Kittel Kroemer Thermal Physics

7/25/2019 Kittel Kroemer Thermal Physics

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NOTES AND SOLUTIONS TO THERMAL PHYSICS BY CHARLES KITTLE AND HERBERT KROEMER

ERNEST YEUNG - LOS ANGELES

ABSTRACT. These are notes and solutions to Kittle and Kroemers Thermal Physics. The solutions are (almost) complete: I will

continuously add to subsections, before the problems in each chapter, my notes that I write down as I read (and continuously reread).I am attempting a manifold formulation of the equilibrium states in the style of SchutzsGeometrical Methods of Mathematical

Physicsand will point out how it applies directly to Thermal Physics. Other useful references along this avenue of investigation is

provided at the very bottom in the references.

Any and all feedback, including negative feedback, is welcomed and you can reach me by email or my wordpress.com blog.

You are free to copy, edit, paste, and add onto the pdf and LaTeX files as you like in the spirit of open-source software. You are

responsible adults to use these notes and solutions as governed by the Caltech Honor Code: No member of the Caltech community

shall take unfair advantage of any other member of the Caltech community and follow the Honor Code in spirit.

SECOND EDITION. Thermal Physics. Charles Kittel. Herbert Kroemer. W. H. Freeman and Company. New York.

QC311.5.K52 1980 536.7 ISBN 0-7167-1088-9

1. STATES OF A M ODELS YSTEM

2. ENTROPY ANDT EMPERATURE

Thermal Equilibrium. EY : 20150821 Based on considering the physical setup of two systems that can only exchange

energy between each other, that are in thermal contact, this is a derivation of temperature.

U=U1+ U2is constant total energy of 2 systems1, 2in thermal contactmultiplicity g(N, U)of combined system is

g(N, U) =U1U

g1(N1, U1)g2(N2, U U1)

The differential ofg(N, U)is

dg= g1U1N1 g2dU+ g1 g2U2N2 dU2 = 0EY : 20150821 This step can be made mathematically sensible by considering the exterior derivative d ofg C(), where is the manifold of states of the system, with local coordinates N , U, whereUhappens to be a global coordinate. Then,consider a curve ins.t. it has no component in N ,

N1

, and this curve is a null curve so that the vector field X X()generated by this curve is s.t. dg(X) = 0.With dU1 = dU2,

1

g1

g1U1

N1

= 1

g2

g2U2

N2

=

ln g1U1

N1

=

ln g2

U2

N2

Define

(N, U) := ln g(N, U)

Then

=

1U1

N1

=

2U2

N2

Date: Fall 2008.

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Temperature. T1= T2 - temperatures of 2 systems in thermal equilibrium are equal.Tmust be a function ofU

N

[?].

= 1T

=kB

U

N

Experimentally, kB = 1.381 1023 J/K= 1.381 1016 ergs/K.Now

1

=

U

N

=kBT

Problems. Solution 1. Entropy and temperature.

(a) Recall that 1U

N,V

and(N, U) log g(N, U). Giveng(U) = CU3N/2,

(N, U) = log CU3N/2 = log C+3N

2 log U

U =

3N

2

1

U =

1

= U=3N

2

(b)2U2

N


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Solution 4. The meaning of never.

Suppose1010 monkeys.

(a) Hamlet represents one specific ordering of1015 with 44 possibilities for each character. The probability of hitting

upon Hamlet from a given, random sequence is 144

100000=

1

44100000 . Given thatlog1044 = 1.64345, then

101.64345 = 44or101.64345 = 441 so then

1

44100000

= 10164345

(b)

(age of universe)

10keys

second

= 1018 s

10keys

second

= 1019 keys

1019 keys 1010 monkeys= 1029 keys typed out

1hamlet

105 characters

= 1024 possible Hamlets

From part (a), the probability that a given, random sequence is Hamlet, 10164345

(1029 characters)(10164345) = 10164316

Note, I think that the probability should be(1029 characters)

1Hamlet105 characters

(10164345) = 10164321

Since we are considering the number of Hamlet,105 character sequences.

3. BOLTZMANN D ISTRIBUTION AND H ELMHOLTZF RE EE NERGY

cf. Example: Energy and heat capacity of a two state system, pp. 62 of Kittel and Kroemer [1]. Kittel and Kroemer

introduces the heat capacity very early, specific to this example.

Definition 1. heat capacityCVat constant volume is defined as

(1) CV :=

V

Recall the thermodynamic identity (which is introduced many equations later):

dU= d pdV 1()whereis a manifold of states of all systems.

Consider local coordinates of,(, V). Consider curve c: R c()

s.t.c generates a vector field c= i.e. no component

in theVdirection. Notice the prescient choice of parameter.Now for internal energyU C(), taking the exterior derivativedresults in

dU=U

d+

U

VdV

Then applyingdUonto vector field ,

dU

=

U

= + 0

Now,

U V V = UV = V Hence,

(2) CV :=

V

=

U

V

EY: 20150825 Why do we need differential geometry? Its because I always wondered why you could do this:

CV :=

V

?=

U

V

with d= dU= ?=U

and talk of differentials.

Definition: Reversible process. EY : 20150824 Mathematically, 1-forms are exact.

3


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Pressure. Consider coordinates(, V) of manifold of thermodynamic states.Imagine a reversible compression of a cube system (so imagine dV 0 whendV


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One atom in a box. one atom of massMin cubical box of volume V =L322M2= p= i p2=

= (x) = A sin nxxL sin nyyL sin nzzL n=

2

2M

L

2(n2x+ n

2y+ n

2z)

Then the partition functionZ1 is

Z1 =n

exp

n

=

(nx,ny,nz)

exp

22M

L

2(n2x+ n

2y+ n

2z)

Let

2 =

22

2ML2 or=

(2M)1/2V2/d

Then

Z1 =

0

dnx

0

dny

0

dnzexp [2(n2x+ n2y+ n2z)] =

0

dnxexp (2n2x)3

=

1

31/2

2

3=

1/2

2

3In general,Z1 =1/2

2

dNow

Z1 = 1/2V1/d

2

(2M)1/2

d

= V

2M d/2 =nQV =

nQn

in terms of concentrationn= 1/V.

nQ:=M22

d/2is thequantum concentration.

Problems. Solution 1. Free energy of a two state system.

(a)

Z= 1 + e/

F = ln Z= ln (1 + e/)(b)

U= 2 (F/)

= e/

1 + e/

= F

= ln(1 + e/) +e/

(1 + e/)

Solution 2. Magnetic susceptibility

(a) Remember to calculate themultiplicityin theN-spin system (its not enough to sum up exp (s/)factors).M= 2sm Us = MB= 2smB N =N++ N

2s= N+ N= N+ (N N+) = 2N+ N

Z=

N/2s=N/2

N

N+

exp

2smB

=

N/2s=N/2

N!N2 + s

!N2 s

!exp

2mBs

=

Ns=0

N!

s!(N s)!exp

2mB

s N

2

=eNmB

(1 + e

2mB

)

N

= 2

N

cosh

NmB where it was crucial to use (1 +x)N =

Nj=1

Nj

xj . Note, in changing the sum index, sinceNis large, we can

neglect dropping thes= 0term.

Z= 2N(N)(coshN1

mB

)sinh

mB

m2

M= 2

ln Z= N m tanh

mB

=

M

B =

N m2

sech2

mB

5


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(b)

F = ln Z= ln

(2 cosh

mB

)N

= N ln(2cosh

mB

)

Forx Mnm = tanhmB

. Now1 tanh2 y= sech2y. F= N ln

21x2

= N2 ln

41x2

.

(c) For mB 1,cosh2mB

1. = m2N

Solution 3. Free energy of a harmonic oscillator

(a)

Z=s=0

exp

s0

=

1

1 e0/ = (1 e0/)1

F = ln Z= ln(1 e0/) ln0

for1 0

(b)

= F

= {ln(1 e0/) + 1 e0/(e

0/)0

} = 0/

e0/ 1 ln(1 e 0 )

Solution 4. Energy fluctuations.

1

=

=

= 1

2

Z=

ses

Z=s

es

2Z=s

2ses

U= s ses/Z

= ln Z

U

= 2

U=2

Z

Z

=2

(2Z)Z (Z)2

Z2

=2

2Z

Z

Z

Z

2=

= 2 U

= 2 2Solution 5. Overhauser effect. System Sin energy eigenstateEn= n.P(E) = (1)gR(E)NoteUR = ( 1). US=. dUSd + dURd = 1 + ( 1) = = dUtotd in a specific energy eigenstate;gS(n) = 1WhilegR(UR) = multiplicity of reservoir R withURenergy.Now

REs

= 1

and

gR(UR) = exp(R(UR))

If dURd = ( 1)= UR small compared toUR.

UR(ES= (n + 1)) = UR(n) +dUR

d = UR(n) + ( 1)

R(UR((n + 1))) R(UR(n)) +1

( 1)P(ES= (n + 1))

P(ES=n) =

exp(R(UR(n)) + 1( 1))exp(R(UR(n)))

= exp (1 )Solution 6. Rotation of diatomic molecules.

(a) (j) = j(j+ 1)0. g(j) = 2j+ 1Remember thatZis a sum over all states, not over all levels.

