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Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA
Assoc. Prof. Dr. Mazlan Abdul WahidFaculty of Mechanical EngineeringUniversiti Teknologi Malaysiawww.fkm.utm.my/~mazlan
Statistical Thermodynamics
ADVANCED THERMODYNAMICS
MMJ 2413
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA
Statistical thermodynamics, or statistical mechanics, is the study of the microscopic behaviors of thermodynamic systems using statistical methods and probability theory .
The essential problem in statistical thermodynamics is to determine the distribution of a given amount of energy Eover N particles in a system. The macroscopic properties, such as thermodynamic energy, heat capacity, etc., can be calculated in terms of partition functions.
Statistical thermodynamics is a bridge of connecting between macroscopic and microscopic properties of a system.
Definition of statistical thermodynamics
Statistical Thermodynamics
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There are two kinds of systems:
•Interacting system
•Non-interacting system
(for instance, ideal gas)
Only the latter will be introduced in this discussion.
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Two kinds of particles:
• Identical particles, or indistinguishable particles (such as gaseous molecules), is also called non-localized particles.
• Distinguishable particles (Such as the atoms in crystal), is also called localized particles.
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Ways of arranging objects:
Problem 1Problem 2Problem 3Problem 4Problem 5
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Combinatorial Analysis
Consider a system of N particles that are allowed to occupy r number of states.
Microstate --- The description of the system that provides the state of each particle.
Number of possible microstates = Nr
Just like Problem 5
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∏=
r
1ii !n
N!
Macrostate--- The description of how many particles,
ni, are in each of the r states.
Number of microstates in a macrostate:
!!...!...!!
!
321 ri
j nnnnn
N=Ω
Just like Problem 2=Ω j
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N=3, r=3E1=1 E2=2 E3=3 Esum Omega %
I 3 0 0 3 1 3.7%II 2 1 0 4 3 11.1%III 2 0 1 5 3 11.1%IV 1 2 0 5 3 11.1%V 0 3 0 6 1 3.7%VI 1 1 1 6 6 22.2%VII 1 0 2 7 3 11.1%VIII 0 2 1 7 3 11.1%IX 0 1 2 8 3 11.1%X 0 0 3 9 1 3.7%
27 100%
N=3, r=3E1 E2 E3
I ABC - - 11 AB C -
AC B - BC A -
III AB - CAC - BBC - A
IV A BC - B AC - C AB -
V - ABC - VI A B C
A C BB A CB C AC A BC B A
VII A - BCB - ACC - AB
VIII - BC A- AC B- AB C
IX - A BC- B AC- C AB
X - - ABC
N=3 and r=3
0
12
3
4
5
67
8
2 4 6 8 10
Sum
Om
eg
a
Combinatorial Analysis
Microstates
Macrostates
DistributionN=3, r=327====Nr
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AssumptionsConsider all particles to be identical.
The net value of a macroscopic property depends onthe number of particles (ni) in each state (i).Exchanging the specific identity of the particles in astate does not change the value of the property.
On average the fraction of time each particle spends inany energy state is the same.
Probability of a macrostate is equal to the
fraction of time the system of particles
spends in that macrostate
Hypothesis
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/2001
Probability of MacrostatesHypothesis
Fraction of time in a macrostate = probability of that macrostate.
smicrostateoftotal
jmacrostateinsmicrostateofPj #
#=
r1
!n
N!
r P Nr
1ii
Nj
j
====ΩΩΩΩ
====∏∏∏∏
====
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Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/2001
Probability of Macrostates
r1
!n
N! P Nr
1ii
j
====∏∏∏∏
====
Sharp distribution --- Most probable state and/or those near
it are observed most of the time.
N=10, r=3
0%
2%
4%
6%
8%
10%
12%
14%
16%
9 11 13 15 17 19 21 23 25 27 29 31
Sum
P
N=10, r=2
0%
5%
10%
15%
20%
25%
30%
9 11 13 15 17 19 21
Sum
PN=3, r=3
0%
5%
10%
15%
20%
25%
30%
2 4 6 8 10
Sum
P
N=6, r=3
0%
5%
10%
15%
20%
25%
5 7 9 11 13 15 17 19
Sum
P
N=4, r=3
0%
5%
10%
15%
20%
25%
3 5 7 9 11 13
Sum
P
N=10, r=4
0%
2%
4%
6%
8%
10%
12%
9 13 17 21 25 29 33 37 41
Sum
P
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N=3, r=3
0%
5%
10%
15%
20%
25%
30%
0% 20% 40% 60% 80% 100%
Sum
P
N=10, r=3
0%
2%
4%
6%
8%
10%
12%
14%
16%
0% 20% 40% 60% 80% 100%
Sum
P
N=6, r=3
0%
5%
10%
15%
20%
25%
0% 20% 40% 60% 80% 100%
Sum
P
N=4, r=3
0%
5%
10%
15%
20%
25%
0% 20% 40% 60% 80% 100%
Sum
P
Plot probability as function of fractional range.
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N-3, r=3E1 E2 E3 Esum Omega %
I 3 0 0 3 1 3.7%II 2 1 0 4 3 11.1%III 2 0 1 5 3 11.1%IV 1 2 0 5 3 11.1%V 0 3 0 6 1 3.7%VI 1 1 1 6 6 22.2%VII 1 0 2 7 3 11.1%VIII 0 2 1 7 3 11.1%IX 0 1 2 8 3 11.1%X 0 0 3 9 1 3.7%
N=4, r=3E1 E2 E3 Esum Omega %
I 4 0 0 4 1 1.2%II 3 1 0 5 4 4.9%III 3 0 1 6 4 4.9%IV 2 2 0 6 6 7.4%V 1 3 0 7 4 4.9%VI 2 1 1 7 12 14.8%VII 0 4 0 8 1 1.2%VIII 2 0 2 8 6 7.4%IX 1 2 1 8 12 14.8%X 0 3 1 9 4 4.9%XI 1 1 2 9 12 14.8%XII 1 0 3 10 4 4.9%XIII 0 2 2 10 6 7.4%XIV 0 1 3 11 4 4.9%XV 0 0 4 12 1 1.2%
N=6, r=3E1 E2 E3 Esum Omega %
I 6 0 0 6 1 0.1%II 5 1 0 7 6 0.8%III 5 0 1 8 6 0.8%IV 4 2 0 8 15 2.1%V 3 3 0 9 20 2.7%VI 4 1 1 9 30 4.1%VII 4 0 2 10 15 2.1%VIII 2 4 0 10 15 2.1%IX 3 2 1 10 60 8.2%X 1 5 0 11 6 0.8%XI 3 1 2 11 60 8.2%XII 2 3 1 11 60 8.2%XIII 0 6 0 12 1 0.1%XIV 3 0 3 12 20 2.7%XV 1 4 1 12 30 4.1%XVI 2 2 2 12 90 12.3%XVII 0 5 1 13 6 0.8%XVIII 1 3 2 13 60 8.2%XIX 2 1 3 13 60 8.2%XX 0 4 2 14 15 2.1%XXI 2 0 4 14 15 2.1%XXII 1 2 3 14 60 8.2%XXIII 0 3 3 15 20 2.7%XXIV 1 1 4 15 30 4.1%XXV 1 0 5 16 6 0.8%XXVI 0 2 4 16 15 2.1%XXVII 0 1 5 17 6 0.8%XXVIII 0 0 6 18 1 0.1%
N=10, r=3E1 E2 E3 Esum Omega %
I 10 0 0 10 1 0.00%II 9 1 0 11 10 0.02%III 9 0 1 12 10 0.02%IV 8 2 0 12 45 0.08%V 8 1 1 13 90 0.15%VI 7 3 0 13 120 0.20%VII 8 0 2 14 45 0.08%VIII 7 2 1 14 360 0.61%IX 6 4 0 14 210 0.36%X 7 1 2 15 360 0.61%XI 6 3 1 15 840 1.42%XII 5 5 0 15 252 0.43%XIII 7 0 3 16 120 0.20%XIV 4 6 0 16 210 0.36%XV 6 2 2 16 1260 2.13%XVI 5 4 1 16 1260 2.13%XVII 3 7 0 17 120 0.20%XVIII 6 1 3 17 840 1.42%XIX 5 3 2 17 2520 4.27%XX 4 5 1 17 1260 2.13%XXI 2 8 0 18 45 0.08%XXII 6 0 4 18 210 0.36%XXIII 3 6 1 18 840 1.42%XXIV 5 2 3 18 2520 4.27%XXV 4 4 2 18 3150 5.33%XXVI 1 9 0 19 10 0.02%XXVII 2 7 1 19 360 0.61%XXVIII 3 5 2 19 2520 4.27%XXIX 5 1 4 19 1260 2.13%XXX 4 3 3 19 4200 7.11%XXXI 0 10 0 20 1 0.00%XXXII 1 8 1 20 90 0.15%XXXIII 2 6 2 20 1260 2.13%XXXIV 5 0 5 20 252 0.43%XXXV 3 4 3 20 4200 7.11%XXXVI 4 2 4 20 3150 5.33%XXXVII 0 9 1 21 10 0.02%XXXVIII 1 7 2 21 360 0.61%XXXIX 2 5 3 21 2520 4.27%
XL 4 1 5 21 1260 2.13%XLI 3 3 4 21 4200 7.11%XLII 0 8 2 22 45 0.08%XLIII 4 0 6 22 210 0.36%XLIV 1 6 3 22 840 1.42%XLV 3 2 5 22 2520 4.27%XLVI 2 4 4 22 3150 5.33%XLVII 0 7 3 23 120 0.20%XLVIII 3 1 6 23 840 1.42%XLIX 2 3 5 23 2520 4.27%
L 1 5 4 23 1260 2.13%LI 3 0 7 24 120 0.20%LII 0 6 4 24 210 0.36%LIII 2 2 6 24 1260 2.13%LIV 1 4 5 24 1260 2.13%LV 2 1 7 25 360 0.61%LVI 1 3 6 25 840 1.42%LVII 0 5 5 25 252 0.43%LVIII 2 0 8 26 45 0.08%LIX 1 2 7 26 360 0.61%LX 0 4 6 26 210 0.36%LXI 1 1 8 27 90 0.15%LXII 0 3 7 27 120 0.20%LXIII 1 0 9 28 10 0.02%LXIV 0 2 8 28 45 0.08%LXV 0 1 9 29 10 0.02%LXVI 0 0 10 30 1 0.00%
27====Nr
81====Nr
729====Nr
049,59====Nr
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N=10, r=4
0%
2%
4%
6%
8%
10%
12%
0% 20% 40% 60% 80% 100%
Sum
P
N=10, r=3
0%
2%
4%
6%
8%
10%
12%
14%
16%
0% 20% 40% 60% 80% 100%
Sum
P
N=10, r=2
0%
5%
10%
15%
20%
25%
30%
0% 20% 40% 60% 80% 100%
Sum
P
Plot probability as
function of
fractional range.
