Intermediate Physical Chemistry

Driving force of chemical reactionsBoltzmann’s distributionLink between quantum mechanics & thermodynamics

Intermediate Physical Chemistry

Contents:Distribution of Molecular States Partition FunctionPerfect Gas  Fundamental Relations Diatomic Molecular Gas  

In 1929, Dirac declared, “The underlying physical laws necessary for the mathematical theory of ...thewhole of chemistry are thus completely know, and the difficulty is only that the exact application of these laws leads to equations much too complicated to be soluble.”

Beginning of Computational Chemistry

Dirac

H E

SchrÖdinger Equation

HamiltonianH = (h2/2m

h2/2me)ii2

+ ZZeri e2/ri

ije2/rij

Wavefunction

Energy

Principle of equal a priori probabilities:

All possibilities for the distribution of energyare equally probable provided the number ofmolecules and the total energy are kept the same

Democracy among microscopic states !!!

a priori: as far as one knows

For instance, four molecules in a three-level system: the following two conformations havethe same probability. 

---------l-l-------- 2 ---------l--------- 2---------l---------- ---------1-1-1---- ---------l---------- 0 ------------------- 0

a demon

THE DISTRIBUTION OFMOLECULAR STATES

Consider a system composed of N molecules,and its total energy E is a constant. These molecules are independent, i.e. no interactions exist among the molecules.

Population of a state: the average number of molecules occupying a state. Denote the energy of the state i

The Question: How to determine the Population ?

Configurations and Weights

Imagine that there are total N molecules amongwhich n0 molecules with energy 0, n1 with energy 1,

n2 with energy 2, and so on, where 0 < 1 < 2 < ....

are the energies of different states. The specific distribution of molecules is called configuration of the system, denoted as { n0, n1,

n2, ......}

{N, 0, 0, ......} corresponds that every molecule is in the ground state, there is only one way to achieve this configuration; {N-2, 2, 0, ......} corresponds that two molecule is in the first excited state, and the rest in the ground state, and can be achieved in N(N-1)/2 ways.

A configuration { n0, n1, n2, ......} can be achieved

in W different ways, where W is called the weight of the configuration. And W can be evaluated as follows,

W = N! / (n0! n1! n2! ...)

2. Calculate the number of ways of distributing20 different molecules among six different states with the configuration {1, 0, 3, 5, 10, 1}.

Answer:20! / 1! 0! 3! 5! 10! 1! = 931170240

note: 0! = 1

Stirling’s Approximation:

When x is large, ln x! x ln x - x x ln x! x ln x - x ln A1 0.000 -1.000 0.0812 0.693 -0.614 0.6524 3.178 1.545 3.1578 10.605 8.636 10.59516 30.672 28.361 30.66620 42.336 39.915 42.33230 74.658 72.036 74.656

note, A = (2)1/2 (x+1/2)x e-x

ln W ( N ln N - N ) - ( ni ln ni - ni )

= N ln N - ni ln ni

The Dominating Configuration

Imagine that N molecules distribute among two states. {N, 0}, {N-1, 1}, ..., {N-k, k}, ... , {1, N-1}, {0, N} are possible configurations, and their weights are 1, N, ... , N! / (N-k)! k!,... , N, 1, respectively. For instance, N=8, the weight distribution is then

0 1 2 3 4 5 6 7 80

20

40

60 N = 8W

k

0 1 2 3 4 5 6 7 8 9 101112131415160

2000

4000

6000

8000

10000

12000N =16

W

k

0 4 8 12 16 20 24 28 320.00E+000

1.00E+008

2.00E+008

3.00E+008

4.00E+008

5.00E+008

6.00E+008

N = 32

k

When N is even, the weight is maximum at k = N/2,

Wk=N/2 = N! / [N/2)!]2.

When N is odd, the maximum is at k = N/2 1

As N increases, the maximum becomes sharper!

