chapter 19 statistical thermodynamics: the concepts statistical thermodynamics kinetics dynamics t,...
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Chapter 19 Statistical thermodynamics: the concepts
Stat
isti
cal T
herm
odyn
amic
sK
inet
ics
Dyn
amic
s
T, P, S, H, U, G, A...
{ri},{pi},{Mi},{Ei}…
How to translate mic into
mac?
The job description
Brute force approach does not work!
2
2
2
2
2
2
2
2
2
2
iii
i
zyxi
iiiiiim EV
222222
22 2
2 EVm
333332
32 3
2 EVm
• 1 mol 100000000000000000000000 particles• ~100000000000000000000000 equations needed to be solved!!!
555552
52 5
2 EVm
111112
12 1
2 EVm
444442
42 4
2 EVm
………
Bad news: We cannot afford it!
Good news:We do not need that detailed description!
How?
Thanks to them
Josiah Willard Gibbs
James Clark Maxwell
Ludwig Boltzmann
Cheng-Ning Yang
Tsung-Dao Lee
Van Hove
Landau
Arrhennius Enrico Fermi
Bose Paul Dirac
Langevin
Einstein
Ising
Mott Anderson Bardeen …
THE magic word
Statistical
We, the observers, are macroscopic. We only need averageof microscopic information.
Spatial and temporal average
T, P, S, H, U, G, A...
{ri},{pi},{Mi},{Ei}...
Still lots of challenges (opportunities) herein!
Contents
The distribution of molecular states
19.1 Configuration and weights
19.2 The molecular partition function
The internal energy and the entropy
19.3 The internal energy
19.4 The statistical entropy
The canonical partition function
19.5 The canonical ensemble
19.6 The thermodynamic information in partition function
19.7 Independent molecules
Assignment for Chapter 19
Exercises:
• 19.1(a), 19.2(b), 19.4(a)
Problems:
• 19.6(a), 19.9(b), 19.11(b), 19.15(a)
• 19.3, 19.7, 19.14, 19.18, 19.22
The distribution of molecular states
EEE E
These particles might be distinguishable
Distribution = Population pattern
•
..............
Enormous possibilities!
E E
E
EE
E
EE
E
E
E
EE
E
..............
Distinguishable particles
E EE EE
Principle of equal a priori probabilities
• All possibilities for the distribution of energy are equally probable.
An assumption and a good assumption.
They are equally probable
E E
E
EE
E
EE
E
E
E
EE
E
..............
E
EE
E
E
They are equally probable
Configuration and weights
{5,0,0,...}
The numbers of particles in the states
{3,2,0,...}
{3,2,0,...}
One configuration may have large number of instantaneous configurations
{N-2,2,0,...}
How many instantaneous configurations?
N(N-1)/2
E
18!/3!/4!/5!/6!
{3,4,5,6}
!...!!!
210 nnnNW
Nnnn
nnn
...
,...},,{
210
210
Configuration and weights
W is huge!
20 particles: {1,0,3,5,10,1} W=931000000
How about 10000 particles with {2000,3000,4000,1000}?
i
iii
iii nnNNnnnNNNW lnln)ln()ln(ln
xxxx ln!ln
ii
nnnN
nN
nnnN
nnnNW
!ln!ln
...)!ln!ln!(ln!ln
!...)!!ln(!lnlnln
210
210!...!!!
210
Stirling’s approximation:
xx exx 21
2!
Wmax
{ni}max {ni}
There is an overwhelming configuration
W
Equilibrium configuration
The dominating configuration is what we actually observe. All other configurations are regarded as fluctuation.
Eni
ii
Nni
i
The dominating configuration is the configuration with largest weight.
