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Molecular Thermodynamics (CH3141)

The ideal gas II: Intramolecular degrees of freedom

N.A.M. (Klaas) Besseling

SandlerCh4

Molecular rotations

Molecular vibrations

Equipartition of energy

2

Summary of what we learned about the ideal gas so far:

The total partition function Qof an ideal gas can be build

up fromsingle-molecule partition functions q:

Q = exp !E

i

kT

"

#$%

&'i(

Q =1

N!qN

Nis the number of molecules

division byN! because of the indistinguishability of theN

atoms of the same kind.

For a one-component ideal gas (Maxwell-Boltzmann statistics)

Divide and Rule! (independent modes or atoms)

q = exp ! !

n

kT

"

#$%

&'n(with

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hence, the single-molecule partition function can be factorised:

q =qtrans

qint

Divide and rule further!

For the mono-atomic ideal gas at not-too-high T,- qintis trivial and only qtransmatters,

- we can write:

q = qtrans

Translational and internal degrees of freedom are independent

(taking the electronic and nuclear partition functions equal to 1)

4

There are 3 independent translational degrees of freedom,

henceztrans itself can be factorised:

qtrans

= qxqyqz with qx = exp !

!lx

kT

"#$

%&'lx

( etc.

. . . and further (independent translational modes)

where are the energy levels of a particle in a 1D box!lx

=

h2lx

2

8mLx

2

Because the energy levels are closely spaced ,and vary smoothly with lx, the sum can be replaced by an

integral, solving the integral yields

qx =

Lx

! where ! =

h

2!mkT

thermal wavelength

(!"

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Hence expressions for all desired thermodynamic propertiescan be obtained:

e.g. pressure (ideal gas law)

energy ( per degree of freedom)

heat capacity

entropy

1

2kT

5

Q =1

N!qN

=

1

N!

VN

!3N

We can use this to construct the total partition function:

for the monoatomic ideal gas

which directly yields an expression for the Helmholtz energy

because

A =!kTlnQ

6

Intramolecular degrees of freedom

So far we discussed the monoatomic ideal gas at not-too-high T,

for which only translational degrees of freedom are relevant.

Then thesingle-molecule partition function essentiallyequals the

translational partition function:

q =qtrans = exp !!ltrans

kT

"

#$

%

&'

ltrans

(

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Hence, we can write the single-molecule partition function as a product

q =qtrans

qint

qtrans = exp !!ltrans

kT

"#$

%&'ltrans

( qint = exp !!iint

kT

"#$

%&'iint

(with

check this; show that indeed

exp !!

ltrans+ !

iint

kT

"#$

%&'iint

(ltrans

( = exp !!

ltrans

kT

"#$

%&'

exp !!

iint

kT

"#$

%&'iint

("

#$%

&'ltrans(

"

#$

%

&'

Oftenintramolecular degrees of freedomare relevant:

q = exp ! !i

kT

"#$

%&'

i

( = exp !!ltrans

+ !iint

kT

"#$

%&'iint

(ltrans

(

Internaland translationaldegrees of freedom are independent;

!Energies are additive

A =!kTlnQ =kTN lnN!lnqtrans! lnqint!1( )

=kTN ln N"3

eV

#$%

&'(! lnq

int

#

$%&

'(8

qint

!qrotqvibqeleqnuc

Internal molecular degrees of freedom

The internal molecular state is a combination of:

the rotational, vibrational, electronic, and nuclear states

Sandler4.1

this presumes that

rotational, vibrational, electronic, and nuclear states are independent

!!! This is only approximately true under certain conditions !!!

(details later)

Q =1

N!qN

=

1

N!qtransqint( )

N

q =qtrans

qint

lnqint

!ln qrot+ lnqvib + lnqele + lnqnuc

check this

check this

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what modification to the expression for chemical potential

that we have seen so far:

arise from the intramolecular degrees of freedom?

A =kTN ln N!3

eV

"#$

%&'(lnq

int

"#$

%&'

=kTln !!3( )

!kTNlnqint

=kT ln !!3( )"lnqint( ) =kTln !!3

qint

#$%

&'(

Term of Helmholtz energy:

!

!NV,T

!kTlnqint

with intramolecular

degrees of freedom:

!!"#

independent ofN

?

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Often ok to assume that only

ground states(states with lowest energy levels) are relevant.

(ok at moderate temperatures, then )!! >>kT

Nuclear and electronic states

There may be more than one state with the same energy level.

the nr. of states with same energy is called

the degeneracyor multiplicityof that energy level

Then !!" = !#$ !

!!"%"

#$

"

#$

%

&'"=&

(

) *"!"%&!#$ !!

!"%&

#$

"

#$

%

&' ="!"%&

choosing the ground state level

to be the zero of energy

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!1enters qjust as a constant factor.