The weight for k = N/4 is Wk=N/4 = N! / [(N/4)! (3N/4)!]

| 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

The ratio of the two weights R(N) Wk=N/2 / Wk=N/4

is equal to (N/4)! (3N/4)! / [(N/2)!]2

Therefore, for a macroscopic molecular system( N ~ 1023 ), there are dominating configurationsso that the system is almost always found in thedominating configurations, i.e. Equilibrium

Dominating Configuration: Equilibrium Configuration

To find the most important configuration, we vary { ni } to seek the maximum value of W.

But how?One-Dimensional Function: F(x) = x2

dF/dx = 0

X

Two-Dimensional Case: for instance, finding theminimum point of the surface of a half water melon F(x,y).

F/x = 0,F/y = 0.

Multi-Dimensional Function: F(x1, x2, …, xn)

F/xi = 0, i = 1,2,…,n

1. The total energy is a constant, i.e.

ni i = E = constant

2. The total number of molecules is conserved,

ni = N = constant

How to maximize W or lnW ? How to maximizea function ?

To find the maximum value of W or lnW,

lnW / ni = 0, i=1,2,3,... ???

Let’s investigate the water melon’s surface:

cutting the watermelon

how to find the minimum or maximum of F(x, y) under a constraint x = a ?

L = F(x, y) - x

L / x = 0 L / y = 0 x = a

Generally, to minimize or maximize a functionF(x1, x2, …, xn) under constraints,

C1(x1, x2, …, xn) = Constant1

C2(x1, x2, …, xn) = Constant2

.

.

.Cm(x1, x2, …, xn) = Constantm

 L = F(x1, x2, …, xn) - i iCi(x1, x2, …, xn)

 L/xi = 0, i=1,2, ..., n

JUSTIFICATION

dL = dF - i i dCi

under the constraints, dCi = 0, thus

dF = 0

i.e., F is at its maximum or minimum.

This method is called the method of LagrangianMultiplier, or Method of undetermined multipliers

Procedure

Construct a new function L,

L = lnW + i ni - i ni i

Finding the maximum of L by varying { ni },

and is equivalent to finding the maximum of W under the two constraints, i.e.,

L/ni = lnW/ni + -i = 0

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) + -i = 0

ni / N = exp( -i)

The Boltzmann Distribution

Pi = exp ( - i )

Interpretation of Boltzmann Distribution

1 = i ni / N = i exp( -i)

exp( ) = 1 / i exp(-i)

= - ln [i exp(-i)]

> 0, more molecules occupying the low energy states.

Physical Meaning of

Imagine a perfect gas of N molecules. Its totalenergy

E = i ni i = i N exp(-i) i / i exp(-i)

= -Nq-1dq /d = -N dlnq / d

where q = i exp(-i), ni and i are the

population and energy of a state i, respectively.It will be shown later that

q = V / 3 and = h (/2m)1/2 [ is called the thermal wavelength]

dlnq/d = -3dln/d = -3/2

Therefore, E = N < > = 3N/2, where < > isthe average kinetic energy of a molecule.Therefore,

< > = <mv2/2> = 3/2.

On the other hand, according to the Maxwelldistribution of speed, the average kinetic energyof a molecule at an equilibrium,

: the reciprocal temperature

1 / = kT

where k is the Boltzmann constant.

Physical Meaning of

Two friends A and B are drinking beer in a pub one night. Out of boredom, the two start to play “fifteen-twenty”. A mutual friend C steps in, and the three play together. Of course, this time one hand is used by each. A is quite smart, and studies chemistry in HKU. He figures out a winning strategy. Before long, B and C are quite drunk while A is still pretty much sober. Now, what is A’s winning strategy? When A plays with B and try the same strategy, he finds that the strategy is not as successful. Why?

Example 1:

Example 2:

Consider a molecular whose ground state energy is-10.0 eV, the first excited state energy -9.5 eV, thesecond excited state energy -1.0 eV, and etc.Calculate the probability of finding the molecule inits first excited state T = 300, 1000, and 5000 K.