Z=j=0

(2j+ 1)ej(j+1)0/ =j=0

d

dj(e(j

2+j)0/)

0

=

0

j=0

d

dj

e0/j2+j

(b) For1 0ZR() =

0

0

d

dx

e0/x2+x

dx= 0

(e0/)x

2+x (e0/)0

=

0

6


7/40

(c) For 0 1ZR() = 1 + 3e

20/

(d)

U=2ln Z

for1 0

U=2

ln

0

=

for

0 1, U=2

1

1 + 3e20/(3e20/)

202

= 60e

20/

1 + 3e20/

CV =

U

V

= 1 when1 0

CV = 60

e20/202

(1 + 3e20/) (3e20/) 202 e20/

(1 + 3e20/)2

= 1220

e20/

2

1

(1 + 3e20/)2

For very small 0 1, CV 12

e20/

( /0)2

(e) See sketch.

Solution 7. Zipper problem.

(a) Nlinks.s= 0closed,open.

Z=Ns=0

exp(s/) = 1 e(N+1)/

1 e/

(b) 1 .Z=Ns=0exp (sB)

1

ln Z=1

Ns=0(s)es

Z = s =

=

1 e/

1 e(N+1)/e(N+1)/((N+ 1))(1 e/) (e/)()(1 e(N+1)//)

(1 e/)2

=

= e/(eN (N+ 1)(1 e) (1 e(N+1)/))

(1 e(N+1)/)(1 e/) e/(eN (N+ 1)(1 e) 1)

(1 e

)This still does not give the desired approximation. Consider the following:

Z=1 e(N+1)

1 e =e eN

e 1Z=

((e + N eN )(e 1) (e)(e eN ))(e 1)2 =

(e(e + N eN e + eN ) e N eN )(e 1)2

Z

Z =

(e(N+ 1)eN e N eN )(e 1)(e eN )

(e(N+ 1)eN e+ N eN )e2

=

= (N e(N1) + e(N1) e N eN )

e2 =

((N+ 1)e(N1) (e + N eN ))e2

=

= (N+1)eeN neN e

e2 =

(N+1)eNeN ee2

=Ne+eeeNeN

e2 =N eN

eN e

(eN )

eN e = e

= s =e/

Solution 8. Quantum concentration. Now(x,y,z) = A sinnxxL

sinnyyL

sinnzzL

.p = 1i, p

2

2m = 12m2.Ground orbital: nx= ny = nz = 1.

T = 3

2m

L

20|0 = 3

2m

L

27


8/40

where 0|0 = 1, normalized. It was normalized in this way:0

sin2nxx

L

dnx=

L0

1 cos 2nxxL 2

dnx=

nx sin 2nxxL L2nx

2

L

0

=L

2

| =A2

L

2

3=

A2L3

8 = 1orA2 =

8

L3

Recall thatnQ= m22 3/2.Consider the condition that there will be a concentration for which the zero-point quantum kinetic energy is equal to thetemperature:

= 32m

2

L2 =

3

2m2n2/3 =orn2/3 =

2m

322 orn =

2m

322

3/2

= n= (

4

3

m

22)3/2 =

4

3

3/2nQ

Solution 9. Partition function for two systems.

Z(1 + 2) =E1+2

g(E1+2)exp

E1+2

=

E1+E2=E0

g(E1+ E2)exp

E1

exp

E2

=

=E1E2

g(E1)g(E2)expE1 expE2 =Z(1)Z(2)since systems are independent.

Solution 10. Elasticity of polymers.

(a) Consider2s= N+ N; N=N++ N, 2s= N+ (N N+) = 2N+ N. N+= 2s+N2 .For2s, consider 2s= N+ (N N+) = 2N+ N. N+= 2s+N2 .

= g(N, s) + g(N, s) = 2N!N2 + s

!N2 s

!

(b)|s| N

(l) = lngN, l2+ gN, l2 = ln 2(N!)N2 +

l2!N2 +l2! =

= ln(2N!) = {

N

2 +

L

2

ln

N

2 +

l

2

N

2 +

l

2

+

N

2 l

2

ln

N

2 l

2

N

2 l

2

}

where we usedln(x + x) ln x + 1xx.

(l) = ln(2N!) {

N

2 +

l

2

ln

N

2

+

1

N/2

l

2

N

2 +

N

2 l

2

ln

N

2

+

1

N/2

l2

} =

= ln(2N!) {N2

lnN

2 N

2 +

N

2 ln

N

2 N

2 +

l

2ln

N

2

+

l

2+

2

N

l

2

2+

l2

l2

ln

N

2

+

l2

N22}

(l) = ln 2N!N2 ! N2 ! l2

N 2

(c)

l =

2lN 2

= f= 2lN 2

Solution 11. One-dimensional gas.

n=

2

2m

L

2n2 in one dimension

Z1 =

n=1

exp

22m

L

2n2/

=0

dne2n2 =

2

8


9/40

where

2 =

2

2m

L

2=

/L2m1/2

.

Recall that =F

,F =U , so that

F= ln Z= Nln

2 = Nln

2

= Nln

2/L

2m1/2

=

= Nln 2mL1/2 = 12 Nln 22mL2F

=

1

2Nln

22

mL2

+

N

2

122

mL2

22mL22

=

=1

2Nln

22

mL2

+

N2

= N

2

ln

22

mL2

1

4. THERMAL R ADIATION ANDP LANCK D ISTRIBUTION

Problems. Solution 1. Number of thermal photons.

We consider a cavity of volumeV, and of edge lengthL(soV L3). So thenn= nc/L.Now 1

exp (n

)1is the thermal average number of photons in a single mode frequency . So then

sn = 1

exp

n

1Consider(nx, ny, nz)on positive octant, and 2 independent polarizationsof em field.

n

1

exp

ncL

1 =(2)80

4n2dn 1

exp

ncL

1 =0

n2dn

exp

ncL

1 ==

L

c

3 0

x2dx

ex 1

= N= V

2

c

3(2.404)

where I used the substitutions

x= cn

L

L xc

=n

(dx)L

c=dn

Now() = (42V /45)( /c)3, so then

N =

1

2.404

44

45

3.602

Now how was0

dx x2dxex1 evaluated?

Solution 2. Surface temperature of a Sun. Given the solar constant of the Earth, the total radiant energy flux density at

the Earth from the Sun normal to the incident rays, integrated over all emission wavelengths,

solar constant = 0.136 J s1 cm2, (49)

(a)

4(1.49 1011 m)2 0.136 J s1 cm2

102 cm

1 m

2= (4)(1.49)21022 0.136 104 = 3.8 1026 J s1

Note that I had used 1.49 1011 mas the distance of the Earth from the Sun.(b) J=energy flux density or rate of energy emission per unit area.

B = 2k4B603c2 = 5.670 108 W m2 K4.

Note thatW= 1Js . I will useR= 6.9599 1010 cmas the radius of the Sun.4 1026 J s1

4(6.9599 1010 cm)2 =J=BT4

9


10/40

4 1026 J s14(6.9599 1010 cm)2

1 m2102 cm1m

25.670 108 J/s

k4 =T4T 5830 K

Solution 3. Average temperature of the interior of the Sun.

(a)

U= R0

G 43r3 (4r2dr)r

= 163

2G2 R0

r4dr=1615

22GR5; M= 43

R3

U=3GM2

5R

U= 12

R0

G43

r3

r (4r2dr) =

83

2G2 R0

r4dr=8

1522GR5 =

=8

152GR5

M43

R3

2=

310

GM2

R = 1.14 1041 J

(b) Using the virial theorem of mechanics, note that

12

U= 3

20

GM2

R

= 3

20

6.67 1011 kg ms2 m2

kg2 2 1033 g

1 kg103 g

7 1010 cm 1m102 cm = 5.72 1040 J

Nows = 1exp(/)1 , is the Planck distribution function, giving the thermal average number of photons in asingle mode frequency.

thermal average energy = s= exp (/)1 for , So then5.72 1040 J=N =N .

= 5.72 1040 J

1 1057(1.381 1023 J/K) = 4.14 106 K

Solution 4. Age of the Sun.

(a) Consider4H42 H e. Then4(1.0078) 4.0026 = 0.0286 amu. Then

(0.0286 amu)1.6726 1027 kg

1.00727647 u (3 108 m/s)2 = 4.27

1012 J

GivenM = 2 1033 g,

(2 1030 kg)(0.10)

1

4 (1.0078 amu)

1.00727647 u

1.6726 1027 kg

(4.27 1012 J) = 1.28 1044J

So 1.28 1044 J energy is available.(b)

1.15 1045 J4 1026J s1

1 hr

3600 s

1 day

24 hr

1 yr

365 days

= 1.02 1010 years

Solution 5. Surface temperature of the Earth.JS=bT4is the radiant power per unit area.

Total emitted radiation energy of the sun isJS4R2.

4R2JS

4R2

ES

= R2

R2

ES

JS= radiation energy hitting1 cm2 of Earths surface in one second

Since the Earth is considered a black-body, the rate of absorption must equal the rate of emission:

R2R2ES

bT4= bT

4e orT

4e =T

4R

2 = Te = 5800 K

7 1010 cm1.5 1013 cm = 396.2 K= 123 C

Solution 6. Pressure of thermal radiation.

(a) sj =number of photons in that mode. Suppose modes ofj ,j = 0, 1, 2, . . . .j =sjj =total energy injth mode,sj photons injth mode.

U=j

j =j

sjj P= UV

= j

sjjV

10


11/40

(b) j =jc/L, V =L3. So thenj =jcV

1/3.

