Solutions of the Exercises - Springer

Appendix ASolutions of the Exercises

Section 1.6

Solution 1.6.1

1.

A.x; y/ D cos x sin y H)�@A

@y

�x

D cos x cos y ;

B.x; y/ D � sin x cos y H)�@B

@x

�y

D � cos x cos y

H) df not a total differential :

2.

A.x; y/ D sin x cos y H)�@A

@y

�x

D � sin x sin y ;

B.x; y/ D cos x sin y H)�@B

@x

�y

D � sin x sin y

H)�@A

@y

�x

D�@B

@x

�y

H) df total differential :

© Springer International Publishing AG 2017W. Nolting, Theoretical Physics 5, DOI 10.1007/978-3-319-47910-1

163

164 A Solutions of the Exercises

3.

A.x; y/ D x3y2 H)�@A

@y

�x

D 2x3y ;

B.x; y/ D �y3x2 H)�@B

@x

�y

D �2y3x

H)�@A

@y

�x

¤�@B

@x

�y

H) df not a total differential :

Solution 1.6.2 One method of solution utilizes the Jacobian determinant (seeexercise 1.7.1, Vol. 1). We choose here an alternative way.

We solve the functional relation for x and y:

x D x.y; z/ I y D y.x; z/

H) dx D�@x

@y

�z

dy C�@x

@z

�y

dz ;

dy D�@y

@x

�z

dx C�@y

@z

�x

dz :

Combining:

dx D�@x

@y

�z

��@y

@x

�z

dx C�@y

@z

�x

dz

�C�@x

@z

�y

dz

H) dx

�1�

�@x

@y

�z

�@y

@x

�z

�D dz

(�@x

@y

�z

�@y

@z

�x

C�@x

@z

�y

):

Two variables are freely selectable H) dx; dz arbitrary H) coefficients mustvanish:

1 ��@x

@y

�z

�@y

@x

�z

D 0 ;

�@x

@y

�z

�@y

@z

�x

C�@x

@z

�y

D 0 :

It follows from that:

1.�@x

@y

�z

D 1�@y@x

�z

:

A Solutions of the Exercises 165

2.�@x

@y

�z

�@y

@z

�x

D ��@x

@z

�y

1:D � 1�@z@x

y

H)�@x

@y

�z

�@y

@z

�x

�@z

@x

�y

D �1 :

Solution 1.6.3

1. Starting point is the two-dimensional path integral

I.C/ DZ B

A.C/

n˛.x; y/ dx C ˇ.x; y/ dy

o:

Choose

Z D ˛.x; y/ ex C ˇ.x; y/ ey

dr D dx ex C dy ey :

Therewith it is:

I.C/ DZ B

A.C/Z � dr :

Let C1 and C2 be two arbitrary paths in the xy-plane between A and B. C1 and�C2 then build a closed path C in the xy-plane:

I.C1/C I.�C2/ D I.C1/� I.C2/ DICZ � dr :

The right-hand side can be evaluated by the use of the Stokes theorem:

ICZ � dr D

ZFC

curl.Z/ � df DZFC

�@̌

@x� @˛

@y

�ez � df :

Hence:

I.C1/� I.C2/ DZFC

�@̌

@x� @˛

@y

�df :

This means:

@̌

@xD @˛

@yH) I.C1/ D I.C2/ :


Since C and therewith also FC are arbitrary, it can further be concluded:

I.C1/ D I.C2/ H) @̌

@xD @˛

@y:

Combining we get the condition for the path-independence of the integral:

I.C1/ D I.C2/ ” @̌

@xD @˛

@y:

That is equivalent to saying

Z � dr D ˛.x; y/ dx C ˇ.x; y/ dy � dF

represents a total differential. To say it differently:

dF D @F

@xdx C @F

@ydy D ˛.x; y/ dx C ˇ.x; y/ dy

is a total differential if and only if

@˛

@yD @̌

@x

2. One recognizes that

dF D ˛.x; y/ dx C ˇ.x; y/ dy

represents a total differential:

@˛

@yD 2yex D @̌

@x:

Therewith the integral IAB is path-independent. The problem formulation istherefore reasonable. dF is the total differential of

F.x; y/ D y2ex

so that it holds

IAB D F.1; 1/� F.0; 0/ D e :


3. Because of

@˛

@yD 2ex ¤ @̌

@xD y2ex

ıF is now not a total differential. The integral IAB is path-dependent. The problemformulation is thus not reasonable since the indication of the path is lacking.

Solution 1.6.4 We use the solution to Exercise 1.6.2:

�T D � 1V

�@V

@p

�T

D � 1V

�@p

@V

�T

��1;

ˇ D 1

V

�@V

@T

�p

D � 1V

1�@T@p

�V

�@p@V

�T

D � 1V

�@p@T

�V�

@p@V

�T

D �T

�@p

@T

�V

:

Solution 1.6.5

�@T

@p

�V

D �"�

@p

@V

�T

�@V

@T

�p

#�1D �

�@V@p

�T�

@V@T

p

D V �TV ˇ

(chain rule)

H)�@T

@p

�V

D V

nRC a p

n R

H) T D T.V; p/ D V p

n RC a

2 n Rp2 C G.V/ :

In addition it must hold:�@T

@V

�p

D 1

V ˇD p

n RD p

n RC 0C G0.V/

H) G0.V/ D 0 ” G.V/ D const D T0 :

Therewith the equation of state reads:

p V C 1

2a p2 D n R .T � T0/ :


Solution 1.6.6

1. The isotherm of the van der Waals equation of state

�p C a

n2

V2

�.V � n b/ D n R T

has an inflection point at the critical point:

�@p

@V

�Tc

D�@2p

@V2

�Tc

ŠD 0 I

p D n R T

V � n b� a

n2

V2;

�@p

@V

�T

D � n R T

.V � n b/2C 2 a

n2

V3;

�@2p

@V2

�T

D 2 n R T

.V � n b/3� 6 a

n2

V4:

This means:

2 an2

V3cD n R Tc.Vc � n b/2

I 6 an2

V4cD 2 n R Tc.Vc � n b/3

:

Division of the left side by the right side:

1

3Vc D 1

2.Vc � n b/ H) Vc D 3 n b ;

n R Tc D 2 an2

27 n3 b34 n2 b2 H) RTc D 8 a

27 b:

pc follows then directly from the van der Waals equation:

pc D 8 a n

27 b

1

2 n b� a

9 b2H) pc D a

27 b2:

Therewith the constants are fixed:

a D 9R

8 nTc Vc I b D Vc

3 n:

2. It follows from 1.:

pc Vc D 3

8n R Tc :


Division of the van der Waals equation by this expression:

�� C a n2

pc V2

� �v � n b

Vc

�D 8

3t ;

a n2

V2 pcD 27 n2b2

V2D 3

v2

H)�� C 3

v2

� �v � 1

3

�D 8

3t W

Law of corresponding states.

3.

�T D � 1V

�@p

@V

�T

��1:

According to 1.:

�@p

@V

�T

D � n R T�V � 1

3Vc2 C 9

4

R

nTc Vc

n2

V3:

At V D Vc:

�@p

@V

�T;VDVc

D �94

n R T

V2cC 9

4

n R TcV2c

H) �T .V D Vc/ D 4Vc

9 n R

1

T � Tc

�T

>! Tc�:

The add on T>! Tc is important since only then the presumption V D

Vc can be realized. �T diverges like .T � Tc/�1. At the critical point anarbitrarily small pressure change suffices to transform a finite volume of gas intoliquid (condensation!). Keywords: Phase transition of second order, universalityhypothesis, critical exponent � D 1 (Fig.A.1).

Fig. A.1


4. We have according to Exercise 1.6.4:

ˇ D �T

�@p

@T

�V

;

�@p

@T

�V

D n R

V � n bH)

�@p

@T

�VDVc

D 3 n R

2Vc:

ˇ shows thus the same critical behavior as �T :

ˇ D ˇ .T;V D Vc/ D 2

3

1

T � Tc:

Solution 1.6.7

1. Put

Nb D b

NAI Na D a

kBT N2A:

Then it holds:

p D kB T ��1 � � Nb�1 e�Na�

D kB T �

1X�D 0

Nb��! 0@ 1X�D 0

1

�Š.�Na/��

1A

D kB T�

1C

1XnD 1

Bn �n

!;

Bn DnX

�D 0

1

�ŠNbn�� .�Na/�

H) B1 D Nb � Na D 1

NA

�b � a

R T

�:

Boyle temperature TB:

B1 .TB/ŠD 0 H) TB D a

R b:


2. Van der Waals:

p D N kB T.V � n b/�1 � an2

V2

D � kB T�1 � Nb ��1 � kB T Na �2

D � kB T

1C

1XnD 1

Bn�n

!

H) B1 D Nb � Na D 1

NA

�b � a

R T

�I Bn D

�b

NA

�n

for n � 2 :

Dieterici:B1 as for the van der Waals model. Deviations appear not before B2:

Bn DnX

�D 0

1

�Š

�b

NA

�n�� a

kB T N2A

��:

At high temperatures only the � D 0-term survives. Then one gets the sameresult as for the van der Waals model.

Meaning of the parameters:Van der Waals:

a �2 � intrinsic pressure;

bN � proper volume:

Dieterici:

vij W interaction potential;

r W distance:

a. b is, as in the van der Waals model, a measure of the proper volumes of themolecules, which are considered in the ideal gas as mathematical points. Theparticle repulsion gives rise to a hard core-potential (Fig.A.2).

Fig. A.2


b. At larger distance an attraction of the particles sets in due to a mutual electricpolarization of the atom shells, being connected to a tendency to a bound state,in which case one would no longer need a wall of the container. That means inany case a pressure decrease. This tendency is described in the Dieterici gasapproximately by the exponential function (Fig.A.2):

a � � average interaction energy.activation energy/:

3. The equation of state

p D kB T ��1 � � Nb�1 e�Na �

becomes of course physically absurd as soon as a particle densification (”increase of �) leads, because of the exponential function, to a pressure reduction.Sign-expectation:

�@p

@�

�T

� 0 :

Because of �T D � 1V

�@V@p

�Tit is (N=const):

�@p

@�

�T

D�@p

@V

�T

�@V

@�

�T

D �1V �T

�@�

@V

�T

��1D �1

V �T

� N

V2

��1

H)�@p

@�

�T

D 1

� �T:

It is to require (stability criterion):

�T � 0 ;

otherwise the system would collapse and therewith would be unstable.4.

�@p

@�

�T

D p

�C Nb1 � � Nb p � Na p D p

1

��1 � � Nb � Na

!

H)�@p

@�

�T0

ŠD 0 H) kB T0.�/ D a��1 � � Nb a� D a

N2A

H) kB T0.�/ W parabola with zeros at � D 0 and � D 1

Nb .Fig:A.3/


Fig. A.3

As unphysical one has to mark the region

� >1

Nb ;

since then T0 < 0 and .@p = @�/T < 0, respectively. Approximately that meansthat � > 1 = Nb, i.e. the proper volume of the molecules would be bigger than thetotal gas volume.

Maximum:

kBdT0d�

D a� �1 � � Nb � a� Nb � D a� �1�2 � Nb ŠD 0 H) �CD 1

2 Nb ;

kB

�d2T0d�2

��D �C

D �2 a� Nb < 0 H) maximum,

kB Tc D kB T0 .�c/ D a�

4 Nb :

The comparison with 1. yields:

TB D 4 Tc :

The material constants a, b are determinable from the critical data!

a� D a

N2AD 2 kB Tc

�c;

Nb D b

NAD 1

2 �c:

5. The region � > 2 �c ” p < 0 is unphysical.We investigate at first

�@p

@�

�T

D p

�1

�.1 � Nb�/ � a�

kBT

�


with respect to zeros, for which it must obviously hold:

�.1 � Nb�/ ŠD kBT

a� D 1

4NbT

Tc:

This quadratic equation has two solutions:

�1;2 D 1

2Nb

1˙

s1 � T

TC

!:

T > Tc

In this case�@p@�

�Thas no real zero. Since otherwise

�@p

@�

�T

.� ! 0/ ! C1 ;

it must be valid for all �:

�@p

@�

�T

> 0 :

p is thus in any case a monotonically increasing function of � with p.� D 0/ D 0.

T < Tc Now�@p@�

�Thas two real(!) zeros at �1;2.

p.�/ is for small � monotonically increasing. Hence the first zero correspondsto a maximum, the second to a minimum of p.�/. In between there is anunphysical region since .@p = @�/T < 0.

T D Tc

�@p

@�

�TC

.� D �c/ D 0 ;

�@2p

@�2

�T

D p

�1

�.1 � � Nb/ � a�

kB T

�2� p

1 � 2 � Nb�2�1 � � Nb2 ;�

@2p

@�2

�TC

.� D �c/ D 0 : inflection point

For T < TC a phase transition appears H) gas liquefaction! The unphysicalregion is corrected by the same Maxwell construction .A D BŠ/ as for the vander Waals gas (Fig.A.4).


Fig. A.4

A

B

gas liquid

p

Cp

×

CTT >

CTT >

CTT =

0T

C C2

Solution 1.6.8

�T D � 1V

�@V

@p

�T

I ˇ D 1

V

�@V

@T

�p

;

ideal gas: p V D n R T

H)�@V

@p

�T

D �n R T

p2D �V

p;

�@V

@T

�p

D n R

pD V

T

H) � idT D 1

pI ˇid D 1

T:

1.

p .V � n b/ D n R T

H)�@V

@p

�T

D �n R T

p2D �1

p.V � n b/ ;

�@V

@T

�p

D n R

pD 1

T.V � n b/

H) �T D 1

p

�1 � n b

V

�D � idT

�1 � n b

V

�;

ˇ D 1

T

�1 � n b

V

�D ˇid

�1 � n b

V

�:


2.

p V D n R T .1C A1.T/ p/

H)�@V

@p

�T

D �n R T

p2D �V

pC n R T A1.T/

p

H) �T D � idT

�1 � n R T A1.T/

V

�;

�@V

@T

�p

D n R

pC n RA1 C n R T

dA1dT

D V

TC n R T

dA1dT

H) ˇ D ˇid

�1C n R T2

V

dA1dT

�:

3.

p V D n R T

�1CB1.T/

V

�;

�@. p V/

@p

�T

D VCp

�@V

@p

�T

D �n R T B1V2

�@V

@p

�T

H)�@V

@p

�T

D �V

p

�1Cn R T B1

p V2

��1

H) �T D � idT

�1Cn R T B1.T/

p V2

��1;

p

�@V

@T

�p

D n R

�1CB1

V

�Cn R T

V

dB1dT

�n R T B1V2

�@V

@T

�p

H)�@V

@T

�p

�pCn R T B1

V2

�D p V

TCn R T

V

dB1dT

H) ˇ D ˇid1C nR T2

pV2dB1dT

1C n RTpV2

B1:

Solution 1.6.9

ıW D B0 dm W m W magnetic moment,

dm D V dM W M W magnetization, V D const

H) ıW D �0 V H dM :


Fig. A.5

Curie law: M D CT H

.ıW/T D �0CV

TH dH

H) �W12 DH2Z

H1

.ıW/T D �0 CV

T

1

2

�H22 � H2

1

D �0V T

2C

�M22 � M2

1

:

Solution 1.6.10 According to equation (2.51), Vol. 3 it holds for the electric fieldinside the capacitor (Fig. A.5)

E D Q

"0F0ex ; F0 W area of the plates.