Answer

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(-2) with 2 = -9.5 eV

And, the probability of finding the second excitedstate is,

p3 = exp(-3) with 3 = -1.0 eV

the ratio of probabilities between second & first excited states is,

exp[-(-1.0+9.5) x 11600/5000] = exp(-20)

[ 1 eV = 1.60 x 10-19 J; k = 1.38 x 10-23 J K-1; 1 eV 11600 k K-1] i.e., compared to the first excited state, 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.

The probability of finding the molecule in the first excited state is

p = exp(-2) / [exp(-1) + exp(-2)]

where, 1 = -10.0 eV, and 2 = -9.5 eV

(1) T = 300 K

300116000.5

exp1

300116000.5

expp

(2) T = 1000 K

(3) T = 5000 K

1000116000.5

exp1

1000116000.5

expp

5000116000.5

exp1

5000116000.5

expp

The Molecular Partition Function

The Boltzmann distribution can be written as

pi = exp(-i) / q

where pi is the probability of a molecule being found in a state iwith energy i. q is called the molecular partition function,

q = i exp(-i)

The summation is over all possible states (not the energy levels).

Molecular Partition Function

Interpretation of the partition function

As T 0, q g0,

i.e. at T = 0, the partition function is equal to the degeneracy ofthe ground state.

As T , q the total number of states.

Therefore, the molecular partition function gives an indicationof the average number of states that are thermally accessible to amolecule at the temperature of the system. The larger the value ofthe partition function is, the more the number of thermallyaccessible states is.

The relationship between q and :

exp() = q-1

Example 3.Consider a proton in a magnetic field B. The proton’s spin(S=1/2) has two states: spin parallel to B and spin anti-parallelto B. The energy difference between the two states is = pBwhere p is proton’s magneton. Calculate the partitionfunction q of the proton.

q = 1 + exp(-)

Example 4

Calculate the partition function for a uniform ladder of energylevels

q = 1 + e-

+ e-2

+ e-3

+ .......

e-

q = e-

+ e-2

+ e-3

+ ...... = q - 1

Therefore,

q = 1 / ( 1 - e-

)

Example 5:

Calculate the proportion of I2 molecules in their ground, firstexcited, and second excited vibrational states at 25oC. Thevibrational wavenumber is 214.6 cm-1.

n = (n+1/2) h

q = n exp(-n) = exp(-h/2) n exp(-n h) = exp(-h/2)/[1- exp(-h)]

Portion of I2 is in n-th vibrational state is

pn = exp(-n)/q = exp(-nh) [1- exp(-h)]

Therefore, at T = 25oC = 298.15 oK,

h = 1.036p0 =0.645 p1 = 0.229 p2 = 0.081

Partition function contains all the thermodynamic information! The relation between U and q If we set the ground state energy

0 to zero, E should be interpreted as the relative energy to the internal energy of the system at T = 0,

E = i ni i = i Nexp(-i) i / q = - Ndlnq/d.

Therefore, the internal energy U may be expressed as

U = U(0) + E = U(0) - N (lnq/)V

Where, U(0) is the internal energy of the system at T = 0. Theabove equation provides the energy as a function of variousproperties of the molecular system (for instance, temperature,volume), and may be used to evaluate the internal energy.

The relation between S and the partition function q

The Statistical Entropy

According to thermodynamics, entropy S is some measurementof heat q. The change of entropy S is proportional to the heatabsorbed by the system:

dS = dq / T

The above expression is the definition of thermodynamicentropy.

Boltzmann Formula for the entropy

S = k lnW

where, W is the weight of the most probable configuration ofthe system.Boltzmann Formula(1) indicates that the entropy is a measurement of theweight (i.e. the number of ways to achieve theequilibrium conformation), and thus a measurement ofrandomness,(2) relates the macroscopic thermodynamic entropy of asystem to its distribution of molecules among itsmicroscopic states,(3) can be used to evaluate the entropy from themicroscopic properties of a system; and(4) is the definition of the Statistical Entropy.