= djdV

=1

3 jcV4/3 =

13

jV

(c) p= U3V(d) We want the Kinetic pressure at a concentration of(1 mol/cm3). RecallingP = NkBTV ,

P =

1 mol

cm3

6.022 1023

1 mol

102 cm

1 m

3 1.381 1023 J

K

(2 107 K) = 1.663 1014 N

m2

Now for the thermal radiation pressure,

p= U

3V =

1

3

2

153c34 = 4.03 1013 N

m2

wheret= 2 107 K.For the pressures to be equal,

1

3

2

153c3k4BT

4 =N kB

V T orT3 =

45(c)3N kB2k4BV

so that T = 3.2 107 K

Solution 7. Free energy of a photon gas

(a) Z=n

11en/ ConsiderZ=s=0 e

s/ = 11e/for a single mode.

(b) F = ln Z=

nln(1 en/) n= ncL .

F=n

ln(1 en/) = 0

4n2

dn8 (2)ln(1 enc/L) = 0 dnn2 ln(1 enc/L) ==

(n3 ln(1 enc/L))0

0

dn n3enc/L

1 enc/Lc

L

=

2c

L

0

dn n3

en/L 1=

=

L

c

3 0

x3

ex 1= ( L)3

(2c3)

2

45

where I usedx= ncL .

Solution 8. Heat shields. ForJu = B(T4u T4l), the thermal flux without the heat shield, in the middle region. Plane mabsorbsBT

4u + BT

4l and emitsJm= B(T

4u+ T

4l ) =T

4m.

Tm= (T4u+ T

4l)

1/4. The key point is that, by symmetry, plane memits Jm2 flux on each side.

Jnet= BT4u (B T4u + T4l2 ) = B T4u T4l2 = Ju2

Jnetis the same for the other side of the heat shield:

Jnet= BT4l + B

T4u + T4l

2

=B

T4u T4l

2

Solution 9. Photon gas in one dimension. E = E0sin (kx)cos(t)is the form of a solution with kL = n or k =

nL ,

since

v2Exx= Ett v2k2 =2 or k

=v, n= vn

L

Zj =s=0

es/ = 1

1 e/whereZj is the partition function for a particular mode frequency.

s =s=0 se

sn

Z =Z1

d

d(n/) (1 expn

)1 =Z1(1 exp

)2 expn

=

= exp(n/)1 exp(n/)

So

sn= n = nexp

n

1 U=n

n =n

vn/L

exp

vnL

1U

=n

vn/Lexp

vnL

12

exp

vn

L

vnL2

=n

vL

2n2 exp

vnL

(exp

nvL

1)2 =n

vL

2n2 exp

vnL

(exp

vL

n 1)2

11


12/40

Nown0

dnfor one-dimensional photon. Let = vL .Lettingx= n,

U

=nv

L

2 dn

n2 exp(n)

(en 1)2 = 1

dx

x2ex

(ex 1)2 = 1

{ x

2

(ex 1)+0

xdx

ex 1}

Coefficient ofk term off(k)isj =0

xex1dx. Nowf(k) =0

sin(kx)ex1 dx,sin (kx) = kx (kx)

3

3! + . . . , so that

2

6 =

0

x

ex

1

dx

U

=

L

v

2

6

=

L

6v =CV

Solution 10. Heat capacity of intergalactic space.

Given the density1 atomm3, considering thermal radiation at 2.9 K, thenkBT = (1.381 1023 J/K)(2.9 K), c =(1.05457 1034 J s)(3 108 ms).Recall for radiation, that the energy per unit volume: U

V = 2

153c34 so that U

= 42

153c33V.

Assume hydrogen atoms modeled as ideal gas: U= 32N , dUd =

32

N.

CVmatterCVradiation

=32N

42

153c33V

=45(c)3(N/V)

82(kBT)3 = 2.8 1010

Solution 11. Heat capacity of solids in high temperature limit.

n

=

vn

L ;n=

vn

L . For

n,0 n nD.

exp

n

1 1 + n

+

n

21

2

+

1

6

n

3+ 1

By doing long division

n

exp

n

1 nn

1 + n2 +

(n)2

62

= 1 + n2 +

(n)2

62

=+ n2

+(n)2

12 + . . .

Fornp = (6N/)1/3

U=3

2

nD0

dnn2( n2

+(n)

2

12 ) =

3

2

nD0

dnn2( 2

vn

L +

22v2n2

12 L2 ) =

= 32{ 1

3n3D v2L n

4D

4 +

2

2

v2

12 L215

n5D} = 2 6N 32

v16L6N

4/3 +323v2(6N/)5/3

120 L2

So

U= 3N 32v

16L

6N

4/3+

323V2(6N/)5/3

120 L2

NowT=, =

vkB

(62N)1/3

L

So then

U= 3Nv(62N)1/3

L 3

2v

16L

61/3N4/36

4/3 =

15

8

(62N)1/3v

L =N kB

15

8

U

=N kB

15

8 =

15

8 (1.381 1023 J/K)

6.022 1023particles1 mol

= 15.59

which is very close to experimental values.

Solution 12. Heat capacity of photons and phonons. For a photon: U= 2V

153c34 U=

42V153c3

3.

phonon:U() = 34N4

5(kB)3. U =

124N3

5(kB)3

So then

CV =1241022

5

1

100

3= 2.3 1018

for a phonon.

For a photon,

42

15

(1.381 1023J/K)3(1.05457 1034J s(3 1010 cm/s))3

3 = 220 /K33

12


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Temperature at which to photon contribution equals to phonon contribution:

(220/K3)3 = 2.3 1018 = = 2.2 105 KSolution 13. Energy fluctuations in a solid at low temperatures.

2

U

V

= ( )2Recall that

UV =125 4N 3

(kB)3 whereU=34N 4

5(kB)3

( )22 =

124N 5

5(kB)3

1

98N28

25(kB)6

=

20

3

1

4

1

N

1

3

(kB)

3 =0.068

N

T

3

F=

0.070

1015

200

102

3= 0.02

Solution 14. Heat capacity of liquid4Heat low temperatures.

(a) Givenv= 2.383 104 cm s1 and accounting for only longitudinal waves (only longitudinal polarization), then theDebye temperature is

= vkB182N

V 1/3

=(1.05457 1034 J s)(2.383 104 cm/s)

1.381 1023 J/K 1820.145 g

cm3 1.00727647 u 1He4.0026u1.67262 1024 g

1/3

=

= 28.6 K

(b) Recall the derivation forUfor phonons in a solid. Account for only longitudinal waves (only longitudinal polariza-tion).

U=

2

nD0

dn n2n

exp(n/) 1=

2

nD0

n2dn

nvL

exp(x) 1= 2v

2L

nD0

n3

exp(x) 1 dn

Withn= nL

v ,x= nvL orLv

x= n, then

U=

2v

2L

L

v

4 nD0

x3

ex 1 dx

For low temperatures,small so takexD = nDvL = vL 18N 1/3 = 181/32/3n1/3v to go to .U

2

2(v)3

4

15V

Recall thatCV =U

V

. ThenCV/V = 215

2

(v)3

3. RecallB =kBT, and givenv= 2.383 104 cm/s, then

kBv

= (1.381 1023 J/K)

(1.05457266 1034 J s)(2.383 104 cm/s) = 5.495 106 (1/K cm)

So ifwe take CV/V and divide by the given density = 0.145 g/cm3 to get the heat capacity per gram, (and multiply

bykB, the Boltzmann constant to get the correct units; Kittel and Kroemer likes using dimensionless formulas ) then

(CV/V)/= (kB)2

152

kBv

3

T3

cm3

0.145 g= 0.0208 T3

Solution 15. Angular distribution of radiant energy flux.

(a) Recall

u =

2c23

exp

1is the radiation energy per unit volume per unit frequency range.

cu =energy per unit time, per cross sectional area per unit frequency range.em waves emitted spherically from pt. Q.

Suppose em wave comes in at a funny angle other than directly inward.

Consider areadathats from the spherical wave from pt. Q. How much of that goes into solid angle d?

= cucos dA13


14/40

So cucos is the energy per unit time, per cross-sectional area, per unit frequency range, that enters into some solidangled.

r2d

4r2 =

d

4is the fraction of the spectral density that if arrives in solid angled.

= cucos d4

is the spectral density of radiant energy flux that arrives in solid angle d.(b)

cos sin dd= 2

sin2

2 d=

cos2

2

/20

= 1 1

2

=

= cu4

Solution 17. Entropy and occupancy.

Z=

s=0

es/ = 1

1 e/

Z=

s=0

es/s2 = (1 e/)2(e/)2 =Z2 s

Then

s = e/

1 e/ s + 1 = 1

1 e/

= (ln Z) = ln Z+ Z

Z = ln s + 1 +

2s = ln s + 1 +

s

Now (nyn)

s ln s + 1

s

=Z e/ ln e/ =

s

= = s + 1 ln s + 1 + s ln s

Solution 18. Isentropic expansion of photon gas.(a) iV

1/3i =fV

1/3f or

ViVf

1/3=

fi

= 2.9 K

3000 K = 103

r= r(t) = t so thatr= tor rr = tt

rf rirf

= 1 rirf

= tf ti

tf= 1 ti

tf

Knwowing that rirf = 103, then titf = 10

3.(b) Now

= V1/3

3 =V

Forconstant entropy expansion.

U

V =

2

153c34 U=

2

15(c)3V

4

V4/3

=

2

15(c)34

V1/3U

V

= 2

15(c)341

3 V4/3

W =

pdV =

2

15(c)34 (V1/3)VfVi

= 24

15(c)3

1

V1/3i

1V

1/3f

=

2V4/3i

4i

15(c)3

1

V1/3i

fiV

1/3i

=

2Vi3i

15(c)3(if

14


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Exercise 19. Reflective heat shield and Kirchhoffs law.