Hence it enters the Helmholtz energy only as

aconstant contribution per molecule:

Irrelevant for measurable thermodynamic properties

(if no chemical reactions occur!)

So it is often taken to be just 1 (even if it is not).

Then terms in the free energy are simply omitted

Similar arguments apply tonuclear states.

qint

!qrotqvibqeleqnuc !qrotqvib

!"!#!$%

lnqel,1

= ln1=0

( )

hence:

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Polyatomic molecules have vibrationaland rotationaldegrees

of freedom.

It is a reasonable approximation to assume that vibrational and

rotational degrees of freedom are independent:

qint

! qvib

qrot

This is not completely exact; it neglects centrifugal effects

Vibrational and rotational states

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qvib

!qvib

1

qvib

2

qvib

3

. . .

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Molecular vibrations

If vibration energies not too high,

molecular vibrations can be approximated as

harmonic oscillation (oscillations in a parabolic potential energy),

and as

separable in independent vibrational normal modes.

The number of modes depends on the molecular structure.

Hence

Sandler4.1c

diagram by Derek Kverno at http://www.cartage.org.lb/en/themes/sciences/Physics/MolecularPhysics/MolecularSpectra/MolecularSpectroscopy/MolecularSpectroscopy.htm

vibration modes of the CO2 molecule

(linear tri-atomic molecule!threemodes)A diatomicmolecule

has only onevibration

mode

more complex

molecules!more

vibrational modes

In the following we

focus on just one mode,

indicated by subscript vib

without a number.14

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Harmonic potential(= parabolic potential):

V(x) = 12 Kx2 ! F(x) = "

dV

dx= "Kx

x= displacement of oscillator from its minimum-energy state

K= spring constant

V= potential energy

F= restoring force

We have chosen

For mass min a parabolic potential

the vibrationfrequencyis

V(0) = 0

! =

1

2!

K

m

from Classical Mechanics:Christiaan Huygens 1673Horologium oscillatorium sive de motu pendulorum

more generally:

mto be replaced by a reduced mass

e.g. diatomic molecule: m1m

2 m

1+ m

2( )

(Hookes law)

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According to the

Schrdinger equation for

a harmonic potential energy well

!vib,j = j+12( )h!

(j = 0, 1, 2, . . . !)

equidistant energy levels: !!vib

=h"

ground state energy:(fromh

ttp://131.1

04.1

56.2

3/Lectures/CHEM_

207/vibrational_spectroscopy)

The temperature for which is!"vib= kT !vib =h" k

the energy levels are

!O 1000K( )

choosing the minimumpotential as the zero of energy:

!0 =

1

2h"!

!

!

h"

j = 0

j = 1

j = 2

j = 3

j =4

j = 5

j = 6

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If that is if T

then only the ground state is occupied,

the higher energy states are virtually unoccupied.

With the Boltzmann distribution law in mind,

we understand immediately, without any further calculations,

that

This is usually approximately true at room temperature since

(this is the low-Tlimit)

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The partition function for one vibrational mode is

!!"#

= $%& !!

!

"#

"

#$

%

&'

!=0

(

) =

=exp ! h!

2kT

"#$

%&'

exp !h!

kT

"#$

%&'

"#$

%&'

j

j=0

(

)using (geometric series)xi

i=0

!

" =1 1#x( )

qvib =exp ! 12h! kT( )1! exp !h! kT( )

=

exp ! 12!vib T( )1! exp !!

vib T( )

= exp ! h!

2kT

"#$

%&'exp !

jh!

kT

"#$

%&'j=0

(

)

!"# ! !+ 1

2( )!!!"

"

#$$

%

&''

#=0

(

)

!!"#

!

$"

%vibrational temperature

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(energy-level spacing )

then ~ only the ground state occupied:

T ! h" k

qvib! exp " h!

2kT

#$%

&'( =exp "

"vib

2T

#$%

&'(

qvib =exp ! 12h! kT( )1! exp !h! kT( )

!!vib

=h" >>kT

Helmholtz energy: avib =!

kTln

qvib =

1

2h!

entropy:

Contributionsper moleculeto the:

energy:(mean energy of vibrational mode)

!!"#

= !"#

$%& '

"1 !=

1

2!"

heat capacity: !!"#

=

!!!"#

!"= 0

!!"#

= !!"#! !

!"#( ) ! = 0

Low temperature limit

(just the first term of partition func.)

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energy-level spacing : classical limit

qvib

!

kT

h!=

T

"vib

!">> #!

Helmholtz energy: avib

=!kTlnqvib =!kTlnkT

h!

Contributions per molecule to the

energy: !vib= k

d avib

T

d1 T= k

dln1 T

d1 T= k

1

1 T= kT

heat capacity: cvib =d!

vib

dT= k

with for |x|

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Graphical summaryof dependence of

"viband cvibon T

!