JUSTIFICATION

The energy of a molecular system U can be expressed as,

U = U(0) + i nii

where, U(0) is the internal energy of the system at T=0, ni is thenumber of molecules which are in the state with its energyequal to i

Now let’s imagine that the system is being heated while thevolume V is kept the same. Then the change of U may bewritten as,

dU = dU(0) + i nidi + i idni = i idni

[dU(0) = 0 because U(0) is a constant; di = 0 because i doesnot change as the temperature of the system arises.]

According to the First Law of thermodynamics, the change of internal energy U is equal to the heat absorbed (q) and workreceived (w), i.e.,

dU = dq + dwdq = TdS (thermodynamic definition of entropy; or more the heat absorbed, the more random the system)dw = -PdV = -Force * distance (as the system shrinks, it receives work from theenvironment)

dU = TdS - PdV = TdS (dV = 0) dS = dU/T = k i

idni

= k i (lnW/ni)dni + k i dni

= k i (lnW/ni)dni = k dlnW (-lnW/ni = ln (ni/N) = - i ) d(S - lnW) = 0 S = k lnW + constant

What is constant?

According to the Third Law of thermodynamics,

as T 0, S 0;

as T 0, W 1 since usually there is only one ground state,and therefore,

constant = 0.

July 1998 GunaHua CHEN @copyright

July 1998 GunaHua CHEN @copyright

Relation between S and the Boltzmann distribution pi

S = k lnW = k ( N lnN -i ni lnni ) = k i ( ni lnN - ni lnni ) = - k i ni ln(ni /N) = - Nk i (ni /N)ln(ni /N) = - Nk i pi ln pi

since the probability pi = ni /N.

The above relation is often used to calculate the entropy of asystem from its distribution function.

The relation between S and the partition function q

According to the Boltzmann distribution,

ln pi = - i - ln q

Therefore,

S = - Nk i pi (- i - ln q) = k i ni i + Nk ln qi pi

= E / T + Nk ln q = [U-U(0)] / T + Nk ln q

This relation may be used to calculate S from the knownentropy q

Hatch of an egg

Independent Molecules

Consider a system which is composed of N identical molecules.We may generalize the molecular partition function q to thepartition function of the system Q

Q = i exp(-Ei)

where Ei is the energy of a state i of the system, and summationis over all the states. Ei can be expressed as assuming there isno interaction among molecules,

Ei = i(1) + i(2) +i(3) + … + i(N)

where i(j) is the energy of molecule j in a molecular state i

The partition function Q

Q = i exp[-i(1) - i(2) - i(3) - … -i(N)] = {i exp[-i(1)]}{i exp[-i(2)]} … {i exp[-i(N)]} = {i exp(-i)}

N

= qN

where q i exp(-i) is the molecular partition function. Thesecond equality is satisfied because the molecules areindependent of each other. The above equation applies only tomolecules that are distinguishable, for instance, localizedmolecules. However, if the molecules are identical and free tomove through space, we cannot distinguish them, and theabove equation is to be modified!

The relation between U and the partition function Q

U = U(0) - (lnQ/)V The relation between S and the partition function Q

S = [U-U(0)] / T + k ln Q The above two equations are general because they not only apply to independent molecules but also general interacting systems.

Perfect Gas

Perfect gas is an idealized gas where an individual molecule istreated as a point mass and no interaction exists amongmolecules. Real gases may be approximated as perfect gaseswhen the temperature is very high or the pressure is very low.

The energy of a molecule i in a perfect gas includes only itskinetic energy, i.e.,

i = iT

q = qT

i.e., there are only translational contribution to the energy andthe partition function.

