molecular principles

7/25/2019 molecular principles

1/23

1

1

Molecular Thermodynamics (CH3141)

First application of Canonical Probability Distribution

The Ideal Gas

N.A.M. (Klaas) Besseling

The Ideal Gas I (monoatomic, not-too-high T )

Independent degrees of freedom (Divide and Rule)

Single-particle translational partition function

Total partition function of the monoatomic ideal gas

Thermodynamic properties of the monoatomic ideal gas

2

SandlerCh3


2/23

2

3

Review of the

Q = exp !Ei kT( )i

"

Boltzmann distribution law:

pi =exp !Ei kT( )

Q

the normalisation factor is

the(canonical) partition function

The probability that a system

is in (quantum) state iis

whereEiis the energy of state i, and

A =!kTlnQ

The canonical partition function

is related to theHelmholtz energythe mean mechanical energy

E = EiPii

! =E

iexp "Ei kT( )

Qi!

is identified with

the thermodynamic

internal energy: E =U

4

First example of Molecular Thermodynamics applied.

Demonstration of the machinery of Molecular Thermodynamics:

Formulation of expression for the partition function,

based on the quantum mechanics of the molecules.

Derivation of expressions for thermodynamic propertiesfrom

these molecular properties.

Ideal gas: molecules do not interact (molecules are point masses)

does not exist in reality (but ideal-gas behaviour is observed at

low densities, when intermolecular encounters are rare)

molecules do have intramolecular degrees of freedom (e.g.

vibrations)


3/23

3

5

For this first example: simplest possible system:

monoatomic ideal gas at not-too-high temperature.

For an ideal gas (independent molecules) the total partition

function can be factorised into

single-molecule partition functions.

Divide and Rule

6

Intermezzo: independent subsystems, modes / degrees of freedom

independent subsystems e.g.

molecules in an ideal gas

binding sites

they do not interact => energy is sum of the energies of subsystems

independent modes / degrees of freedom e.g.

molecular translations and vibrations

energy contributions of separate modes can be added up toobtain the total energy

afmaken


4/23

4

The single-molecule partition function:

7

q = exp

!

!n

kT

"

#$

%

&'n(

!

nis the energy level of quantum state n

According to quantum mechanics

a molecule has discrete (quantum) states

Sandler2.3

these quantum states are numbered: by quantum numbers (e.g. n)

8

-

For mono-atomic molecules, e.g. Argon,

there are novibrationsand no rotations.

- at moderate temperatures

electronic and nuclear states are not thermally active;

(only ground states populated because )!! >>kT

First examine the simplest case: mono-atomic ideal gas

generally, molecular quantum states involve

- molecular translations (in three independent directions)

- molecular vibrations (depending on the molecular structure)

-

molecular rotations (in maximal three independent directions)

-

molecular electronic states

-

nuclear states


5/23

5

Lx

Lz

Ly

9

-

specifies the translational quantum statesof a particle in a 3D rectangular container of size

- Quantum number lxspecifies the quantum states

(with energy level ) for a particle in a 1D box of sizeLx.

-The energy levels for the 3D box are

V=LxLyLzltrans = (lx, ly, lz )

!lx

!ltrans= !lx

+ !ly+ !lz

Sandler1.3, 3.1 ,

A molecule has

threetranslational degrees of freedom.

These are independent.

lx = 1, 2, 3, ... !

q =qtrans = exp !!ltrans

kT

"#$

%&'ltrans

(only translationaldegrees of freedomare relevant:

sum over all translational states

Energies of quantum states follow from the Schrdinger equation:

10

!lx

=

h2lx

2

8mLx

2 (l

x =1, 2, 3, . . . !)

(= 6.627 !10"34 Js)

plx =hlx

2Lx! mvx

!lx=

plx2

2m!

px2

2m=

1

2mvx

2

The momentum levels are

h= Plancks constant

m= the mass of the particle

For a particle in a 1D box of sizeLxthe energy levels are

ideal gas:

no potential energy!

classical limit


6/23

6

11

qtrans = exp !!ltrans

kT

"#$

%&'

()*

+,-ltrans

. = exp !!lx

+ !ly+ !lz

kT

"

#$%

&'()/

*/

+,/

-/lz.

ly

.lx

.

= exp !!lx

kT

"#$

%&'exp !