Capacity ((2.55), Vol. 3):

C D "0F02 a

:

Energy within the capacitor ((2.58), Vol. 3):

W D 1

2

Q2

CD Q2

2 0

2 a

F0:

Force on the plates of the capacitor (section 2.2.1, Vol. 3):

F.CQ/ D QE.x�/ D Q2

2 "0 F0ex ;

F.�Q/ D �QE.xC/ D � Q2

2 "0 F0ex :

Change of the capacity:

C D "0 F0xC � x�

:


Let x� be variable:

dC

dx�D "0 F0.xC � x�/2

D C

xC � x�H) dx� D dC

C.xC � x�/

dx� > 0 means distance-reduction leading to dC > 0. Analogously one finds dxC.

dx˙ D �dC

C.xC � x�/ :

1. Mechanical work due to distance-reduction:

ıW D �F.CQ/ dx� D �F.CQ/ ex dx

D � Q2

2 "0 F0

.xC � x�/C

dC D �12

Q2

C2dC ;

ıA D �ıW D 1

2

Q2

C2dC :

That is the work done on the system from the outside.2. Change of the field energy:

before:

Wb D 1

2

Q2

C;

after:

Wa D 1

2

Q2

C C dC;

dW D Wa � Wb D Q2

2

�1

C C dC� 1

C

�� 1

2

Q2

C2dC :

The change of the field energy thus corresponds to the work done from the outsideon the system:

ıA C ıW D 0 :

3. The last relation is always valid, i.e., for dC > 0 as well as for dC < 0. The statechange is therefore reversible!


Solution 1.6.11

1. We have�@B0@m

�T

D ˛T

�1

m0 C mC 1

m0 � m

��

D ˛T2m0

m20 � m2� � D 2˛

m0T

1 � m2

m20

��1� �

D 2˛

m0

T

1 � m2

m20

��1� TC

!:

Instabilities follow therewith for

T

1 � m2

m20

��1< TC ” jmj < m0

s1 � T

TC:

This means that for T < TC and jmj < m0q1 � T

TCthe isotherms of the model-

equation of state become unphysical. For T > TC and arbitrary .�m0 � m �Cm0/ there are basically no instabilities. The isotherms are then physical in thewhole allowed space.

2. Limiting curve:

mS.T/ D ˙m0

s1 � T

TC:

3. The ferromagnet is characterized by a ‘spontaneous’ magnetic moment, whichis not enforced by an external field. We have to therefore seek solutions of theequation

B0.T;m/ D 0 :

One recognizes directly from the equation of state that the non-magnetic casem D 0 always represents a possible solution. That corresponds to paramag-netism. Of special interest is therefore the question whether there does exist anadditional solution

m D mS ¤ 0 :


(a) T > TC

�@B0@m

�T

D 2˛

m0

T

1 � m2

m20

��1� TC

!

>2˛

m0

T

1 � m2

m20

��1� T

!> 0 :

B0 D B0.T;m/ is therewith for T > TC a bijective monotonic increasingfunction of m, and m D 0 is the only zero of B0.T;m/ for T > TC .

(b) T < TCAt first it is clear that even now m D 0 represents a possible solution. The

equation of state shows in addition that, if mS ¤ 0 is a further solution, then�mS is also a zero. B0-zeros have to fulfill the equation (Fig.A.6):

0ŠD T

2TCln

�m0 C m

m0 � m

�„ ƒ‚ …

f .m/

� m

m0„ƒ‚…g.m/

As to be seen in Fig.A.6 g.m/ is a straight line through the origin with theslope 1=m0 and the end points g.˙m0/ D ˙1. f .m/ has the properties f .0/ D0; f .˙m0/ D ˙1. The slope is given by:

f 0.m/ D T

2TC

�1

m0 C mC 1

m0 � m

�Õ f 0.0/ D T

TC 1m0

:

Because of T < TC the slope of f at the origin is smaller than that of g.Fig. A.6 illustrates that then two additional intersection points of f and gmust exist and therefore zeros of B0.T;m/ at m D ˙mS.

4. Isotherms:The very simple model already yields a qualitatively accurate description of

the ferro-/paramagnet (Fig.A.7). However, below TC we observe an unphysicalbehavior (see 1.). There the curves are to be replaced by a linear piece of the

Fig. A.6

·

·m

f g,

–m0 f

gT= const

×××–mS +m0

+mS×


Fig. A.7

m-axis from �mS to CmS, similar to the Maxwell construction for the van derWaals gas (Fig. 1.4).

Solution 2.9.1

1. First law of thermodynamics (gas!):

dU D ıQ C ıW D ıQ � p dV ;

U D U.T;V/ H) dU D�@U

@T

�V

dT C�@U

@V

�T

dV

H) ıQ D�@U

@T

�V

dT C�@U

@V

�T

C p

�dV :

Integrability condition:

@

@V

�@U

@T

�V

�T

ŠD�@

@T

�@U

@V

�T

C p

��V

H)@

@V

�@U

@T

�V

�T

ŠD@

@T

�@U

@V

�T

�V

C�@p

@T

�V

:

dU total, therefore

0ŠD�@p

@T

�V

:

That is a contradiction, hence ıQ can not be a total differential!2. Ideal gas:

�@U

@V

�T

D 0 (Gay-Lussac),

�@U

@T

�V

D CV D const

H) ıQ D CV dT C p dV :


a) � D �.T/ to be chosen so that

dy D �ıQ total differential

”@

@V.CV�/

�T

ŠD@

@T.� p/

�V

” 0 D@

@T.� p/

�V

D d�

dTp C �

�@p

@T

�V

” �n R

VD �n R T

V

d�

dT

” �C Td�

dTD d.�T/ D 0 ” �T D const :

The constant is arbitrarily chosen equal to 1 H) integrating factor: �.T/ D1 = T. We have therewith

dy D ıQ

TD dS total differential.

Entropy of the ideal gas:

S.T;V/� S .T0; V0/ D CV

TZT0

dT 0

T 0 CVZ

V0

p

TdV 0

D CV lnT

T0C n R ln

V

V0:

Test:�@S

@T

�V

D CV

TI

�@S

@V

�T

D n R

VD p

T:

b) � D �.V/ to be chosen so that

dy D �ıQ D .�CV / dT C .� p/ dV

becomes a total differential.


Integrability condition:

CVd�

dVD �

�@p

@T

�V

D �n R

V

” CV

n R

d�

�D dV

V:

Ideal gas:

n R D Cp � CV D CV .� � 1/ I � D Cp

CV

H) d ln� D .� � 1/ d lnV D d lnV� � 1

” d ln��V1� �

D 0 ” �V1� � D const ; e.g. D 1

H) �.V/ D V� � 1 H) dy D V� � 1 ıQ I

dy D CV V� � 1

�dT C .� � 1/T

VdV

�:

Solution 2.9.2

�W D �2Z1

p.V/ dV D �const

2Z1

dV

Vn; .n ¤ 1/

D const

1

Vn�12

� 1

Vn�11

�1

n � 1 ;

const D p1 Vn1 D p2V

n2

H) �W D 1

n � 1. p2V2 � p1V1/ :

Ideal gas:

�U D CV .T2 � T1/ D CV

N kB. p2V2 � p1V1/

H) �Q

��WD �U ��W

��WD 1 � �U

�WD 1C .1 � n/

CV

n kBD const :


Special case n D 1:

p V D const ” isotherm of the ideal gas

H) �U D 0 H) �Q

��WD 1 :

Solution 2.9.3 Integration:

ln p0ˇ̌ˇpp0

D a lnV 0ˇ̌ˇVV0:

This means:

p.V/ D p0

�V

V0

�a

Õ p V D p0Va0

VaC1 :

With the equation of state of the ideal gas

pV

TD p0V0

T0

it follows

V D p0p

VaC1

Va0

D V

V0

T0T

VaC1

Va0

Õ VaC1 D VaC10

T

T0

and therewith

V.T/ D V0

�T

T0

� 1aC1

.V0 D V.T0// :

First law of thermodynamics for closed systems:

ıQ D dU C p dV D�@U

@T

�V

dT C�@U

@V

�T

dV C p dV :

Ideal gas, Gay-Lussac:�@U@V

T

D 0:

ıQ D CVdT C p dV :

Heat capacity:

Ca D�ıQ

dT

�a

D CV C p

�@V

@T

�a

D CV C p

a C 1

V

TD CV C nR

a C 1:


Special cases:

a D 0 W p.V/ D p0 D const. isobaric

Ca D CV C nR D Cp

a ! 1 W V.T/ D V0 D const. isochoric

Ca D CV

a ! �1 W pV D p0V0 D const. isothermal

a ! �� W pV� D p0V�0 D const. adiabatic

Ca D CV � nR

� � 1

D CV � CVnR

Cp � CV

D CV � CV D 0

Solution 2.9.4

1. According to (2.59) it is generally valid:

�@U

@V

�T

D T

�@p

@T

�V

� p :

Therewith:�@U

@V

�T

D 0 Õ�@p

@T

�V

D p

T:

This means:

p D p.T;V/ D T f .V/ :

Valid in particular for the ideal gas!2.

�@U

@V

�T

D bp D T

�@p

@T

�V

� p

Õ b

Vf .T/ D T

Vf 0.T/ � 1

Vf .T/

Õ 1

V.b C 1/ f .T/ D T

Vf 0.T/

Õ df

fD .b C 1/

dT

T


Õ ln f .T/ D .1C b/ lnT C c

Õ f .T/ / T1Cb or f .T/ D p0V0

�T

T0

�1Cb

:

Solution 2.9.5 Adiabatic means ıQ D 0 and therewith dS D 0 (isentropic). For theisotherm it is dT D 0. If there were, as plotted in Fig. A.8, two intersection pointsA and B then it would hold obviously:

IABA

ıW D �IABA

p dV ¤ 0 :

First law of thermodynamics:

ıW D dU � ıQ D dU � T dS :

It follows therewith:IABA

ıW DIABA

.dU � T dS/ D �IABA

T dS (since dU total differential)

D �ZC1

T dS �ZC2

T dS D �ZC1

T dS (since C2 adiabatic, isentropic)

D �T�ZC1

dS (since C1 isotherm)

D �T� .S.A/� S.B// (since dS total differential).

A and B lie on the same isotropic, i.e., S.A/ D S.B/. The integral thus vanishes:

IABA

ıW D 0 :

This conflicts with Fig.A.8. The figure must therefore be wrong. Isotherms andadiabatics never intersect twice!

Fig. A.8


Solution 2.9.6

1.

Cm D�@U

@T

�m

follows directly from the first law, see (2.19).We exploit (2.20):

CH D Cm C�@U

@m

�T

� �0H

��@m

@T

�H

:

Curie law:

m D V M D VC

TH

H)�@m

@T

�H

D �VC

T2H D �V

M2

CH:

Above inserted, this yields:

CH D�@U

@T

�m

C�@U

@m

�T

�@m

@T

�H

C �0V

CM2 ;

U D U.T;m/ H) dU D�@U

@T

�m

dT C�@U

@m

�T

dm :

This means:�@U

@T

�H

D�@U

@T

�m

C�@U

@m

�T

�@m

@T

�H

:

Hence we are left with the assertion:

CH D�@U

@T

�H

C �0V

CM2 :

2. We can write:�@m

@H

�ad

D�@m

@T

�ad

�@T

@H

�ad:

We determine the two factors separately:


a) First law of thermodynamics:

dU D ıQ C �0 H dm

D�@U

@T

�m

dT C�@U

@m

�T

dm ;

ıQ D 0, because the state change is adiabatic!

H)��

@U

@T

�m

dT

�ad

D��0H �

�@U

@m

�T

�dm

�ad

:

This means:�@m

@T

�ad

D Cm

�0H � �@U@m

T

:

b) First law of thermodynamics:

0 D ıQ D dU � �0H dm ;

now: U D U.T;H/I m D m .T;H/.It follows:

0 D��

@U

@T

�H

� �0H

�@m

@T

�H

�dT C

��@U

@H

�T

� �0 H�@m

@H

�T

�dH :

According to part 1. the first bracket is equal to CH :

CH dT D��0H V

�@M

@H

�T

��@U

@H

�T

�dH

D��0 V M.T;H/ �

�@U

@H

�T

�dH :

From this it can be read off:

�dT

dH

�ad

D �0 m .T;H/� �@U@H

T

CH:

The combination of a) and b) leads to the assertion:

�@m

@H

�ad

D�@m

@T

�ad

�@T

@H

�ad

D Cm

CH

�0 m � �@U@H

T

�0H � �@U@m

T

:


Solution 2.9.7

1. The wall is thermally insulating, so the process running in the right chamber isadiabatic. By use of the adiabatic equations of the ideal gas (2.24), (2.25) we get(� D Cp =CV given!):

�Wr D �VrZ

V0

p dV D �const1

VrZV0

dV

V�D �const1

1 � ��V1� �r � V1� �

0

�;

const1 D p0 V�0 D pr V

�r

H) �Wr D � 1

1 � � . prVr � p0V0/ D �N kB1 � � .Tr � T0/ :

We still fix Tr by the initial data:

T�r p1��r D T�0 p

1� �0 H) Tr D T0

�p0pr

�.1� �/ = �

H) �Wr D �N kB T01 � �

"�p0pr

�.1� �/ = �

� 1#

H) �Wr D N kB T0� � 1

�3.� � 1/ = � � 1

> 0 .because � > 1/ Š

Hence work is done on the right system! Trivially:�Qr D 0.2.

