molecular principles
TRANSCRIPT
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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
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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)
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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
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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
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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
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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
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(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
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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!
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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
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"
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)
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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.
!"
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! " #( )( ) !" #( )$##1
#2
" = ! "( )$"" #
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.
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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( )
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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
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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!
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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
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! =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)
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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
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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
>>"
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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)
!!"#""
$"%"=!!"
#"%"!!"
$"%" =
$
"#%"
$%'#
"#'
$
"$
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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
#
$%&
'(
hint: use Stirling approximation !"!#!!!"! "!
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Thepressure
from thermodynamics p =! "A
"V
#$%
&'(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
+ ...
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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
!"#
$%&V,N
3
2k
1
2k
3
2R
1
2R
per atom per mole of atoms
per degree of freedom per mole of degrees
of freedom
These are just the translational contributions.
For polyatomic molecules there are additional contributions
CV=
3
2NkhenceU= 3
2kTNwith
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45
The chemical potential
=!A
!N
"#$
%&'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