For a left plane sheet at utemperature, right plane sheet atl temperature

+Ju= b4u

reflection rJu = (1 a)b4uabsorb (left)aJu= ab

4u

Jl = b4lreflectionrJl = (1 a)b4labsorb (left)aJl= ab

4l

total absorption:a(Ju+ Jl) = ab(4u+

4l)

total emission:ab(4u+

4l). By symmetry,

a(Ju+Jl)2 emitted to the left, and to right.

eemissivity, where emissivity is defined so radiation flux emitted by the object is e times the flux emitted by a blackbody atthe same temperature.

By Kirchhoff law, for equilibrium,a = e; object must emit at same rate as it absorbs.

Jnet=a(Ju+ Jl)

2 rJu+ Ju=(1 r)(Ju+ Jl)

2 + (1 r)Ju= (1 r)Ju

2 (1 r)Jl

2 = (1 r) (Ju Jl)

2 =

= (1 r)

b(4u 4l)

2

=

a(Ju+ Jl)

2 + rJl Jl = (1 r)(Ju Jl)

2

5. CHEMICAL P OTENTIAL ANDG IBBSD ISTRIBUTION

Solution 1. Centrifuge.

T= 12mr2 + 12mr

22.

Considerext = 12Mr22 (negative so for bigger r away from r = 0 axis, lower chemical potential , so to showcentrifugal force outwards.tot = ln (n/nQ) 12M r22.tot(r) = tot(0)for diffusion equilibrium.

ln

n

nQ

1

2Mr22 =ln

n(0)

nQ

= ln

n(r)

n(0)

=

1

2Mr22

n(r)

n(0)= exp

Mr22

2

or n(r) = n(0)exp

Mr22

2

Solution 2. Molecules in the Earths atmosphere.

Recall that for ideal gas,F= [Nln Z1 ln N!];Z1 = nQV = M22

3/2V.

= FN,V

= [log Z1 ddN

ln N!] =ln nnQ

wherenQ=M22

3/2.

Now =log nnQ

ext=

GMer = gR

2m

r sinceg = GMeR2e

.

tot = ln (n/nQ) + M gh

In equilibrium, this must be independent ofr:tot(r) = tot(R).

ln (n(r)/nQ) g R2M

r =ln (n(R)/nQ) M gR

2

R

ln n(r)n(R) =M gR2r R or expMg R2r R = n(r)n(R)= n(r) = n(R)exp

M g

R2

r 1

R

so that

N= 4n(R)expMgR

R

r2dr exp

MgR2

r

Solution 6. Gibbs sum for a two level system. Recall that

Z(, ) =N=0

s(N)

exp[(N s(N))/] =ASN

exp[(N s(N))/]; = exp

15


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(a)Z= 1 + + exp ,1forN= 0; forN= 1, + exp . for= 0.(b)N = 0(1)+(1)(+ exp (

))

Z = (1+exp(/))

Z(c)N() = 0(1)+(0)()+ exp(/)Z = exp(/)Z(d) =N() = exp (/)Z(e)

Z= 1 + + exp(/) + 2 exp(/) = (1 + )(1 + exp(/)where we considered the possibility the orbital at 0and atare each occupied by one particle at the same time.

So that, for total energy being,exp [(2 )/] = (exp )2e/ =2e/.Solution 7. States of positive and negative ionization.

Z=e 2 + e 2 + e2 + 2e2Z= 0 + e/2 + e

2 + 2e

2

ln Z= (e/2 + e/2 + 2e/2)

e 2 + e

2 + e

2 + 2e

2

= 1

e/2 + e/2 + 22e/2 =e/2 =e/2 + e/2 + 2e/2 =2e/2 =e/2

2 =e/ or2 ln = /

2ln =

Solution 8. Carbon monoxide poisoning.

(a)Z= 1 + (O2)eA/

P(O2) = (O2)e

A1 + (O2)eA/

= 0.9or0.9 = 0.1(O2)exp

A

= ln

9

(O2)

=

A

; orln

(O2)

9

=A

A= (kBT) ln

(O2)

9

= (8.617 105 eV

k )(273 + 39) ln

105

9

= 0.3686 eV

(b)Z= 1 + (O2)eA/

+ (CO)eB/

P(O2) = (O2)eA/

1 + (O2)eA/ + (CO)eB/ = 0.1 = 0.1 + 0.1(CO)eB/ = 0.9(O2)eA/

ln

9(O2)eA/ 1

(CO)

=

B

or B =ln

(CO)

9(O2)eA/ 1

= 0.5511 eV

Solution 9. Absorption ofO2 in a magnetic field. Recall that2j+ 1 =total number of spin states.2(1) + 1 = 3.NowU= m B.

Z= 1 + (O2)e

+BB

+ (O2)e A + (O2)e

(A+BB)

=

Z= 1 + (O2)eA/(1 + 2 coshBB

)

0.91 =(O2)e

A/(1 + 2 coshBB

)

1 + (O2)eA/(1 + 2 coshBB

)

or0.91 = 0.09(O2)eA/(1 + 2 cosh

BB

)

1

2

91

9

1

(O2)eA/ 1

= cosh

BB

orB =

Barccosh

1

2

91

9

1

(O2)eA/ 1

The Gibbs sum in the limit of zero magnetic field will differ from that of Problem 8 because there the spin multiplicity of the

bound state was neglected.

16


17/40

Z= 1 + 3(O2)eA/

P(O2) = 3(O2)e

A/

1 + 3(O2)eA/ = 0.9or0.9 = 0.3(O2)e

A/

ln 3

(O2)= A/orA= ln

(O2)

3

= 0.6227

forT= 300 K, so that= 0.049375 eV.

= B= B

0.59927 = 0.049375 0.599275.7884 1011 106 eV/T= 511.1 T

Solution 10. Concentration fluctuations.

(a) Recall that Z=N=0s(N)exp [(N s(N))/].2Z2

=ASN

N

2exp[(N s(N))/] N2 =

2

Z2Z2

(b)

N =

Z

Z

,V

=

Z

2,V

Z2 +

2Z2

,V

Z

=

N = N2 N2 = (N)2

Solution 11. Equivalent definition of chemical potential.

Recall that

d =

U

V,N

dU+

V

U,N

dV +

N

U,V

dN (31)

=

N

U,V

(35)

Consider whend = dV = 0.

=

dN+

U

V,N

dU= 0

U

V,N

dU

dN =

Note that dUdN =UN

,V

.

Using the definition,U

N,V

1, so then

=

U

N

,V

NowF =U , by definition. Consider the thermodynamic identity, dU= d pdV + dN.dF =dU d d= d pdV + dN d d =

= pdV + dN d

= FN,V

=( , V , N )

Likewise, =UN

,V

is equivalent to( , V , N ) =FN

,V

through the thermodynamic identity as well.

dU= d pdV + dN, V constant.d, dV = 0.

=

U

N

,V

=

Solution 12. Ascent of sap in trees. Given: relative humidityr = 0.9.T= 25 C

17


18/40

n0 = concentration in saturated air that stands immediately above pool of water of water vapor in air.rn0 = actual concentration of water vapor in air at uppermost leaves is rn0.

At pool,H2O = vapor for diffusion equilibrium.

Same condition at uppermost leaves, otherwise theres evaporation:

sap= vapor(h).sap= H2O (no flow going on in water). Thus, vapor(h) = vapor(0)and treat water vapor as an ideal gas.

lnn(h)nQ+ mgh = lnn(0)

nQ

ln

rn0n0

= mgh = ln

1

r

=mgh orh=

mgln

1

r

h=

(1.381 1023 1K)(298 K)(kg m2

s2 ) ln109

(18 amu)1.671027 kg

1amu

(9.8 m/s2)

= 147.1 m

Solution 13. Isentropic expansion.

(a)

(b) Recall thatF = [Nln Z1 ln N!]whereZ1 = nQV =M22

3/2V and

= F V,W

.

F

= [Nln Z1 ln N!] [N 1

Z1

M223/2 3

21/2V] =

= [Nln Z1 ln N!] [N 1Z1

nQ3

2V] = [Nln Z1 ln N!] 3

2N

WithNconstant andZ1 = M22

3/23/2V, then for an isentropic expansion, V2/3 must remain constant.

Solution 14. Multiple binding ofO2.

(a) Be wary of the multiplicity, how you count, each of the energy states.

Z= 1 + 4e/ +

4

2

2e2/ +

4

3

3e3/ + 4e4/ = (1 + e/)4

P() =

4e/

(1 + e/)4

(b)

P(4) = 4e/

(1 + e/)4 =

e/ 1+e/

46. IDEAL G AS

Reversible Isothermal Expansion. Q= 0, insulated gas, no heat flow to or from the gas (adiabatic)constant in system isolated from reservoir, if expansion reverisble (slowly)

What is the pressure after expansion? Remember

Cp

CV=

52

N32N

=5

3

=

So let1

1=3

2

(7) (, V) = N(ln 11 + ln V + constant) (61)

(8) ln 11 V = constant or

11 V constant (62)

(9) = 1

1

1 V1 = 1

1

2 V2 (63)18


19/40

for ideal monoatomic gas.

UsepV =N

(10) =

1

1

p1=

1

2

p2(64)

orp1

1

1 V

1

1 =p1

1

2 V

1

2 orp1V1 =p2V

2

I will recap Problem 10 and my solution.