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/2001
N=10, r=2I E1 E2 Esum Omega %II 10 0 10 1 0.10%III 9 1 11 10 0.98%IV 8 2 12 45 4.39%V 7 3 13 120 11.72%VI 6 4 14 210 20.51%VII 5 5 15 252 24.61%VIII 4 6 16 210 20.51%IX 3 7 17 120 11.72%X 2 8 18 45 4.39%XI 1 9 19 10 0.98%XII 0 10 20 1 0.10%
024,1====Nr
N=10, r=4E1 E2 E3 E4 Esum Omega %
I 10 0 0 0 10 1 0.000%II 9 1 0 0 11 10 0.001%III 9 0 1 0 12 10 0.001%IV 8 2 0 0 12 45 0.004%V 9 0 0 1 13 10 0.001%VI 8 1 1 0 13 90 0.009%VII 7 3 0 0 13 120 0.011%VIII 8 0 2 0 14 45 0.004%IX 8 1 0 1 14 90 0.009%X 7 2 1 0 14 360 0.034%XI 6 4 0 0 14 210 0.020%XII 8 0 1 1 15 90 0.009%XIII 7 2 0 1 15 360 0.034%XIV 7 1 2 0 15 360 0.034%XV 6 3 1 0 15 840 0.080%XVI 5 5 0 0 15 252 0.024%XVII 8 0 0 2 16 45 0.004%XVIII 7 0 3 0 16 120 0.011%XIX 7 1 1 1 16 720 0.069%XX 4 6 0 0 16 210 0.020%XXI 6 3 0 1 16 840 0.080%XXII 6 2 2 0 16 1,260 0.120%XXIII 5 4 1 0 16 1,260 0.120%XXIV 3 7 0 0 17 120 0.011%XXV 7 0 2 1 17 360 0.034%XXVI 7 1 0 2 17 360 0.034%XXVII 6 1 3 0 17 840 0.080%XXVIII 6 2 1 1 17 2,520 0.240%XXIX 5 4 0 1 17 1,260 0.120%XXX 4 5 1 0 17 1,260 0.120%XXXI 5 3 2 0 17 2,520 0.240%XXXII 2 8 0 0 18 45 0.004%XXXIII 7 0 1 2 18 360 0.034%XXXIV 6 0 4 0 18 210 0.020%XXXV 3 6 1 0 18 840 0.080%XXXVI 6 2 0 2 18 1,260 0.120%XXXVII 6 1 2 1 18 2,520 0.240%XXXVIII 4 5 0 1 18 1,260 0.120%XXXIX 5 2 3 0 18 2,520 0.240%XL 5 3 1 1 18 5,040 0.481%
XLI 4 4 2 0 18 3,150 0.300%XLII 1 9 0 0 19 10 0.001%XLIII 7 0 0 3 19 120 0.011%XLIV 2 7 1 0 19 360 0.034%XLV 6 0 3 1 19 840 0.080%XLVI 3 6 0 1 19 840 0.080%XLVII 6 1 1 2 19 2,520 0.240%XLVIII 5 1 4 0 19 1,260 0.120%XLIX 5 3 0 2 19 2,520 0.240%L 3 5 2 0 19 2,520 0.240%LI 5 2 2 1 19 7,560 0.721%LII 4 4 1 1 19 6,300 0.601%LIII 4 3 3 0 19 4,200 0.401%LIV 0 10 0 0 20 1 0.000%LV 1 8 1 0 20 90 0.009%LVI 2 7 0 1 20 360 0.034%LVII 6 1 0 3 20 840 0.080%LVIII 6 0 2 2 20 1,260 0.120%LIX 2 6 2 0 20 1,260 0.120%LX 5 0 5 0 20 252 0.024%LXI 5 1 3 1 20 5,040 0.481%LXII 3 5 1 1 20 5,040 0.481%LXIII 5 2 1 2 20 7,560 0.721%LXIV 4 4 0 2 20 3,150 0.300%LXV 4 2 4 0 20 3,150 0.300%LXVI 3 4 3 0 20 4,200 0.401%LXVII 4 3 2 1 20 12,600 1.202%LXVIII 0 9 1 0 21 10 0.001%LXIX 1 8 0 1 21 90 0.009%LXX 1 7 2 0 21 360 0.034%LXXI 6 0 1 3 21 840 0.080%LXXII 2 6 1 1 21 2,520 0.240%LXXIII 5 0 4 1 21 1,260 0.120%LXXIV 4 1 5 0 21 1,260 0.120%LXXV 5 2 0 3 21 2,520 0.240%LXXVI 3 5 0 2 21 2,520 0.240%LXXVII 2 5 3 0 21 2,520 0.240%LXXVIII 5 1 2 2 21 7,560 0.721%LXXIX 3 3 4 0 21 4,200 0.401%LXXX 4 3 1 2 21 12,600 1.202%LXXXI 4 2 3 1 21 12,600 1.202%LXXXII 3 4 2 1 21 12,600 1.202%LXXXIII 0 9 0 1 22 10 0.001%LXXXIV 0 8 2 0 22 45 0.004%LXXXV 1 7 1 1 22 720 0.069%LXXXVI 6 0 0 4 22 210 0.020%LXXXVII 4 0 6 0 22 210 0.020%LXXXVIII 1 6 3 0 22 840 0.080%LXXXIX 2 6 0 2 22 1,260 0.120%XC 5 0 3 2 22 2,520 0.240%XCI 3 2 5 0 22 2,520 0.240%XCII 5 1 1 3 22 5,040 0.481%XCIII 2 5 2 1 22 7,560 0.721%XCIV 2 4 4 0 22 3,150 0.300%XCV 4 1 4 1 22 6,300 0.601%XCVI 4 3 0 3 22 4,200 0.401%XCVII 3 4 1 2 22 12,600 1.202%XCVIII 4 2 2 2 22 18,900 1.802%XCIX 3 3 3 1 22 16,800 1.602%C 0 8 1 1 23 90 0.009%
C I 0 7 3 0 23 120 0.011%C II 1 7 0 2 23 360 0.034%C III 3 1 6 0 23 840 0.080%C IV 1 6 2 1 23 2,520 0.240%C V 5 1 0 4 23 1,260 0.120%C VI 1 5 4 0 23 1,260 0.120%C VII 4 0 5 1 23 1,260 0.120%C VIII 5 0 2 3 23 2,520 0.240%C IX 2 3 5 0 23 2,520 0.240%C X 2 5 1 2 23 7,560 0.721%C XI 3 4 0 3 23 4,200 0.401%C XII 4 2 1 3 23 12,600 1.202%C XIII 4 1 3 2 23 12,600 1.202%C XIV 2 4 3 1 23 12,600 1.202%C XV 3 2 4 1 23 12,600 1.202%C XVI 3 3 2 2 23 25,200 2.403%C XVII 0 8 0 2 24 45 0.004%C XVIII 3 0 7 0 24 120 0.011%C XIX 0 7 2 1 24 360 0.034%C XX 0 6 4 0 24 210 0.020%C XXI 2 2 6 0 24 1,260 0.120%C XXII 1 6 1 2 24 2,520 0.240%C XXIII 5 0 1 4 24 1,260 0.120%C XXIV 1 4 5 0 24 1,260 0.120%C XXV 2 5 0 3 24 2,520 0.240%C XXVI 1 5 3 1 24 5,040 0.481%C XXVII 3 1 5 1 24 5,040 0.481%C XXVIII 4 0 4 2 24 3,150 0.300%C XXIX 4 2 0 4 24 3,150 0.300%C XXX 4 1 2 3 24 12,600 1.202%C XXXI 2 3 4 1 24 12,600 1.202%C XXXII 2 4 2 2 24 18,900 1.802%C XXXIII 3 3 1 3 24 16,800 1.602%C XXXIV 3 2 3 2 24 25,200 2.403%C XXXV 0 7 1 2 25 360 0.034%C XXXVI 2 1 7 0 25 360 0.034%C XXXVII 0 6 3 1 25 840 0.080%C XXXVIII 1 6 0 3 25 840 0.080%C XXXIX 3 0 6 1 25 840 0.080%C XL 1 3 6 0 25 840 0.080%C XLI 5 0 0 5 25 252 0.024%C XLII 0 5 5 0 25 252 0.024%C XLIII 1 5 2 2 25 7,560 0.721%C XLIV 2 2 5 1 25 7,560 0.721%C XLV 4 1 1 4 25 6,300 0.601%C XLVI 1 4 4 1 25 6,300 0.601%C XLVII 4 0 3 3 25 4,200 0.401%C XLVIII 3 3 0 4 25 4,200 0.401%C XLIX 2 4 1 3 25 12,600 1.202%C L 3 1 4 2 25 12,600 1.202%
576,048,1====Nr