T�r p1��r D T�0 p

1� �0 I pr D 3 p0 H) Tr D T0 3

.�� 1/ = � > T0 :

Equation of state:

Tl D pl Vl

N kBD pl

N kB.2V0 � Vr/ :

Equilibrium:

pl D pr D 3 p0

H) Tl D 3p0 2V0N kB

� Tr D 6 T0 � Tr

H) Tl D T0�6 � 3.� � 1/ = �

:


3.

First law: �Ql D �Ul ��Wl ;

ideales Gas: �Ul D CV .Tl � T0/ :

�Wl takes care of the energy change on the right-hand side, i.e.:

�Wl D ��Wr D ��Ur .adiabatic on the right side/

H) �Wl D �CV .Tr � T0/

H) �Ql D CV .Tl � T0 C Tr � T0/ D CV.Tl C Tr � 2 T0/ ;

Tl C Tr D 6 T0 H) �Ql D 4CV T0 :

Solution 2.9.8

1.

ıW D �p dV; V D RT

p; dV D �RT

p2dp

H) �W D RT

p1Zp0

dp

pD RT ln

p1p0

D �RT ln 20 < 0 :

The system therefore carries out work:

R D 8:315J

centigrade moleH) �W D �7:298 103 J :

2.

T D const ; ideal gas H) dU D 0

H) ıQ D �ıW H) �Q D j�Wj :

3. State change now adiabatic:

p V� D C H) dV D � 1�C1 = �

1

p1 = �C1 dp

H) �W D C1 = �

�

p1Zp0

dp

p1 = �D C1 = �

��1 � 1

�

� hp1� 1 = �1 � p1� 1 = �

0

i:


Now we have

C1 = � p�1 = �1; 0 D V1; 0

H) �W D p1V1 � p0V0� � 1

;

p1V�1 D p0V

�0 H) V1 D V0

�p0p1

�1 = �

H) �W D V0� � 1

"p1

�p0p1

�1 = �� p0

#D RT0

� � 1

"�p0p1

�1 = � � 1

� 1

#;

CV D 5

2R H) � D Cp

CVD 1C R

CVD 1:4 ;

T0 D 293K I p0 D 20 p1

H) �W D �3:503 103 J :

4. Adiabatic process:

�W D �U D CV .T1 � T0/

H) T1 D T0 C �W

CVD 124:5K :

Solution 2.9.9 A displacement by z means a volume change of

�V D z F .F D cross section area:/

By that a difference �p between external and internal pressure arises which takescare for a repelling force in z-direction:

K D F�p :

The state change of the ideal gas takes place adiabatically:

p V� D const :

By that we calculate �p:

d . p V� / D 0 D dp V� C � p V� � 1 dV H) dp D �� pV

dV :

This means

�p D �� pV�V D �� p

Vz F


and therewith

K D �� pV

F2z D �k z :

If m is the mass of the sphere then the oscillation period reads:

D 2�

rm

kD 2�

smV

� p F2H) � D 4�2mV

pF2 2:

Solution 2.9.10 First we determine

1

TıQrev

for both the systems as functions of T and V . For that first we use therefore the firstlaw of thermodynamics,

ıQ D dU C p dV ;

and the presumption,

U D U.T/ H) dU D CV.T/ dT ;

for a reversible state change:

1

TıQrev D CV .T/

dT

TC p

TdV ;

p.A/ D ˛N T

V2I p.B/ D

�ˇN

VT

�1 = 2

H) 1

TıQrev .A/ D CV .T/

dT

TC ˛ N

dV

V2;

1

TıQrev .B/ D CV .T/

dT

TC�ˇ

N

T V

�1 = 2dV :

For the entropy to be a state quantity,

dS D 1

TıQrev

must be a total differential. For that we check the integrability conditions:

(A)

@

@V

�CV.T/

T

�ŠD @

@T

�˛N

V2

�I obviously fulfilled!

H) entropy as state quantity definable!


(B)

@

@V

�CV.T/

T

�ŠD @

@T

�ˇ

N

V T

�1 = 2

” 0 D � 1

2 T

�ˇ N

T V

�1 = 2:

This is a contradiction. An entropy is thus not definable for system B. System Btherefore can not exist!

Solution 2.9.11

1. According to (2.59) it is:

�@U

@V

�T

C p

�D T

�@p

@T

�V

:

This means according to (2.58):

dS D CV.T;V/

TdT C

�@p

@T

�V

dV :

Integrability condition for dS:

1

T

�@

@VCV .T;V/

�T

ŠD�@2p

@T2

�V

D�@˛.V/

@T

�V

D 0

” CV .T;V/ D CV .T/ :

2.

dS D CV .T/

TdT C

�@p

@T

�V

dV :

The van der Waals gas fulfills the presumptions of part 1.:

�p C a

n2

V2

�.V � n b/ D n R T

H) p D Tn R

V � n b� a

n2

V2H)

�@p

@T

�V

D n R

V � n b

H) dS D CV

TdT C n R

V � n bdV

CV according to the presumption T-independent


H) �S D S.T;V/ � S.T0;V0/

D CV lnT

T0C n R ln

V � n b

V0 � n b:

3. U D U.T;V/

dU D�@U

@T

�V

dT C�@U

@V

�T

dV ;

.see 1:/

�@U

@V

�T

D T

�@p

@T

�V

� p .D 0 for the ideal gas/

2:D p C an2

V2� p D a

n2

V2

H) dU D CV dT C a n2dV

V2

H) U D U.T;V/ D CV T � an2

VC const :

The interaction of the gas-particles thus provides a volume-dependence of theinternal energy, which, in the last analysis, is responsible for the temperaturechange due to the expansion:

U .T2;V2/� U .T1;V1/ D CV .T2 � T1/ � a n2�1

V2� 1

V1

�ŠD 0

H) �T D a n2

CV

�1

V2� 1

V1

�:

4. According to 2. it holds for reversible adiabatic state changes of the van derWaalsgas .�S D 0/:

CV lnT C n R ln.V � n b/ D const1

” lnhT .V � n b/

n RCV

iD const2

H) T .V � n b/n RCV D const3 :

Compare this adiabatic-equation with that of the ideal gas T V� � 1 D const (� DCp

CVD 1C n R

CV).

Insertion into the equation of state:

�p C a

n2

V2

�.V � n b/

n RCCVCV D const4 :


For the ideal gas these adiabatic-equations read: p V� D const.—Finally theequation of state still yields:

TnRCCV

CV

�p C a

n2

V2

�� n RCV D const5 :

This is to be compared with T� p1� � D const, valid for the ideal gas!

Solution 2.9.12 Thermal equation of state:

V D V0 � ˛p C �T :

Cp; ˛; � are known, as material-specific parameters.�@p

@T

�V

D �

˛I�@V

@T

�p

D � :

Equation (2.65):

Cp � CV D�@U

@V

�T

C p

��@V

@T

�p

D T

�@p

@T

�V

�@V

@T

�p

means in our case here:

Cp � CV D �2

˛T

and therewith for the heat capacity CV :

CV D Cp � �2

˛T :

Internal energy:

dU D CV dT C�T

�@p

@T

�V

� p

�dV

D�Cp � �2

˛T

�dT C V � V0

˛dV :

This can easily be integrated:

U.T;V/ D U0 C CpT � �2

2˛T2 C .V � V0/2

2˛

U.T; p/ D U0 C CpT C ˛

2p2 � �Tp :

In the last step we merely inserted the thermal equation of state.


Solution 2.9.13

1. Thermal equation of state:

p D nRT

V � nb� a

n2

V2:

It follows therewith:�@p

@T

�V

D nR

V � nb:

Furthermore we utilize:

�@V

@T

�p

D ��@T

@p

�V

�@p

@V

�T

��1D �

�@p

@T

�V

�

@p

@V

�T

��1:

It holds for the van der Waals gas:

�@p

@V

�T

D � nRT

.V � nb/2C 2a

n2

V3:

It follows then:�@V

@T

�p

D nR

nRT

V � nb� 2a

n2

V3.V � nb/

��1:

Difference of the heat capacities:

Cp � CV(2.65)D T

�@p

@T

�V

�@V

@T

�p

D n2R2T

nRT � 2a n2

V3.V � nb/2

:

Correction with respect to the ideal gas (a and b small):

Cp � CV D nR

1 � 2a n2

V3.V�nb/2

nRT

� nR

�1C 2a

n2

V3.V � nb/2

nRT

�

� nR�1C 2a

n

VRT

�:

For the ideal gas it is Cp � CV D nR.2. Thermal equation of state:

p D nRT

V � nb� a

n2

V2:


We apply (2.58), (2.59):

dS D 1

TCV dT C

�@p

@T

�V

dV :

Adiabatic-reversible means dS D 0. Therefore:�@T

@V

�S

D ��@p

@T

�V

T

CVD �nRT

CV.V � nb/:

Separation of the variables:

dT

TD � nR

CV

dV

V � nbÕ d ln T D � nR

CVd ln.V � nb/ :

Solution:

T D T0

�V � nb

V0 � nb

�� nRCV

:

Solution 2.9.14 First law of thermodynamics

ıQ D dU � ıW D dU � ' dq :

Choose U D U.T; q/, i.e.

dU D�@U

@T

�q

dT C�@U

@q

�T

dq :

Therewith:

ıQ D�@U

@T

�q

dT C��

@U

@q

�T

� '�dq :

Second law of thermodynamicsreversible process: ıQ D T dS, hence:

dS D 1

T

�@U

@T

�q

dT C 1

T

��@U

@q

�T

� '�dq :

dS and dU are total differentials. Maxwell relation:

0 D � 1T

�@'

@T

�q

� 1

T2

��@U

@q

�T

� '

�:


That means:�@U

@q

�T

D ' � T

�@'

@T

�q

:

Heat quantity:

ıQ D�@U

@T

�q

dT � T

�@'

@T

�q

dq :

Isothermal processing:

.ıQ/T D 0 � T

�@'

@T

�q

dq :

It then remains:

.�Q/is DZ qe

qa

ıQ D �Td'

dT

Z qe

qa

dq D �Td'

dT.qe � qa/ :

Solution 2.9.15

1. Basic relation of thermodynamics:

dS D 1

TdU C p

TdV D CV

dT

TC n R

dV

V

H) S D S0 C CV lnT C n R lnV :

2. We determine by use of 1. the temperature as function of S and V:

lnT D 1

CV.S � S0/ � n R

CVlnV

H) T Dexp

�1CV.S � S0/

�

VnRCV

H) U D CV

VnRCV

exp

1

CV.S � S0/

�C U0 D U.S;V/ :

U as function of the variables S and V even for the ideal gas depends on thevolume V . Note that this is not a contradiction to the Gay-Lussac experiment.


3.

free expansion: U .T2;V2/ D U .T1;V1/ ;

ideal gas: U D U.T/ H) �T D 0

H) with 1.: �S D n R lnV2V1

:

Solution 2.9.16 Internal energy:

U.T;V/ DTZdT 0 CV.T

0/C '.V/ D 3

2N kB T � N kB

N

V

�T2

df

dT

�C '.V/

H)�@U

@V

�T

D N2 kB T2

V2df

dTC ' 0.V/ :

It follows according to (2.59):

�@U

@V

�T

D T

�@p

@T

�V

� p ;

�@p

@T

�V

D N kBV

�1C N

Vf .T/

�C N2 kB T

V2df

dT

H)�@U

@V

�T

D N2 kB T2

V2df

dT:

Comparison with the above expression:

' 0.V/ D 0 ” '.V/ D const D U0

H) U D 3

2N kB T � N kB

N

V

�T2

df

dT

�C U0 :

We are left with the determination of the entropy!

(2.58) C (2.59) H) dS D CV

TdT C

�@p

@T

�V

dV :

Hence:�@S

@T

�V

D CV

TI

�@S

@V

�T

D�@p

@T

�V

H) S.T;V/ DTZdT 0CV .T 0/

T 0 C .V/ ;


CV

TD 3

2

N kBT

� N kBN

V

1

T

d

dT

�T2

df

dT

�

D 3

2

N kBT

� N kBN

V

d

dT

�f C T

df

dT

�

H) S.T;V/ D 3

2N kB ln T � N2 kB

V

�f .T/C T

df

dT

�C .V/

H)�@S

@V

�T

D N2 kBV2

�f .T/C T

df

dT

�C 0.V/ :

Otherwise it holds also:

�@S

@V

�T

D�@p

@T

�V

D N kBV

�1C N

Vf .T/

�C N2 kB T

V2df

dT:

The comparison yields:

0.V/ D N kBV

H) .V/ D N kB lnV C S0 :

This means eventually:

S.T;V/ D 3

2N kB lnT C N kB lnV � N2 kB

V

�f .T/C T

df

dT

�C S0 :

Solution 2.9.17

1. p D const D p0

�W D �V2Z

V1

p dV D �p0 .V2 � V1/ ;

ıQ D CV dT C p dV I dT D p0n R

dV

H) ıQ D�CV

n RC 1

�p0 dV D �

� � 1p0 dV

H) �Q D �

� � 1p0 .V2 � V1/ :


Reversible state change:

dS D CVdT

TC p

TdV I dT

TD dV

V

H) dS D .CV C n R/dV

VD Cp

dV

V

H) �S D Cp lnV2V1:

2. T D const D T0

�W D �n R T0

V2ZV1

dV

VD �n R T0 ln

V2V1

�U D 0 ; since isotherm H) �Q D ��W ;

reversible with T D const H) �S D 1

T0�Q D n R ln

V2V1

:

3. Adiabatic:

in addition reversible H) �S D �Q D 0 ;

�W D �V2Z

V1

p dV D �C

V2ZV1

dV

V�D 1

� � 1

C

V� � 12

� C

V� � 11

!

