GHC@copyright G. H. CHEN Department of Chemistry University of Hong Kong Intermediate Physical Chemistry

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G. H. CHENDepartment of Chemistry

University of Hong Kong

Intermediate Physical Chemistry

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Intermediate Physical Chemistry

Contents:Distribution of Molecular States Perfect Gas Fundamental Relations Diatomic Molecular Gas

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Distribution of Molecular States

Configurations and Weights Boltzmann Distribution, and Physical Meanings of Molecular Partition Function and its Interpretation The Internal Energy and the Entropy Independent Molecular and their Partition Function

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Perfect Gas

Partition FunctionEnergyHeat CapacityPressure and the gas law

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Diatomic Molecular Gas

Factorization of Partition Function Rotational Partition FunctionVibrational Partition FunctionElectronic Partition Function Mean Energy and Heat Capacity

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Fundamental Relations

Helmholtz Energy Pressure Enthalpy Gibbs Energy

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Statistical Mechanics provides the link between the microscopic properties of matter and its bulk properties.

Statistical Mechanics:

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THE DISTRIBUTION OF MOLECULAR 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. Countless collisions occur. It is hopeless to keep track positions, momenta, and internal energies of all molecules.

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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.

That is, we assume that vibrational states of acertain energy, for instance, are as likely to bepopulated as rotational states of the same energy.


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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


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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, ......}


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For instance, a system with 17 molecules, and each molecule has four states.

1

2

3

4

5

6

7

89

10

1112

13

14

15

16 17

1

2

3

4

STATE

The above configuration is thus, { 4, 6, 4, 3 }


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{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! ...)


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Justification

1. N! different ways to arrange N molecules;

2. ni! arrangements of ni molecules with energy i

correspond to the same configuration;


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1. Calculate the number of ways of distributing3 objects a, b and c into two boxes with thearrangement {1, 2}.

Answer:| a | b c |, | b | c a |, | c | a b |.

Therefore, there are three ways 3! / 1! 2!

Example:


| a | c b |, | b | a c |, | c | b a |

To eliminate overcounting of these configurations

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2. Calculate the number of ways of distributing20 objects into six boxes with the arrangement {1, 0, 3, 5, 10, 1}.

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

note: 0! = 1

Example:


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Stirlings 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 3.1576 6.579 4.751 6.5668 10.605 8.636 10.59510 15.104 13.026 15.096


Note: A = (2)1/2 (x+1/2)x e-x

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Therefore,

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

= N ln N - ni ln ni

x ln x! x ln x - x ln A12 19.987 17.819 19.98016 30.672 28.361 30.66620 42.336 39.915 42.33230 74.658 72.036 74.656

Stirlings approximation(cont’d):


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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

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t0 1 2 3 4 5 6 7 80

20

40

60 N = 8

W

k


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t0 1 2 3 4 5 6 7 8 9 101112131415160

2000

4000

6000

8000

10000

12000N =16

W

k


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t0 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


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

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)!]


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| 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

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

The ratio of the two weights is equal to

R(N) Wk=N/2 / Wk=N/4

=(N/4)! (3N/4)! / [(N/2)!]2


If the system has more states, would we reach the same conclusion as above? Why?

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

ni i = E = constant

2. The total number of molecules is conserved,i.e.

ni = N = constant

How to maximize W or lnW under these constraints?

Two constraints for the system

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The Boltzmann Distribution

ni / N = Pi = exp ( - i )

Interpretation of Boltzmann Distribution

Meaning of : ensure total probability is ONE

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

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

= - ln [i exp(-i)]

1 / = kT

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Starting from the principle of equal a priori probabilities, we evaluate the probabilities of different configurations by simple counting, and find that the dominating configuration whose population obeys the Boltzmann distribution which relates the macroscopic observables to the microscopic molecular properties, and is capable of explaining the equilibrium properties of all materials.


Summary

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To find the most important configuration, we vary { ni } to

seek the maximum value of W. But how?

E.g. One-Dimensional Function: F(x) = x2

dF/dx = 0


Enough?

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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

To find the maximum value of W or lnW,

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


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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

The method of Lagrange Multipliers

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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

The Dominating ConfigurationThe method of Lagrange Multipliers

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JUSTIFICATION

dL = dF - i i dCi

under the constraints, dCi = 0, thus

dF = 0

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

The Dominating ConfigurationThe method of Lagrange Multipliers

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The method of Lagrangian Multiplier

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

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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 method of Lagrangian Multiplier

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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,


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This is the physical meaning of , the reciprocaltemperature.