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C LI 3 2 2 3 25 25,200 2.403%C LII 2 3 3 2 25 25,200 2.403%C LIII 2 0 8 0 26 45 0.004%C LIV 0 7 0 3 26 120 0.011%C LV 1 2 7 0 26 360 0.034%C LVI 0 4 6 0 26 210 0.020%C LVII 0 6 2 2 26 1,260 0.120%C LVIII 2 1 6 1 26 2,520 0.240%C LIX 0 5 4 1 26 1,260 0.120%C LX 4 1 0 5 26 1,260 0.120%C LXI 3 0 5 2 26 2,520 0.240%C LXII 1 5 1 3 26 5,040 0.481%C LXIII 1 3 5 1 26 5,040 0.481%C LXIV 4 0 2 4 26 3,150 0.300%C LXV 2 4 0 4 26 3,150 0.300%C LXVI 1 4 3 2 26 12,600 1.202%C LXVII 3 2 1 4 26 12,600 1.202%C LXVIII 2 2 4 2 26 18,900 1.802%C LXIX 3 1 3 3 26 16,800 1.602%C LXX 2 3 2 3 26 25,200 2.403%C LXXI 1 1 8 0 27 90 0.009%C LXXII 0 3 7 0 27 120 0.011%C LXXIII 2 0 7 1 27 360 0.034%C LXXIV 0 6 1 3 27 840 0.080%C LXXV 1 2 6 1 27 2,520 0.240%C LXXVI 1 5 0 4 27 1,260 0.120%C LXXVII 0 4 5 1 27 1,260 0.120%C LXXVIII 4 0 1 5 27 1,260 0.120%C LXXIX 0 5 3 2 27 2,520 0.240%C LXXX 3 2 0 5 27 2,520 0.240%C LXXXI 2 1 5 2 27 7,560 0.721%C LXXXII 3 0 4 3 27 4,200 0.401%C LXXXIII 1 4 2 3 27 12,600 1.202%C LXXXIV 1 3 4 2 27 12,600 1.202%C LXXXV 3 1 2 4 27 12,600 1.202%C LXXXVI 2 3 1 4 27 12,600 1.202%C LXXXVII 2 2 3 3 27 25,200 2.403%C LXXXVIII 1 0 9 0 28 10 0.001%C LXXXIX 0 2 8 0 28 45 0.004%C XC 1 1 7 1 28 720 0.069%C XCI 0 6 0 4 28 210 0.020%C XCII 4 0 0 6 28 210 0.020%C XCIII 0 3 6 1 28 840 0.080%C XCIV 2 0 6 2 28 1,260 0.120%C XCV 0 5 2 3 28 2,520 0.240%C XCVI 2 3 0 5 28 2,520 0.240%C XCVII 3 1 1 5 28 5,040 0.481%C XCVIII 1 2 5 2 28 7,560 0.721%C XCIX 0 4 4 2 28 3,150 0.300%C C 1 4 1 4 28 6,300 0.601%
CC I 3 0 3 4 28 4,200 0.401%CC II 2 1 4 3 28 12,600 1.202%CC III 2 2 2 4 28 18,900 1.802%CC IV 1 3 3 3 28 16,800 1.602%CC V 0 1 9 0 29 10 0.001%CC VI 1 0 8 1 29 90 0.009%CC VII 0 2 7 1 29 360 0.034%CC VIII 3 1 0 6 29 840 0.080%CC IX 1 1 6 2 29 2,520 0.240%CC X 0 5 1 4 29 1,260 0.120%CC XI 1 4 0 5 29 1,260 0.120%CC XII 0 3 5 2 29 2,520 0.240%CC XIII 2 0 5 3 29 2,520 0.240%CC XIV 3 0 2 5 29 2,520 0.240%CC XV 2 2 1 5 29 7,560 0.721%CC XVI 0 4 3 3 29 4,200 0.401%CC XVII 1 2 4 3 29 12,600 1.202%CC XVIII 2 1 3 4 29 12,600 1.202%CC XIX 1 3 2 4 29 12,600 1.202%CC XX 0 0 10 0 30 1 0.000%CC XXI 0 1 8 1 30 90 0.009%CC XXII 1 0 7 2 30 360 0.034%CC XXIII 3 0 1 6 30 840 0.080%CC XXIV 0 2 6 2 30 1,260 0.120%CC XXV 2 2 0 6 30 1,260 0.120%CC XXVI 0 5 0 5 30 252 0.024%CC XXVII 1 1 5 3 30 5,040 0.481%CC XXVIII 1 3 1 5 30 5,040 0.481%CC XXIX 2 1 2 5 30 7,560 0.721%CC XXX 2 0 4 4 30 3,150 0.300%CC XXXI 0 4 2 4 30 3,150 0.300%CC XXXII 0 3 4 3 30 4,200 0.401%CC XXXIII 1 2 3 4 30 12,600 1.202%CC XXXIV 0 0 9 1 31 10 0.001%CC XXXV 3 0 0 7 31 120 0.011%CC XXXVI 0 1 7 2 31 360 0.034%CC XXXVII 1 0 6 3 31 840 0.080%CC XXXVIII 1 3 0 6 31 840 0.080%CC XXXIX 2 1 1 6 31 2,520 0.240%CC XL 0 4 1 5 31 1,260 0.120%CC XLI 0 2 5 3 31 2,520 0.240%CC XLII 2 0 3 5 31 2,520 0.240%CC XLIII 1 2 2 5 31 7,560 0.721%CC XLIV 1 1 4 4 31 6,300 0.601%CC XLV 0 3 3 4 31 4,200 0.401%CC XLVI 0 0 8 2 32 45 0.004%CC XLVII 2 1 0 7 32 360 0.034%CC XLVIII 0 4 0 6 32 210 0.020%CC XLIX 0 1 6 3 32 840 0.080%CC L 2 0 2 6 32 1,260 0.120%
CC LI 1 2 1 6 32 2,520 0.240%CC LII 1 0 5 4 32 1,260 0.120%CC LIII 0 3 2 5 32 2,520 0.240%CC LIV 1 1 3 5 32 5,040 0.481%CC LV 0 2 4 4 32 3,150 0.300%CC LVI 0 0 7 3 33 120 0.011%CC LVII 2 0 1 7 33 360 0.034%CC LVIII 1 2 0 7 33 360 0.034%CC LIX 0 3 1 6 33 840 0.080%CC LX 1 1 2 6 33 2,520 0.240%CC LXI 0 1 5 4 33 1,260 0.120%CC LXII 1 0 4 5 33 1,260 0.120%CC LXIII 0 2 3 5 33 2,520 0.240%CC LXIV 2 0 0 8 34 45 0.004%CC LXV 0 3 0 7 34 120 0.011%CC LXVI 1 1 1 7 34 720 0.069%CC LXVII 0 0 6 4 34 210 0.020%CC LXVIII 1 0 3 6 34 840 0.080%CC LXIX 0 2 2 6 34 1,260 0.120%CC LXX 0 1 4 5 34 1,260 0.120%CC LXXI 1 1 0 8 35 90 0.009%CC LXXII 0 2 1 7 35 360 0.034%CC LXXIII 1 0 2 7 35 360 0.034%CC LXXIV 0 1 3 6 35 840 0.080%CC LXXV 0 0 5 5 35 252 0.024%CC LXXVI 0 2 0 8 36 45 0.004%CC LXXVII 1 0 1 8 36 90 0.009%CC LXXVIII 0 1 2 7 36 360 0.034%CC LXXIX 0 0 4 6 36 210 0.020%CC LXXX 1 0 0 9 37 10 0.001%CC LXXXI 0 1 1 8 37 90 0.009%CC LXXXII 0 0 3 7 37 120 0.011%CC LXXXIII 0 1 0 9 38 10 0.001%CC LXXXIV 0 0 2 8 38 45 0.004%CC LXXXV 0 0 1 9 39 10 0.001%CC LXXXVI 0 0 0 10 40 1 0.000%
576,048,1====Nr
4,10 ======== rN
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Energy level and its degeneracy
∑=i
inN ∑=i
iinU ε
0 1 2 3 4 51ε 2ε 3ε
Energy levels are said to bedegenerate, if the same energy
level is obtained by more than one quantum mechanical state. They
are then called degenerate energy levels.
The number of quantum states at the same energy level is called
thedegree of degeneracy.