D 1

� � 1 . p2V2 � p1V1/ I p2 D p1

�V1V2

��

H) �W D p1� � 1

V2

�V1V2

�� V1

�:

Solution 2.9.18

�W D 0 ; because �V D 0 ;

ıQ D CV dT C p dV D CV dT D CV

n RV0 dp

H) �Q D 1

� � 1V0 . p2 � p1/ ;

dS D CV

TdT H) �S D

p2Zp1

CVdp

pD CV ln

p2p1:


Solution 2.9.19

1. The general equation (2.59) is valid:

�@U

@V

�T

D T

�@p

@T

�V

� p ;

U.T;V/ D V ".T/

H) ".T/ D ˛

�Td".T/

dT� ".T/

�

” .1C ˛/".T/ D ˛ Td"

dT

” 1C ˛

˛

dT

TD d"

"H) lnT

.1C ˛/˛ D ln "C C0

H) U.T;V/ D AV T.1C ˛/˛ :

2.

dS D 1

T.dUCp dV/ D 1

T

�@U

@T

�V

dTC 1

T

pC

�@U

@V

�T

�dV

H)�@S

@T

�V

D 1

T

�@U

@T

�V

D 1C˛˛

AV T.1˛ /� 1

H) S.T;V/ D .1C˛/AV T1˛ Cf .V/ ;�

@S

@V

�T

D 1

T

pC

�@U

@V

�T

�D�˛ ".T/CAT

.1C˛/˛

� 1

TD .˛C1/AT

1˛

ŠD .1C˛/AT1˛ Cf 0.V/ H) f .V/ D const

H) S.T;V/ D S0C.1C˛/AV T1˛ :

Solution 2.9.20

1. Mixing temperature Tm:The pressure does not change when the separating wall is removed. Thus we

have as equations of state:

before: p V1; 2 D n1; 2 RT1; 2 ;

after: p V D n R Tm

V D V1 C V2 ; n D n1 C n2


H) n R Tm D R.n1T1 C n2T2/ ;

Tm D n1nT1 C n2

nT2 D n1T1 C n2T2

n1 C n2:

So Tm is known.2. The intermixing is irreversible. The entropy has therefore to be determined via a

reversible auxiliary process.

a) Isothermal expansion (reversible, see section 2.7) of each of the partial gases:

V1; 2 �! V ; �U1; 2 D 0 ; since isotherm,

�W1; 2 D �n1; 2 RT1; 2

VZV1; 2

dV

VD �n1; 2 RT1; 2 ln

V

V1; 2;

�S.a/1; 2 D ��W1; 2

T1; 2D n1; 2 R ln

V

V1; 2:

The pressure of the two partial gases will have changed.b) Isochoric temperature change (reversible):

T1; 2 �! Tm ;

�U D �Q ; since isochoric,

�S.b/1; 2 D n1; 2 CV

TmZT1; 2

dT

TD n1; 2 CV ln

TmT1; 2

:

c) Net balance: Entropy of mixing

�S D �S.a/1 C�S.a/2 C�S.b/1 C�S.b/2 ;

�S D R

n1 ln

V

V1C n2 ln

V

V2

�C CV

n1 ln

TmT1

C n2 lnTmT2

�:

3. Let us take T1 D T2. In the case of identical gases the state of the total systemcan not have changed after the intermixing because of equal pressures and equaltemperatures in both the chambers. �S D 0 is to be expected. Our calculation,however, yields:

�S .T1 D T2 D Tm/ D R

�n1 ln

V

V1C n2 ln

V

V2

�:

That is paradoxical because then the entropy of the system could be madearbitrarily small by squeezing in arbitrarily many separation walls. The entropy


would then be a function of the prehistory of the system and therefore not a statequantity!

Let bCV be the heat capacity per particle. Then it is generally valid for theentropy of the ideal gas:

S D N�bCV ln T C kB lnV C C

�:

The constant C is independent of T and V , must, however, obviously depend onN. But then we can avoid the paradox by the definition

S D N

�bCV ln T C kB lnV

NC C0

�:

The entropy of mixing is unambiguous:

�S D S.N;V;T/ � ŒS .N1;V1;T1/C S .N2;V2;T2/�

Solution 2.9.21

1.

�Q1 DZ T0

Ta

Cp dT D Cp .T0 � Ta/ :

2. The Carnot-machine receives ıQ0 from the heat bath, converts from that ıW intowork, and gives ıQ� back to the steel block at the intermediate temperature T�(Fig. A.9). Efficiency of the Carnot machine:

�C D 1 � T�

T0D 1C ıQ�

ıQ0Õ 0 D ıQ�

T� C ıQ0T0

Õ ıQ� D T�

T0.�ıQ0/ D �Cp dT

� :

Fig. A.9 T0HB

C

BlockδQ*

δQ0

δW

T*

<0

<0

>0


In the last step we have exploited that the process, all in all, shall be performedisobaricly, and that ıQ� is to be counted negative. We can now integrate:

�Q2 DZ T0

TU

ıQ0 D �T0

Z T0

TU

ıQ�

T� D CpT0

Z T0

TU

dT�

T� :

That yields:

�Q2 D CpT0 lnT0TU

:

3. (a) Entropy change for process 1) At T D T0 the bath gives away heat in areversible manner:

.�S/.1/HB D ��Q1T0

D �Cp

�1 � Ta

T0

�:

The steel block receives irreversibly heat to raise its temperature from Tato T0. In order to calculate its entropy change we need a reversible auxiliaryprocess. This, however, is just the process 2). Therefore

.�S/.1/block D �Q2T0

D Cp lnT0TU

:

Total entropy change:

�S.1/ D Cp

��1C TU

T0C ln

T0TU

�> 0

since ln x � 1C 1=x > 0 for x > 1.

Irreversibility creates entropy!

(b) Entropy change for process 2)All sub-steps are now reversible:

.�S/HB D ��Q2T0

D �Cp lnT0TU

.�S/Carnot D 0

.�S/Block DZ T0

TU

�ıQ�

T� D Cp lnT0TU

:


That yields for the total entropy change the expected result:

�S.2/ D 0 :

Solution 2.9.22

�C D T1 � T2T1

D 1

6H) ��W D �C�Q1 D 1

6kJ

Solution 2.9.23

1. Segment a ! b W

p � V H) p D paVa

V H) Vb D Va

papb ;

ideal gas: p V D n R T

H) paVa

pbVbD Ta

TbD p2a

p2bH) Tb D Ta

�pbpa

�2:

Segment b ! c W

V D const H) Vc D Vb D Va

papb ;

pc Vc D n R Tc D pa Vb H) Tc D pa VbTb

pb Vb

H) Tc D Tapbpa:

2. Work done:

�Wab D �bZ

a

p dV D � paVa

bZa

V dV D � pa2Va

�V2b � V2a

H) �Wab D �12paVa

"�pbpa

�2� 1

#< 0 ;

�Wbc D 0 ; da dV D 0,

�Wca D �pa.Va � Vc/ D �pa Va

�1 � pb

pa

�> 0 :


Internal energy:The internal energy of the ideal gas depends only on the temperature. From

that it follows:

�Uab D CV.Tb � Ta/ D CV Ta

"�pbpa

�2� 1

#;

�Ubc D CV.Tc � Tb/ D CV Tapbpa

�1 � pb

pa

�;

�Uca D CV.Ta � Tc/ D CV Ta

�1 � pb

pa

�:

Heat quantities:

�Qab D �Uab ��Wab D�CV Ta C 1

2pa Va

�"�pbpa

�2� 1

#;

�Qbc D �Ubc ��Wbc D CV Tapbpa

�1 � pb

pa

�;

�Qca D �Uca ��Wca D .CV Ta C pa Va/

�1 � pb

pa

�;

�Qab > 0 ;

�Qbc < 0 I �Qca < 0 :

Entropy changes:

�Sb! c DcZ

b

ıQ

TD CV

cZb

dT

TD CV ln

TcTb

D CV lnpapb;

�Sc! a DaZ

c

ıQ

TD Cp ln

TaTc

D Cp lnpapb;

S is state quantity:HdS D 0

H) �Sa! b D � .�Sb! c C�Sc! a/ D �.CV C Cp/ lnpapb;

Cp D CV C n R D CV C pa Va

Ta

H) �Sa! b D ��2CV C pa Va

Ta

�ln

papb:


3.

� D total work done

accepted heat quantityD ��W

�Qab;

�W D �12pa Va

"�pbpa

�2� 1C 2 � 2

pbpa

#

D �12pa Va

�pbpa

� 1

�2< 0

H) � D12pa Va

CV Ta C 12pa Va

pbpa

� 1pbpa

C 1D pa Va

2CV Ta C pa Va

pb � papb C pa

:

Solution 2.9.24

1. Temperatures:On the adiabatics: T p.1� �/ = � D const

H) Tc p.1� �/ = �2 D Td p

.1� �/ = �1 I Tb p

.1� �/ = �2 D Ta p

.1� �/ = �1

H) Ta � TdTb � Tc

D�p2p1

�.1� �/ = �

:

2. Heat quantities:

�Qab D �Qcd D 0 ;

�Qbc D Cp .Tc � Tb/ > 0 .Tc > Tb follows from the equation of state!/ ;

�Qda D Cp .Ta � Td/ < 0 :

3. Efficiency:

� D ��W

�QbcD 1C �Qda

�QbcD 1 � Ta � Td

Tb � TcD 1 �

�p1p2

�.� � 1/ = �

:

Solution 2.9.25

1. ıQ D T dS

1 ! 2 W T D const D T2 H) �Q12 D T2 .S2 � S1/ > 0 ;

2 ! 3 W adiabatic, isentropic H) �Q23 D 0 ;

3 ! 4 W T D const D T1 H) �Q34 D T1 .S1 � S2/ < 0 ;

4 ! 1 W adiabatic, isentropic H) �Q41 D 0 :


2. 0 D HdU, because (thermodynamic) cycle

H) ��W DIıQ D .T2 � T1/ .S2 � S1/

H) � D ��W

�Q12D 1 � T1

T2D �c :

3. The thermodynamic cycle is actually the Carnot process!

Solution 2.9.26

Path (A):

T is a linear function of S:

T.S/ D �T2 � T1S2 � S1

S C b :

With T.S2/ D T1 it follows:

b D T1 C T2 � T1S2 � S1

S2 D T2S2 � T1S1S2 � S1

:

Hence it holds on the path (A):

T.S/ D �T2 � T1S2 � S1

S C T2S2 � T1S1S2 � S1

:

This yields the heat-exchange contribution:

�QA DZ.A/

dS T.S/ D �12

T2 � T1S2 � S1

�S22 � S21

C T2S2 � T1S1S2 � S1

.S2 � S1/

Õ �QA D 1

2.T2 C T1/.S2 � S1/ > 0 :

On .A/ the system therefore absorbs heat!Work done (first law of thermodynamics):

�WA D �UA ��QA D U.T1/� U.T2/��QA :

Here we have exploited that the internal energy of the ideal gas depends only on thetemperature.


Path (B)

The process runs isothermally:

�UB D 0 Õ �WB D ��QB D �T1.S1 � S2/ Õ �QB < 0 :

Path (C):

�QC D 0 Õ �WC D �UC D U.T2/� U.T1/ :

Sum of all the work done:

�W D �WA C�WB C�WC

D U.T1/ � U.T2/ � 1

2.T2 C T1/.S2 � S1/� T1.S1 � S2/C U.T2/� U.T1/

Õ �W D �12.T2 � T1/ .S2 � S1/ :

Efficiency

� D ��W

�QAD T2 � T1

T2 C T1:

Solution 2.9.27

Segment 1 ! 2 WAdiabatic compression of the gas H) temperature rise (via the ignition temperatureof the fuel mixture!):

�W12 D �V2Z

V1

p dV D �V2Z

V1

C1V�

dV D C1� � 1

�V1��2 � V1� �

1

�

D 1

� � 1. p2 V2 � p1 V1/ ;

�Q12 D 0 :

Segment 2 ! 3 WInjection of the fuel (isobaric!):

�W23 D �p2 .V3 � V2/ ;

�Q23 D Cp .T3 � T2/ ;

p2 V2 D n R T2 ;


p2 V3 D n R T3

H) T3 � T2 D p2n R

.V3 � V2/ ;

�Q23 D Cp

n Rp2 .V3 � V2/

D �

� � 1p2 .V3 � V2/ > 0 :

Segment 3 ! 4 WExpansion along an adiabatic curve (work done):

�W34 D C2� � 1

�V1� �4 � V1� �

3

�D 1

� � 1. p4 V1 � p2 V3/ ;

�Q34 D 0 :

Segment 4 ! 1 WEjection of the residual gas:

�W41 D 0 ;

�Q41 D CV .T1 � T4/ ;

p4 V1 D n R T4 ;

p1 V1 D n R T1

H) T1 � T4 D V1n R

. p1 � p4/ ;

�Q41 D CV

n RV1 . p1 � p4/

D 1

� � 1V1 . p1 � p4/ < 0 :

Net balance:

�W DIıW D 1

� � 1. p2 V2 � p1 V1 C p4 V1 � p2 V3/� p2 .V3 � V2/

D p2 .V3 � V2/�

1 � � C V1� � 1

. p4 � p1/

D 1

� � 1ŒV1 . p4 � p1/ � � p2 .V3 � V2/� :


Solution 2.9.28

1. 0 ! 1 W aspiration, piston is shifted, volume-expansion at constant pressure,1 ! 2 W densification, adiabatic compression, thereby pressure rise,2 ! 3 W ignition, pressure rise at constant volume,3 ! 4 W adiabatic expansion, thereby work done,4 ! 1 W opening of the outlet valves, pressure decrease at constant volume,1 ! 0 W expulsion of the residual gas.

(0 ! 1; 1 ! 0 do not belong to the thermodynamic cycle!)2. �U D �Q C�W D 0, since cycle

H) �W D ��Q ;

�Q D CV .T3 � T2/C CV .T1 � T4/ D �Q23 C�Q41 :

Equation of state:

p2 V2 D n R T2 I p3 V2 D n R T3

H) T3 > T2 ; because p3 > p2 ;

p1 V1 D n R T1 I p4 V1 D n R T4

H) T4 > T1 ; because p4 > p1

H) �Q23 D CV .T3 � T2/ > 0 ;

�Q41 D CV .T1 � T4/ < 0 :

Adiabatic equations:

T1 V� � 11 D T2 V

� � 12 H) T1 < T2 ;

T3 V� � 12 D T4 V

� � 11 H) T4 < T3 :

T3 is thus the highest, T1 the lowest temperature of the cycle!