Constant total number of molecules:
Constant total energy :
ji
jij
i
i nn
nN
if 0
if 1,1
Find the distribution with largest lnWi
i i
dnn
WWd
lnln
iiidn 0
iidn 0
ii
ii
i iiiii
i i
dnn
W
dndndnn
W
ln
ln
0ln
i
in
W
j i
jj
ii n
nn
n
NN
n
W lnlnln 1lnln
ln
N
n
NN
n
N
n
NN
iii
Nni
i 1ln1lnln
lnln
i
jj
i
j
j i
jjj
i
j
j i
jj nnn
n
n
nnn
n
n
n
nn
N
nNn
n
W ii
i
ln1ln1lnln
0ln i
i
N
n ieN
ni
Lagrange’s method of undetermined multipliers
z=f(x,y) with g(x,y)=c1, h(x,y)=c2
To find the maximum of z with constraints g and h, we May use
0** dhbdgadz
Boltzmann distribution
i
i
i
i
e
e
N
n
kT
1
Boltzmann constant
ieN
ni jj
jjeNenN
j
iee
1
The molecular partition function(nondegenerate case)
q
ep
i
i
Nn
i
j
ieq
E
1
54
3
2
G
11 g
15 g
14 g13 g
12 g
The molecular partition function(degenerate case)
j levels
jiegq
E
,...,, 3,52,51,5
,...,, 3,32,31,3
,...,, 3,42,41,4
,...,, 3,22,21,2
,...,, 3,12,11,1
G
1g
5g
4g
3g
2g
Angular momentum
...3,2,1,0 2
)1( 2
22
lI
llI
E
...3,2,1,0 2
)1( 2
22
lI
llI
JE
...3,2,1,0 ,)1( || lllJ
llllmmJ llz ,1,...1, ,
2zJ
1zJ
00 zJ
zJ2zJ
6)12(2 || J
The Rotational Energy Levels (Ch 16)
2aaa I
2
1E Around a fixed-axis
222
2
1
2
1
2
1ccbbaa IIIE Around a fixed-point
aaa IJ c
c
b
b
a
aI
J
I
J
I
JE
222
222
I2
J
I2
JJJE
22c
2b
2a
(Spherical Rotors)
22 1JJJ ,...2,1,0J
2
j I21JJE
I2hcB
2
cI4B
1JhcBJE j
BJJFJF 21
Example: Linear Molecules (rigid rotor)
0
)1(
j levels
)12(j
JhcBJj eJegq i
E
9,53,52,51,5 ,...,,
5,33,32,31,3 ,..,, 7,43,42,41,4 ,...,,
3,22,21,2 ,,
1
G
11 g
95 g74 g53 g32 g
E1
E2
E3
E4
E5
Exercises
E
,...,, 3,52,51,5
,...,, 3,32,31,3
,...,, 3,42,41,4
,...,, 3,22,21,2
,...,, 3,12,11,1
G
11 g
25 g34 g13 g22 g
E1
E2
E3
E4
E5
q=?
E=0,g=1
E=ε,g=2q=?
eq 21
The physical interpretation of the molecular partition function
j levels
jiegq
00
gqlimT
qlimT
q is an indication of the average number of states that are thermally accessible to a molecule at the temperatureof the system.
E, TE, T=0 E, ∞
eq
1
1
ieepi 1
Example: Uniform ladder of energy levels
(e.g., harmonic vibrator)...1 32
0
eeeegqj
jj
xxxx 1132 ...1
ieepi 1
eq
1
1
eq 1
E=0,g=1
E=ε,g=1
1,,0
1
qT
eq
E=0,g=1
E=ε,g=1
2,0,
1
qT
eq
E=0,g=1
E=ε,g=1
ep
1
10
e
ep
11
E=0,g=1
E=ε,g=1
E=0,g=1
E=ε,g=1
11
1 00
T
ep
01
01
T
e
ep
E=0,g=1
E=ε,g=1
2/11
10
T
ep
2/111
T
e
ep
E=0,g=1
E=ε,g=1
E=0,g=1
E=ε,g=1
Approximations and factorizations
• Exact, analytical partition functions are rare.
• Various kinds of approximations are employed:
dense energy levels
independent states (factorization of q)
…
Dense energy levels
Xh
mqX
21
2
2
2
22
8mX
hnEn One dimensional box: ,...,n 21
12 nEn 2
2
8mX
h
1
12
n
nX eq dneq n
X
1
12
Xh
mdxeq x
X
21
2
21
21
0
21
2
2
11 2
Independent states (factorization of q)
Zn
Yn
Xnnnn 321321
ZYX qqq
q
Zn
Yn
Xn
Zn
Yn
Xn
Zn
Yn
Xn
3
-
2
-
1
-
--
n all
-
n all
---
321
321321
eee
eeee
XYZh
mq
23
2
2
3V
q
21
21
22 mkT
h
mh
Three-dimensional box:
Thermal wavelength
(Translational partition function)
Why q, the molecular partition function, so important?