Isentropic relations of ideal gas.

(a) = CpCV

. For isentropic process,pV =piVi .

Then, essentially and equivalently, take the exterior derivative:

Vdp + pV1dV = 0 = dpp

+ dV

V = 0

For

V1 =iV1i

then takingd:

d V1 + ( 1)V2dV = 0 = d

+ 1

V dV = 0

For

1p=

1i pi

then takingd:

dp

1 +

1 211 d p= 0 = dp

p +

1 d

= 0

(b) isentropic bulk moduli

B = V

p

V

=piV

i

V =p

sincep= piV

i

V

B = p

EY : 20150606 I think when one consider small, linear longitudinal perturbations of the gas system, with pressure

being the external restoring force, then sound waves propagate (correct me if Im wrong) and this is the way to deriveB.

isothermal bulk moduliB= VpV

= nV =p

velocity of sound in gas isc=B

1/2=p

1/2for ideal gas of molecules of massM,pV =N

Manifold interpretation; Im using Chapter 5 Applications in Physics, Section A Thermodynamics of [2]. Consider a 2-

dimensional manifold of equilibrium statesM, s.t. dimM = 2. Local coordinates are (U, V) with U being a globalcoordinate.

NowU=CVso useas a global coordinate, i.e. the local coordinates ofM can be(, V).Consider a curve in M parametrized by:((), V()). The corresponding tangent vectorX= + V V X(M)has

components in coordinate vector basis = ()V = V()

Recalld

+

1V

dV = 0

Consider applying this 1-form ontoX:d

1

V dV

(X) =

+

1V

V = 0 = dd

( V1) = 0

Then the equalities of the endpoints of this curve ((), V()) are the equalities above. The interpretation is that the isentropicprocess draws out a curve in M and can be written as a curve or as a tangent vector field (specifically a section ofTM).

19


20/40

1V11 =2V

12 (66)(11)

1

1 P1 =

1

2 P2 (67)(12)

P1V1 =P2V

2 (68)(13)

T1 = 300 K , V1/V2 = 12

,

(14) T2 = 122/3

(300 K) = 189 K (69)

The gas is cooled in expansion process by

(15) T1 T2 = 300 K 189 K= 111 K (70)Expansion at constant entropy is important e.g. methods of refridgeration.

What is the change in energy in the expansion?

U2 U1 = CV(2 1) = 1 1 N(2 1)

Problems. Solution 1. Derivative of Fermi-Dirac function. Recallf= 1exp[()/]+1 = (e()/ + 1)1

f= 1(e()/ + 1)2(e()/)1

f(= ) =

14

Solution 2. Symmetry of filled and vacant orbitals. = +

f() = f( + ) = 1

e/ + 1=

e/

1 + e/ = 1 +

1e/ + 1

= 1 f( )Solution 3. Distribution function for double occupancy statistics.

(a) = 1 + e/ + 2e2/ where= e/.

N =ln =

(e/+ 2e2/) =e/ + 22e2/

(b) = 1 + 2e/ + 2e2/

N =ln = 2e/

+ 22

e2/

= 2(e

/

)(1 + e/

)

Solution 4. Energy of gas of extreme relativistic particles. Forpp,s ep/tau =Z.

With the factor 2 for the 2 possible polarizations,

Z= (2)4

8

0

p2dpep/ ={p2ep/()0

0

2pep/()} = 2 0

pep/dp=

== 2 {pep/()0

0

ep/()dp} = 23

U=2ln Z

=2{ln(23)} =2(3ln ) = U= 3

Solution 5. Integration of the thermodynamic identity for an ideal gas. For constantN, recall

d= dU

+ pdV

= 1U

V

d+ 1U

V

dV + pdV

CV =U

V

, and for an ideal gas pV =N . d= = CV ln + Nln V +

1

U

dV + 1

NowU= 32for an ideal gas, soUV

= 0.

= = CV ln + Nln V + 11independent constant of andV.Solution 6. Entropy of mixing.

20


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Solution 7. Relation of pressure and energy density.

(a) Recall thatU=U(,V,N). p = UV N, andU=

sse

s/

Z

U

V =s

sV

N

es/ + s

1

sV

N

es/

/Zs

s

V

1

es/s2

s2es//Z2

Nowp=

UV N.So if the system is in state s; thenps= sV N.

= p =p =ssV

N

es/

Z

(b) Nows= L2n22

2M = V2/3

n22

2M sV

N

= 23

V5/3n22

2M = 2

3Vs

(c)

p=

s2s3V

es/

Z =

2

3

1

V

U

Solution 8. Time for a large fluctuation.(a) Recall = N[lnnQn

+ 52 ].nQ=M22

3/2.

f=N[lnnQA2VN

+ 52 ]

i= N[lnnQAVN

+ 52 ]

Now

g e exp

lnnQ

n

N+

5N

2

=e

5N2

NQ

n

NnQ= 4 (938 MeV/c

2)((0.8617 104 eV)/K)(300 K)2(6.582122 1022 MeV s)2

1 M eV

106 eV

1 c

3 108 m/s2

= 7.88 1030/m3

NowP V =N . Then

P

= NV

= (1 atm)(

1.013105N/m2

1atm )kgm/s21N 1.381 1023 J/K(300 K) = 2.445 10

25/m3

Now1 L= 103 m3 and so for0.1 L,

(2.445 1025 1m3

)(104 m3) = 2.445 1021

With 7.88 1030/m32.445 1025 /m3

= 3.22 105

= g e 52 (2.4451021)(3.22 105)2.4451021(b)

(c)

Solution 9. Gas of atoms with internal degree of freedom.

For an ideal monatomic gas, assume noninteracting.

(a) int= . = exp (/)ideal gas. = ln (n/nQ); nQ=M22

3/2.

ext= exp (/)or1Z1is the usual canonical partition function, Z1 = nQV wherenQ=

M22

3/2Z1 =s

( exp(s/)(1) + exps

exp

)Z1 = (1 + e

/)Z1

(Z1)N = ((1 + e/)Z1)N21


22/40

(b)

(c)

Solution 10. Isentropic relations of ideal gas.

(a) Isentropic process, sopV =piVi .

Vdp + pV1dV = 0 = dpp

+

VdV = 0

Dealing with an ideal gas,pV =N still applies.

V1 =iV1i

d V1 + ( 1)V2dV = 0

= d

+( 1)

V dV = 0

UsingpV =N again, note thatp1 = constant .

(1 )pdp+p11d= 0 = dpp

+

1 d

= 0

(b) Usingp= piV

i

V ,

p

V =

piViV+1

= V pV

=piV

i

V =p

So thatB = V(p/V) =p, the isentropic bulk moduli.

B= V

P

V

=n

V =p

sincepV =n

p=n

Vp

V =

nV2

7. FERMI AND B OS EG ASES

Problems. Problem 1. Density of orbitals in one and two dimensions.

(a) Show that the density of orbitals of a free electron in one dimension is

(16) D1() = (L/)(2m/2)1/2,whereLis the length of the line.

(b) Show that in two dimensions, for a square of areaA,

(17) D2() = Am/2independent of.

Solution 1. Recall, for the free electron:H= p2

2m =22m 2

= s =

2

2mn2

x

2

L2 for 1 dim.,s =

2

2m

(n2x+n2y)

2

L2 for 2-dim.IfF =Fermi energy, energy of the highest filled orbital,

1-dim: N= 2nF. 2-dim.:N= (2)14

(n2F) =

n2F2

2 factor for 2 possible spin states.

1-dim: F =

2

2m

N2

2 2L2 =

12m

N2

2 1V2

= 12m

m2

2n2 N=

2mF2V

22-dim: F =

2

2m

2N

L

2=

2

mN2

V =

2

mNV =

2m

n N= mV F2

1-dim.: D() = dNd =

2m2V

2 12(F)

1/2 = N2 =L

2m

21/2

2-dim.:N = Am/2

22


23/40

Problem 2. Energy of relativistic Fermi gas. For electrons with an energy mc2, where m is the rest mass of theelectron, the energy is given by pc, wherepis the momentum. For electrons in a cube of volume V =L3 the momentumis of the form(/L), multiplied by(n2x+ n

2y+ n

2z)

1/2, exactly as for the nonrelativistic limit.

(a) Show that in this extreme relativistic limit the Fermi energy of a gas ofNelectrons is given by

(18) F = c(3n/)1/3,

wheren = N/V.(b) Show that the total energy of the ground state of the gas is

(19) U0 =34

N F.

The general problem is treated by F. Juttner, Zeitschrift fur Physik47, 542 (1928).

Solution 2.

(a) pc= nL c, F = nFL c.Recall, for 3-dim.: N= (2)18

43

n3F

= 3 n3F. nF =3N

1/3.

F = L

3N

1/3= c3n

1/3(b)

U0 = 2

nnFn= 2 1

8 4 nF0

dnn2nc

L =

2c

L

nF0

dnn3 = 2c

4L n4F =

= 2c

4L

FLc

4= 2c

4L

3N

FLc

=

3

4N F

Problem 3. Pressure and entropy of degenerate Fermi gas.

(a) Show that a Fermi electron gas in the ground state exerts a pressure

(20) p=(32)2/3

5

2

m

N

V

5/3In a uniform decrease of the volume of a cube every orbital has its energy raised: The energy of an orbital is

proportional to1/L2 or to1/V2/3.(b) Find an expression for the entropy of a Fermi electron gas in the region F. Notice that 0as 0.

Solution 3.

