ChemE_2200_lecture_T1.ppt weryuiutewryyuuu

1. ChemE 2200 – Chemical Thermodynamics Lecture 1
Today:
Introduction to Statistical Thermodynamics
Defining Question:
What is a partition function?
Reading for Today’s Lecture:
McQuarrie & Simon, Chp 17.1-17.3, MathChapters H & J.
Reading for Thermodynamics Lecture 2:
McQuarrie & Simon, Chp 17.4, 17.6, 18.1-18.2.

2. Physical Chemistry II for Engineers
Syllabus:
Applied Quantum Chemistry
Classical Thermodynamics
Chemical Kinetics
Statistical Thermodynamics

3. PChem II for Engineers – Part 2.
Statistical Thermodynamics:
Classical Thermodynamics:
physical observables
classical
thermodynamics
abstract quantities
P, V, T, N, and chemical identity U – internal energy (1st Law)
Description of the Equilibrium State.
S – entropy (2nd Law)
Conversions of Energy: Heat – thermal equilibrium
Work – mechanical equilibrium
Chemical Reaction – 2H2 + CO  CH3OH
Physical Reaction – liquid  gas
Start with molecular energy levels from quantum chemistry.
Use statistics to predict populations of energy levels.
Given Nmolecules, Etotal, and V, predict P, T, CV, CP, and entropy.
Empirical; not based on microscopic (molecular) description.

4. Thermodynamics and Chemical Kinetics
in the Chemical & Biomolecular Engineering Curriculum
Mathematics & Science
Calculus
Physics
Chemisty
Biology
Engineering Sciences
Mass & Energy Balances
Fluid Mechanics
Thermodynamics
Chemical Kinetics
Chemical & Biomolecular Process Units
Heat & Mass Transfer
Separation Processes
Chemical Reactor Design
Unit Operations Laboratory
Chemical & Biomolecular Processes
Process Design
Process Optimization & Control

5. Understanding Thermodynamics
“Thermodynamics is a funny subject. The first time you go through it,
you don't understand it at all. The second time you go through it, you think
you understand it, except for one or two points. The third time you go through
it, you know you don't understand it, but by that time you are so used to the
subject, it doesn't bother you anymore.”
Arnold Sommerfeld (1868-1951)
Ordinarius Professor of Physics
Director, Theoretical Physics Institute
University of Munich

6. Understanding Thermodynamics
“The first time you go through it (thermodynamics),
you don't understand it at all.”
Why?
Copious mathematics (mostly The Calculus)
Myriad variables.
A functional command of thermodynamics.
Calculate heat flow and work done for a process.
Our Goals:
Abstract quantities: energy and entropy.
Predict the spontaneity of a process, such as a chemical reaction,
and its dependence on concentration, temperature, and pressure.
An understanding of thermodynamic phenomena.
A molecular basis for thermodynamic quantities.

7. Introduction to Statistical Thermodynamics
What is the Equilibrium State of a System (of Molecules)?
Nmolecules, V, T
empirical
postulates
Esystem, Ssystem
Macroscopic Description
can replace either with P
Microscopic Description
position (x, y, z) of every molecule
momentum (px, py, pz) of every molecule
Esystem, Ssystem
statistics

8. Introduction to Statistical Thermodynamics
Consider a system of N molecules, i = 1, 2, 3, … N
Example: An Ideal Gas in a cubic box, V = a3
no interaction between molecules.
indistinguishable molecules.
Energy states are given by Particles in a Box
Define Ni  number of molecules in state ei Ni is a State Population number



i
i
N
Ntotal
molecules
total
 e


i
i
i
N
Etotal
energy
total
The State of the System is the set of population numbers = {Ni} = (N1, N2, N3, …)
1
,
1
,
1
:
1 


e z
y
x n
n
n
1
,
1
,
2
:
2 


e z
y
x n
n
n
 
2
2
2
2
2
,
,
8
z
y
x
n
n
n n
n
n
ma
h
z
y
x



e

9. The State of the System is the set of population numbers = {Ni} = (N1, N2, N3, …)
Example 1: 10 molecules, e1 = 0, e2 = 1, e3 = 3
e = 0
1
3
State A
10
2
3
5
3
2
1
total






 N
N
N
N
State B
10
1
6
3
total 



N
9
3
2
1
3
0
5
3
3
2
2
1
1
total







e

e

e
 N
N
N
E 9
3
1
1
6
0
3
total 






E
State C
10
3
0
7
total 



N
9
3
3
1
0
0
7
total 






E
Which state is more probable?
For a set of indistinguishable molecules, Number of permutations  W