!ly

kT

"

#$%

&'exp !

!lz

kT

"#$

%&'

()*

+*

,-*

.*lz/

ly

/lx

/

= exp !!lx

kT

"#$

%&'lx

()*+

,-.

exp !!ly

kT

"

#$%

&'ly(

)*/

+/

,-/

./exp !

!lz

kT

"#$

%&'lz

()*/

+/

,-/

./

qtrans

=qxqyqz with etc.qx = exp !

!lx

kT

"#$

%&'lx

(

hence

!ltrans= !lx

+ !ly+ !lz

The translational partition function for a particle in a 3D box

can be written as a product of 3 partition functions for the

three independent translational degrees of freedom.

!"+#

= !"!#( )!

12

exp !!nx

kT

"#$

%&'exp !

!ny

kT

"

#$%

&'exp !

!nz

kT

"#$

%&'

(

)**

+

,--nz

.ny

.nx

. =

= exp !!nx

kT

"#$

%&'nx

()

*+

,

-. exp !

!ny

kT

"

#$%

&'ny(

)

*++

,

-..

exp !!nz

kT

"#$

%&'nz

()

*++

,

-..

On the previous slide we saw

If you dont see this right away, write out for yourself:

FnG

m[ ]m=1

2

!n=1

2

! and to see that they are equalFnn=1

2

!"

#$

%

&' Gm

m=1

2

!"

#$

%

&'

Compare with the above equation


7/23

7

(as illustrated above, for the three translational degrees of

freedom of a particle in a 3D box: )

13

When a system has independent degrees of freedom

(also called independent modes),

meaning that the energy contributions of the modes are additive,

then the partition function can be written as aproductof the

partition functions for each of the degrees of freedom

qtrans

=qxqyqz

Divide and Rule

Generally:

- lmaxindicates the state for which

-

decreases rapidly for

- states with are relatively sparsely occupied14

Terms of for whichqx = exp !!lx

kT

"#$

%&'lx

( !lx

=h2lx

2

8mLx

2 >>kT

!lx

kT=

h2lx

2

8mLx

2kT

>>1

lx >> 8mkTL

x h ! l

max

!"# !!

!"

#$

"

#$$

%

&'' = !"# !

!"

!$%"

"

#$

%

&'

2"

#$$

%

&''

lx lmax

>1

! = kT

lx > lmax (that is !>kT)

Single-particle translational partition function

contribute little to the partition function qx.

that is for which

that is for which


8/23

8

15

For examplewhen

- size of the 1D boxLx= 1 cm = 0.01 m

- particle mass

-

-

-

then

number of easily accessible states in a macroscopic system is huge!!!

m =1.67 !10"27Mkg

mass of proton or neutron relative atomic mass (e.g. 40 for 40Ar)

h = 6.627!10"34

J.s

k= 1.38!10"23 J K

room temperature T= 300 K

#$%

kT= 4.14 !10"21J

lmax

=

8mkTLx

h!10

9M

Typical value of lmax???

!

1

exp !!lx

kT( )exp !!

lx+1

kT( )! exp !!lx

kT( )(l

x+1)! l

x

"

16

! " 1

exp "!lx

kT( )d

dlx

exp "!

lx

kT

#$%

&'( ="

d

dlx

lnexp "!

lx

kT

#$%

&'( =

= d

dlx

!lx

kT= d

dlx

lx

lmax

!"#

$%&

2

=2lx

lmax

2

the relative decrease of

upon increase of lxby 1

!"#!!!"

#$( ) = !"# ! !" !$%"( )2"

#$%

!lx+1

! !lx

( ) kT the energy difference between adjacent states /kT!


9/23

9

For all relevant translational energy levels

(energy levels for which not )

the relative decrease of

upon increase of lxby 1 is extremely small

17

It increases with lx, but even for lx= lmax

2lmax

! 110

9M

=

2lx

lmax

2

exp !!lx

kT( )

it is only

lx >>l

max! !

x >> kT

18

"

effectively constant

"

" relative decrease of becomes significant when

but then the value of

!" =#!1" ! $%& "

!!"

$%

#

$%%

&

'(( = $%& "

!"