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A molecular energy state is the sum of an
electronic (e), nuclear (n), vibrational (v),
rotational (r) and translational (t)
component, such that:
t r v e nε ε ε ε ε ε= + + + +
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The degree of freedom of movement
• Translation: x,y,z F=3
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Rotation
• For linear molecules, F=2
• For non-linear molecules, F=3
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Vibration
• A polyatomic molecule containing natoms has 3n degrees of freedom totally. Three of these degrees of freedom can be assigned to translational motion of the center of mass, two or three to rotational motion.
• 3n-5 for a linear molecule;
• 3n-6 for a nonlinear molecule
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• CO2 has 3×3-5 = 4 degrees of freedom of vibration; nonlinear molecule of H2O has 3×3-6 = 3 degrees of freedom of vibration.
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Translational particle
The expression for the allowed translational energy levels of a particle of mass m confined within a 3-dimensional box with sides of length a, b, c is
22 22
t 2 2 2( )
8yx z
nn nh
m a b cε = + +
Where h=6.626×10-34J·s,nx, ny, nz are integrals called
quantum numbers. The number of them is 1,2,…∞ .
22 2 2
t x y z3/ 2( )
8
hn n n
mVε = + +
If a=b=c, equation becomes
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All energy levels except ground energy level
are degenerate.
Example
At 300K, 101.325 kPa, 1 mol of H2 was
added into a cubic box. Calculate the energy
level εt,0 at ground state, and the energy
difference between the first excited state and
ground state.
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SolutionTake the H2 at the condition as an ideal gas, then the
volume of it is
The mass of hydrogen molecule is
31 8.3145 3000.02462 m
101325
nRTV
p
× ×= = =
3 23 27/ 2.0158 10 / 6.022 10 3.347 10 kgm M L − −= = × × = ×
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2 34 240
t,0 2/3 27 2/3
3 (6.626 10 )3 5.811 10 J
8 8 3.347 10 0.02462
h
mVε
−−
−
× ×= × = = ×× × ×
240
t,1 2/36 11.622 10 J
8
h
mVε −= × = ×
40 40t,1 t,0 (11.622 5.811) 10 5.811 10 Jε ε ε − −∆ = − = − × = ×
the energy difference is so small that the translational particles
are excited easily to populate on different excited states, and
that the energy changes of different energy levels can be think
of as a continuous change approximately.
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Rigid rotator (diatomic)
The equation for rotational energy level of diatomic molecules is :2
r 2( 1) 0 1 2
8
hJ J J
Iε
π= + = ⋅ ⋅ ⋅,,,
2 21 20 0
1 2
( )m m
I R Rm m
µ= =+
where J is rotational quantum number, I is the moment of
inertia
µ is the reduced mass, The degree of degeneracy is
r, 2 1 Jg J= +
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One-dimensional harmonic oscillator
v
1( ) 0,1,2,
2hε ν= + = ⋅⋅ ⋅v v
v,0
1
2hε ν=
Where v quantum number,when v=0,the energy is called zero point energy.
v,1
3
2hε ν=
v,2
5
2hε ν=
v,3
7
2hε ν=
One dimensional harmonic vibration is non-degenerate.
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Electron and atomic nucleus
The differences between energy levels of electron
motion and nucleus motion are big enough to keep the
electrons and nuclei stay at their ground states.
Both degree of degeneracy, ge,0, for electron motion
at ground state and degree of degeneracy, gn,0, for
nucleus motion at ground state are different for
different substances, but they are constant for a given
substance.
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Distribution and microstate
Distribution of energy levels
We call the occupation number ni the number of distribution in
energy level εi.
For example, a distribution of 6 identical particles among 9
units of energy must satisfy with the conditions
6ii
N n= =∑
9i ii
U n ε= =∑
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The total number of ways of distribution is 26.
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Distribution of states
The number of particles occupied in a microscopic quantum state
is called the number of distribution of states.
One distribution D of energy levels has a certain number of microstatesWD, the sum of all WD is the total number of
microstates Ω of a system. That is.
DD
WΩ =∑
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• There are 16 ways of distribution of four distinguishable particles in two identical boxes.
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Distinguishable particles
• Consider N distinguishable or localized particles distribute into N nondegenerate energy levels.
( 1)( 2) (2)(1) !DW N N N N= − − ⋅⋅⋅ =
Now consider another kind of distribution that the numbers of
particles occupied in different energy levels are denoted as n1,
n2, ⋅⋅⋅, ni. All the energy levels are still nondegenerate.
1 2
! !
! ! ! !Di i
i
N NW
n n n n= =
⋅⋅⋅ ∏!
!DD D i
i
NW
nΩ = =∑ ∑∏
Just like Problem 1
Just like Problem 2
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• three different distributions of six particles. The exchanges of particles in the same energy level do not create new microstate because every energy level has only one quantum state. The numbers of microstates for three distributions are
6! 6! 6!6 60 180
1!5! 3!2!1! 2!2!1!1!A B CW W W= = = = = =
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• We now consider that the degree of degeneracy of energy levels is g1, g2, ⋅⋅⋅, gi. Suppose number of quantum states is unconstrained.
• Consider ni particles occupy energy level i, every particle can chose one from all quantum states in the energy level. Hence the ways of selection for ni particles are
inig
inig
For all energy levels, the number of microstates caused by
the degeneracy of levels is in
ii
g∏
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• the number of microstates for a certain distribution D can be written as
!!
! !
i
i
nn i
D ii ii i
i
gNW g N
n n= × =∏ ∏∏
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Identical particles
• assume that there is no restriction on the number of particles which can occupy a given energy level and that energy levels is nondegenerate.
• there is only one way for ni particles to occupy the energy level εi. Therefore, the number of microstates of a distribution D for a system is WD =1.
• If energy level is degenerate, It is easy to see that there is (2 +1) ways of distributing 2 particles in two quantum states which can be written as
(2 1)!3
2!1!W
+= =
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Suppose 8 identical particles populate in 4 quantum states in an
energy level. This is equivalent to the permutation of the sum
of 8 persons and (4-1) dividing walls, both persons and
dividing walls are indistinguishable.
(8 4 1)!165
8!(4 1)!W
+ −= =− Just like
Problem 4
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• the number of microstates for ni particles distributing in gi quantum states in an energy level is ( 1)!
!( 1)!i i
i i
n g
n g
+ −−
one kind of distribution is the products of the number of
microstates for every level multiplied by one another. That is
( 1)!
!( 1)!i i
DI i i
n gW
n g
+ −=−∏
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• If ni<<gi, this equation can be simplified into
( 1)( 2) ( )( 1)( 2)
!( 1)( 2)
!
i
i i i i i i i i iD
I i i i
ni
i i
n g n g n g n g gW
n g g
g
n
+ − + − ⋅⋅⋅ + − − − ⋅⋅⋅=− − ⋅⋅⋅
≈
∏
∏Compare this equation with equation, we can see that under
the same conditions of N, ni, and gi, the number of microstates
of a distinguishable-particle system is N! times that of an
identical-particle system.
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The most probable distribution, equilibrium distribution, and Boltzmann distribution
• The principle of equal a priori probabilitiesStatistical thermodynamics is based on the
fundamental assumption that all possible configurations of a given system, which satisfy the given boundary conditions such as temperature, volume and number of particles, are equally likely to occur.
( , , ) DD
N U V WΩ = Ω =∑
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• ExampleConsider the orientations of three unconstrained and distinguishable spin-1/2 particles. What is the probability that two are spin up and one spin down at any instant?
• Solution Of the eight possible spin configurations for the system,
• ↑↑↑ ↑↑↓ ↑↓↑ ↓↑↑ ↑↓↓ ↓↑↓ ↓↓↑ ↓↓↓• The second, third, and fourth comprise the subset
"two up andone down". Therefore, the probability for this particular configuration is
P = 3/8
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The most probable distribution
• The probability for distribution D is
DD
WP
Ω=
the microstates of three harmonic oscillators which are
distinguishable particles with total vibrational energy of 92U hν=
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I
1 13! 3
2! 1!W = ⋅ =
II
1 1 13! 6
1! 1! 1!W = ⋅ ⋅ =
III
13! 1
3!W = =
Ω=WI + WII + WIII = 3 + 6 + 1 = 10
Which distribution
is the most probable
distribution?
WD is called
thermodynamic
probability of
distribution D
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Equilibrium distribution
• In a system with large number of N, the most probable distribution may represent all distributions.
• Stirling's approximation
• a more accurate form
ln ! lnN N N N≈ −
!lim 1
2 ( / )NN
N
N N eπ→∞=
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( )! !
!
MNM
NWD −
=
Consider a system consisting of N localized particles which
distribute over two degenerate quantum states, A and B. M denotes
for the number of particles in state A and (N-M) in state B. the
number of microstates for this distribution can be expressed as
When M=N/2, WD has a maximum value.
( ) ( )!2/!2/
!
NN
NWB =
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• Every particle has two possibilities to populate on the quantum states, state A or state B. The total number of microstates for the system would be 2N.
• The probability for the most probable distribution is
0 0
!2
!( )!
N NN
DM M
NW
M N MΩ
= =
= = =−∑ ∑
( ) max2
2N WP
NΩ π= =
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• When the number of particles in a system is about 1024, the probability is then
• We consider another distribution that has a distribution number deviating m from N/2, its probability would be
( ) 132 24
28 10
10NP
π−= ≈ ×
×
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( ) ( )22
1 !