• It contains all information needed to calculate the thermodynamic properties of a system of independent particles (e.g., U, S, H, G, A, p, Cp, Cv …)
• It is a kind of “thermal wavefunction”. (Remember the wavefunction in quantum mechanics which contains all information about a system we can possibly acquire)
Find the internal energy U from q
i
iinE
i
iie
q
NE
ii ed
dei
d
dq
q
Ne
d
d
q
Ne
d
d
q
NE
ii
ii
EUU )0(
V
q
q
NUU
)0(
V
qNUU
ln
)0(
Total energy of the system:
q
ep
i
i
Nn
i
At T=0, U=U(0)
A two-level system
e
N
e
eNe
d
d
e
NE
111
1
E=0,g=1
E=ε,g=1 eq 1
U=?
eN
eeN
q
Vq
V
UU
eNU
q
qNU
qNUU
11
1
)0()0(
)()0(
))(ln
()0(
)ln
()0(
2
2
)1(
1
)()(
)0(
e
eNkTv
UvT
Uv
eN
C
UU
A two-level system eq 1
)1ln(ln
]lnlnln[ln
]lnlnln[lnln
lnlnln
lnln
lnlnln
11
1
1
1
ln
eqe
qeeqNqN
qeeNeqNN
N
N
NNW
eq
q
q
qNe
qe
qN
q
qNe
iq
eNW
iq
Neq
Ne
ii
ii
E=0,g=1
E=ε,g=1
W=?
i
ii nnNNW lnlnln
Exercise
A two-level system
e
NE
1
E=0,g=1
E=ε,g=1
E=0,g=1
E=ε,g=1
01
1/ 0/
kT
eNE
E=0,g=1
E=ε,g=1
5.01
1/ 2
1/
kT
eNE
The value of βV
qNUU
ln
)0(
nRTUU2
3)0(
2
3)0(
NUU kTkTnN
nN
nRT
N
A
A 1
For monatomic perfect gas,
d
dV
d
dV
Vq
VV433
31
222
1
2 21
21
21
21
m
h
m
h
d
d
d
d32
3
Vq
V
2
3)0(
2
3)0( 3
3 NU
V
VNUU
3V
q Translational partition function:2
1
2
mh
The statistical entropy
0ln
i
in
W
i
iinUU )0( ii
ii
ii dndndUdU )0(
ii
ii dndUd 0
TdSdqdU rev When no work is done,
ii
idnkT
dUdS
For the most probable configuration:
ii n
Wln
i i
iii
dnkdnn
WkdS ln
i
ii
Wdkdnn
WkdS ln
ln
WkS ln
dN=0
Heat does not change energy levels
Work changes energy levels
Boltzmann formula
WkS lnT0, W=1, S0 (Third law of thermodynamics )
)]1ln([ln
)1ln(
1
1ln
ekNWkS
e
e
eNW
E=0,g=1
E=ε,g=1
The two-level system
)]1ln([1
eNkSe
Negative temperature!
q
ep
i
i
Nn
i
E=0,g=1
E=ε,g=1
eNN
E=0,g=1
E=ε,g=1
T>0, N+<N-
T<0, N+>N-Examples: laser, maser, NMR etc.