.T1 � T4/ V� � 11 D .T2 � T3/ V

� � 12

H) �Q D ��W D CV .T3 � T2/

"1 �

�V2V1

�� 1#:

3. On the segment 2 ! 3 the system absorbs heat, therefore:

� D ��W

�Q23D 1 �

�V2V1

�� 1

D 1 � T1T2:


4. The Carnot machine between the heat baths HB.T1/ and HB.T3/ has theefficiency

�C D 1 � T1T3

H) �otto < �C :

T2 ! T3 is not possible with the Otto motor, because then the ignition would beleft out.

Solution 2.9.29

1 ! 2 W Heat QD D Q.1/D CQ.2/D is taken from the heat bath HB.T/ and applied to

overcome the cohesive forces .Q.1/D / and for the expansion .Q.2/D /:

�W12 D �. p C�p/ .V2 � V1/ :

2 ! 3 W Hardly any work is done because of negligible volume-change:

�W23 � 0 :

3 ! 4 W Isothermal condensation:

�W34 D �p .V1 � V2/ :

Q.1/D goes via condensation into the heat bath HB.T ��T/.4 ! 1 W �W41 � 0, because of an only unimportant volume-change.

H) �W DIıW D ��p .V2 � V1/ < 0 :

Efficiency:

.Carnot/ � D 1 � T ��T

TD �T

TŠD ��W

QDD �p .V2 � V1/

QD

H) �p

�TD QD

T .V2 � V1/

Solution 2.9.30 According to (2.73) it holds for adiabatic state changes:

.dT/ad D �Tˇ.T/

CV �T.dV/ad :


Because of ˇ.T D 4 ıC/ D 0 the adiabatic cooling down of water from 6 to 4 ıCis not possible. The described Carnot process is actually not realizable. Therefore itcan not be about a contradiction.

Solution 2.9.31 Works done:

�W12 D �V2Z

V1

p.V/ dV D �n R T2 lnV2V1

< 0 ;

�W23 D 0 ; da dV D 0 ;

�W34 D �V1Z

V2

p.V/ dV D �n R T1 lnV1V2

> 0 ;

�W41 D 0

H) �W D �n R .T2 � T1/ lnV2V1

:

Equation of state:

p4 V1 D n R T1p1 V1 D n R T2

H) T2 > T1 H) �W < 0 :

Heat quantities:(during isothermal state changes the internal energy of the ideal gas remainsconstant!)

�Q12 D ��W12

�Q23 D CV .T1 � T2/ < 0 ;

�Q34 D ��W34 ;

�Q41 D CV .T2 � T1/ > 0 :

Efficiency:

� D ��W

�Q:

�Q is the amount of heat supplied to the system:

�Q D �Q12 C�Q41 ¤ �Q12 ;

�Q D n R T2 lnV2V1

C CV .T2 � T1/

H) � D T2 � T1

T2 C CV .T2�T1/n R lnV2 =V1

< �C :


Fig. A.10

Fig. A.11

Fig. A.12

Is that a contradiction to the statement, proven in Sect. 2.5, that �C is reached by allmachines which work reversibly and periodically between two heat baths?

Solution 2.9.32

1. Isotherms: p / 1 =V (equation of state!)

a) p1 V1 D n R T1;b) p1 V2 D n R T2;c) p2 V2 D n R T3;d) p2 V1 D n R T4:

Because of p1 V2 D p2 V1 we have T2 D T4. There are thus only threedifferent temperatures T1, T2, T3 (Fig. A.10). We need three isotherms for therepresentation.

2. Isobares: T / V (Fig. A.11)

p.1/ D p.2/ D p1 I p.3/ D p.4/ D p2 ;

V.1/ D V.4/ D V1 I V.2/ D V.3/ D V2 :

3. Isochores: p / T (Fig. A.12)


Solution 2.9.33

1. An elongation of the thread means work is done on the system which is thereforeto be counted as positive. It comes out therewith the following analogy to the gas:

Z dL ” �p dV ;

Z ” �p ;

L ” V :

First law of thermodynamics:

dU D ıQ C Z dL ;

dS D dU

T� Z

TdL ;

U D U.T;L/ H) dS D 1

T

�@U

@T

�L

dT C1

T

�@U

@L

�T

� Z

T

�dL :

dS is a total differential. From that it follows:

1

T

@

@L

�@U

@T

�L

�T

ŠD�@

@T

1

T

�@U

@L

�T

� Z

T

��L

.integrability conditions/

H) 1

T2

�@U

@L

�T

D �@

@T

�Z

T

��L

;

@

@T

�Z

T

��L

D�@

@T

L � L0˛

�L

D 0

H)�@U

@L

�T

D 0 ” U.T;L/ � U.T/ :

Heat capacity:

CL.T;L/ D�ıQ

dT

�L

D�@U

@T

�L

� CL.T/ :

According to the scope of the exercise we have especially for L D L0:

CL0 .T/ D CL0 D C > 0 :

Because of CL.T/ D CL0 .T/ it follows then for arbitrary L the assertion:

CL.T/ � C > 0 :


2. From part 1. it follows already:

U.T/ D C T C U0 ;

dS D dU

T� Z

TdL D C

dT

T� L � L0

˛dL

H) entropy:

S.T;L/ D C lnT � 1

2˛.L � L0/

2 C S0 :

Adiabatic equations:

dS D 0 H) CdT

TD L � L0

˛dL

” d lnT D L � L0˛ C

dL

” T.L/ D D exp

".L � L0/

2

2 ˛C

#;

The constant D takes different values on different adiabatic curves!

D D T.L0/ D exp Œ.S � S0/ =C� D D.S/ :

Z.L/ D 1

˛.L � L0/ T.L/ D D

˛.L � L0/ exp

".L � L0/

2

2 ˛C

#:

3. See Fig.A.13

Z

L

T

S

adiabatic curves

isotherms

0L

)( 0LLZ −=T

Fig. A.13


4.

CZ D T

�@S

@T

�Z

:

Thus we need S D S.T;Z/:

dS D C

TdT � Z

TdL ;

L D L.T;Z/ D L0 C ˛Z

TH) dL D �˛ Z

T2dT C ˛

TdZ

H) dS D�C

TC ˛

Z2

T3

�dT � ˛ Z

T2dZ

H) CZ D C C ˛Z2

T2:

5.

�L D ˛ Z

�1

T2� 1

T1

�< 0 ; falls ˛ > 0 ;

ˇ D ��W

�QD 1 � �U

�Q< 1 ;

since during the heating alsothe internal energy will change,

�U D C .T2 � T1/ ;

�Q DT2Z

T1

CZ dT D C .T2 � T1/ � ˛ Z2�1

T2� 1

T1

�

H) ˇ D 1

1C C˛ Z2

T1 T2< 1 :

The supplied heat quantity is therefore not exclusively used for mechanicalwork!

6. Thermally isolated H) adiabatic equation applicable (see 2.).

T .L1; 2/ D D exp

".L1; 2 � L0/

2

2 ˛C

#;

L2 > L1 H) T .L2/ > T .L1/ :

The thread therefore heats up.


Fig. A.14

Solution 2.9.34

1. The Carnot process consists of two isotherms and two adiabatic curves(Fig.A.14). It works as heat engine if it extracts heat from a heat reservoirand gives away a part of it for doing work:

IıW

Š< 0 I

IıQ > 0 :

Because of ıW D Z dL the direction of the path is as sketched in Fig.A.14.2. First law of thermodynamics:

ıQ D dU � Z dL I dU D C dT ;

�Q2 DbZ

a

ıQ D �bZ

a

Z.T;L/ dL D .isotherm!/

D �T2˛

LbZLa

.L � L0/ dL D � T22 ˛

h.Lb � L0/

2 � .La � L0/2i;

La > Lb > L0 H) �Q2 > 0 :

The system absorbs heat on the segment a ! b. Analogously one finds:

�Q1 D � T12 ˛

h.Ld � L0/

2 � .Lc � L0/2i

L0 < Lc < Ld H) �Q1 < 0 :

On the segment c ! d the system transfers heat into the heat bathHB.T1/. Usingpart 2. of exercise 2.9.33 one gets:

.La; b; c; d � L0/2 D 2 ˛ ŒC ln Ta; b; c; d � .Sa; b; c; d � S0/� :


Fig. A.15

L

Z1T

2T

isotherm

adiabatic curve

0L 1L

On the adiabatic curves we have:

Sa D Sd I Sb D Sc

H) .Lb � L0/2 � .La � L0/

2 D 2 ˛ ŒSa � Sb� ;

.Ld � L0/2 � .Lc � L0/

2 D 2 ˛ ŒSc � Sd� D 2 ˛ ŒSb � Sa� :

Efficiency:

� D �Q1 C�Q2�Q2

D 1 � T1T2

D �C :

3. The point at L0 is ambiguous on the isotherm as well as on the adiabatic curve(Fig.A.15). Away from this point it is on the isotherm T D T1. For the adiabaticcurve we have according to part 2. of exercise 2.9.33:

T D D.S/ exp

".L � L0/

2

2 ˛C

#D T.L/

H) T2 D .T .L0//ad D D ;

T1 D .T .L1//ad D D exp

".L1 � L0/

2

2 ˛C

#

D T2 exp

".L1 � L0/

2

2 ˛C

#> T2 :

At L D L0 the system must be cooled down from T1 to T2. This process isirreversible if one uses the heat bath HB.T2/ for that. We therefore expect

� < �C D 1 � T2T1:

Work done:

�W DZis

Z dL CZad

Z dL

D T1˛

L0ZL1

.L � L0/ dL C D

˛

L1ZL0

.L � L0/ e.L�L0/

2

2˛C dL


D � T12˛.L1 � L0/

2 C T2˛˛C

L1ZL0

dLd

dLe.L�L0/

2

2˛C ;

�W D � T12 ˛

.L1 � L0/2 C C T2

(exp

".L1 � L0/

2

2 ˛C

#� 1

):

Amount of heat:

�Q D .�Q/is DZis

.dU � Z dL/ D �T1˛

L0ZL1

.L � L0/ dL

H) �Q D T12 ˛

.L1 � L0/2 > 0 ; is taken by the system!

Efficiency:

� D ��W

�QD 1 � T2

T1

e.L1�L0/

2

2˛C � 1.L1�L0/2

2˛C

;

1

x.ex � 1/ D 1C 1

2Šx > 1 ; .x > 0/ :

Thus it is as expected:

� < �C D 1 � T2T1:

Solution 2.9.35

�W D 0; since the change of the length takes place against Z D 0,�Q D 0; since no heat exchange with the surroundings occurs.

A reversible auxiliary process would yield the following entropy change:Part 2. of exercise 2.9.33 had given:

S.T;L/ D C lnT � 1

2 ˛.L � L0/

2 C S0

H) �S D S .T;L0/� S.T;L/ D 1

2 ˛.L � L0/

2 D 1

2˛Z2

T2:

Reversible auxiliary process:


Isothermal state change by contacting a heat bath HB.T/, quasi-static procedure.Thereby Z D 0, i.e. L ! L0:

dU D 0 H) ıQ D �ıW D �Z dL D T dS

H) �S D � 1T

L0ZL

Z dL0 D � 1˛

L0ZL

.L0 � L0/ dL0

D .L � L0/2

2 ˛D 1

2˛Z2

T2

That was to be shown!

Solution 2.9.36

1.

ıQ.m/ D dU.m/ � ıW.m/ ;

ıW.m/ D �0 V H dM

H) C.m/M D�ıQ.m/

dT

�M

D�@U.m/

@T

�M

;

C.m/H D C.m/M C�@U.m/

@M

�T

� �0H V

��@M

@T

�H

:

It holds for the entropy of the paramagnetic moment system:

dS.m/ D 1

T

�@U.m/

@T

�M

dT C 1

T

�@U.m/

@M

�T

� �0VH�dM :

Integrability condition for the total differential dS.m/:

1

T

�@2U.m/

@M@T

�ŠD � 1

T2

�@U.m/

@M

�T

� �0VH

�

C 1

T

�@2U.m/

@T@M

�� 0V

�@H

@T

�M

�:

We utilize now the integrability condition for dU.m/ and insert the equation ofstate (Curie law):

�@U.m/

@M

�T

D �0VH � T�0V

�@H

@T

�M

D 0 :


This means

dU.m/ D�@U.m/

@T

�M

dT D C.m/M dT

or for the heat quantity:

ıQ.m/ D C.m/M dT � �0VH

�@M

@T

�H

dT C�@M

@H

�T

dH

�

D C.m/H dT � �0VH

�@M

@H

�T

dH :

That shows the assertion.2a.

ıQ.m/ D 0

H) dT

dHD �0 V H

�@M@H

T

C.m/H

DbC 1

T HbC H2CH2r

T2

H) dT

dHD T H

H2 C H2r

:

This can easily be integrated:

d lnT D dT

TD H dH

H2 C H2r

D 1

2

d

dHln�H2 C H2

r

dH :

It follows with T0 D T.H D 0/:

lnT

T0D 1

2ln

H2 C H2r

H2r

H) T.H/ D T0

sH2 C H2

r

H2r

:

2b. Thermal equilibrium means:

ıQ.m/ D �ıQL D �CL dT ;


CL is thought to be known,

H) ��CL C C.m/H

�dT D ��0 V H

�@M

@H

�T

dH

H) dT

dHD �0 V

bC HT

CL C C.m/H

� �0 VbCCL

H

T:

since CL C.m/H

This is again easily integrated:

1

2

�T2 � T20

D �0 VbCCL

1

2H2

H) T.H/ Ds�0 V

bCCL

H2 C T20 :

3a. That is the situation of 2a.:

T0 D T�s

H2r

H�2 C H2r< T� (cooling down!).

3b.