!'(%

#

$%

&

'(

2#

$%%

&

'(() $%&0 =1

lx = O(lmax ) ! relative decrease of exp "

!lx

kT

#$%

&'(

is O(1 lmax )

exp !!

lx

kT

"#$

%&'

lx! l

max

2 exp !!

lx

kT

"#$

%&'(0

(previous slide)


10/23

10

19

We see that

the number of thermally accessible translational states is huge

energy levels very close (for adjacent levels: )

varies smoothly with lx

So we may treat lx, and as continuousvariables.

#

This implies we could have started just as well from

classical mechanicsrather than from quantum mechanics.

!"


11/23

11

! " #( )( ) !" #( )$##1

#2

" = ! "( )$"" #

1( )

" #2( )

"

21

substitution rule:

special case:

f x t( )( )Cdtt1

t2

! = f x( )dxx t

1( )

x t2( )

! " f x t( )( )dtt1

t2

! =1

Cf x( )dx

x t1( )

x t2( )

!

! "( ) = #"! !#$"%= #

Hence, with

!"# ! !

"

!$%"

"

#$

%

&'

2"

#$$

%

&''#!

"

1

(

) = !$%" !"#!"2#"1 !

$%"

(

)

!x t( ) = dx dt so !x t( )dt= dx( )

x = lx

lmax

(as on the previous slide)

C = 1

lmax

(If not in classical limit then the shape of the box matters,

because the values for the energy levels depend on the shape)

22

qtrans

=qxqyqz =

Lx

!

Ly

!

Lz

!=

V

!3

As said before,

for an ideal monoatomicgas at not-too-high T

this is essentially the complete molecular partition function.

q =qtrans

For the 3D box

In the classical limit the shape of the box does not matter,

only the volume.


12/23

12

23

Rough intuitive meaning of lmaxand"

lmax

=

8mkTLx

h

!number of accessible quantum states of a

particle in 1D box of sizeLxat temperature T

(as you probably expected)

qx = 1

2 !l

max =

Lx

!=

2!mkTLx

hidem

qtrans

=

V

!3

!the number of accessible quantum states of a

particle in 3D box of size Vat temperature T

! = h2!mkT

thermal wavelength~ the de Broglie wavelengthof a particle with kinetic energy kT

!length element in a 1D box that corresponds to

1 accessible quantum state

!3!volume element in a 3D box corresponding to 1 accessible

quantum state

!!"

24

The multiparticle partition function

(for a one-atomic ideal gas at not-too-high temperatures)

If there areNindependent, distinguishableparticles in the box:

microstate of the whole system defined by the states of all particles:

and the energy of a state of the total system is

sum of all single-particle energiesE

i = !

l1

+ !l2

+ !l3

+ . . . + !lN

iindicates the state of the total system

lmindicates the state of particle nr. m

Sandler2.4

i = l1,l

2,l

3,...l

N( )


13/23

13

25

Since Ei = !

l1

+ !l2

+ !l3

+ . . . + !lN

Q = exp !E

i

kT

"#$

%&' = ... exp !

!l1+ !l2

+ . . . + !lN

kT

"#$

%&'lN

(l2

(l1

(i

(

Q = q1q2q3...qN

For distinguishableindependent particles, the total partition

function can be written as aproduct of single-particle partition

functions (Divide and Rule)

(check this, similar procedure as with )qtrans =qxqyqz

26

What if the particles are indistinguishable

(all atoms of the same kind e.g. 40Ar)

then

# the single-particle states nm

# the single-particle energy levels

#

the single particle partition functions qmare all the same for each particle m

Can we then just write ?Q = q1q2q3...q

N = q

N

!nm

DistinguishabilityandIndistinguishability


14/23

14

i = I : l1= iv l2 = iii l3 = ii . . . . . . . . . . . lN= vi

i = II: l1= iii l2 = iv l3 = ii . . . . . . . . . . . lN= vi

27

examine e.g. the following two different states i= I and i= II

for the whole system containingNdistinguishableparticles:

if particle 1 and 2 are indistinguishablethen

I and II are notdifferent states of the total system

I and II do not deserve two terms in the partition function!

(2permutationsfor particle 1 and 2)

DistinguishabilityandIndistinguishability

28

How many permutation are there forNdistinguishable particles

that are each in a different state?formulated differently:

How many ways are there to distributeNdistinguishable

particles overNdifferent states?