E=0,g=1
E=ε,g=1
eq 1
eNUU
1
)0(
)(0/
)(0/ 1
kT
kT
q
q
2)0(/ kTq
E=0,g=1
E=ε,g=1
eq 1
0/ 0/ kTNkS
2ln/ / kTkNS
2
2
)1(
e
eNkvC
0)(0/ kTvC
0)0(/ kTvC
)]1ln([1
eNkSe
E=0,g=1
E=ε,g=1
eq 1
eNUU
1
)0(
)(0/
)(0/ 1
kT
kT
q
q
E=0,g=1
E=ε,g=1
2ln/ 0 kNS
0/ NkS
2
2
)1(
e
eNkvC
0)0(/ kTvC
0)(0/ kTvC
)]1ln([1
eNkSe
E=0,g=1
E=ε,g=1
VT U
S)(1
eNUU
1
)0(
Positive temperature
Negative temperature
)]1ln([1
eNkSe
U and S
qNk
T
UUS ln
0
ii i
ii
ii
iii ppNkN
nnknnNnkS lnlnlnln
qp ii lnln
qNkUUk
qNknk
qpNkNpk
qppNkS
iii
ii
iii
i iiii
ln0
ln
ln
ln
i
ii nnNNW lnlnln WkS ln Nni
i
=1
q
ep
i
i
eq
1
1
Example: Simple harmonic oscillator
V
q
q
NUU
)0(
eN
eeN
Vee
11
])1
1[)(1(
Example: Simple harmonic oscillator
qNk
T
UUS ln
0
)]1ln([
1
1ln/)(
1
1
eNk
eNkT
e
eN
E=0,g=1
E=ε,g=1Exercise: The two-level system
2ln/ )0(/ kTkNS0/ )(0/ kTNkS
)]1ln([1
eNkSe
U=?, S=?
eN
V
ee
N
q
q
NUU
1
)(1
)0(
)]1ln([
)1ln(/)(
ln0
1
1
eNk
eNkT
qNkT
UUS
e
eN
0/ )(0/ kTNkS
2ln/ )0(/ kTkNS
)]1ln([
)1ln(/)(
ln0
1
1
eNk
eNkT
qNkT
UUS
e
eN
The canonical partition function QIndependent system vs interacting system
Molecular partition function q
Ensemble: an imaginary collection of replicationof the actual system with a common macroscopicparameter.
Canonical ensemble: an imaginary collection of replications of the actual system with a common temperature. (N, V,T)
Microcanonical ensemble: an imaginary collection of replications of the actual system with a common energy. (N, V,E)Grand canonical ensemble: an imaginary collection of replicationsof the actual system with a common chemical potential. (μ, V,T)
(close system)
(isolated system)
(open system)
!...~!~!
~~
10 nn
NW
ionsconfigurat
ousinstantane possible ofnumber
the,...},~,~{on distributi
with system, theof copies~
10 nn
N
,...}~,~{ 10 nn
in~ The number of states with energybetween Ei and Ei+1
Q
e
N
n iEi
~~
i
EieQ
The canonical partition function Q
The most probable configuration:
Find internal energy U from Q
NE~
/~
NEUEUU~
/~
00 N~
Q
ep
iE
i
~ iE
ii
iii eE
QUEpUU
10~0
VV
QU
Q
QUU
ln
01
0
The total energy of the ensemble: E~
The average energy:
The fraction of members of the ensemble in a state with energy Ei:
(classroom exercise)
WN
kWkWkS N ~
ln~~
lnln~1
Qk
T
UUS ln
0
Find entropy S from Q
NWW~~ The total weight :
(5 points bonus!!!)
i
NiiieQ 21
NE iiii 21
NqQ
Independent molecules
N
i
N
ii
qeeeQ iii ...21
(classroom exercise)
(a)For distinguishable independent molecules:
NqQ
(b) For indistinguishable independent molecules:
!NqQ
N
E
E
Monatomic gas(Sackur-Tetrode equation)
3V
q
3
25
lnAnN
VenRS
21
2 mkT
h
3
25
lnp
kTenRS
!lnln
0NkqNk
T
UUS
nRNnRqnR
T
UUS
lnln
0
)ln(
lnlnlnln1lnln2
3
3
25
32
3
3
A
AA
nN
VenR
enNV
enRnNV
nRnRS
AnNN
xxxx ln!ln
Using the Sackur-Tetrode equation
• Calculate the standard molar entropy of gaseous argon at 25C
30
25
3
025
3
025
0
ln
lnln0
p
kTenR
N
VeR
nN
VeRS
A
m
An
Sm
11
)106.1(101012.40
molJK 1666.18
}ln{ 31125
212/5
R
RSmNm
Jem
Sackur-Tetrode equation
i
f
V
V
if
nR
aVnRaVnRS
ln
lnln
2
22
8mX
hnEn
As the container expands, X increeasesmore states accessiblefor the systemS increases.
aVnRnN
VenRS
A
lnln3
25