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1

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

28

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

Table below shows the electronic energies for a number of atoms

38

!/ 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

Modes of motions; translation, vibration, rotation

40

)()()()()( 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

� Table below shows the molecular constants for several diatomic

molecules.

50

� 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