(a) RecallU0 = 35N

2

2m(32N)2/3V2/3

U

V =

3

5N

2

2m(32N)2/32

3

V5/3 =

15

2

m(32)2/3

N

V

5/3So then

p=U0

V =

1

5

2

m(32)2/3

N

V

5/3(b) Recall thatF F = 22m32NV

2/3and that the heat capacity of an electron gas is Cel =

12

2N F

= U, which

helps directlywith finding the entropy.

() (0) = 0

1dU= 0 1 12 2N Fd= 2

N2F

Let(0 = 0) = 0,

() =2N

2F

Problem 4. Chemical potential versus temperature. Explain graphically why the initial curvature of versusis upwardfor a fermion gas in one dimension and downward in three dimensions (Figure 7.7). Hint: The D1()and D3()curves aredifferent, where D1is given in Problem 1. It will be found useful to set up the integral forN, the number of particles, and toconsider from the graphs the behavior of the integrand between zero temperature and a finite temperature.

Solution 4. Recall,N=F0

dD().23


24/40

1-dim:

N=

F0

dD1() = F0

d

L

2m

2

1/2=

L

(2m)1/2

F0

d1/2 =

=

L

(2m)1/2

(21/2)

F0

=

L

(2m)1/2

2

1/2F

3-dim:

N= F

0

d V

22 2m

2 3/2

1/2 = V

22 2m

2 3/2

2

33/2

F

0

= V

32 2m

2 3/2

3/2F

Note the difference in the concavity of the N()curves.Solution 5.

(a) For 3He, givenI= 1/2, density of liquid0.081 g cm3, we want to findvF,F,F.

F =

2

2m

(32n)2/3 =

= (6.582 1022 MeV s)2

2 3 938 MeV/c2

1 c31010 cm/s2 32 0.081 gcm3

1 kg

103 g

1 u

1.67 1027 kg

1 3He

3 u

2/3=

= 4.24 1010 MeV = 4.24 104 eVNow suppose we have a nonrelativistic gas. Then 12mv

2F orv

2F =

2Fm .

vF = 1.675 104 cmsec

TF = 4.24 104eV

0.8619 104 eV/K = 4.92 K(b)

Cel =2

3D(F)=

2

3

3N

2F=

2

2N

F= 1.003kBT N

Solution 6. Mass-radius relationship for white dwarfs.

(a)

U= (r)(r)r24dr =

4

G 43

r3

r

r2dr= 4 M4

3R3

G4

3

1

5

R5 M4

3R3

= 3GM2

5R(b)

F =

2

2m

(32n)2/3

is the Fermi energy.

WithV = 43R3,

Ttot = N1

2mv2 =N F =N

h2

2m

32N

V

2/3=

(32)2/3

2

2

m

N5/3

V2/3

=

(32)2/3

2

2

m

N5/3432/3

R2=

=

94

2/32

2N5/3

mR2 2 (M/MH)

5/3

mR2

sinceN= MMH

sinceMH

m.(c)

2M5/3

mM5/3H R

2=

GM2

R = M1/3R

2/G

mM5/3H

(6.582 1022 M eV s)(1.05457 1034 J s)0.511 MeV/c2((1.67 1027 kg)103 g1 kg

)5/3

103 g

1 kg

2/(6.67 1011 m

3/s2

kg )

(3 108 m/s)2

1 c2

Dealing with units and dimensions,

J=kg m2

s2 ;

N m2kg2

=kg m3/s2

kg2 =

m3/s2

kg24


25/40

(6.582 1022 MeV s)(1.05457 1034 kg m2s )102 cm1m

2 103 g1 kg

2(0.511 MeV/c2)(1.67 1024 g)5/3(6.67 1011 m3/s2kg

102 cm1m

3)

(3 1010 cm/s)2

1 c2

1020 g1/3 cm

(d)

= M43

R3 =

M

43

1020 g1/3 cmM1/3

3 = M243

1060 g cm3 =

(2 1033 g)243

1060 g cm3 =

3 1066 g21060 g cm3

106 g

cm3

(e)

M1/3R 2/G

mM5/3H

1017 g1/3 cm = R= 1017g1/3 cm

(2 1033 g)1/3 = 7.937 km

Solution 7. Photon condensation.Ne = 2.404V 3/23c3.

The condition is thatN=Ne:

N=Ne =2.404V 3

23c3

=

23c3

2.404

N

V

1/3With a concentration of1020 cm3,T = 1.7

106 K(the critical temperature inKbelow whichNe < N.

Solution 8. Energy, heat capacity, and entropy of degenerate boson gas.Recall that the distribution function for bosons is

f(, ) = 1

exp[( )/] 1ConsiderNnoninteracting bosons of spin zero.= 0for ground state. Thus,f(0, ) = 1

exp ( )1. Recall that

N() =

0

dD()f(, ) =0

d V

42

2M

2

3/21/2

1

1 exp

1Recall,

U= 0 1exp(/) 1+

0

e()/ 1D()d=

0

e()/ 1V

42 2M2 3/2

1/2

d=

= V

42

2M

2

3/2 0

3/2

e()/ 1 < , so 1, forN0to be sufficiently large.

= U= V42

2M

2

3/2 0

3/2

e/ 1

Usingx= /

dx= d/

= U= V42

2M

2 3/2

5/2

0

x3/2dx

ex

1

=B05/2C0

Its true that(e/) = e/ 2

CV =

5

2

V

42

2M

2

3/23/20

x3/2dx

ex 1Now 1 =U

V

.

(U) (U0) =

1

dU=

V B0

52

3/2C0d

=V B0

5

2C0

1/2d=

dU=VB052

3/2C0d= V B0 52

C02

33/2

025


26/40

(U) = V B05

3C0

3/2 =V B05

3C0

U3/5

(V B0C0)3/5 = (V B0C0)

2/5 5

3U3/5 =

=

V

4

2M

2

3/2 0

x3/2dx

ex 1

2/55

3U3/5

where we had usedU=V B05/2C0or

UV B0C0

3/5

=3/2.

Solution 9. Boson gas in one dimension.In one-dim.,

s=

2

2m

2n2

L2 =

2

2m

2n2

V2 or

2mV2

22 s= n

2

= n=

2mV2

22

1/21/2 = dn

d =

2mV2

22

1/21

21/2

Note the difference with 3-dim.: For spinless bosons,

D(n)dn= 4n2dn

8 =

2

2mL2

22 s

2mL2

22

1/21

21/2d=

= 4 2mL222 3/2

1/2d= D()d

N() =

0

dD()f(, ) =0

d

2mV2

22

1/21

21/21

1exp

1 =

=1

2

2mV2

22

1/2 0

d

1/2(1 exp

1)

For 0,N() which is not characteristic of a Boson.Solution 10. Relativistic white dwarf stars. pc = 2/p.Virial theorem:

2T = U 2T =kU

2

3

4N F

=

3

2N4/3c

3

1/31

L=

3GM2

5R

=

3GM2

5R

F = c

3n

1/3Approximating the sphere as a box,

L3 =4

3R3

L=

4

3

1/31/3R

3

4

1/3L= R

N4/3 =

GM2

5R

2L

2/3c31/3

= N=

GM2

5R

2L

2/3c31/3

3/4=

2GM2

5

41/3

32/31/3c

3/4NowM=N mH, wheremHis the mass of hydrogen, so

1 =

2Gm2H

5

41/3

32/31/3c

3/4N1/2

= N=

5(32/3)1/3c

41/3(2Gm2H)

3/2=

(1.05457 1034 J s)(3 108 m/s)(6.67 1011 m3kgs2 )(1.67 1027 kg)2

3/2

5(32/3)1/3

41/32

3/2= 2.2 1058

26


27/40

8. HEAT AND WOR K

Energy and Entropy transfer: Definition of Heat and Work. Consider, a manifold consisting of points representingthermodynamic states of a single system. For instance, for global coordinates (U, V),(U, V) .ConsiderW, Q 1(), 1-forms on.Now defineQas

Q d

with = (U, V) C()=(U, V) C()

Recall energy conservation in this form:

dU=W+ Q

Consider pure heat and, so, no work. NowQ = d

Heat Engines: Conversion of Heat into Work. Consider curve c: R c(t)

s.t.c generates vector field c= (Uis suited

for this).

Act on this vector field

X()withQ, i.e.

Q =

This is whats meant when its said reversible heat transfer accompanying 1 unit of entropy is given by temperature [1].Consider Figure 8.1 on page 229 of Kittel and Kroemer [1]. Roughly it looks like this:

=h dh= Qh/h Qh

W

=l dl= Ql/l Ql

But whats really going on?

ConsiderQh= dU, the initial heat input at high temperature (Ill show that later) h.Consider a curvec0 s.t. c0= 0 . Then

Qh(c0) = dU(c0) =

U

V

=0 h0

We can integrate the 1-formdU 1()for 2 reasons: mathematically, it is an exact form. Physically, we are consideringa reversible process, passing through thermodynamic states of the system, starting with the system being in energy U0 andending up with energyU1, and all the energy states ( R) in between.

= U1 U0 = 10

hd

If this is conducted all at temperaturehduring the whole process, thenU1 U0 = h 1

0d= h(1 0). Its in this case

thatd is an exact form and can be integrated over that curvec0.Legendre transforms revisited. Lets recall 2 of our favorite thermodynamic potentials,U, and Helmholtz free energy F.They are related by Legendre transformations that transform 1 coordinate into its conjugate coordinate, somewhat like how

the Legendre transform transforms that Lagrangian in canonical coordinates into a Hamiltonian written with the conjugate

momentum. However, I do want to point out that, for Lagrangians and Hamiltonians, the Legendre transformation is a fiber

derivative between tangent bundle to the cotangent bundle on the manifold. In our current case, we want a mundane Legendre

transformation between convex function to another convex function, a coordinate transformation by a C function, not amorphism between vector spaces.