1 / = kT

where k is the Boltzmann constant. Thus,

<mv2>/2 = 3kT/2

(This is actually the definition of the temperature)


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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:


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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.


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The summation is over all possible states (not the energy levels). If the energy level is gi-fold degenerate, then the

molecular partition function can be rewritten as

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 i with energy i. q is called the molecular

partition function, q = i exp(-i)

q = i gi exp(-i)

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As T 0, q g0,

i.e. 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.

The relationship between q and :

exp() = q-1

Interpretation of the partition function

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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-parallel to B. The energy difference between the two states is = pB where p is proton’s magneton.

Calculate the partition function q of the proton.


Example 3

Example 4Calculate the partition function for a uniform ladder of energy levels

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.

Example 5

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Partition function contains all the thermodynamic information!

Statistical Thermodynamics

The Internal Energy, the Heat Capacity & the Entropy

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.

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Therefore, the internal energy U may be expressed as

The Internal Energy, the Heat Capacity & the Entropy


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

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

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The relation between CV and q

Cv is the constant-volume heat capacity which measures

the ability of a system to store energy. It is defined as the rate of internal energy change as the temperature T varies while the volume is kept constant:

Cv (U/T)V = N(1/kT2) (2lnq/2)V

Note that, d/dT = (d/dT) d/d = -(1/kT2) d/d

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Example 6:

Calculate the constant-volume heat capacity of a monatomic gas assuming that the gas is an ideal gas.

U = U(0) + 3N / 2 = U(0) + 3NkT / 2


where N is the number of atoms, and k is the Boltzmann constant.

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

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The relation between S and the partition function q


The Statistical Entropy

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

dS = dq / T

The above expression is the definition of thermodynamic entropy.

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Boltzmann Formula for the entropy


S = k lnW

where, W is the weight of the most probable configuration of the system.

Boltzmann Formula(1) indicates that the entropy is a measurement of the weight (i.e. the number of ways to achieve the equilibrium conformation), and thus a measurement of randomness,(2) relates the macroscopic thermodynamic entropy of a system to its distribution of molecules among its microscopic states,(3) can be used to evaluate the entropy from the microscopic properties of a system; and(4) is the definition of the Statistical Entropy.

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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 the number of molecules which are in the state with

its energy equal to i

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

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

[dU(0) = 0 because U(0) is a constant; di = 0 because i

does not change as the temperature of the system arises.]

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According to the First Law of thermodynamics, the change of internal energy U is equal to the heat absorbed (q) and work received (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 the environment)


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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) = 0S = k lnW + constant

What is the constant?


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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.


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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 a system from its distribution function.

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The relation between S and the partition function q


According to the Boltzmann distribution,

ln pi = - i - ln qTherefore,

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 known entropy q

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Independent Molecules


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

Q = i exp(-Ei)

where Ei is the energy of a state i of the system, and summation

is over all the states. Ei can be expressed as assuming there is no

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

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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. The

second equality is satisfied because the molecules are independent of each other.


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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.


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Perfect Gas

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

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

i = iT

q = qT


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

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Translational Partition Function of a molecule qT

Although usually a molecule moves in a three-dimensional space, we consider first one-dimensional case. Imagine a molecule of mass m. It is free to move along the x direction between 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, …


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

qx = 1 dn exp [ -(n2-1) ] = 1 dn exp [ -(n2-1) ]

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

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

qx = n exp [ -(n2-1) ]

is very small, then


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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


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Partition function of a perfect gas,

Q = (qT) N = V N / 3N

Energy

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

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

Heat Capacity

Cv = (E/T)V = 3/2 nR


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Fundamental Thermodynamic Relationships

Consider an equilibrium system which is consistent of N interacting molecules. These molecules may or may not be the same.

Relation between energy and partition function


U = U(0) + E = U(0) - (lnQ/)V

Relation between the entropy S and the partition function Q

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

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Helmholtz energy

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

A - A(0) = -kT ln Q


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Pressure

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

Therefore, pressure may be evaluated by the following expression,

p = -(A/V)T

= kT( lnQ/V)T

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


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

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

the pressure p is then

p = 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.