! N"1( )! N" 2( )

DistributeNparticles overNstates:

N possible states for the first particle,

for the second,for the third,

and so on)

=N!


15/23

15

29

the number of accessible states >> number of particleslmax

>> N

Under what condition will particles, all be in a differentstate?formulated differently:

Under what condition is it very unlikely that some particles will

be in the same state?

The previous slide asked: How many permutation are there

forNdistinguishable particlesthat are each in a different state?

30

!!! "#$ !!

!%

+!!&

+ ! ! ! +!!"

#$

"

#$$

%

&''!

"

(!&

(!%

( = %%%&%(!!!%" = %"- so the sum

includes many terms that should not be included when particles areindistinguishable!

- each term in the correct partition function Q, occursN! times in

the above sum

- the above sum should be divided byN!

If the particles are indistinguishable

all thoseN!permutationscorrespond to

only one quantum stateof the system

Q =1

N!qN


16/23

16

31

! =1

"!#" This is called

Maxwell-Boltzmann statistics

lmax

>>N

typical distance between the particles >> thermal wavelength

1

!=V

N>> !

3

the nr. of accessible states >> nr. of particles

or, equivalently

division byN! correct if

volume per particle >> (thermal wavelength) cubedor, equivalently

Conclusion:

the canonical partition function of a one-component ideal gas is

Why is this Boltzmann statistics not exact?

also includes terms for which notallNparticles are in a differentstate.

e.g. l1= iii l2 = iii l3 = ii . . . . . . . . . . . lN= vi

32

!!! "#$ !!

!%

+!!&

+ ! ! ! + !!"

#$

"

#$$

%

&''!

"

(!&

(!%

( = %"

Hence, division byN! (the number of ways to distribute

Ndistinguishable particles overNdifferentstates)

not exact.

(as indicated before the number of such terms is relatively very small

when the nr. of accessible states >> nr. of particles)


17/23

17

For most casesBoltzmann statistics works fine

because for most cases

nr. of accessible

single-particle states

>> the nr. of particles

33

(check this)

lmax

>> N

the typical distance between the particles >> thermal wavelength

lmax

states

!

!=N

V


18/23

18

35

Simple illustration / example:system consisting of two independent distinguishableparticles N = 2

whole system (of 2 particles)

l1/l2 1 2 31 1:11 2:123:13

2 4:21 5:226:23

3 7:31 8:329:33

system states i

i = 1: l1= 1, l2= 1, E1= !1+ !1

i = 2: l1= 1, l2= 2, E1= !1+ !2etc.

Q = exp !E

i

kT

"

#$%

&'i( = xp !

!l1

+ !l2

kT

"

#$

%

&'

l2

(l1

(

=exp !!

1+ !

1

kT

"#$

%&'+ exp !

!1+ !

2

kT

"#$

%&'+ exp !

!1+ !

3

kT

"#$

%&'

+exp !!

2+ !

1

kT

"

#$%

&'+ exp !

!2+ !

2

kT

"

#$%

&'+ exp !

!2+ !

3

kT

"

#$%

&'

+exp !!

3+ !

1

kT

"

#$%

&'+ exp !

!3+ !

2

kT

"

#$%

&'+ exp !

!3+ !

3

kT

"

#$%

&'

=q1q2 =q

2

Maxwell-Boltzmann

Statistics

when particle 1, 2

indistinguishable

then divide byN! = 2!

i:l1l2

36

Else

Bose-Einsteinstatistics should be used forBosons

Fermi-Diracstatistics should be used forFermions

Fermions(particles consisting of odd nr. of spin-#particles)

are subject to thePauli exclusion principle:

not more than 1 particle can be in the same quantum state.

Bosons(spinless particles and particles consisting of even nr.

of spin-#particles)

more than 1 particle allowed to be in the same state

(This will not be discussed further in this course.)

Maxwell-Boltzmannstatistics applies when

typical distance between particles >> thermal wavelength !!1 3

>>"


19/23

19

37

Harvest time

We have now all the ingredients to find expressions for

the thermodynamic properties

of the ideal monoatomic gas at not-too-high T.