Cvib

k!

low-Tlimit:

just the

ground-state energy

low-Tlimit:

exac

tresul

t

high-Tlimit:

high

-Tlimit

!vib

!vib

NB:

!vib "O(1000K)

!vib

=h"

eh" kT

!1+h"

2

!!

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Molecular rotations

Assuming that vibrations and rotations are independent:

qint

= qvib

qrot

implies that rotations are treated as if molecules are rigid rotors

(neglecting centrifugal effects)

Sandler4.1b

- mono-atomic molecules: 0 rotational modes

- linear molecules: 2 rotational modes

- non-linear molecules: 3 rotational modes

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Hetero-nucleic diatomic molecule (e.g. CO)

R R

mA

mB

=mAmB

mA+m

B

reduced mass

rotation of di-atom AB with AB distance =R

equivalent withmovement of a mass on the surface of a sphere with radiusR

Hence in the classical limit the partition function

is the same as that for translational movement

of a particle with massin a 2D box of area :4!R2

~

2424

qrot =2!kT

h2!"#

$%&

2 2

4!R2

= moment of inertiaI= R2

qrot =8!

2IkT

h2

=

T

!rot

= rotational temperature!rot

=

h2

8!2Ik

= translational single-particle partition function

of a particle with massin a 2D box of size

in the classical limit (check this)

4!R2

So we can use expression ~ translational motion

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If the classical limit does not apply

(low T, or smaller)

the actual values of the lowest energy levels are important.

These do not just depend on the size, but also on the shape

of a 2D box.

Then the correct rotational energy levels have to be used.

Makes things more complicated (see Sandler)

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kT ! "!

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Homo-nuclear diatomic molecule (e.g. N2)

The two nuclei are indistinguishable

Rotational partition function as given above

needs to be devided by 2:

Generally:

where thesymmetry number ! =1 heteronuclear

2 homonuclear

"#$

qrot =8!

2IkT

"h2

=

T

"!r

Molecular symmetry

each orientation of a homonuclear = symmetric linear molecule

corresponds to

twoorientations of a heteronuclear = asymmetric linear molecule

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qrot =8!

2IkT

"h2

The contribution to the Helmoltz energy per molecule is

arot =!kTln qrot with

The contribution to the energy per molecule is

!!"#

=

$%!"#

&

$1 &= '

$!"1 &

$1 &= '&

That is per degree of freedom

(a diatomic molecule has two rotational degrees of freedom)

1

2kT

For non-linear molecules: !rot =3

2kT

crot=

d!rot

dTThe contribution to the heat capacity per molecule is

rotations (classical limit applies)

translations (classical limit applies)

vibrations (classical limit may be used )

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Overview for molecular systems at room temperature:

!"

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The Law of Equipartition of Energies

The Equipartition Theorem

for translations and rotations and vibrations

the mean energy per molecule, per degree of freedom is always

1

2kT

For vibrations it is kTper molecule for each normal mode.

This is in the classical limit

for potential energy and

for kinetic energy.

1

2kT

1

2kT

In the classical limit(high temperatures),

Equipartition = division in equal parts

each degree of freedom contributes an equal amount to energy12kT

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if the energy for a degree of freedom of a particle can be written

as Cy2

where Cis some constant coefficient

ydefines the state for a degree of freedom

ycan be,

-

in classical mechanical terms: a (angular) velocity

component. Then Cy2is a kinetic energy.

- a position coordinate in the case of a harmonic potential.

Then Cy2is a potential energy.

- in quantum mechanical terms: a quantum number, e.g. as for

translation (check e.g. in previous handouts)

Why so often this ?12kT

Because of the quadratic expressions for the energy:

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for each translational degree of freedom,

in classical mechanical terms: 12mv

x

2

for each rotational degree of freedom,

in classical mechanical terms: 12I!

2

I= moment of inertia, != angular velocity

for vibrational modes, kinetic energy:

+ potential energy; if vibration is harmonic: 12kx

2

(in quantum mechanical terms this is )! lx

2

1

2mv

x

2

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q = exp !Cy2

kT

"

#$%

&'dy

!(

(

) =!kT

C

the partition function

classical expression

y(treated as) continuous

the corresponding Helmholtz energy contribution

a =!kTln q =!kTln !kT

C=! 1

2kT ln

!k

C+ lnT

"

#

$%

&

'

with the Gibbs-Helmholtz relation:

! =d a T

d1 T=

1

2k

d ln1 T

d1 T=

1

2kT

This result does not depend on C,

only on the quadratic character of !y =Cy

2

ydefines microstate of some mode

if energy of the state can be written as !y = Cy

2

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Give a rough estimate of the heat capacity of a mole of CObased on the equipartition theorem

at ~ room temperature (assuming low-Tlimit can be used for vibr)

at ~ 5000 K (assuming that it does not decompose)

vibrations (low-Tlimit)

!"