!
!
!
!
3
2
1
total
N
N
N
N
2520
!
2
!
3
!
5
!
10


W 840
!
1
!
6
!
3
!
10


W 120
!
3
!
0
!
7
!
10


W
State D: (N1, N2, N3) = (1,9,0) 10
!
0
!
9
!
1
!
10


W
most
probable

10. The State of the System is the set of population numbers = {Ni} = (N1, N2, N3, …)
Example 2: e1 = 0, e2 = 1, e3 = 3, Ntotal = 20,000, Etotal = 10,000
3 degrees of freedom (e1, e2, e3) with 2 constraints (Ntotal, Etotal)  1 equation
000
,
20
3
2
1
total 


 N
N
N
N
000
,
10
3
3
2
2
1
1
total 
e

e

e
 N
N
N
E
000
,
10
3
1
0 3
2
1 





 N
N
N
substitute
2
1
3 000
,
20 N
N
N 


000
,
10
3
)
000
,
20
( 2
1
2 



 N
N
N
1
2 3
000
,
50
2 N
N 



solve for N3
Set N1, calculate N2 and N3, and then calculate
!
!
!
!
3
2
1
total
N
N
N
N
W 
But calculating N! is intractable for N > 100. 10! = 3,628,800 100! = 9.3  10157
How to calculate 20,000! ? 1023! ?
1
2
2
3
000
,
25 N
N 


11. Stirling’s Approximation for N!
But calculating N! is intractable for N > 100. How to calculate 20,000! ? 1023! ?













 )
4
ln(
)
3
ln(
)
2
ln(
)
1
ln(
)
4
3
2
1
ln(
!
ln N


N
x
1
ln



N
x
x
1
ln
McQuarrie & Simon, Figure J-1, p. 813


N
x
N
1
ln
!
ln
N
x
x
x 1
)
ln
( 

)
1
1
ln
1
(
ln 



 N
N
N
0
neglect for
N » 1
N
N
N
e
N
N
N
N
N




!
ln
!
ln Stirling’s
Approximation
McQuarrie & Simon, p. 812

12. The State of the System is the set of population numbers = {Ni} = (N1, N2, N3, …)
Example 2: e1 = 0, e2 = 1, e3 = 3, Ntotal = 20,000, Etotal = 10,000
!
!
!
!
3
2
1
total
N
N
N
N
W 
2
1
3 000
,
20 N
N
N 


Set N1 
0
0.2
0.4
0.6
0.8
1
0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000
N1
max
W
W
N3 < 0 for N1 < 10,000
N2 < 0 for
N1 > 16,666
Maximum probability at N1 = 12,674, N2 = 5,989, N3 = 1,337
Full Width at
Half Maximum
(FWHM) = 100
1
2
2
3
000
,
25 N
N 


13. The relative width of the probability distribution decreases with Ntotal
Example 3: Consider a box of molecules, each of which can be on the left or right side.
Left Right
What are the
Left-Right
configurations?
Case 1: 4 molecules: 4 in Left side: LLLL
3 in Left side: LLLR LLRL LRLL RLLL
2 in Left side: LLRR LRLR LRRL RRLL RLRL RLLR
1 in Left side: LRRR RLRR RRLR RRRL
0 in Left side: RRRR
Number of Configurations = W(Ntotal, NLeft)
)!
(
!
!
Left
total
Left
total
N
N
N
N


1
4
6
4
1
W

14. Example 3: A Box of Molecules: Probabilities of Left-Right Distributions
width of distribution scales as
1/2
total
N
0 1 2 3 4 5 6 7 8 9 10
10 molecules
0 10 20 30 40 50 60 70 80 90 100
100 molecules
0 100 200 300 400 500 600 700 800 900 1000
1,000 molecules
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
10,000 molecules
NLeft NLeft
NLeft NLeft

15. Example 4: A mole of an Ideal Gas
Infinite energy levels. 6  1023 molecules.
Numerical Modeling (as with Examples 1, 2, and 3) is not practical.
How to find the maximum of the function that gives the number of configurations?





e
!
!
!
!
)
,
(
3
2
1
total
N
N
N
N
W
N
f i
i
with two constraints:










i
i
N
N
N
N
N 3
2
1
total
i
i
i
N
N
N
N
E e





e

e

e
 

3
3
2
2
1
1
total
Use Lagrange’s Method of Undetermined Multipliers (Math 1920).
f = function to be (conditionally) maximized
g1 = constraint 1
g2 = constraint 2
Define K = f + ag1 + bg2
The constrained maximum of f is at 2
1
0 g
g
f
K 
b


a





0
)
,
(
,
0
)
,
(
such that 2
1 
 y
x
g
y
x
g

16. Lagrange’s Method of Undetermined Multipliers – one constraint
f(x, y) = y2 + 4y  x2y constraint: g(x, y) = 0 = 9  x2  y2
The gradient of f(x, y) is parallel to
the gradient of constraint g(x, y)
at the constrained maximum of f(x, y).
The constrained maximum is
f(±2.49, 1.67)  6.4
source: Norm Prokup, www.geogebra