Translational Partition Function of a molecule qT

Although usually a molecule moves in a three-dimensionalspace, we consider first one-dimensional case. Imagine amolecule of mass m. It is free to move along the x directionbetween x = 0 and x = X, but confined in the y- and z-direction.We are to calculate its partition function qx.

The energy levels are given by the following expression,En = n2h2 / (8mX2) n = 1, 2, …

Setting the lowest energy to zero, the relative energies can thenbe expressed as,

n = (n2-1) with = h2 / (8mX2)

qx = n exp [ -(n2-1) ] is very small, then

qx = 1 dn exp [ -(n2-1) ] = 1 dn exp [ -(n2-1) ] = 0 dn exp [ -n2 ] = (2m/h22)1/2 X

Now consider a molecule of mass m free to move in a container of volume V=XYZ. Its partition function qT may be expressed as qT = qx qy qz

= (2m/h22)1/2 X (2m/h22)1/2 Y (2m/h22)1/2 Z = (2m/h22)3/2 XYZ = (2m/h22)3/2 V = V/3 where, = h(/2m)1/2, the thermal wavelength. The thermal wavelength is small compared with the linear dimension of the container. Noted that qT as T . qT 2 x 1028 for an O2 in a vessel of volume 100 cm3, = 71 x 10-12 m @ T=300 K

Partition function of a perfect gas,

Q = (qT) N / N! = V N / [3N N!]

EnergyE = - (lnQ/)V = 3/2 nRT

where n is the number of moles, and R is the gas constant

Heat CapacityCv = (E/T)V = 3/2 nR

Relation between the entropy S and the partition function Q

S = [U-U(0)] / T + k lnQ

Helmholtz energy

The Helmholtz free energy A U - TS. At constant temperatureand volume, a chemical system changes spontaneously to the statesof lower Helmholtz free energy, i.e., dA 0, if possible. Therefore,the Helmholtz free energy can be employed to assess whether achemical reaction may occur spontaneously. A system at constanttemperature and volume reaches its equilibrium when A isminimum, i.e., dA=0. The relation between the Helmholtz energyand the partition function may be expressed as,

A - A(0) = -kT ln Q

Fundmental Thermodynamic Relationships

Pressure

dA = dU - d(TS) = dU - TdS - SdTdU = dq + dwdq = TdSdw = -pdV dA = - pdV - SdT

Therefore, pressure may be evaluated by the following expression,p = -(A/V)T

p = kT( lnQ/V)T

This expression may be used to derive the equation of state for achemical system.

the entropy may also be expressed as

S = -(A/T)V

= klnQ + kT(lnQ/T)V

= klnQ -(lnQ/)V / T = klnQ + [U-U(0)] / T

Consider a perfect gas with N molecules. Its partition function Q isevaluate as

Q = (1/N!) (V / 3)N

the pressure p is thenp = kT( lnQ/V)T

= kT N ( lnV/V)T

= NkT / V

pV = NkT = nNAkT = nRT

which is the equation of the state for the perfect gas.

The enthalpy

During a chemical reaction, the change in internal energy is notonly equal to the heat absorbed or released. Usually, there is avolume change when the reaction occurs, which leads workperformed on or by the surroundings. To quantify the heatinvolved in the reaction, a thermodynamic function, the enthalpyH, is introduced as follows,

H U + pV

Therefore,H - H(0) = -( lnQ/)V + kTV( lnQ/V)T

The Gibbs energy

Usually chemical reactions occur under constant temperature. Anew thermodynamic function, the Gibbs energy, is introduced.

G A + pVAt constant temperature and pressure, a chemical system changesspontaneously to the states of lower Gibbs energy, i.e., dG 0, ifpossible. Therefore, the Gibbs free energy can be employed toaccess whether a chemical reaction may occur spontaneously. Asystem at constant temperature and pressure reaches its equilibriumwhen G is minimum, i.e., dG = 0. The relation between theHelmholtz energy and the partition function may be expressed as,

G - G(0) = - kT ln Q + kTV( lnQ/V)T

Example 6:Calculate the constant-volume heat capacity of a monatomicgas assuming that the gas is an ideal gas.