Sandler3.3, 3.4

38

As derived before:

for mono-atomic ideal gas at not-too-high T

q =qtrans =V

!3 =

2!mkT

h2"#$

%&'

3 2

Vthe single-particle

partition function

assuming Maxwell-Boltzmann statistics

!!""#"$#=$

"%

%"

=

$

"%

#"

!&"

=

$

"%

'!&'$

('

"

#$

%

&'

&" '

#"

generalise this for a two-components ideal-gas mixture

(taking electronic and nuclear partition functions to be 1)

!!"#""

$"%"=!!"

#"%"!!"

$"%" =

$

"#%"

$%'#

"#'

$

"$


20/23

20

39

TheHelmholtz energy

= !!"!" 1

##

$#

"3##

$%&

'(

A =!kTlnQ

=!kT ln1

N!+Nln

V

"3#$%

&'(

using the Stirling approximationfor largeN:

lnN!!NlnN "N+ ln 2#N !NlnN "N

A =kTN ln !!3e( ) =kTN ln !!3( )!1( )

!=

N

Vwith

check this (see next slide)

40

Check

!

"#= !"#!"

1

$#

%$

"3$#

$%&

'( =$!"

!"3

&

#

$%&

'( =$ !" !"3( )!1( )

A

kT=! ln

1

N!+Nln

V

"3

#

$%

&

'( = lnN!+Nln

"3

V

= NlnN!N+Nln"3

V= N lnN!1+ ln

"3

V

#

$%&

'(

= N ln !"3( )!1( ) = Nln !"3

e

#

$%&

'(

hint: use Stirling approximation !"!#!!!"! "!


21/23

21

41

Thepressure

from thermodynamics p =! "A

"V

#$%

&'(T,N

The only V-dependent term in is !kTNlnV

Hence p =! "A

"V

#$%

&'(T,N

= kTN "lnV

"V

#$%

&'(T,N

p = kTN V = kT!

p =RT n V=RTcTada

the ideal gas law

A = kTN ln !!3( ) "1( )

42

The ideal gas law holds just as well for poly-atomic molecules

because molecular vibrations and rotations are independent of V,

and hence do not introduce extra V-dependent contributions in the

partition function.

The ideal gas law also holds for a mixed ideal gas with

(in an ideal gas mixture pressures are simply additive)

N=N1

+N2

+N3

+ ...


22/23

22

43

The energy

from thermodynamics

(Gibbs-Helmholtz relation)

U=!A T

!1 T

"#$

%&'V,N

For the mono-atomic ideal gas:A

T= kNln !"

3e( )

! =h

2"mkT

The only T-dependent term is kNlnT!3 2

=3

2kNln1 T

Hence U= 32kN

!ln1 T

!1 T

"

#

$%

&

'V,N

U= 3

2kTN

The average kinetic energyper moleculeis ( per mol)

( per degree of freedom, per mol of degrees of freedom)

NB for polyatomic molecules extra terms are added

3

2kT

1

2kT

!

"!"

!

"!"

44

The heat capacity

The constant-volume heat capacity

for the monoatomic gas at not-too-high Tis

CV =

!U

!T

!"#

$%&V,N

3

2k

1

2k

3

2R

1

2R

per atom per mole of atoms

per degree of freedom per mole of degrees

of freedom

These are just the translational contributions.

For polyatomic molecules there are additional contributions

CV=

3

2NkhenceU= 3

2kTNwith


23/23

23

45

The chemical potential

=!A

!N

"#$

%&'T,V

=

!

!N

"#$

%&'T,V

kTNlnN(3

Ve

"

#$%

&'

= kTln !"3( )

check this

46

The entropy

S=! "A

"T

#$%

&'(N,V

or S= U! A( ) T

S= kNlne5 2

!3"

#

$%&

'(= kNln

e5 2

!3kT

p

#

$%&

'(

ideal gas law

Sackur-Tetrode equation (1912)

(http://nl.wikisage.org/wiki/Hugo_Martin_T

etrode)

what is the limiting behaviour for T$0 according to this equation?

why does this go wrong?

Check that for an ideal gas the entropy change upon a volume change is

The absolute valueof the entropy of a monoatomic ideal gas!

!S = S(N,V2,T)"S(N,V

1,T) = kN lnV

2"lnV

1( ) = kN lnV

2

V1

#

$%&

'(

http://www.aps.org/publications/apsnew

s/200908/physicshistory.cfm

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molecular principles