17. Lagrange’s Method of Undetermined Multipliers – one constraint
f(x, y) = x2  y constraint: g(x, y) = 0 = 16  x2  y2
The gradient of f(x, y) is parallel to
the gradient of constraint g(x, y)
at the constrained maximum of f(x, y).
The constrained maximum is
f(3.97, 0.5)  16.25

18. Lagrange’s Method of Undetermined Multipliers for a mole of an Ideal Gas
The function to maximize is
2
1
0 g
g
f
K 
b


a





We seek the population distribution among particle-in-a-box energy states
with the maximum number of configurations, W.
 Ntotal
For convenience we will find the maximun of the function ln W.











e
!
!
!
!
ln
ln
)
,
(
3
2
1
total
N
N
N
N
W
N
f i
i






 !
ln
!
ln
!
ln
!
ln 3
2
1
total N
N
N
N









 




 1
1
total
total
total ln
ln
i
i
i
i
i N
N
N
N
N
N
i
i
i
i N
N
N
N
N
f ln
ln
)
,
( total
total 


e
Constraints: total molecules: 





e
1
total
1 0
)
,
(
i
i
i
i N
N
N
g
total energy: 


e



e
1
total
2 0
)
,
(
i
i
i
i
i N
E
N
g
K  f + ag2 + bg2 





e

b








a


 






1
total
1
total
total
total ln
ln
i
i
i
i
i
i
i N
E
N
N
N
N
N
N
Next step: find a and b such that




1
total !
ln
!
ln
i
i
N
N
apply Stirling’s
approximation

19. Lagrange’s Method of Undetermined Multipliers for a mole of an Ideal Gas
K  f + ag2 + bg2 





e

b








a


 






1
total
1
total
total
total ln
ln
i
i
i
i
i
i
i N
E
N
N
N
N
N
N
2
1
0 g
g
f
K 
b


a





Next step: find a and b such that
The function K must be at a maximum with respect to all variables N1, N2, N3, …






e





















b




























a














































i
i
i
i
i
i
i
i
i
i
i
i
i
i
N
i
N
N
N
E
N
N
N
N
N
N
N
N
N
N
N
N
N
N
K
i
j
total
total
total
total ln
ln
ln
0
0
0 0
i
i
N
i
N
N
K
i
j
be

a
















1
ln
0
i
i
N be

a


 )
1
(
ln
i
e
e
Ni
be

a


 )
1
(




be

a






1
)
1
(
1
total
i
i
i
i
e
e
N
N 


be

a



1
)
1
(
i
i
e
e
i
e
N
e be

a



 total
)
1
(
i
i
e
e
N
Ni be

be


 total
1
1 1
i
N
1


20. Statistical Thermodynamics for a mole of an Ideal Gas
i
i
e
e
N
Ni be

be


 total
For a fixed number of molecules and a fixed total energy
the population of energy level ei is
The population distribution decreases exponentially at a rate set by b.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ni Ni
energy level ei energy level ei
large b small b
5
.
0

be

 i
e 1
.
3

be

 i
e
Later we will derive
kT
1

b
low T high T
k is Boltzmann’s constant.
The exponential population decrease is a Boltzmann distribution.

21. Statistical Thermodynamics for a mole of an Ideal Gas
i
i
e
e
N
Ni be

be


 total
For a fixed number of molecules and a fixed total energy
the population of energy level ei is
The sum i
e be

 is a key descriptor of the equilibrium state of a system of molecules.
kT
i
i
e
e
Q /
Define e

be



  Q is called the Partition Function.
Q reflects how the molecules are partitioned into the molecular energy levels ei.
Q is determined by the molecular energy levels ei and the temperature T.
Aside: Some textbooks use Z for the partition function for the German Zustrandssomme.
state sum

22. Statistical Thermodynamics for a mole of an Ideal Gas
i
i
e
e
N
Ni be

be


 total
For a fixed number of molecules and a fixed total energy
the population of energy level ei is
The sum i
e be

 is a key descriptor of the equilibrium state of a system of molecules.
kT
i
i
e
e
Q /
Define e

be



  Q is called the Partition Function.
1 1.4 3
Q = 1.05 1.58 1.99 3.86
1
lim
0


Q
T




Q
T
lim
Figure from Atkins’ Physical Chemistry, p. 566.
spacing
between
levels
 elevel
3
.
0
level

e
kT