RecallF. Its defined as such:F U , sodF =dU d d= dpdV

27


28/40

Consider curve c: R c(t)

Consider 2 curves that generate vector fields:

c=

or c= V

V

Now, in general, mathematically,

dF= +FV d+ FV dVThus,

F

V

= F

V

= p

Maxwell relations are easily derived:

2F

V =

2F

V so

V

=

p

V

So-called natural coordinates for F are , V. So (, V) (i.e. after a Legendre transformation, the coordinates become(, V)for each thermodynamic state.RecallUas a thermodynamic potential. Using energy conservation and howQis defined,

dU=Q + W= d+pdV

Natural coordinates are, V forU. So (, V).heat engine

ideal heat engine:

1

0 2

Qh= hdh Ql= ldl W

(1, V0)

(0, V0) (2, V1)

Qh= hdh Ql= ldl

W+ Ql= Qh or W =Qh QlW=Qh Ql h l

hQh= CQh

l= h soQh

h=

Qll

Carnot efficiencyC hlh is the ratio of the work generated to the heat added, in the reversible process.

Carnot cycle.

3 2

4 1

W21 =Qh

W32

W43

W14

(H, l) (H, h)

(L, l) (L, h)

W21 =Qh

W32

W43

W14

The total work is as such:

dU= 0for 2 reasons: mathematically, the integration of an exact 1-form around a closed curveis0, and physically, we return the system back to its original state, as this is a reversible process.

dU= 0 =

d

pdV =

W =

d= [h(H L) + 0 + l(L H) + 0] = (h l)(H L)

28


29/40

Example: Carnot cycle for an ideal gas.

3 2

4 1

W21 =Qh

W32

W43 = Ql

W1

4

(H, l, V3) (H, h, V2)

(L, l, V4) (L, h, V1)

W21 =Qh

W32

W43 = Ql

W1

4

with

isothermal expansionQh= W21 = 21

pdV =N hln

V2V1

adiabatic expansionW32 = 32

dU=U(h) U(l) = CV(h l)

isothermal compression Ql= W43 = 43

pdV =N llnV3V4

adiabatic compressionW14 =CV(h l)

lV13 =hV

12 or

V3V2

=

hl

11

V4V1

= (hl

) 11

EY : 20150911 I dont have a good reason why CVwhich is defined for constant V, thatCV U V, can be used in theisentropic (i.e. adiabatic) expansion from2 3.

The total work done is

W =N(h l) ln

V2V1

Energy Conversion and the Second Law of Thermodynamics.

11 12

01 02

Qh Qh

W1 = 1Qh Wout= 2Qh 1Qh

Ql2

Ql2 = (1 2)Qh

QhQl1 = (1 1)Qh

Q(in) = (2 1)Qh

Ql2 waste heat

11 12

01 02

Qh

W1 Wout

Ql2

Q(in)

So withQ(in)heat in,Woutnet work can be done. But thats a decrease in overall entropy. This violates the law of increasingentropy.

29


30/40

DefineH= U+ pV.H C(), whereis the manifold of equilibrium (and non-equilibrium) states of the system.Path Dependence of Heat and Work. Mathematically, Q and W are not necessarily exact1-forms. So they are path-dependent.EY : 20150911 ThatQ,Ware not necessarily exact 1-forms would imply that has some nontrivial, interestingtopologicalfeatures.

Heat and Work at Constant Temperature or Constant Pressure.

isothermal work.dU=W+ Q= W+ d

F=U dF =dU d d=W d

Ifd= 0, on an isothermal curve,dF =W,Wbecomes an exact 1-form, with potential functionF, the Helmholtz free energy.

isobaric heat and work. e.g. boiling of liquid. When liquid boils under atmospheric pressure, vapor pressure displacing

atmospheric odes work against atmospheric pressure. isobaric process.

Consider this change of volume:dx= dVA . Now

peq= vapor pressure.

F =peqA= patmA(force equilibrium)

(1, V1)

(0, V0)

W = patmAdx= patmdV

W = patmdV pdV = d(pV)is part of total work done on system.

Ifd(pV)> 0, work provided by environment and is free.Ifd(pV)< 0, work delivered to environment and not extractable from system for other purposes.

W+ d(pV) = dU Q + d(pV) = dH QRecall that for enthalpyH=U+ pV,

dH= dU+ V dp +pdV =dU W+ V dp= d+ V dp, pare natural coordinates ofH.

dH Q= W+ d(pV)An isobaric curve s.t. dp = 0,dH= Q + W+ d(pV)soQ + Wis an exact 1-form ofHpV = d(HpV) = W+ Q.2 classes of constant pressure processes:

(a)W+ d(pV) = 0

dH=Q

e.g. liquid evaporation from open vessel, because no effective work is done.

heat of evaporation is enthalpy difference between vapor phase and liquid phase

(b) constant temperature and constant pressure.

G= F+pV =U +pVdG= dF+ V dp +pdV =dU d d+ V dp +pdV =V dp ddG= W d+ d(pV) = W+ d(pV) d

with natural variables arep, at constant temperature,W+ d(pV)is exact 1-form,dG

30


31/40

FIGURE 1. Problem 8.1(c)

Problems. Solution 1. Heat pump.

(a) For a heat pump,

input:h= Qhh

output:l= Qll

Reversible condition:h= l= Qhh

= Qll so thatQh= hl

Ql.

Qh

Ql= Qh

lh

Qh= hlh

Qhnet heat inputted to pump heat.

Thus,W

Qh=c =

h lh

If heat pump is not reversible, h> l, so that Qhh

> Qll

or lh

Qh> Ql,

W

Qh=

Qh QlQh

h,by energy conservation:Qhh+ Ql Qh = 0reversible refrigerator:hh+ l h = 0,

= Qhhhh

+Ql

l Qh

h= 0

Qhhhh

+Ql

l=

Qhh+ Qlh

orQhh

1

hh 1

h

=Ql

1

h 1

l

Ql

Qhh=

1

hh 1

h

/

1

h 1

l

=

h hh

hhh

/

l h

hl

=

hh hh l

lhh

=

QlQhh

Note thatQl Qh= Ql (Qhh+ Ql) = Qhh; weve removedQhhheat from refrigerators inside.31


32/40

FIGURE 2. Problem 8.2(a)

Solution 3. Photon Carnot engine. Recall, photons are relativistic: = pc. Recallp = i. = s = pc = ksc =nsL

c.

Recalling that there are 2 polarization states for a photon in 3-dim. space,

U= (2)1

8 (4)

0

n2dnn

L

c enL c/ =

2c

L

0

n3dnexpc

L

n ==

2c

L

{n3expc

L n

L

c

0

0

3n2expc

L n

L

c

dn} =

=

2c

L

{(1)K

c

3

0

n2exp

cL

n

dn} =

=

2c

L

{(1)L

c

3{n2expc

L n

L

c

0

0

2nexp

cL

n

L

c

dn=

=

2c

L

(1)2L

c

23(2)

0

nexp

cL

n

L

c

dn=

=

2c

L (1)3

Lc

3

3(2)(1)

0

exp

c

L n

dn=

2c

L (1)3

Lc

4

3(2)1

exp

c

L n

0

=

= 6 L

2c

34 = 6

V

(2c)34 =U

To get the entropy, recall,U

V

= 1, and using this is usually the most direct way to obtain entropy.

= d =

dU

=

6V

(22c)343

d

=

6V

(22c)31

33 = () = 8V

3

(22c)3

Consider

Isothermal expansion: Helmholtz free energyFis needed.

F =U = 6V(22c)3

4 8V 3

(22c)3 = 2V

4

(22c)3

Then

p=

F

V

, N

= 24

(22c)3

W12 = p(V2 V1) = 24h

(22c)3(V2 V1)

12 = 83h

(22c)3(V2 V1)

Q12 =

=

84h(22c)3

(V2 V1)

Isentropic expansion: = V23h =V33l orV3 = V2hl

3.

So for this isentropic process,V23h =V

3,

U= 6V

(22c)3

V2

3h

V

4/3=

6(V23h)

4/3

(22c)3 V1/3

32


33/40

p=U

V = 6(V2

3h)

4/3

(22c)3

13

V4/3

=2(V2

3h)

4/3

(22c)3 V4/3

W23 =

pdV =

2(V2

3h)

4/3

(22c)3 V4/3dV =

2(V23h)

4/3

(22c)3 (3V1/3)V3V2

=6(V23h)4/3

(22c)3

1

V1/33

1V

1/32

=

= 6V24h(22c)3

1 l

h

Isothermal compression: W34 =

24l

(22c)3 (V4 V3) = 23hl

(22c)3 (V1 V2).34 =

83l(2c)3 (V4 V3) = 8

3h

(2c)3 (V1 V2).Isentropic compressiong:V4

3l =V1

3h or V4 = V1hl

3.

W41 =6(V4

3l)

4/3

(22c)3

1

V1/34

1V

1/31

=

6V14h(22c)3

1 l

h

W = 24h

(22c)3(V2V1)+ 6V2

4h

(22c)3

1 l

h

+

23hl(22c)3

(V1V2)+6V14h

(22c)3

1 l

h

=

84h(V2 V1)(22c)3

1 l

h

Qh=84h(V2 V1)

(2

c)3

=

W

Qh= 1

l

hSolution 4. Heat engine-refrigerator cascade. Consider the heat engine as a Carnot cycle.