2 ! !2 2
NN
N
W m NP m
N Nm m
Ω±
± = = − +
When m<<N, in terms of Stirling’s approximation this equation
can be converted into
( )22
2
2 mN NP m e
Nπ−
± =
26
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• The probability for all distributions ranging from
• is the summation of their probabilities. By using error function
• we obtain
2 to 22 2
N NM N M N= − = +
2 2
0
2 1erf ( )
x xt t
x
x e dt e dtπ π
+− −
−
= =∫ ∫
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• the distribution at equilibrium is most certainly going to be the most probable distribution, or at the very least, with these kind of numbers, something very close to it.
( ) ( )22 2 22
2 22 2 2
20.9999
N N mNN N N
m N N N
P m P m dm e dmNπ
+ ++ −
=− − −
− ≈ − = >∑ ∫ ∫
27
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Boltzmann distribution
For a large number of noninteracting particles
/j
j
kTen ελ −=κ = 1.38×10-23 J K-1, Boltzmann constant. λ is proportional
coefficient. The population can also be expressed in the form of
energy level distribution /i kT
i in g e ελ −=The total number is
/j kT
jj j
N n e ελ −= =∑ ∑ /i kTi i
i i
N n g e ελ −= =∑ ∑
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then/ /j ikT kT
iij
N N
g ee ε ελ − −= =∑∑
•Define the particle partition function
/j kT
j
q e ε−=∑then
/j kT
j
Nn e
qε−=
/i kTi
i
q g e ε−=∑
/i kTi i
Nn g e
qε−=
The distribution that obeys these equations is called the Boltzmann distribution . The equations are also known as Boltzmann distribution law .
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for any two levels:/
/
i
k
kTi i
kTk k
n g e
n g e
ε
ε
−
−=
The ratio to total number:/ /
/
i i
i
kT kTi i i
kTi
i
n g e g e
N g e q
ε ε
ε
− −
−= =∑
Boltzmann distribution is the most probable distribution. The
maximum value can be derived by using Lagrange’s method of
undetermined multipliers
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Computations of the partition function
• Some features of partition functions• (1) at T=0, the partition function is equal to the
degeneracy of the ground state.
• (2) When T is so high that for each term εi/kT=0,
• (3) factorization property If the energy is a sum of those from independent modes of motion, then
00
limT
q g→
=i
kTi
i
q g eε
−=∑
limT
q→∞
= ∞
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t r e nq q q q q q=v
, , , , ,i t i r i i e i n iε ε ε ε ε ε= + + + +v
, , , , ,i t i r i i e i n ig g g g g g=v
The partition functions for 5 mode motions are expressed as
, , ,
, ,
, , ,
, ,
; ;
;
t i r i v i
e i e i
kT kT kTt t i r r i v v i
i i i
kT kTe e i e e i
i i
q g e q g e q g e
q g e q g e
ε ε ε
ε ε
− − −
− −
= = =
= =
∑ ∑ ∑
∑ ∑
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, r,it,i r,i
v,i e,iv,i e,i
n,in.i
[ exp( )] [ exp( )]
[ exp( )] [ exp( )]
[ exp( )]
t i
i i
i i
i
q g gkT kT
g gkT kT
gkT
ε ε
ε ε
ε
= − ⋅ − ⋅
− ⋅ − ⋅
−
∑ ∑
∑ ∑
∑
t r v e nq q q q q= ⋅ ⋅ ⋅ ⋅
30
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Zero-point energy
• zero-point energyis the energy at ground state or the energy as the temperature is lowered to absolute zero.
• Suppose some energy level of ground state is ε0, and the value of energy at level i is εi, the energy value of level i relative to ground state is
• Taking the energy value at ground state as zero, we can denote the partition function as q0.
00i iε ε ε= −
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0
0i
kTi
i
q g eε−
=∑0
0 kTq e qε
=
,0 ,0 ,0
,0 ,0
/ / /0 0 0
/ /0 0
; ;
;
t r v
e n
kT kT kTt t r r v v
kT kTe e n n
q e q q e q q e q
q e q q e q
ε ε ε
ε ε
= = =
= =
Since εt,0≈0, εr,0=0, at ordinary temperatures.
0 0,t t r rq q q q≈ =
31
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• The vibrational energy at ground state is
• therefore
• the number of distribution in any levels does not depend on the selection of zero-point energy.
1,0 2 hε ν=v
0 /2h kTq e qν=v v
0 00
0
/ ( )/ //0 0
i i ikT kT kTi i i ikT
N N Nn g e g e g e
q q e qε ε ε ε
ε− − + −
−= = =×
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Translational partition function
222 2
,t 2 2 2( )
8yx z
i
nnh n
m a b cε = + +
,it ,i exp( )t
ti
q gkT
ε= −∑
Energy level for translation
The partition function
32
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22 22
2 2 21 1 1
2 2 22 2 2
2 2 21 1 1
, , ,
exp /8
exp exp exp8 8 8
x y z
x y z
yx zt
n n n
x y zn n n
t x t y t z
nn nhq kT
m a b c
h h hn n n
mkTa mkTb mkTc
q q q
∞ ∞ ∞
= = =
∞ ∞ ∞
= = =
= − + +
= − × − × −
=
∑∑∑
∑ ∑ ∑
22
, 21
exp8
x
t x xn
hq n
mkTa
∞
=
= −
∑
22
, 21
exp8
y
t y yn
hq n
mkTb
∞
=
= −
∑
22
, 21
exp8
z
t z zn
hq n
mkTc
∞
=
= −
∑
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Take qt,x as an example
22
t, 21
exp( )8
x
xx
n
nhq
mkT a
∞
=
= − ⋅∑
2 2t, 0
exp( )dx x xq n nα∞
= −∫
For a gas at ordinary temperature α2<<1, the summation
converts into an integral.
2α
22 2 2
21
exp( ) (8
x
xn
hn
mkTaα α
∞
=
= − =∑ 设 )
33
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2 12
0
1d ( )
2xe xα π
α∞ − =∫
12
12
t, 2
1 2( ) ( )
2x
mkTq a
h
ππα
= = ⋅
32
t 2
2( )
mkTq a b c
h
π= ⋅ ⋅ ⋅
In like manner,t,yq t,zq
32
2
2 ( )
mkTV
h
π= ⋅
From mathematic relations in Appendix
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• ExampleCalculate the molecular partition function q for He in a cubical box with sides 10cm at 298K.
• Solution The volume of the box is V=0.001m3. The mass of the He molecule is 0.004/(6.022×1023)=6.6466×10-27kg. Substituting these numbers and the proper natural constants, we have
3/227 2327
34 2
2 6.6466 10 1.38 10 2980.001 7.820 10
(6.626 10 )q
π − −
−
⋅ × × × ×= × = × ×
34
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L
Mm =
p
NkT
p
nRTV ==
( )
⋅×
=−
Pa
Kmolkg102052.8 2
523
17
t p
TMNq
For ideal gas,
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Rotational partition function
The rotational energy of a linear molecule is given by εr = J(J+1)h2/8π2I and each J level is 2J+1 degenerate.
2
r 2( 1) 0 1 2
8
hJ J J
Iε
π= + = ⋅ ⋅ ⋅,,,
, 2
, 20
(2 1)exp ( 1)8
r i
kTr r i
i J
hq g e J J J
IkT
ε
π
∞−
=
= = + − +
∑ ∑
define the characteristic rotational temperature
2
28r
h
IkΘ
π=
35
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rr
0
( 1)(2 1)exp( )
J
J Jq J
T
Θ∞
=
+= + −∑Θr<<T at ordinary temperature, The summation can be
approximated by an integral
[ ]r0(2 1)exp ( 1) / drq J J J T JΘ
∞≈ + − +∫
Let J(J+1)=x, hence J(2J+1)dJ=dx, then
2 2r r0
exp( / )d 8rq x T x T IkT hΘ Θ π∞
≈ − = =∫
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2
r 2r
8T IkTq
h
πΘ σ σ
= =
For a homonuclear diatomic molecule, such as O2, it comes back
to the same state after only 180o rotation.
where σ is called the symmetry number. σ is the number of
indistinguishable orientations that a molecule can exhibit by
being rotated around symmetry axis. It is equal to unity for
heteronuclear diatomic molecules and is equal to 2 for
mononuclear diatomic molecules.
For HCl, σ = 1; and for Cl2, σ = 2.
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Vibrational partition function
v
1( ) 0,1,2,
2v h vε ν= + = ⋅ ⋅ ⋅
Vibrational energies for one dimensional oscillator are
Vibration is non-degenerate, g=1. The partition function is
,
,0
1exp /
2
i
kTi
i
q g e h kTε
ν∞−
=
= = − +
∑ ∑v
v v
v
v
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v v, h
k
νΘ Θ=
vv
v v3 5exp( ) exp( ) exp( )
2 2 2q
T T T
Θ Θ Θ−= + − + − + ⋅⋅⋅
v v v2 exp( ) [1 exp( ) exp( ) ]
2T T T
Θ Θ Θ= − ⋅ + − + − + ⋅⋅⋅
Define the characteristic vibrational temperature
37
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• Characteristic vibrational temperatures are usually several thousands of Kelvins except for very low frequency vibrational modes.
2v,H 5986KΘ =
v,CO 3084KΘ =
2v,O 2239KΘ =we cannot use integral instead of summation in the calculation
of vibrational partition function.