” �TfZ

T0

ıQ.m/ŠD

TfZT�

ıQL .H D const D 0/

” �TfZ

T0

C.m/H .H D 0/ dT DTfZ

T�

CL dT

” �bC�0 V H2r

TfZT0

dT

T2D CL

�Tf � T�

” bC�0 V H2r

�1

Tf� 1

T0

�D CL

�Tf � T� :


The temperature of the lattice will change only very little with the above heatexchange because of its large heat capacity. Thus we can replace on the left-hand side approximately Tf by T�:

Tf � T� DbC�0 VCL

H2r

�1

T� � 1

T0

�

D bC�0 VCL

H2r

T�

1 �

pH2

r C H�2Hr

!< 0 :

4. That is now the situation of 2b., for which it remains to solve because of CL C.m/H :

dT

dH� �0 V

bCCL

H

T

H) bT2f � T�2 D ��0 VbCCL

H�2

”�bT f � T�

� �bT f C T��

D ��0 VbCCL

H�2 :

From the same reasons as in 3b. we can take

bT f C T� � 2 T� ;

getting therewith:

H) bT f � T� � ��0 VbC2CL

H2�T� < 0 :

Even now it gives rise to a cooling!5a. The process 3. is irreversible, namely because of the heat exchange between the

system of moments and the crystal (part 2.). The process 4. is reversible.The total system is thermally isolated:

�Q D �Q.m/ C�QL D 0 :

Generally it holds in such a case:

dS � ıQ

TD 0 ;


‘D’ for reversible; ‘>’ for irreversible H) process 3.: �S > 0; process 4.:�S D 0

H) Sf >bSf :5b. By resolving S D S.T;H/ for T,

T D T.S;H/ ;

one finds for the final states:

Tf D T .Sf; 0/ ¤ bT f D T.bSf; 0/ :We assert that

bT f � Tf

That is indeed so if it holds:

0 � �0 VbCCL

1

T�

��12H�2 � H2

r C Hr

qH2

r C H�2�

” H�2 � 2

�Hr

qH2

r C H�2 � H2r

�

” H�2 C 2H2r � 2Hr

qH2

r C H�2

” H�4 C 4H�2H2r C 4H4

r � 4H4r C 4H2

r H�2

” H�4 � 0 q. e. d.

The equal sign .bT f D Tf/ is valid only for H� D 0.

Section 3.9

Solution 3.9.1

1. Free energy: F D F.T;V/

�@F

@T

�V

D �S.T;V/ I�@F

@V

�T

D �p.T;V/ :


We integrate the first equation:

F.T;V/ D �TZ

T0

dT 0 S.T 0;V/C f .V/ D �RV0V

1

Ta0

TZT0

dT 0 T 0a C f .V/

.a¤ �1/D �RV0V Ta

0

1

a C 1

�TaC 1 � TaC 1

0

C f .V/ :

Intermediate result:

F.T;V/ D �RV0V

T

a C 1

�T

T0

�a

C RV0V

T0a C 1

C f .V/ :

Because of

dF D �S dT � p dV D �S dT C ıW

we get for isothermal state changes:

.dF/T D .ıW/T :

That we utilize for T D T0:

�@F

@V

�T0

D�ıW

@V

�T0

D RT0V

D�RV0V2

T

a C 1

�T

T0

�a

� RT0 V0V2.a C 1/

C f 0.V/�T DT0

D f 0.V/

H) f .V/ D RT0 lnV

V0C f .V0/ :

This fixes eventually the free energy:

F.T;V/ D RV0V

T0a C 1

"1 �

�T

T0

�aC 1#

C RT0 lnV

V0C F .T0;V0/ :

2. Equation of state:

p D ��@F

@V

�T

D RV0V2

T0a C 1

"1 �

�T

T0

�aC 1#

� RT0V

:


3. Work done:

�WT D �VZ

V0

p.T;V 0/ dV 0

D RV0T0

a C 1

"1 �

�T

T0

�aC 1#�

1

V� 1

V0

�C RT0 ln

V

V0

H) �WT D RT0a C 1

"1 �

�T

T0

�aC 1#

V0 � V

VC RT0 ln

V

V0

D F.T;V/ � F .T;V0/ :

Solution 3.9.2

2.59 H)�@U

@V

�T

D T

�@p

@T

�V

� p :

That means for the photon gas:

".T/ D 1

3

�Td

dT� "

�” 4 ".T/ D T

d"

dT

H) ".T/ D T4 :

It follows therewith the caloric equation of state:

U.T;V/ D V T4 :

We determine the entropy as follows:

T

�@S

@T

�V

D�@U

@T

�V

D 4 V T3

H)�@S

@T

�V

D 4 V T2 :

A first integration yields:

S.T;V/ D 4

3 V T3 C f .V/ :


We apply now the following Maxwell relation of the free energy:

�@S

@V

�T

D�@p

@T

�V

D 1

3

d

dT

� T4

D 4

3 T3

D 4

3 T3 C f 0.V/

H) f .V/ D const :

We know therewith the entropy as function of T and V:

S.T;V/ D 4

3 T3 V C const :

The third law of thermodynamics says that the constant must be zero. We solve for T

T D�3

4

�1 = 3S1= 3 V�1 = 3

and insert the result into the caloric equation of state:

U.S;V/ D"3

4

�3

4

�1 = 3#V�1 = 3 S4 = 3 :

That is the internal energy of the photon gas as function of its natural variables Sand V .

The free energy is simpler to be calculated:

F.T;V/ D U.T;V/� T S.T;V/ D �13 V T4 :

For the free enthalpy an especially simple expression comes out:

G D F C pV D �13 V T4 C 1

3 V T4 D 0 :

According to the Gibbs-Duhem relationG D �N we find therewith for the chemicalpotential of the photon gas:

� � 0 :

The enthalpy H is still left:

H D U C p V H) H D 4

3 V T4 D S T :


It follows with T D�3p

�1 = 4:

H.S; p/ D�3

�1 = 4p1 = 4 S :

Solution 3.9.3 Work:

ıW D �Fk dx D Ck.T/ x dx :

Analogy to the gas:

ıW D �p dV H) p ” �.k x/ ;V ” x :

For a state change characterized by x and T the free energy F is the properthermodynamic potential:

dF D �S dT C k x dx :

For an isothermal elongation of the spring it holds:

F.x;T D const/ DxZ

x0

kx0 dx0 C F .x0;T/

D 1

2k�x2 � x20

C F .x0;T/ ;

�F.x;T D const/ D 1

2k�x2 � x20

:

The entropy change results from the Maxwell relation of F:

�@S

@x

�T

D �@

@T.kx/

�x

D �x

�@k

@T

�x

H) .�S/TDconst D �xZ

x0

x0 dkdT

dx0 D �12

dk

dT

�x2 � x20

:

We calculate therewith the change of the internal energy:

.�U/T D .�F/T C T.�S/T

D 1

2

�x2 � x20

�k � T

dk

dT

�D k

�x2 � x20

:


Solution 3.9.4

1. Work:

ıW D dL :

When stretching the band, work is executed on the system (sign convention!).Free energy:

F.T;L/ D �S dT C dL :

Maxwell relation:

��@S

@L

�T

D�@

@T

�L

D ˛ :

Internal energy:

U.T;L/ D F.T;L/C T S.T;L/ :

We show that U does depend only on T, but not on L:

�@U

@L

�T

D�@F

@L

�T

C T

�@S

@L

�T

D � ˛ T D 0 :

2.�@S

@L

�T

D �˛ < 0 :

The entropy of the band decreases when the band is elongated!3. Required: .@T = @L/S.

Chain rule:�@T

@L

�S

�@L

@S

�T

�@S

@T

�L

D �1 ;

CL D T

�@S

@T

�L

W heat capacity at constant length; always positive

H)�@T

@L

�S

D ˛

CLT > 0 :

The temperature raises for an adiabatic elongation!


Solution 3.9.5 Entropy:

S D S.T H/ H) dS D�@S

@T

�H

dT C�@S

@H

�T

dHŠD 0

H)�@T

@H

�S

D ��@S@H

T�

@S@T

H

D � T

CH

�@S

@H

�T

:

Free enthalpy:

dG D �S dT � mdB0

B0 D �0H I m D MV I V D const ; no variable

H)�@S

@B0

�T

D�@m

@T

�B0

H)�@S

@H

�T

D �0 V

�@M

@T

�H

D ��0 V C

T2H

C W Curie constant

H)�@T

@H

�S

D �0 VCH

CH T:

Solution 3.9.6

1. Basic relation:

dS D 1

TdU � 1

TQdL D 1

T

�@U

@T

�L

dT C1

T

�@U

@L

�T

� Q

T

�dL :

Integrability condition for S:

1

T

�@

@L

�@U

@T

�L

�T

ŠD � 1

T2

�@U

@L

�T

� Q

�

C 1

T

�@

@T

�@U

@L

�T

�L

��@Q

@T

�L

�:

Because of the integrability condition for U the two double derivatives canceleach other and it remains:

�@U

@L

�T

D Q � T

�@Q

@T

�L

D �T2�@

@T

Q

T

�L�

@U

@T

�L

D CL.T;L/ :


• Internal energyWe integrate along the path .T0;L0/ ! .T;L0/ ! .T;L/:

U.T;L/ D U.T0;L0/ CZ T

T0

dT 0 CL.T0;L0/

� T2�@

@T

�1

T

Z L

L0

dL0 Q.T;L0/��

L

:

• Heat capacity

CL.T;L/ D�@U

@T

�L

D CL.T;L0/ ��@

@TT2�@

@T

�1

T

Z L

L0

dL0 Q.T;L0/��

L

�L

:

• EntropyIt follows with the above relations

�@S

@T

�L

D 1

TCL.T;L/

�@S

@L

�T

D1

T

�@U

@L

�T

� Q

T

�D �

�@Q

@T

�L

by integration along the path .T0;L0/ ! .T;L0/ ! .T;L/:

S.T;L/ D S.T0;L0/CZ T

T0

dT 0 1T 0 CL.T

0;L0/�Z L

L0

dL0�@Q.T;L0/@T

�L0

:

• Free energyT;L are the ‘natural’ variables of the free energy:

F D U � TS D F.T;L/ :

By combining the results for U and S we get:

F.T;L/ D U.T0;L0/� TS.T0;L0/

CZ T

T0

dT 0�1 � T

T 0

�CL.T

0;L0/CZ L

L0

dL0 Q.T;L0/ :


2.• Internal energy

With the special ansatz one easily calculates:

Z T

T0

dT 0 CL.T0;L0/ D 1

2b.T2 � T20 /

Z L

L0

dL0 Q.T;L0/ D 1

2aT2 .L � L0/

2

�@

@T

�1

T

Z L

L0

dL0 Q.T;L0/��

L

D 1

2a .L � L0/

2 :

For the internal energy that leads to:

U.T;L/ D U.T0;L0/C 1

2b.T2 � T20 /� 1

2aT2.L � L0/

2 :

• Heat capacity

CL.T;L/ D�@U

@T

�L

D bT � aT.L � L0/2 :

• Entropy

Z T

T0

dT 0 1T 0 CL.T

0;L0/ DZ T

T0

dT 0 b D b.T � T0/

�@

@T

Z L

L0

L.0 Q.T;L0/

�L

D 2aTZ L

L0

dL0 .L0 � L0/ D aT.L � L0/2 :

It follows therewith:

S.T;L/ D S.T0;L0/C b.T � T0/ � aT.L � L0/2 :

• Free energy

F D U.T0;L0/� TS.T0;L0/C 1

2aT2.L � L0/

2 � 1

2b.T � T0/

2 :

3. For the thermal expansion coefficient we need:

L D Q

aT2C L0

Õ�@L

@T

�Q

D � 2Q

aT3:


Therewith:

˛ D � 2Q

aLT3D � 2

T

�1 � L0

L

�:

4. Adiabatic-reversible means:

S.T1;L/ŠD S.T2;L0/

and therewith

b.T1 � T0/� aT1.L � L0/2 D b.T2 � T0/ :

This eventually yields

T2 D T1�1 � a

b.L � L0/

2�:

Solution 3.9.7

1.

S.T;V/ D ��@F

@T

�V

D N kB.˛ C lnC0 V/C N kB lnC1.kB T/˛ :

2.

p D ��@F

@V

�T

D N kB T

V:

3.

U D F.T;V/C T S.T;V/ D N kB T ˛ :

4.

CV D�@U

@T

�V

D T

�@S

@T

�V

D N kB ˛ :

5.

�T D � 1V

�@V

@p

�T

D � 1V

�@p

@V

�T

��1D V

N kB TD 1

p:


Solution 3.9.8 Internal energy before the mixing:

U D U1 C U2 I U1 D 3

2NkBT1 I U2 D 3

2NkBT2 :

Since no work is needed to be done and no heat transfer takes place, it yields afterthe mixing:

U D 3

2.2N/kBT D 3

2NkB.T1 C T2/ H) T D 1

2.T1 C T2/ :

By use of the thermal equation of state one gets additionally:

U D 3

2pV D 3

2p0.V1 C V2/ D 3

2p0V H) p D p0 :

According to (3.43) it holds for the entropy of the ideal gas:

S.U;V;N/ D Nc C 3

2NkB ln

U

NC NkB ln

V

N

D Nc C 3

2NkB ln

�3

2kBT

�C NkB ln

�kBT

p

�:

Entropy of mixing:

�S D S.T; p0; 2N/� S.T1; p0;N/ � S.T2; p0;N/

D 3NkB ln

�3

2kB1

2.T1 C T2/

�C 2NkB ln

kB 12 .T1 C T2/

p0

!

�32NkB ln

�3

2kBT1

�� NkB ln

�kBT1p0

�

�32NkB ln

�3

2kBT2

�� NkB ln

�kBT2p0

�

D 3

2NkB

"ln

�3

2kB

�21

4.T1 C T2/

2

!� ln

�3

2kBT1

�� ln

�3

2kBT2

�#

CNkB

"ln

�kBp0

�21

4.T1 C T2/

2

!� ln

�kBp0

T1

�� ln

�kBp0

T2

�#

D 3

2NkB ln

1

4

.T1 C T2/2

T1 T2

!C NkB ln

1

4

.T1 C T2/2

T1 T2

!

Õ �S D 5

2NkB ln

1

4

.T1 C T2/2

T1 T2

!:


For the special case T1 D T2 it obviously results �S D 0, i.e., the Gibb’s paradoxdoes not appear.