W+ Wr = (h l)hwhereWr = work consumed by refrigerator.

h =Qh

h=

Qll

=l

This must be true for any heat engine undergoing Carnot cycle; furthermore, we can say its the most efficient heat engine

possible.

reversible refrigerator:QL+ Wr = QH, (byE-consv.)L= H=

QLL

= QHH , (by reversible condition)Note,Qlis energy transfer from heat engine tol reservoir.QLis energy transfer froml reservoir to refrigerator.QL Ql,otherwise, no cooling, no thermal energy extracted from l resevoir to lower its temperature. QL = Ql at equilibrium; nofurther cooling,r reached.Note that l is given as the environmental temperature. Assume refrigerator throws out QH heat into the environment. H=l. SinceQLheat inputed into refrigerator from a l reservoir now lowered tor,l r.

Wr = lr

QL QL=

lr

1

QL

since for a reversible refrigerator,L= H= QLL

= QHH .

= WQh

=

1 r

h

lr

1

QLQh

=

1 r

h

lr

1

rh

= 1 l

h

Combinations of reversible systems=reversible system.

Solution 5. Thermal pollution. Given Tl = 20 CTh= 500

C. Consider a Carnot cycle.

W= (h l)l = (h l) Qll

=

hl

1

Ql=

500

20 1

1500 MW = 36000MW

If improvements in hot-steam technology would permit raising Thby 100 C,

W =

600

20 1

1500 MW= (29)(1500 MW) = 43500 MW

There was a17.2 %increase in output.Solution 6. Room air conditioner.

33


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(a)

W = (h l) Qll

=

hl

1

Ql

P=

hl

1

dQldt

=

hl

1

A(h l) = PA

l= (h l)(h l) = 2h 2hl+ 2l

= 2l 2hl P

Al+

2h = 0

l= h+ P

2A

(h+ P

2A)2 2h

(b) ForTl= 17 C= 290 K,Th= 310 K,

A= P l

(h l)2 =(2 kW)(290 K)

(310 290)2 =580 103 W

400 K = 1450

W

K

Solution 7. Light bulb in a refrigerator

Carnot refrigerator draws 100 W. For any Carnot cycle,

W = (h l) Qll

=

hl

1

Ql=

1 l

h

Qh

Carnot refrigerator expelsQh thermal energy to hothenvironment andinputsQl thermal energy froml reservoir.Ql+ W =Qh

WorkWmust be drawn by Carnot refrigerator to do work. Suppose Carnot cycle part of the refrigerator must input in heatfrom light bulb to cool down its inside, i.e. consider Carnot refrigerator in equilibrium with light bulb, now inputting in heat

from light bulbQext, and drawing in work to expend out Qhthermal energy into the environment.

= Qext= QlW = Ql in this case, so

hl

1

Ql Ql= 0or

hl

2

Ql = 0

= l = h

2 =

300 K

2 = 150 K

Solution 8. Geothermal energy.

GivenQh= MCdTh,Tllower reservoir temperature stays constant. h decreasing,dh < 0.

W= (h l) Qhh

=

1 l

h

(M C) dh

kB

= W =

MC

kB

(h lln h)|fi =

MC

kB

lln

if

(i f)

ForM= 1017 g,C= 1 J/g K,Tl = 20 C= 293 K,Ti = 600 C= 873 K,Tf= 110 C= 383 KW= 2.486

1019 J

Note that1014 kW h= 1017 Js h 3600 sec1h = 3.6 1020 J.Solution 9. Cooling of nonmetallic solid toT = 0. Recall thatC= aT3 =

UT

V

. ThendQl = aT3ldTl. Nowdl < 0

sincel decreasing.For the refrigerator: Ql+ W =Qh.

dW = (h l) Qll

=

hl

1

(a3ldl)

1

k4B

=

ak4B

hl

1

3ldl=a

k4B(h

2l 3l)dl

W =a

k4B

h

1

33l

1

44l

0h

= aT3h12kB

=W

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9. GIBBSF RE EE NERGY AND C HEMICAL R EACTIONS

Solution 1. Thermal expansion near absolute zero

(a) G

N, p

= 2G

p

=

p

G

p

=V2G

p

p

=

V

p

=

V

p

=

p

GNp

= 2G

pN

N

=

p

N

Gp

=V 2G

Np

p

=

V

N

p

=

V

N

p

=

p

N

2G

N

N

=

N

2G

N

=

N

=

N

=

N

(b) = 1VV

p

= 1Vp

= 0 as 0since constant as 0by third law of thermodynamics.Solution 2. Thermal ionization of hydrogen.

(a) Givene + H+ H, note thate + H+ H= 0. Recall[e][H+]

[H] =K() =j njQj exp[jFj(int)/]

wherenQ=M22

3/2V.

For dissocation ofH intoe + H+ choose zero of internal energy of each composite particle (hereH) to concidewith energy of dissociated particles (hereH+, e) at rest; place energy of ground state of composite particle HatI,Iis energy required in reaction to dissociate composite particle into its constituents and is taken to be positive,i.e. the ionization energy.

K() = (ne)1 exp[Fint(e)/] (nH+)1 exp[Fint(H+)/](nH)1 exp((1)(Fint(H)/))

Note thatnH+ nH. Letne =nQ. Importantly, noteFint(e

) + Fint(H+) Fint(H) = I

Fint(H)is at a lower free energy thaneand H+.

= K() = nQeI/ = [e][H+]

[H] =nQe

I/

(b) By charge conservation,[e] = [H+], so that

[e] = [H]1/2n1/2Q exp(I/2)

Given[H] 1023 cm3,me = 0.511 MeV/c2,T= 5000 K,I= 13.6 eV ionization energy,[e] = (1023 cm3)1/2(2.92 10101/cm3/2)1/2 exp(13.6 eV /2kB5000 K) = 1.3 1015 cm3

Note thatH(exc)andHare just two different states of atomic hydrogen. Their concentrations must therefore beproportional to the probability of occurrence of these states, and the ratio of probabilities is the ratio of the respective

Boltzmann[H(exc)]

[H] =

p(H(exc))p(H)

IfH(exc)is the internal energy of the first excited state and His the internal energy of the ground state of atomic

hydrogen, we are given thatH(exc) H= 34I. We also need to take into account the fact that the first excitedelectronic state of hydrogen is 4-fold degenerate i.e. one 2s-orbital and three 2p-orbitals. 1 Therefore,

[H(exc)]

[H] =

p(H(exc))

p(H) =

4eH(exc)/

eH/ = 4e3I/4

[H(exc)] = 4[H]e3I/4 = 2.092 1013 cm3

1(from solutions to Homework 8, Ph12c, Caltech, June 6, 2008, by Prabha Mandayam, Heywood Tam)35


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[e]

[H(exc)]= 62

Solution 4. Biopolymer growth.

Recall thatG(N,p,) = N (p, ), sinceG was chosen to be an extensivequantity (it scales with size). For more than onechemical speciesG=jNjj .

dF = 0 for equilibrium, for constantP, .

j =chemical potential of species j ,j = (G/Nj),p.

Giveni jAj , e.g. H2+ Cl2 = 2HCl ,

dG= (jjj)d

NwheredNj =jdN,dG= 0 jjj = 0.

Recall the mass action law derivation: assume constituents act as ideal gases; j = (ln nj ln cj), nj concentration ofspeciesj;cj nQj Zj(int).

j

jln nj =j

jln cj =j

ln njj =j

ln cjj = lnj

njj = ln K()

j

njj =K() mass action law

(a) By mass action law, [monomer][Nmer]

[(N+1)mer] = [1][N][N+1] =KN.

[1][1]

[2] =K1

[1]2

[2]

[1][2]

[3] =

[1]3

[3] =K1K2

[1]j+1

[j+ 1]

[1][j+ 1]

[j+ 2] =

jl=1

KlKj+1 = [1]j+2

[j+ 2]=

j+1l=1

Kl

= [N+ 1] = [1]N+1/K1K2K3 . . . K N

(b) Recall thatK() =jnjQj

exp[jFj(int)/]

KN =nQ(N)nQ(1)

nQ(N+ 1) exp

FN

F1

exp

FN+1

=

nQ(N)nQ(1)

nQ(N+ 1) exp

(FN+ F1 FN+1)

wherenQ(N) =MN22

3/2andMNis the mass ofNmer molecules,FNis the free energy of oneNmer molecule.

(c) AssumeN 1 so nQ(N) nQ(N+ 1). Assume [1] = 1020 cm3. AssumeF = FN+1 FN F1 = 0,meaning zero free energy change in the basic reaction step. Were given the molecular weight of the monomer to be

200.

We want [N+1][N] at room temperature. NowKN nQ(1) =

M122

3/2.

[1][N]

[N+ 1]=nQ(1) =

M1

22

3/2or

[N+ 1]

[N] =

[1]

nQ(1)=

22

M1

3/21020 cm3

Note that

2(6.5821022MeV s)2

200(938MeV/c2)(0.8617104eV/K)(298K) 31010 cm/s

1 c 2

3/2

= 0.3627 1027 cm.

= [N+ 1][N]

= 3.627 108

(d) We want the condition

1< [N+ 1]

[N] =

[1]

nQ(1)exp

F

or ln

nQ(1)

[1]

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Kittel Kroemer Thermal Physics

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