and degeneracy's
for rotational and vibrational motion.
( )2 10,1,2,
8J
J JhJ
Iε
π+
= = K
Rigid Rotor Harmonic Oscillator
2 1J Jω = +
I→ moment of inertia 2
eI rµ=
re→ bond length
m→ reduced mass1 2
1 2
mm
m mµ =
+
( )12 0,1,2,v h n nε ν= + = K
1nω =
v→ frequency
1 21
2
kv
π µ
=
k→ force constant
→ wave numberv
cω =ω→ rotational const. 28B h Icπ=B
45
5--------------5hv
4--------------4hv
3--------------3hv
2--------------2hv
1--------------hv
0--------------0
hvnV
n )2/1( +=ε
nhvV
n =ε
hv=ε
Vibrational Partition Function
� Two atoms vibrate along an axis connecting the two atoms. The
vibrational energy levels; n= 0, 1, 2, …….
� If we set the ground state energy to zero or measure energy from the
ground state energy level, the relative energy levels can be expressed as;
46
�Then, the molecular partition function can be evaluated;
∑∑ −−=−=−=nn n
v hvhvnq )]exp(1/[1)exp()exp( βββε
...1 32 βεβεβε −−− +++= eeeq v1....32 −=+++= −−−− vv qeeeqe βεβεβεβε
hv
v
eeq β
βε−
−
−=−=1
1)1/(1
�At very high temperature where kT >>hv, i.e.,
1, q 1vhvβ << ≈
� The vibrational temperature, θv ≡ hv/k,
is often used instead of the vibrational
frequencies.
1hve hβ β ν− ≈ −
ΙΙΙΙ2222 F2 HCl H2θθθθv/K 309 1280 4300 6330v/cm-1 215 892 2990 4400
Assignment 2;
47
Q. Calculate the proportion of I2 molecules in their ground, first excited,
and second excited vibrational states at 25oC. The vibrational
wavenumber is 214.6 cm-1.
εεεεn = (n+1/2) hνννν
q = ∑∑∑∑n exp(-ββββεεεεn) = exp (-ββββhνννν/2) ∑∑∑∑n exp (-nββββ hνννν)
= exp (-ββββhνννν/2) / [1- exp (-ββββhνννν)]
Portion of I2 is in n-th vibrational state is
pn = exp (-ββββεεεεn) / q = exp (-nββββhνννν) [1- exp (-ββββhνννν)]
Therefore, at T = 25oC = 298.15 oK,
ββββhνννν = 1.036
p0 =0.645 p1 = 0.229 p2 = 0.081
48
49
� For Example, evaluate the vibrational partition functions at 1500 K. We
know that there are three normal modes of H2O vibrational partition
function.
1594.8 cm-1 3656.7 cm-1 3755.8 cm-1
At 1500 K: 16.1042 −= cmhckT
Mode 1 2 3
3656.7 1594.8 3755.8
3.507 1.530 3.602
1.031 1.276 1.028Vq
kThc
cm
ν
ν~
~ 1−
353.1028.1276.1031.1 =××=Vq
� Thermodynamics is the science of energy conversion involving heat and
other forms of energy.
�A thermodynamic system is defined by a macroscopic region of the universe,
often called a physical system studied by using the principles of
thermodynamics.
� Statistical thermodynamics, or statistical mechanics, is the study of the
microscopic behaviors of thermodynamic systems using statistical methods and
probability.
� The essential problem in statistical thermodynamics is to determine the
distribution of a given amount of energy E over N particles in a system.
� The macroscopic properties such as internal energy, heat capacity, & … can
be calculated in terms of partition functions.
� By this, statistical thermodynamics is a bridge between macroscopic and
microscopic properties of a system.
Connection between Thermodynamic & Statistical Quantities
51
Top Related