U = U(0) + 3N / 2 = U(0) + 3NkT / 2where N is the number of atoms, and k is the Boltzmannconstant.

Cv (U/T)V = 3Nk / 2 =3nNAk/2= 3nR/2

where, n is the number of moles, R NAk is the gas constant,and NA = 6.02 x 1023 mol-1 is the Avogadro constant

Example 7Calculate the translational partition function of an H2 molecule confined to a 100-cm3 container at 25oC

Example 8Calculate the entropy of a collection of N independent harmonicoscillators, and evaluate the molar vibraitional partition function of I2 at 25oC. The vibrational wavenumber of I2 is 214.6 cm-1

Example 9What are the relative populations of the states of a two-level systemwhen the temperature is infinite?

Example 10Evaluate the entropy of N two-level systems. What is the entropywhen the two states are equally thermally accessible?

Example 11Calculate the ratio of the translational partition functions of D2 andH2 at the same temperature and volume.

Example 12A sample consisting of five molecules has a total energy 5. Eachmolecule is able to occupy states of energy j with j = 0, 1, 2, ….(a) Calculate the weight of the configuration in which the moleculesshare the energy equally. (b) Draw up a table with columns headedby the energy of the states and write beneath then all configurationsthat are consistent with the total energy. Calculate the weight of each configuration and identify the most probable configuration.

Example 14Given that a typical value of the vibrational partition function of onenormal mode is about 1.1, estimate the overall vibrational partition function of a NH3.

Diatomic Gas

Consider a diatomic gas with N identical molecules. A molecule is made of two atoms A and B. A and B may be the same or different. When A and B are he same, the molecule is a homonuclear diatomic molecule; when A and B are different, the molecule is a heteronuclear diatomic molecule. The mass of a diatomic molecule is M. These molecules are indistinguishable. Thus, the partition function of the gas Q may be expressed in terms of the molecular partition function q,

The molecular partition q

where, i is the energy of a molecular state I, β=1/kT, and ì is the

summation over all the molecular states.

!/ NqQ N

i

iq )exp(

Factorization of Molecular Partition Function

The energy of a molecule j is the sum of contributions from its different modes of motion:

where T denotes translation, R rotation, V vibration, and E the electronic contribution. Translation is decoupled from other modes. The separation of the electronic and vibrational motions is justified by different time scales of electronic and atomic dynamics. The separation of the vibrational and rotational modes is valid to the extent that the molecule can be treated as a rigid rotor.

)()()()()( jjjjj EVRT

i

Ei

Vi

Ri

Tii iq )](exp[)exp(

i i i i

Ei

Vi

Ri

Ti )]exp()][exp()][exp()][exp([ EVRT qqqq

The translational partition function of a molecule

ì sums over all the translational states of a molecule.

The rotational partition function of a molecule

ì sums over all the rotational states of a molecule.

The vibrational partition function of a molecule

ì sums over all the vibrational states of a molecule.

The electronic partition function of a molecule

ì sums over all the electronic states of a molecule.

i

Ti

Tq )exp(

i

Ri

Rq )exp(

i

Vi

Vq )exp(

i

Ei

Eq )exp(

RVT qqqq 1/ Eqw

3/ VqT2/1)2/( Mh

kT/1where

Vibrational Partition Function

Two atoms vibrate along an axis connecting the two atoms. The vibrational energy levels:

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

5--------------5hv4--------------4hv3--------------3hv2--------------2hv1--------------hv0--------------0

hvnVn )2/1(

nhvVn

n= 0, 1, 2, …….