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The enthalpy

During a chemical reaction, the change in internal energy is not only equal to the heat absorbed or released. Usually, there is a volume change when the reaction occurs, which leads work performed on or by the surroundings. To quantify the heat involved in the reaction, a thermodynamic function, the enthalpy H, is introduced as follows,

H U + pVTherefore,

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


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The Gibbs energy

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

G A + pV

At constant temperature and pressure, a chemical system changes spontaneously to the states of lower Gibbs energy, i.e., dG 0, if possible. Therefore, the Gibbs free energy can be employed to access whether a chemical reaction may occur spontaneously. A system at constant temperature and pressure reaches its equilibrium when G is minimum, i.e., dG = 0. The relation between the Helmholtz energy and the partition function may be expressed as,

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


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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?


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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.


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Example 15Consider Stirling’s approximation for lnN! In the derivation of the Boltzmann distribution. What difference would it make if the following improved approximation is used ?

x! = (2)1/2 (x+1/2)x e-x

Example 16Consider N molecules in a cube of size a.(a) Assuming molecules are moving with the same speed v along one of three axises (x-, y-, or z-axis). The motion may be along the positive or negative direction of an axis. On the average, how many molecules move parallel to the positive x direction?


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(b)When a molecule collides with one of six sides of the cube, it is reflected. The reflected molecule has the same speed v but moves in the opposite direction. How many molecules are reflected from the side within a time interval t ? What is the corresponding momentum change of these molecules?(c) Force equals the rate of momentum change. Calculate the force that one side of the cube experiences.Pressure is simply the force acting on a unit area. What is the pressure of the gas? Assuming the average kinetic energy of a molecule is 3kT/2, derive the equation of state for the system.


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!/ NqQ N

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.

i iq )exp(


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B

A


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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

Factorization of Molecular Partition Function

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


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i

Ei

Vi

Ri

Tii iq )](exp[)exp(

i i i i

Ei

Vi

Ri

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

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RVT qqqq

1/ Eqw

3/VqT

2/1)2/( Mh

kT/1

where


The translational partition function of a molecule

ì sums over all the translational 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(

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.

i

Ri

Rq )exp(

i

Vi

Vq )exp(

i

Ei

Eq )exp(

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

hvnVn )2/1(

nhvVn

n= 0, 1, 2, …….

kT hv

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

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Then the molecular partition function can be evaluated

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

Therefore,

Vibrational Partition Function

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

)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

where gJ is the degeneracy of rotational energy level

εJR Usually hcB is much less than kT,

Note: kT>>hcB

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

=kT/hcB

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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

Rotational Partition Function

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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

Electronic Partition Function

=g0

=gE

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

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

Mean Energy and Heat Capacity

(T>>1)

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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


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Translational energy

22

22

2

1

8mVx

mx

hnTn

mx

hnVx

2

where n = 1, 2, …n is a measurement of the speed of the molecule

mx

50h

mx

h

2mx

hVx

.... 100 .... 2 1 n

sec/109.12

5 mmx

h

Quantum Classical

for a H in a one-dimonsional box x= 1cm


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Vibrational energy

2

2

1 )

2

1( kAhVn

v

n

Quantum Classical

A is the amplitude of the vibration

k

hvn

k

hvnA )12()

2

1(2

Here the vibrational quantum number n is a measurement of the vibrational amplitude.


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mjJ J

R

eeqR

J

R

J J,

)12(

mj=-J,-J+1…….


Rotational energy

2

2

1 )1( IJhcBJ

R

J

Quantum Classical is the rotational angulor velocity

)1(2

JJI

h

J is a measurement of angular velocity

mJ is a measurement of the projection of the angular velocity of the z-

axis. i.e. a measurement of the rotation’s orientation.

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Molecule with N atoms Degree of freedom: Translation: 3Rotation: 3 nonlinear2 linearvibration: 3N – 6 nonlinear

3N – 5 linear Diatomic Molecule A-B

Symmetry =

Supplementary note

0 if A B1 if A = B

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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 configuration

Dominating Configuration vs Equilibrium


Pi = exp (-i ) / q

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

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HC

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

Partition Function

Energy

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

Heat Capacity

CV = (E/T)V = k2 (2lnQ/2)V

Entropy

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

A= A(0) - kT lnQ

Helmholtz energy

Summary

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

Q = qN or (1/N!)qN

q = qTqRqVqE

Enthalpy

Molecular partition function


Summary

Documents

GHC@copyright G. H. CHEN Department of Chemistry University of Hong Kong Intermediate Physical Chemistry