Solution 3.9.9

1.

dF D �S dT C B0 dm D �S dT C �0 V H dM

H)�@F

@M

�T

D �0 V H :

Susceptibility:

�T D�@M

@H

�T

D�@H

@M

�T

��1D �0 V�

@2F@M2

�T

:

Free energy:

�@2F

@M2

�T

D �0 V

�T

H)�@F

@M

�T

D �0 V

MZ0

��1T.T;M0/ dM0 C f .T/

H) F.T;M/ D F.T; 0/C �0 V

MZ0

dM0M0Z0

dM00 ��1T .T;M

00/C f .T/M :

This is the most general solution!Special case:

�T.T;M/ � �T.T/ (e.g. Curie law)

H) �T D M

H:

It follows then:�@F

@M

�T

D �0 VM

�TC f .T/ H) f .T/ � 0

H) F.T;M/ D F.T; 0/C �0 V1

2

M2

�T:


2. Entropy:

S D � � @F@T

M

D S.T; 0/� Mdf .T/dT � �0 V

MR0

dM0 M0R0

dM00 � @@T�

�1T .T;M

00/M00 :

The above special case:

S.T;M/ D S.T; 0/ � �0 V 12M2

�d

dT��1T .T/

�:

Internal energy:

U D F C T S ;

U.T;M/ D U.T; 0/C M

�f .T/ � T

df

dT

�

C�0 VMZ0

dM0M0Z0

dM00��1T .T;M

00/ � T@

@T��1T .T;M

00/�:

Special case:

U.T;M/ D U.T; 0/C �0 V1

2M2

��1T � T

@��1T

@T

�:

Thereby:

U.T; 0/ D F.T; 0/C T S.T; 0/ :

Solution 3.9.10 We read off from dF D �S dTCB0 dm the integrability condition:

�@S

@m

�T

D ��@B0@T

�m

D ��0�@H

@T

�m:

Curie-Weiss law:

M D C

T � TcH D V m .V D const/ :

Heat capacity:

�@Cm

@m

�T

D T

@

@m

�@S

@T

�m

�T

D T

@

@T

�@S

@m

�T

�m


D ��0 T�@2H

@T2

�m

D 0

H) Cm.T;M/ � Cm.T/ :

Internal energy:

dU D T dS C �0 V H dM

H)�@U

@T

�M

D T

�@S

@T

�M

D CM D Cm ;

�@U

@M

�T

D T

�@S

@M

�T

C �0 V H

D T V

�@S

@m

�T

C �0 V H

D ��0 T V

�@H

@T

�m

C �0 V H

D ��0 T VM

CC �0 V

M

C.T � Tc/

D ��0 V M

CTc

H) U.T;M/ D ��0 V TcM2

2CC g.T/ :

Because of �@U

@T

�M

D g0.T/ D Cm

so that finally:

U.T;M/ DTZ0

Cm.T0/ dT 0 � �0 V Tc

M2

2CC U0 :

Entropy:

S.T;M/ DTZ0

Cm.T 0/T 0 dT 0 C f .M/ ;

�@S

@m

�T

D 1

V

�@S

@M

�T

D ��0�@H

@T

�m

D ��0CM


H) f 0.M/ ŠD ��0VC

M

H) S.T;M/ D S0 CTZ0

Cm.T 0/T 0 dT 0 � �0 V

2CM2 :

Free energy:

F.T;M/ D U.T;M/ � T S.T;M/

D F0 CTZ0

Cm.T0/�1 � T

T 0

�dT 0 C �0 V

2CM2 .T � Tc/ :

Free enthalpy:

G D F � mB0 D F � �0 V MH D F � �0 V

C.T � Tc/ M

2

D F0 CTZ0

Cm.T0/�1 � T

T 0

�dT 0 � �0 V

2CM2 .T � Tc/

H) G.T;B0/ D F0 CTZ0

Cm.T0/�1 � T

T 0

�dT 0 � V C

2�0B20

1

T � Tc:

Solution 3.9.11

H D U C p V D H.S; p/ I dH D T dS C V dp ;�@H

@p

�V

D T

�@S

@p

�V

C V :

Maxwell relation for U:

dU D T dS � p dV H)�@p

@S

�V

D ��@T

@V

�S

H)�@H

@p

�D V � T

�@V

@T

�S

:

Solution 3.9.12

1. First law of thermodynamics:

dU D ıQ C ıW :


The work is composed of an electrical and a mechanical part:

ıWe D V E dP (the volume V is again to be seen as constant ,

does not belong to the thermodynamic variables),

ıWm D dL :

Thus we have for reversible state changes:

dU D T dS C V E dP C dL :

The differential of the free enthalpy,

G D U � T S � V EP � L ;

thus reads:

dG D �S dT � V P dE � L d :

It is a total differential from which we get the Maxwell relation

V

�@P

@

�T; E

D�@L

@E

�T;

:

2. There are as many thermodynamic potentials as one can create from U byLegendre transformations:

U IU � T S I U � V PE I U � L

” transformation in one variable,

U � T S � V PE I U � T S � L I U � V PE � L

” transformation in two variables,

U � T S � V PE � L

” transformation in three variables.

There are therefore altogether eight different thermodynamic potentials.3. Each potential depends on three variables. That leads in each case to three

integrability conditions. Altogether that are then twenty four!


Solution 3.9.13

1.

U.T;V;N/ D F C TS D F � T

�@F

@T

�V;N

H)�@U

@N

�T; V

D�@F

@N

�T; V

� T

"@

@N

�@F

@T

�V;N

#T; V

:

Note that the derivative with respect to the particle number yields the chemicalpotential � only then when the differentiated quantity is a thermodynamicpotential:

�@U

@N

�T;V

¤ � I but:

�@U

@N

�S;V

D �.S;V;N/ ;

�@F

@N

�T; V

D �.T;V;N/ :

This means for the above relation:

�@U

@N

�T; V

D �.T;V;N/ � T

"@

@T

�@F

@N

�T;V

#V;N

H)�@U

@N

�T; V

� �.T;V;N/ D �T

�@�

@T

�V;N

:

2.

N D N.T;V; �/

H) dN D�@N

@T

�V; �

dT C�@N

@V

�T; �

dV C�@N

@�

�T; V

d� :

Take:

x D �

T

H)�@N

@T

�V; x

D�@N

@T

�V; �

C 0C�@N

@�

�T; V

�@�

@T

�V; x

;

�@�

@T

�V; x

D@

@T.T x/

�V; x

D x D �

T:


This yields the intermediate result:

�@N

@T

�V; x

D�@N

@�

�T; V

"�@N

@T

�V; �

�@�

@N

�T; V

C �

T

#:

Chain rule:�@N

@T

��;V

�@T

@�

�N;V

�@�

@N

�T;V

D �1

H)�@N

@T

�V; x

D�@N

@�

�T; V

"�

T��@�

@T

�N;V

#

1:D 1

T

�@N

@�

�T;V

�@U

@N

�T; V

3.

U D U.T;V;N/

H) dU D�@U

@T

�V;N

dT C�@U

@V

�T;N

dV C�@U

@N

�T; V

dN :

From that we read off:�@U

@T

�V; x

D�@U

@T

�V;N

C 0C�@U

@N

�T; V

�@N

@T

�V; x

:

After insertion of the result of part 2. the assertion comes out!

Solution 3.9.14

1.

dU D T dS � p dV C B0 dm

dF D �S dT � p dV C B0 dm :

2. (a) Integrability relation for F Õ�@S

@V

�T;m

D�@p

@T

�V;m

�! NkBV

:

(b) Integrability condition for F Õ�@S

@m

�T;V

D ��@B0@T

�V;m

�! � m

˛V:


(c) With 1.) and 2.a) one finds

�@U

@V

�T;m

D T

�@S

@V

�T;m

� p D T

�@p

@T

�V;m

� p :

Because of

T

�@p

@T

�V;m

D NkBT

VD p

that means for the ideal paramagnetic gas:

�@U

@V

�T;m

D 0 :

(d) Wit 1.) and 2.b) it holds now:

�@U

@m

�T;V

D T

�@S

@m

�T;V

C B0 D �T

�@B0@T

�V;m

C B0 :

Because of

T

�@B0@T

�V;m

D Tm

˛VD B0

it is also here�@U

@m

�T;V

D 0 :

3.• Since the entropy S is a state quantity, the integration path between two

points in the state space can be chosen arbitrarily. Here it appears convenient:.T0;V0;m0/ �! .T0;V0;m/ �! .T0;V;m/ �! .T;V;m/, where the particlenumber N is kept constant:

S.T;V;m;N/ D S.T0;V0;m0;N/CZ m

m0

�@S

@m0

�T0;V0

dm0

CZ V

V0

�@S

@V 0

�T0;m

dV 0 CZ T

T0

�@S

@T 0

�V;m

dT 0

D � 1

˛V0

Z m

m0

m0 dm0 C NkB

Z V

V0

dV 0

V 0 C CV;m

Z T

T0

dT 0

T 0 :


It thus results for the entropy:

S.T;V;m;N/ D S.T0;V0;m0;N/� 1

2˛V0.m2�m20/CNkB ln

V

V0C3

2NkB ln

T

T0:

• Internal energy:Because of

�@U

@V

�T;m

D�@U

@m

�T;V

D 0

it remains

U.T;V;m;N/ D 3

2NkBT C U.T0;V0;m0;N/ :

4. The above result for the entropy does, at first glance, not look as if thehomogeneity relation (�: arbitrary real number)

S.T; �V; �m; �N/ D �S.T;V;m;N/

were fulfilled. But one has to take into consideration that the integration-constantS0.N/ � S.T0;V0;m0;N/ still depends on the particle number N, which is anextensive variable and has actually been considered during the derivations in part3.) as constant. The homogeneity is guaranteed if the following requirement canbe fulfilled:

S0.�N/� �2

2˛V0m2 C 1

2˛V0m20 C �NkB ln

�V

V0C 3

2�NkB ln

T

T0

ŠD �S0.N/ � �

2˛V0m2 C �

2˛V0m20 C �NkB ln

V

V0C 3

2�NkB ln

T

T0:

Hence it is to require:

�S0.N/ D S0.�N/ � 1

2˛V0�.� � 1/m2 C 1

2˛V0.1 � �/m20 C �NkB ln� :

� is still free. We choose � D N0N :

S0.N/ D N

N0S0.N0/ � 1

2˛V0.N0N

� 1/m2 C 1

2˛V0.N

N0� 1/m20 C NkB ln

N0N:


S0.N0/=N0 � � is a constant. We insert S0.N/ into the expression for the entropyfrom part 3.):

S.T;V;m;N/ D N� � 1

2˛V0

�N0N

�1�m2C 1

2˛V0

�N

N0� 1

�m20CNkB ln

N0N

� 1

2˛V0

�m2 � m20

C NkB lnV

V0C 3

2NkB ln

T

T0

D N� � 1

2˛V0

N0N

m2 C 1

2˛V0

N

N0m20

CNkB lnV=N

V0=N0C 3

2NkB ln

T

T0:

The final result shows that the entropy is indeed extensive:

S.T;V;m;N/ D N

�� N0

2˛V0

m2

N2C 1

2˛V0

m20N0

C kB lnV=N

V0=N0C 3

2kB ln

T

T0

�:

Solution 3.9.15

1. Free energy:According to Exercise 3.9.9 we have:

F.T;m/ D F.T; 0/C �0

2V

m2

�T:

This means:

F.T;m/ D F.T; 0/C �0T � Tc2V C

m2 :

Internal energy:According to Exercise 3.9.9 we have:

U.T;m/ D U.T; 0/C �0

2Vm2��1T � T

@��1T

@T

�;

��1T � T

@��1T

@TD 1

C.T � Tc/� T

CD �Tc

C:

This means:

U.T;m/ D U.T; 0/� �0 Tc2V C

m2 :


Entropy:According to Exercise 3.9.9 we have:

S.T;m/ D S.T; 0/� �0

2Vm2�d

dt��1T

�:

This means here:

S.T;m/ D S.T; 0/� �0

2V Cm2 :

2. Entropy:

Cm.T;m D 0/ D T

�@S

@T

�mD0

H) S.T; 0/ DTZ0

Cm.T 0; 0/T 0 dT 0 D � T :

With the partial result from 1. we get:

S.T;m/ D � T � �0

2V Cm2 (take part 4. into consideration!)

Because of

m D CV

T � TcH

it follows immediately:

S.T;H/ D � T � 1

2�0 CV

H2

.T � Tc/2:

Free energy:

�@F

@T

�m

D �S.T;m/

H) F.T; 0/ D F0 �TZ0

� T 0 dT 0 D F0 � 1

2� T2 :


With the result of part 1. we then get:

F.T;m/ D F0 � 1

2� T2 C �0

T � Tc2V C

m2 :

Internal energy:

CmD0 D�@U

@T

�mD0

” U.T;m D 0/ D 1

2� T2 C U0 :

With the result of part 1. we have:

U.T;m/ D U0 C 1

2� T2 � �0 Tc

2V Cm2 :

3. Heat capacities:

Cm D T

�@S

@T

�m

D � T D Cm.T;m D 0/ ;

CH D T

�@S

@T

�H

D � T C �0 CVT H2

.T � Tc/3:

Because of T > Tc it follows: CH � Cm.Adiabatic susceptibility:According to (2.84) it is:

�S D �TCm

CH:

Insertion into the above results leads to:

�S.T;H/ D C

T � Tc

� T

� T C �0 CV T H2

.T�Tc/3

H) �S.T;H/ D C

T � Tc C �0 C V�

H2

.T�Tc/2

:

4. Tc D 0 H) according to part 2.:

S.T;H/ D � T � 1

2�0 CV

H2

T2:


The third law of thermodynamics requires:

limT ! 0

S.T;H/ D 0 :

Our above result comes for H ¤ 0 to a contradiction. Thus the Curie law cannotbe correct for arbitrarily low temperatures!

We had found in Exercise 3.9.9:

S.T;m/ D S.T; 0/� �0

2Vm2�d

dt��1T

�:

If it holds �T.T;m/ � �T.T/, i.e., m D V �T H, then it follows:

S.T;m/ D S.T; 0/C �0

2Vm2

1

�2T

d�TdT

H) S.T;H/ D S.T; 0/C 1

2�0 V H2 d�T

dT:

In order to fulfill the third law of thermodynamics we thus have to require

limT!0

d�TdT

D 0 ” �T D const C O.T2/ ;

i.e., �T remains finite for T ! 0!