kT

hv

Then the molecular partition function can be evaluated

Therefore,

nn n

v hvhvnq )]exp(1/[1)exp()exp(

...1 32 eeeqv

1....32 vv qeeeqe

hvv

eeq

1

1)1/(1

Consider the high temperature situation where kT >>hv, i.e.,

hvkThvhv v //1q ,1

Vibrational temperature v

High temperature means that T>>v

hvk v

he hv 1

F2 HCl H2

v/K 309 1280 4300 6330v/cm-1 215 892 2990 4400m

kvwhere

Rotational Partition Function

If we may treat a heteronulcear diatomic molecule as a rigid rod, besides its vibration the two atoms rotates. The rotational energy

where B is the rotational constant. J =0, 1, 2, 3,…

where gJ is the degeneracy of rotational energy level εJR

Usually hcB is much less than kT,

=kT/hcB

)1( JhcBJRJ

states rotational all

]exp[ RJ

Rq

levelsenergy rotational all

]exp[ RJJg

J

)]1(exp[)12( JhcBJJ

0

)]1(exp[)12( dJJhcBJJqR

dJ /)]}1({exp[)/1(0

dJJhcBJdhcB

0]}1(){exp[/1( lJhcBJhcB

h/8cI2

c: speed of light I: moment of Inertia

i

i rmI i

2

hcB<<1

Note: kT>>hcB

For a homonuclear diatomic molecule

Generally, the rotational contribution to the molecular partition function,

Where is the symmetry number.

Rotational temperature R

hcBkTqR 2/

hcBkTqR /

12

CH

3

NH

2432OH

hcBk R

Electronic Partition Function

=g0

=gE

where, gE = g0 is the degeneracy of the electronic ground state, and the

ground state energy 0E is set to zero.

If there is only one electronic ground state qE = 1, the partition function of a diatomic gas,

]exp[]exp[states electronic all energies electronic all

Ejj

Ej

E gq

]exp[ 00Eg

NhvNN ehcBkTVNQ )1()/()/)(!/1( 3

At room temperature, the molecule is always in its ground state

Mean Energy and Heat Capacity

The internal energy of a diatomic gas (with N molecules)

(T>>1)

Contribution of a molecular to the total energyTranslational contribution(1/2)kT x 3 = (3/2)kTRotational contribution(1/2)kT x 2 = kTVibrational contribution(1/2)kT + (1/2)kT = kT kinetic potentialthe total contribution is (7/2)kT

venNNnNUU hvvv ]/)1(1[)/ln()/1(3)0(

)1/(/1/1)2/3( hveNhvNN )1/()2/5( hveNhvNkT

kTN )2/7(

qV = kT/hvqR = kT/hcB

The rule: at high temperature, the contribution of one degree of freedom to the kinetic energy of a molecule (1/2)kT

the constant-volume heat capacity

(T>>1)Contribution of a molecular to the heat capacityTranslational contribution

(1/2) k x 3 = (3/2) kRotational contribution

(1/2) k x 3 = kVibrational contribution

(1/2) k + (1/2) k = kkinetic potential

Thus, the total contribution of a molecule to the heat capacity is (7/2) k

vv TUC )/( 22 )1/(hv)(K N k )2/5( hvhv eeN

k )2/7( N

Summary

Principle of equal a priori probabilities: All possibilities for the distribution of energy are equally probable provided the number of molecules and the total energy are kept the same.

A configuration { n0, n1, n2, ......} can be achieved in W

different ways or the weight of the configurationW = N! / (n0! n1! n2! ...)

Dominating Configuration vs Equilibrium

The Boltzmann DistributionPi = exp (-i ) / q

Partition Function

q = i exp(-i) = j gjexp(-j) Q = i exp(-Ei)

Energy E= N i pi i = U - U(0) = - (lnQ/)V

Entropy S = k lnW = - Nk i pi ln pi = k lnQ + E / T

A= A(0) - kT lnQ

H = H(0) - (lnQ/)V + kTV (lnQ/V)T

Q = qN or (1/N!)qN

EVRT qqqqq