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At low T, , ,according
to mathematics
v 1T
Θ >> vexp( ) 1T
Θ− <<
21
when 1 11
x x xx
<< + + + ⋅ ⋅ ⋅ ≈−
,
/2/2
/ /2 /2
1 1
1 1
TT
T T T
eq e
x e e e
ΘΘ
Θ Θ Θ
−−
− −= ⋅ = =− − −v
v
v
v v v
38
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( )v,0 v0v /
1exp1 exp( )
kT qqhkT
ε ν× ==− −
take the ground energy level as zero,
For NO, the characteristic vibrational temperature is
2690K. At room temperature Θv/T is about 9;
the , indicating that the vibration is almost in the
ground state.
0 1q ≈v
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Electronic and nuclear partition function
e,0 e,1e e,0 e,1exp( ) exp( )q g g
kT kT
ε ε= − + − + ⋅⋅⋅
e,0 e,1 e,1 e,0e,0
e,0
exp( )[1 exp( ) ]g
gkT g kT
ε ε ε−= − + − + ⋅⋅⋅
-1e,1 e,0( ) 400 kJ mol ,ε ε− = ⋅
e,0e e,0exp( )q g
kT
ε= −
Energy difference is large, so electrons are generally at ground
state, all terms except first one in the summation expression is
negligible.
,0 /0,0
e kTe e eq e q gε= =
39
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• If the quantum number of total angular momentum for electronic motion is j, the degeneracy is (2j+1). Then the electronic partition function can be written as
• A rare exception is halide atoms and NO molecule. The difference between the ground state and the first excited state of them are not so large, the second term in the summation has to be considered.
0 2 1eq j= +
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n,0n n,0 exp( )q g
kT
ε= −
Nuclear motion is always in the ground state at ordinary
chemical and physical process because of large energy
difference between ground and first excited state. Its
partition function has the form of
Nuclear motion
0,0 2 1n nq g I= = +
where I is a quantum number of nuclear spin.
40
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Thermodynamic energy and partition function
/
/
Independent particle system:
;
(8.48)
i
i
i ii
kTi i
kTi i
i
U n
Nn g e
q
NU g e
q
ε
ε
ε
ε
−
−
=
=
=
∑
∑
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/
/2
/2
/2
1
1
i
i
i
i
kTi
iV V
kT ii
i
kTi i
i
kTi i
iV
qg e
T T
g ek T
g ekT
qkT g e
T
ε
ε
ε
ε
ε
ε
ε
−
−
−
−
∂ ∂ = ∂ ∂
= − −
=
∂ = ∂
∑
∑
∑
∑Substitute this equation into equation (8.48), we have
Thermodynamic energy and partition function
41
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2 2 ln
V V
N q qU kT NkT
q T T
∂ ∂ = = ∂ ∂
Substitute the factorization of partition function for q
2 ln t r v e n
V
q q q q qU NkT
T
∂ = ∂ Only qt is the function of volume, therefore
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2 2 2
2 2
ln lnln
ln ln
t vr
V
e n
t r v e n
q d qd qU NkT NkT NkT
T dT dT
d q d qNkT NkT
dT dTU U U U U
∂ = + + ∂
+ +
= + + + +
If the ground energy is specified to be zero, then0
0 2 ln
V
qU NkT
T
∂= ∂
0 /0Substitute = into this equation, it follows thatkTq qeε
42
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00U U Nε= −
It tells us that the thermodynamic energy depends
on the zero point energy. Nε0 is the total energy of
system when all particles are localized in ground
state. It (denoted as U0) can also be thought of as
the energy of system at 0K. Then,
00U U U= −
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• U0 can be expressed as the sum of different energies
0 0 0 0 0 0
0 0 0
0 0
20 0
t r v e n
t t r r v v
e n
U U U U U U
NhvU U U U U U
U U
= + + + +
≈ = = −
= =
43
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The calculation of The calculation of
• (1) The calculation of
0 0 0, ,t r vU U U
0tU
0 2
3/2
22
ln
2ln
3
2
tt t
V
V
qU U NkT
T
mkTV
hNkT NkT
T
π
∂ ≈ = ∂
∂ =
∂
=
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The calculation of The calculation of
0 2
2
ln
ln
rr r
V
r
qU U NkT
T
Td
NkT NkTdT
σ
∂ = = ∂
Θ= =
0rU
The degree of freedom of rotation for diatomic or
linear molecules is 2, the contribution to the energy of
every degree is also ½ RT for a mole substance.
44
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The calculation ofThe calculation of
0vU 0
vU
0vU
0 /0 2 2
/
1lnln 1
1
1
v
v
Tv
v
v T
dd q eU NkT NkTdT dT
Nke
−Θ
Θ
−= =
= Θ−
Usually, Θv is far greater than T, the quantum effect
of vibration is very obvious. When Θv/T>>1,
Showing that the vibration does not have contribution
to thermodynamic energy relative to ground state.
0 0vU ≈
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• If the temperature is very high or theΘv is very small, thenΘv/T<<1, the exponential function can be expressed as
/ 1TeT
Θ Θ≈ +v v
0v /
1 1
1 1 1T
U Nk Nk NkTe
T
ΘΘ Θ Θ= ≈ =− + −
vv v
v
45
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• For monatomic gaseous molecules we do not need to consider the rotation and vibration, and the electronic and nuclear motions are supposed to be in their ground states. The molar thermodynamic energy is
0,
3
2m mU RT U= +
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• For diatomic gaseous molecules vibration and rotation must be considered. If only lowest vibrational levels are occupied, the molar thermodynamic energy is
00, v
5( 0)
2m mU RT U U= + ≈
46
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• If all vibrational energies are equally accessible, the molar thermodynamic energy for vibration is
• The molar thermodynamic energy for diatomic molecules is then
0vU RT=
00, v
7( )
2m mU RT U U RT= + =
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HeatHeat capacitycapacity andand partitionpartition functionfunction
• The molar heat capacity, CV,m, can be derived from the partition function.
,UmCV m T V
∂=
∂
ln2,
qRT
T VV m T V
C ∂ ∂
∂=∂
Replace q with 0 /0 kTq q e ε−=0ln2
,q
RTT
VV m T
V
C ∂ ∂
∂=∂
We can see from above equations that heat capacity does not
depends on the selection of zero point of energy.
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• Electrons and nucleus are in ground state
002 2
, , ,
0ln2,
lnln vr
V V
V t V r V v
qtRTT
VV m T
V
qqC RT RT
T T T T
C C C
∂ + + ∂
∂=∂
∂∂∂ ∂ ∂ ∂ ∂ ∂
= + +
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The calculation of CThe calculation of CV,tV,t, C, CV,rV,r and Cand CV,vV,v
• (1) The calculation of CV,t
32
2
2( )t
mkTq V
h
π=
0 3ln22,
qtRT RT
VV m T
V
C ∂ = ∂
∂=∂
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• (2) The calculation of CV,r
2
2
8(linear molecules)r
r
IkT Tq
h
πσ σ
= =Θ
02
,
ln rV r
V
qC RT R
T T
∂∂= = ∂ ∂
If the temperature is very low, only the lowest rotation state is
occupied and then rotation does not contribute to the heat
capacity.
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The calculation of The calculation of CCV,vV,v
T
kT
eqeq
v
0,v
1
1v
0v Θ−
−==
ε
22
vv, 1
vv−ΘΘ
−
Θ= TTV ee
TRC
02 v
,v
lnV
V V
qdC RT
dT T
∂= ∂
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v v v v
v
2 22 2v v
,v
2
v
1
0
T T T TV
T
C R e e R e eT T
R eT
− −Θ Θ Θ Θ
Θ−
Θ Θ = − ≈
Θ = ≈
It shows that under general conditions, the contribution to heat
capacity of vibration is approximately zero.
Generally, Θv/T>>1, equation becomes
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v
1 vTeT
Θ Θ ≈ +
When temperature is high enough,
v v2 2
v v,v
T TVC R e Re R
T T
−Θ ΘΘ Θ ≈ = ≈
50
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3
2VC R=
, ,
50
2V V t V vC C C R= + + =
In gases, all three translational modes are active and their
contribution to molar heat capacity is
The number of active rotational modes for most linear
molecules at normal temperature is 2
122VC R R= × =
In most cases, vibration has no contribution to the heat capacity,
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Entropy and partition function
Entropy and microstate
Boltzmann formula
k = 1.38062×10-23 J K-1
As the temperature is lowered, theΩ, and hence theS of the
system decreases. In the limitT→0,Ω=1, so lnΩ=0, because
only one configuration is compatible withE=0. It follows
thatS→0 asT→0, which is compatible with the third law of
thermodynamics.
lnS k Ω=
51
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• For example
maxln ln lnDD
W WΩ = ≈∑
maxlnS k W=
48
50
100.01
10=
48
50
ln100.96 1
ln10= ≈
When N approaches infinity, maxln1
ln
W
Ω=
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Entropy and partition function
• For a non-localized system, the most probable distribution number is
• Using Stirling equation ln N!=N ln N - N and Boltzmann distribution expression
•
!
ini
Di i
gW
n=∏
ln ( ln ln !)D i i ii
W n g n= −∑
/i kTi i
Nn g e
qε−=
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• We have,
0 0
ln ( ln ln )
( ln ln ln )
ln
ln ln (non-localised system)
or
ln (non-localised system)
B i i i i ii
i ii i i i i i
i
B
W n g n n n
nNn g n n g n
q kT
q UN N
N kTq U
S k W Nk NkN T
q US Nk Nk
N T
ε
= − +
= − − + +
= + +
= = + +
= + +
∑
∑
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• For localized system
• Entropy does not depend on the selection of zero point energy .