Solution 3.9.16 Because of Tc D 0:

U.T;m/ D U0 C 1

2� T2 � U.T/

(cf. Gay-Lussac experiment for the ideal gas).

1. Isotherm: 0 ! H H) dU D 0

This means:

�Q D ��W D ��0HZ0

H dm ;

dm D CV

T1dH

H) �Q D ��0 CV

2 T1H2 < 0 :

heat is dissipated!


2. Adiabatic-reversible ” dS D 0

” S .T1;H/ D S .Tf; 0/ :

We use the result

S.T;H/ D � T � 1

2�0 CV

H2

T2

from part 2. of the preceding exercise.

� T1 � 1

2�0 CV

H2

T21

ŠD � Tf ” Tf D T1 � �0 CV

2 �

H2

T21„ ƒ‚ …>0

< T1 :

One therefore achieves a cooling effect! Compare this result with part 3. ofExercise 2.9.36!

Solution 3.9.17

1. Pressure:

p D ��@F

@V

�T

D �d F0dV

� AT1

kB TdE.V/dV

1 � e� E.V/kB T

e� E.V/kB T

H) p D � B

V0.V � V0/C AE1

kB V0n.T;V/ :

Entropy:

S D ��@F

@T

�V

D �A ln

�1 � e� E.V/

kB T

�C AT

E.V/kB T2

e� E.V/kB T

1 � e� E.V/kB T

D �A ln

�1 � e� E.V/

kB T

�C A

E.V/

kB Tn.T;V/ :

One easily verifies:

eE.V/kB T D n C 1

n;

1 � e� E.V/kB T D 1 � n

n C 1D 1

n C 1;

E.V/

kB TD ln.n C 1/� ln n :


The entropy reads therewith:

S D A f.n C 1/ ln.n C 1/� n ln ng :

Internal energy:

U D F C T S D F0.V/C AT ln

�1 � e� E.V/

kB T

�

�AT ln

�1 � e� E.V/

kB T

�C A

E.V/

kBn.T;V/

H) U D F0.V/C AE.V/

kBn.T;V/ :

2. From 1. it follows for p D 0:

B

V0.Vm � V0/

ŠD AE1kB V0

n .T;Vm/

H) Vm D V. p D 0/ D AE1kB B

n .T;Vm/C V0 :

This is an implicit conditional equation for Vm, which, for instance, can beiterated:

n .T;Vm/ D�exp

1

kB T

�E0 � E1

Vm � V0V0

�� 1

��1:

That has to be expanded in powers of E1 around E1 D 0. Since E1 appears alsoas factor, the zeroth order suffices for n.T;Vm/:

Vm � V0 C AE1kB B

n .T;V0/ :

Expansion coefficient:

ˇ D 1

V

�@V

@T

�p

:

Estimation:

1

VmD 1

V0 C AE1kB B n .T;Vm/

� 1

V0

�1 � AE1

kB BV0n .T;V0/

�;

@Vm

@TD AE1

kB B

@n

@TD AE1

kB B

E.V/kB T2

eE.V/kBT

�e

E.V/kB T � 1

�2


D AE1kB B

1

Tn.n C 1/ ln

n C 1

n

H) ˇ � 1

V0

A

B

E1kB T

n0 .n0 C 1/ lnn0 C 1

n0;

n0 D n .T;V0/ D�e

E0kB T � 1

��1:

T ! 0

n0 � e� E0kB T ! 0 ;

n0 C 1 ! 1 I lnn0 C 1

n0� � ln n0 � E0

kB T

H) 1

kB Tn0 .n0 C 1/ ln

n0 C 1

n0� E0.kB T/2

e� E0kB T �!

T!00

H) ˇ.T D 0/ D 0 I Vm D V0 :

kB T E.V/

n.T;V/ � kB T

E.V/I @n

@T� kB

E.V/D kB

E0C O.E1/

H) ˇ � 1

V0

AE1BE0

I Vm � V0 C AE1BE0

T :

3. Equation (2.65) is a convenient starting point:

Cp � CV D T

�@p

@T

�V

�@V

@T

�p

:

It is difficult to keep the pressure constant in our equations:

�@V

@T

�p

D �1�@T@p

�V

�@p@V

�T

D ��@p@T

�V�

@p@V

�T

:

This means:

Cp � CV D �T

h�@p@T

�V

i2�@p@V

�T

;


�@p

@T

�V

D AE1kB V0

�@n

@T

�V

� AE1kB V0

�@n0@T

�V

;

�@p

@V

�T

D � B

V0C AE1

kB V0

�@n

@V

�T

;

�@n

@V

�T

D � 1kB T e

E.V/kB T

�e

E.V/kB T � 1

�2 @E.V/@V;

@E.V/

@VD �E1

V0

H)�@p

@V

�T

D � B

V0C 0.E21/

H) Cp � CV � TA2 E21k2B BV0

�@n0@T

�2V

:

Solution 3.9.18

1.

U D U.S;V;A/ ;

dU D ıQ C ıW ;

ıW D ıWV C ıWA ;

ıWV D �p dV ;

ıWA D dA :

If the area A is enlarged by dA work is done on the system:

dU D T dS � p dV C dA :

2. Maxwell relation for dU:�@T

@A

�S; V

D�@

@S

�V;A

D�@

@T

�V;A

�@T

@S

�V;A

Dd dT�

@S@T

V;A

D T

CV;A

d

dT

3. Let us abbreviate

� D ˛

Tc CV;A;


then it is to be integrated:

dT

TD �� dA H) ln T D �� A C ˇ :

Initial values:

ˇ D � A0 C lnT0

H) lnT

T0D �� .A � A0/ H) T D T0 e

��.A�A0/ :

The temperature decreases with an adiabatic-isochoric enlargement of thesurface!

4.

dF D d.U � T S/ D dU � T dS � S dT

H) dF D �S dT � p dV C dA :

5. Independent variables: T;V;A:

�@F

@A

�T; V

D .T/ ; independent of V

H) F.T;V;A/ D .T/A C FV.T;V/ :

In addition we have:

@

@A

�@F

@V

�T;A

D @

@V

�@F

@A

�T; V

D @

@V .T/ D 0

H)�@F

@V

�T;A

D f .T;V/ ; independent of A

H) F.T;V;A/ DVZf .T;V 0/ dV 0 C FA.T;A/ :

Obviously we can write:

F.T;V;A/ D FV.T;V/C FA.T;A/ :

FA can be specified explicitly:

FA.T;A/ D .T/A :


6. Maxwell relation for F:�@S

@A

�T; V

D ��@

@T

�V;A

D ˛

Tc> 0 :

The entropy S increases when the surface is enlarged!7. dU D T dS C dA; if isochoric

H)�@U

@A

�T; V

D T

�@S

@A

�T;V

C 6:D ˛

T

TcC ˛

�1 � T

Tc

�D ˛ > 0 :

8.

S D ��@F

@T

�V;A

D SV.T;V/C SA.T;A/

H) SA.T;A/ D ��@FA

@T

�V;A

5:D �Ad

dTD CA

˛

Tc:

A1 ! A2: isotherm-isochoric ” SV D const

�Q D T .SA .T;A2/� SA .T;A1// D ˛T

Tc.A2 � A1/ ;

�Q > 0, if A2 > A1.9.

dG D d.F C p V/ D �S dT C V dp C dA :

10.�@G

@A

�T; p

D .T/ ; independent of p

H) @

@p

�@G

@A

�T; p

D 0 D @

@A

�@G

@p

�T;A

H)�@G

@p

�T; A

D V.T; p/ ; independent of A:

This means:

G.T; p;A/ D GV.T; p/C GA.T;A/ :


Surface-part:

GA.T;A/ D .T/A ;

V D�@G

@p

�T; A

D�@GV

@p

�T; A

:

Solution 3.9.19

1.

G.1/V .T; p/ D M1 g1.T; p/ ;G.1/A .T;A/ D .T/A1 D .T/ 4� r2 ;

)(drop)

G.2/.T; p/ D M2 g2.T; p/ .vapor; no surface/

H) G.T; p;A/ D M1 g1.T; p/C .T/ 4� r2 C M2 g2.T; p/ :

2. Equilibrium means: dG D 0

Since T and p are fixed, onlyM1; M2 and r are variable:

M1 C M2 D M D const H) dM1 D �dM2 :

Part 1. then yields:

0 D dG D dM1 .g1 � g2/C 8� r dr

H) g2 � g1 D 8� rdr

dM1

:

The mass density �1 of the liquid drop,

�1 D M1

4�3r3;

is to be considered as constant:

H) M1 D �14�

3r3 H) dM1

drD �1 4� r2 :


Therewith the assertion

g2 � g1 D 2

r �1

is proven.3. From the general relation

�@G

@p

�T

D V

it follows here:

V D V1 C V2 D M1

�@g1@p

�T

C 0C M2

�@g2@p

�T

:

This means obviously:

�@gi@p

�T

D Vi

MiI i D 1; 2

H) V2M2

� V1M1

D 1

�2� 1

�1D@

@p.g2 � g1/

�T

D �2 .T/r2 �1

dr

dp:

�1 �2 W1

�2� �2 .T/

r2 �1

dr

dp:

Vapor D ideal gas:

�2 D M2

V2D M2

N2 kB T 1p

D mp

kB T.m W mass of a molecule/ ;

H) kB T

mp� � 2

r2 �1

dr

dpH) dp

pD 2 m

�1 kB T

��dr

r2

�

H) ln p D 2 m

�1 kB T

1

rC ˛ :


p1.T/ W vapor pressure at infinite radius of the drop:

H) ˛ D ln p1

H) lnp

p1D 2 m

�1 kB T

1

r:

Vapor pressure of the drop:

p.r;T/ D p1.T/ exp�2m .T/

�1 kB T

1

r

�:

Solution 3.9.20

1. First law of thermodynamics: dU D ıQ C �0 V H dM

H) CM D�@U

@T

�M

;

ıQ D CM dT C�@U

@M

�T

� �0 V H

�dM

H) CH D�ıQ

dT

�H

D CM C�@U

@M

�T

� �0 V H

��@M

@T

�H

H) CM � CH D�0 V H �

�@U

@M

�T

��@M

@T

�H

:

2.�@M

@T

�H

D � C

T2H

H) CM � CH D ��0 VC

M2 :

3a. Maxwell relation of the free energy:

dF D �S dT C �0 V H dM

H)�@S

@M

�T

D ��0 V�@H

@T

�M

:

3b. Maxwell relation of the free enthalpy:

dG D �S dT � �0 V M dH

H)�@S

@H

�T

D �0 V

�@M

@T

�H

:


3c. The assertion follows immediately from 1. for ıQ D T dS.4.

CM � CH1:D �T

�@S

@M

�T

�@M

@T

�H

D �0 V T

�@H

@T

�M

�@M

@T

�H

:

5.�@H

@T

�M

D M

C;

dH D M

CdT C 1

C.T � Tc/ dM C 3 bM2 dM

H)�@M

@T

�H

3 bM2 C 1

C.T � Tc/

�D �M

C

H)�@M

@T

�H

D �M

3 bM2 C C .T � Tc/

H) CM � CH D ��0 V T M2

3 bM2 C2 C C .T � Tc/:

6.

@

@MCM D

�@

@M

T

�@S

@T

�M

��T

D T

@

@T

�@S

@M

�T

�M

.3a:/D T .��0 V/�@2H

@T2

�M

D 0 :

7.�@U

@T

�M

D CM.T/ :

According to part 3. it is also valid:

�@U

@M

�T

D T

�@S

@M

�T

C �0 V H D ��0 V T

�@H

@T

�M

C �0 V H

D ��0 V TM

CC �0 V

1

C.T � Tc/ M C �0 V bM3

D �0 V

�bM3 � Tc

CM

�:


From that it follows by integration:

U.T;M/ D �0 V

�1

4bM4 � Tc

2CM2

�C f .T/ ;

�@U

@T

�M

D CM.T/ D f 0.T/

H) U.T;M/ D �0 V

�1

4bM4 � Tc

2CM2

�C

TZ0

CM.T0/ dT 0 :

Analogously one finds the entropy:

�@S

@T

�M

D 1

TCM.T/ I

�@S

@M

�T

D ��0 V�@H

@T

�M

D ��0 VM

C

H) S.T;M/ D ��0 V M2

2CC

TZ0

CM.T 0/T 0 dT 0 C S.0; 0/„ƒ‚…

D0 (2.82):

That means finally for the free energy:

F D U � T S D F0 C �0 V1

2C.T � Tc/ M

2 C �0 V1

4bM4

CTZ0

CM.T0/�1 � T

T 0

�dT 0 :

8.

H D M

1

C.T � Tc/C bM2

�:

H D 0 thus possesses the solutions:

a/ M D 0 ;

b/ MS D ˙r

1

bC.Tc � T/ :


We have for the free energy according to part 7.:

F D f .T/C �0 V

2C.T � Tc/ M

2 C 1

4�0 V bM4

H) F.T;M D 0/ D f .T/

F .T;M D ˙MS/ D f .T/C �0 V

2C.T � Tc/

Tc � T

bC

C1

4�0 V b

1

.bC/2.Tc � T/2

D f .T/ � 1

4

�0 V

bC2.Tc � T/2 :

So it is:

F .T;M D ˙MS/ < F.T;M D 0/ :

Consequently, the ferromagnetic solutionMS ¤ 0 is the stable one. It exists asreal solution only for T � Tc.

9. Magnetic susceptibility:

�T D�@M

@H

�T

D�@H

@M

�T

��1D 1

1C .T � Tc/C 3 bM2

H) limH!0

�T D 11C .T � Tc/C 3 bM2

S

D C

2.Tc � T/:

�T diverges in the zero-field for T ! Tc!For the difference of the heat capacities we use the result of part 5.:

limH!0

.CM � CH/ D ��0 V T M2S

3 bM2SC

2 C C .T � Tc/

D ��0 V T 1bC .Tc � T/

3 bC2 1bC .Tc � T/C C .T � