Adv Thermo Chapter 6 Statistical Thermodynamicsmazlan/?download=Adv Thermo Chapter 7 -...

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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


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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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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%


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%


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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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%

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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


FKMFKM


Rotation

• For linear molecules, F=2

• For non-linear molecules, F=3

11


FKMFKM


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


FKMFKM


• 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.

12


FKMFKM


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


FKMFKM


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.

13


FKMFKM


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 − −= = × × = ×


FKMFKM


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.

14


FKMFKM


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= +


FKMFKM


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.

15


FKMFKM


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.


FKMFKM


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= =∑

16


FKMFKM


The total number of ways of distribution is 26.


FKMFKM


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 =∑

17


FKMFKM


• There are 16 ways of distribution of four distinguishable particles in two identical boxes.


FKMFKM


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

18


FKMFKM


• 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= = = = = =


FKMFKM


• 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∏

19


FKMFKM


• 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= × =∏ ∏∏


FKMFKM


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

+= =

20


FKMFKM


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


FKMFKM


• 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

+ −=−∏

21


FKMFKM


• 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.


FKMFKM


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 =∑

22


FKMFKM


• 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


FKMFKM


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=

23


FKMFKM


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


FKMFKM


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→∞=

24


FKMFKM


( )! !

!

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 =


FKMFKM


• 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= =

25


FKMFKM


• 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−= ≈ ×

×


FKMFKM


( ) ( )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


FKMFKM


• 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

+− −

−

= =∫ ∫


FKMFKM


• 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


FKMFKM


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 −= =∑ ∑


FKMFKM


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 .

28


FKMFKM


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


FKMFKM


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→∞

= ∞

29


FKMFKM


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

− − −

− −

= = =

= =

∑ ∑ ∑

∑ ∑


FKMFKM


, 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


FKMFKM


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= −


FKMFKM


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


FKMFKM


• 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− − + −

−= = =×


FKMFKM


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


FKMFKM


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

∞

=

= −

∑


FKMFKM


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


FKMFKM


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


FKMFKM


• 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


FKMFKM


L

Mm =

p

NkT

p

nRTV ==

( )

⋅×

=−

Pa

Kmolkg102052.8 2

523

17

t p

TMNq

For ideal gas,


FKMFKM


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


FKMFKM


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∞

≈ − = =∫


FKMFKM


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.

36


FKMFKM


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


FKMFKM


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


FKMFKM


• 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.


FKMFKM


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


FKMFKM


( )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


FKMFKM


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


FKMFKM


• 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= +


FKMFKM


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


FKMFKM


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

−

−

=

=

=

∑

∑


FKMFKM


/

/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


FKMFKM


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


FKMFKM


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


FKMFKM


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= −


FKMFKM


• 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


FKMFKM


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

∂ ≈ = ∂

∂ =

∂

＝


FKMFKM


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


FKMFKM


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 ≈


FKMFKM


• 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


FKMFKM


• 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= +


FKMFKM


• 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


FKMFKM


• 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= + =


FKMFKM


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.

47


FKMFKM


• 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

∂ + + ∂

∂=∂

∂∂∂ ∂ ∂ ∂ ∂ ∂

= + +


FKMFKM


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 ∂ = ∂

∂=∂

48


FKMFKM


• (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.


FKMFKM


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

∂= ∂

49


FKMFKM


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


FKMFKM


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


FKMFKM


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,


FKMFKM


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


FKMFKM


• 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=


FKMFKM


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−=

52


FKMFKM


• 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

= − +

= − − + +

= + +

= = + +

= + +

∑

∑


FKMFKM


• 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

= = +

= +

53


FKMFKM


• 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= + + + +


FKMFKM


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

54


FKMFKM


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= + +


FKMFKM


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=

55


FKMFKM


( )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.


FKMFKM


(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= +

56


FKMFKM


(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− −− −= − + −


FKMFKM


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.

57


FKMFKM


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

= − = − + +

= − − = − − +

= − − − = − −

= −

= −


FKMFKM


( )

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

∂ ∂ = + = − = ∂ ∂

∂ = − + ∂

∂ = − + ∂

= +∂ ∂ = + ∂ ∂

58


FKMFKM


• 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.


FKMFKM


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

59


FKMFKM


• 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= − +


FKMFKM


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 −

= = − = −

60


FKMFKM


• 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− −

= − =

∑ ∏ ∏


FKMFKM


• 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∆ = − − + +

61


FKMFKM


• 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−∆ =

∏


FKMFKM


• 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−∆=

62


FKMFKM


• 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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FKMFKM


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

63

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FKMFKM


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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FKMFKM


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

64

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FKMFKM


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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FKMFKM


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 ======== ∑∑∑∑

====

65

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FKMFKM


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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FKMFKM


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

66

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FKMFKM


∑∑∑∑++++−−−−==== idnPdVTdSdU µµµµ

By analogy

Compare phenomenological & statistical expressions for dS to evaluate α & β:


∑∑∑∑∑∑∑∑========

++++−−−−====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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FKMFKM


All equilibrium thermodynamic functions can be derived if the partition function is known.

Complete expression equilibrium distribution of particles over energy levels:


====kTe-

exp1

Nn i

sys

i

P

kTe

exp Function Partitionr

1i

i∑∑∑∑====

−−−−======== P

67

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FKMFKM


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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FKMFKM



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====

68

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FKMFKM



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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FKMFKM


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

69

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FKMFKM


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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FKMFKM


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 ====

70

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FKMFKM


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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FKMFKM


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:

71

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R R

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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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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.

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