00
ln ln (localised system)
or
ln (localised system)
B
US k W Nk q
T
US Nk q
T
= = +
= +
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• Factorizing the partition function into different modes of motions and using
• We can give
0 0 0 0 0 0t r v e nU U U U U U= + + + +
t r v e nS S S S S S= + + + +
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0 0
ln t tt
q US Nk Nk
N T= + +
0 0
ln v vv
q US Nk
N T= +
0 0
ln e ee
q US Nk
N T= +
0 0
ln e et
q US Nk
N T= +
0 0
ln r rr
q US Nk
N T= +
For identical particle system, entropies for every mode of
motion can be expressed as
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Calculation of statistical entropy
• At normal condition electronic and nuclear motions are in ground state, and in general physical and chemical process the contribution to the entropy by two modes of motion keeps constant. Therefore only translational, rotational and vibrational entropies are involved in computation of statistical entropy.
vt rS S S S= + +
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Calculation of statistical entropy
30 22
2( )t t
mkTq q V
h
π= = ⋅
0 0
ln t tt
q US Nk Nk
N T= + +
( )3/2
3
2 5ln
2t
mkT VS Nk Nk
Nh
π= +
(1) Calculation of St
0 3
2tU NkT=
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( )1,
3 5ln / kg mol ln( / K) ln( / Pa) 20.723
2 2m tS R M T p− = ⋅ + − +
• For ideal gases, the Sackur–Tetrode equationis used to calculate the molar translational entropy.
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(2) Calculation of (2) Calculation of SSrr
• For linear molecules
• When all rotational energy levels are accessible
• We obtain
0 0
ln r rr
q US Nk
N T= +
0 0/r r r rq q T U NkTσ= = Θ =
ln( / )r rS Nk T Nkσ= Θ +
, lnm rr
TS R R
Θ σ= +
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(3) Calculation of (3) Calculation of SSvv
• Substitute
• Into the following equation
( ) ( )1 1/ /0 01 and 1v vT Tv v rq e U Nk e
− −−Θ Θ= − = Θ −
( ) ( )0 0
v v
1 1/ /1v
ln /
ln 1 1v v
v
T T
S Nk q U T
Nk e Nk T e− −−Θ Θ−
= +
= − + Θ −
( ) ( )v v1 1/ /1
m,v vln 1 1T TS R e R T eΘ ΘΘ− −− −= − + −
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Residual EntropyResidual Entropy
• in some the experimental entropy is less than the calculated value. One explanation to this discrepancy is that the experimental system does not reach a real state of equilibrium. In other words, some disorder is present in the solid even at T = 0 K. In this case, the entropy at T = 0 is then greater than zero. This difference in entropy is called the residual entropy.
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Other thermodynamic functions and partition functions
• 1 A, G, H and q
[ ]( ) ( )
ln
ln ln ln
ln ln ln ln !
ln (for identical particles)!
ln (localized system)
N
N
q UA U TS U T Nk Nk
N T
qNkT NkT kT N q N N N
N
kT N q N N N kT N q N
qA kT
N
A kT q
= − = − + +
= − − = − − +
= − − − = − −
= −
= −
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( )
2
ln
lnln / ! (non-localized system)
lnln (localized system)
ln ln
T T
N
T
N
T
V T
A qG A pV p NkT
V V
qG kT q N NkTV
V
qG kT q NkTV
V
H U pV
q qNkT NkTV
T V
∂ ∂ = + = − = ∂ ∂
∂ = − + ∂
∂ = − + ∂
= +∂ ∂ = + ∂ ∂
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• Note
• Compound functions, A, G, and H are defined form U, they are depend on the zero-point energy.
• A and G are depend on the entropy, therefore, localized and non-localized particles are different.
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of ideal gases
• Except translational partition function, all the other partition functions are independent of volume.
• Stirling:lnN! = N lnN – N
1lnln t
T T
qqV
V V−∂∂ = = ∂ ∂
( ) lnln / !
ln ln
ln( / )
NT
T
qG kT q N NkTV
V
NkT q NkT N NkT NkT
NkT q N
∂ = − + ∂
= − + − += −
,m TG
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• At standard state,when N=L
• If q is expressed asq0
, ln( / )m TG RT q N= −
0, 0,ln( / )m T mG RT q N U= − +
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Standard equilibrium constant for ideal-gas reactions
• We define the equilibrium constant in the number of molecules as
• The Gibbs function of one particle
BN B
B
K Nν= ∏
0,
0/ln ln B kTB B B
BB B B
G q qkT kT e
N N Nεµ −
= = − = −
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• When chemical reaction reaches equilibrium
r m , 0B m BB
G Gν∆ = =∑
0BBB
ν µ =∑
0,
0/ln 0B kTB
BB BB B B
qkT e
Nεν µ ν −
= − =
∑ ∑
( )0, 0,
0/ /0ln ln ( ) ln 0
B
B B BB BkT kTBB B
B B BB
qe q e N
N
νε ν εν ν− −
= − =
∑ ∏ ∏
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• Where
• for a reaction aA + bB = lL + mM
0 /0( ) rB BkTB B N
B B
q e N Kεν ν−∆ = = ∏ ∏
r 0 0,B BB
ε ν ε∆ =∑
0
0 0/
0 0r
l m l mkTL M L M
N a b a bA B A B
N N q qK e
N N q qε−∆= =
r 0 0,A 0,B 0,L 0,Ma b l mε ε ε ε ε∆ = − − + +
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• The number of molecules per unit volume is defined as the molecular concentration
• The equilibrium constant expressed in the molecular concentration is then
BB
NC
V=
BC B
B
K Cν= ∏( ) 0 /0 /
Br kT
C BB
K q V eν ε−∆ =
∏
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• is called partition function of B per unit volume at equilibrium condition, and denoted by
0 /Bq V*Bq
( ) 0 /* BrB kT
C B BB B
K C q eν εν −∆ = =
∏ ∏
0
* */
* *r
l mkTL M
C a bA B
q qK e
q qε−∆=
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• Relation between concentration cB and molecular concentration CB is cB = CB /L
• For a general reaction of ideal gases
0 0
* * * */ /
* * * *
( / ) ( / )
( / ) ( / )Br r
l m l mkT kTL M L M
C a b a bA B A B
q L q L q qK e L e
q L q L q qνε ε−−∆ −∆∑= =
0,0
* *//
B B
r mr U RTkTB BC
B B
q qK e e
L L
ν νε −∆−∆
= = ∏ ∏
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Boltzman Hypothesis
where S is the entropy.ΩΩΩΩ is the # of microstates in a macrostate.The Boltzman constant, k = R/NO.NO is Avogardo’s number.R is the ideal gas constant.
Provides a sharp extremum.Range is compressed by assuming logarithmic relation.
Average energy of particles is fixed.
ΩΩΩΩ==== ln k S
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Find Conditions for Equilibrium
• Find an expression for change in entropy of the system.
• Determine the constraints.
• Apply the constraints and the extremum criterion: .0)( ====indS
• Solve the remaining equations for the
conditions for equilibrium.
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Find an expression for dS(ni)Substitute for ΩΩΩΩ:
====∏∏∏∏
====
r
i 1i !n
N!ln k S
Expand:
(((( )))) (((( ))))
==== ∑∑∑∑====
r
1ii !nln-N!ln k S
Note the Stirling approximation: x-x lnx x! ln ====
(((( )))) (((( ))))
++++−−−−==== ∑∑∑∑∑∑∑∑========
r
1ii
r
1iii nnlnnN-NlnN k S
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Find an expression for dS(ni)
Rearranging:
Note: ∑∑∑∑====
====r
1iin N
(((( )))) (((( ))))
−−−−−−−−==== ∑∑∑∑====
r
1iii NlnNnlnn k S
and x ln - x1
ln ====
−−−−==== ∑∑∑∑====
r
1i
ii N
nlnn k S
Taking the
derivative:
−−−−==== ∑∑∑∑====
r
1i
i
Nn
ln k dS idn
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Isolation ConstraintsConsider an isolated system.
Closed system ---
∑∑∑∑====
====r
1iisys n N ∑∑∑∑
====
====r
1iiisys ne U
Insulated system ---
∑∑∑∑====
====r
1ii? V
Rigid system ---
0 d? dVr
1ii ======== ∑∑∑∑
====
Closed system ---
Insulated system ---
Rigid system ---
(((( )))) 0 dnened dUr
1iii
r
1iiisys ============ ∑∑∑∑∑∑∑∑
========
0 dn dNr
1iisys ======== ∑∑∑∑
====
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Constrained Maximum EntropyApply Lagrange multipliers to constraints &
add to condition for entropy maximum.
Rearrange, raise to power of e to yield r equations:
0 dU dN dS syssys ====++++++++ ββββαααα
r),1,2, (i]ke
exp[ ]k
exp[ Nn i
sys
i …………======== ββββαααα
Substitute for entropy and constraints:
0dnednNn
ln k r
1i
r
1iiii
r
1i
i ∑∑∑∑ ∑∑∑∑∑∑∑∑==== ========
====++++++++
−−−− ββββααααidn
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Constrained Maximum Entropy
∑∑∑∑====
====r
1iisys n N ∑∑∑∑
====
====r
1i sys
i
Nn
1Apply: and
Solve for αααα:
P
1ke
exp k
exp1r
1i
i ====
====
−−−−
====∑∑∑∑
ββββαααα
ke
exp Function Partitionr
1i
i∑∑∑∑====
======== ββββP
Define:
Yielding:
P
==== ke
exp
Nn
i
sys
i
ββββ
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∑∑∑∑++++−−−−==== idnPdVTdSdU µµµµ
By analogy
Compare phenomenological & statistical expressions for dS to evaluate α & β:
Constrained Maximum Entropy
∑∑∑∑∑∑∑∑========
++++−−−−====r
1i
idnP
r
iii kdnedS
1
lnββββ
sysNkdUdS Pdln++++−−−−==== ββββ
sysNT
dVTp
dUT
dS dµµµµ−−−−++++==== 1
T1−−−−====ββββ Plnk
T−−−−====µµµµ
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All equilibrium thermodynamic functions can be derived if the partition function is known.
Complete expression equilibrium distribution of particles over energy levels:
Constrained Maximum Entropy
====kTe-
exp1
Nn i
sys
i
P
kTe
exp Function Partitionr
1i
i∑∑∑∑====
−−−−======== P
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Thermodynamic Functions in
Terms of Partition Function
−−−−==== ∑∑∑∑
====
r
1i sys
ii N
nlnn k S
====kTe-
exp1
Nn i
sys
i
P
∑∑∑∑∑∑∑∑========
++++====r
1i
P 11
lnT1
S nkner
iii
sysNkU PlnT1
S ++++====
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Thermodynamic Functions in
Terms of Partition FunctionDeduce F from S:
ln kT N- F sys P====
Apply: TF
- SV
∂∂∂∂∂∂∂∂====
VTkT
∂∂∂∂∂∂∂∂++++==== Pln
N ln k N S syssys P
TS- F U====
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Thermodynamic Functions in
Terms of Partition Function
Apply: TU
CV
V
∂∂∂∂∂∂∂∂====
VT
kT
∂∂∂∂∂∂∂∂++++
∂∂∂∂∂∂∂∂==== 2
22
sysVsysV
lnN
Tln
T k 2N CPP
Apply: TS U ++++==== F
VTkT
∂∂∂∂∂∂∂∂==== Pln
N U 2sys
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Monatomic Gas ModelAssumptions:
All particles are identical.Volume = lx x ly x lzEnergy of the system is not quantized & is equal to Σ kinetic energies of the particles.
3/2
mkT2
V
==== ππππP
2vm21
KE ====
1/2
x-
2
mkT2
dv2
exp
====
−−−−∫∫∫∫∞∞∞∞++++
∞∞∞∞
ππππxv
kTm
∫∞
∞−−∫
∞
∞−−∫
∞
∞−−∫∫∫= zzyyxx dvkTmvdvkTmvdvkTmv
lxdxdydz
lylz)2/
2(exp[)2/
2(exp[)2/
2(exp[
000P
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Thermodynamic Properties
of Ideal Monatomic Gases
Apply
====3/2
mkT2
V ln lnππππ
PT1
23
T
ln
V
====
∂∂∂∂∂∂∂∂ P
ln kT N- F sys P====
====3/2
sys mkT2
V ln kT N- Fππππ
ApplyVT
kT
∂∂∂∂∂∂∂∂++++==== Pln
N ln k N S syssys P
kN 23
mkT2
Vln kN S sys
3/2
sys ++++
==== ππππ
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kTN2 1
U O====
R23
kN 23
C OV ========
Thermodynamic Propertiesof Ideal Monatomic Gases
VTkT
∂∂∂∂∂∂∂∂==== Pln
N U 2sysApply:
Apply: TU
CV
V
∂∂∂∂∂∂∂∂====
Equipartition of energy:freedomofdegreeperkT
2 1
U ====
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Einstein’s Model of a CrystalConsider a simple cubic crystal --- 6 nearest neighbors,
1 atom & 3 bonds per unit cell. Hypothesis --- Energy of crystal is the sum of the
energies of its bonds. The atoms vibrate around equilibrium positions as if bound by vibrating springs. Only certain vibrational frequencies are allowed in coupled springs. The energies (εi) of the bonds are proportional to their vibrational frequencies (ν).
εi = (i + 1/2) hννννwhere h = Planck’s constant.
The adjustable parameter νννν is set by assuming an Einstein temperature: θE = hνννν/k
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Einstein’s Model of a Crystal
∑∑∑∑====
++++====
r
0i kT
h )21
(i-exp
ννννP
Evaluate the partition function:
ir
i∑∑∑∑
====
====0 kT
h-exp
kTh
21
-exp νννννννν
PFactor:
Approximate as infinite
series:
i
i∑∑∑∑
∞∞∞∞
====
====kTh
-expkTh
21
-exp νννννννν
P
−−−−−−−−
====
kTh νννν
νννν
exp1
1kTh
21
-exp PSubstitute for
infinite series:
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====2kTh
-exp-1ln - kTh
21
- lnνννννννν
P
Einstein’s Model of a CrystalTake ln of both sides:
For simple cubic: Osys NN 3====
Apply ln kT N- F sys P====
−−−−−−−−++++====kTh
kTN O
νννννννν exp1ln3hN23
F O
ApplyVT
kT
∂∂∂∂∂∂∂∂++++==== Pln
N ln k N S syssys P
exp-1ln k3N - exp-1
exp k3N S OO
====kTh
kTh
kTh
kTh νννν
νννν
νννννννν
09/19/20013-142
CC
HH
AA
PP
TT
EE
R R
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2
2
O
exp1
exp
kTh
3N C
−−−−−−−−
−−−−
====
kTh
kTh
kVνννν
νννννννν
VTkT
∂∂∂∂∂∂∂∂==== Pln
N U 2sysApply
Einstein’s Model of a Crystal
−−−−−−−−
−−−−++++====
kThkTh
νννν
νννν
ννννexp1
exp1hN
23
U O
Apply: TU
CV
V
∂∂∂∂∂∂∂∂====
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA
JAMES CLERK MAXWELL
JAMES CLERK MAXWELL (1831-1879)
British physicist, presented his first scientific paper
to the Royal Society of Edihburgh at the age of 15. In
chemistry he is best known for his Maxwell distribution
and his contributions to the kinetic theory of gases. In
physics his name is most often associated with his
Maxwell equations for electromagnetic fields.
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LUDWIG BOLTZMANN
LUDWIG BOLTZMANN (1844-1906)
Austrian scientist, is best known for his work in
the kinetic theory of gases and in thermodynamics and
statistical mechanics. His suicide in 1906 is attributed by
some to a state of depression resulting from the intense
scientific war between the atomists and the energists at
the turn of the century. On his tombstone is the
inscription S = k ln W.
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ALBERT EINSTEIN
ALBERT EINSTEIN (1879-1955)
was born in Germany and educated in Switzerland; and
he died in the United States. He was refused a position as
assistant in the physics department in the Zurich
Polytechnical institute on his graduation, and he settled
for position as an examiner in the Swiss Patent Office in
1900.
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ALBERT EINSTEIN
In a few short years he produced three theories, each of
which was fundamentally important in different branches
of physics and chemistry: the theory of the photoelectric
effect, the theory of Brownian motion, and the theory of
relativity. Einstein was one of the few scientists to
achieve worldwide stature in nonscientific circles for his
scientific work.
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ALBERT EINSTEIN
The name Einstein is a household word, and has been
introduced as a word in the English language. The
expression “He’s a regular Einstein” is often applied to
bright children. When I was a schoolboy, it was accepted
fact among my associates that Einstein was the smartest
man who ever lived, and that his theory of relativity was
so complicated that only three people understood it, one
of whom was Einstein himself.
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ALBERT EINSTEIN
Einstein was forced out of Nazi Germany in the early
1930s along with Fritz Haber (1868-1934) and others,
and came to the United States, where he spent the rest of
his life at the Institute for Advanced Study at Princeton.
Einstein received the Nobel Prize in physics in 1921 for
his work on the photoelectric effect.
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ENRICO FERMI
ENRICO FERMI (1901-1954)Italian physicist, was actively engaged in many
branches of physics during his career. His trip to Sweden to accept the Nobel Prize in physics in 1938 was used as a cover to flee Italy, and his intention not to return was known only to a few of his most intimate friends. He came to the United States, where he accepted a position on the faculty of columbia University. Later developments in the Axis nations rendered this decision a very fortunate one, especially since his wife was Jewish.
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ENRICO FERMI
It was also lucky for the United States, since Enrico
Fermi directed the research that led to the first successful
chain reaction at the University of Chicago in 1942 and
pointed to the feasibility of the atomic bomb. His Nobel
Prize was for “ the discovery of new radioactive elements
produced by neutron irradiation, and for the discovery of
nuclear reactions brought about by slow electrons.”
Fermi had devoted the years before 1938 to studying
radioactivity induced by neutron bombardment.
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ENRICO FERMI
He thought that he had produced transuranic elements by
bombarding uranium, and all workers in the field at that
time accepted this explanation. It remained for Hahn
(1879-1968) and Strassman to show that the measured
radioactivity was produced because of isotopes of much
lighter eldments, and that Fermi had actually produced
nuclear fission instead of nuclear transmutation. It was a
case of the right man getting the Nobel Prize, but for the
wrong reason.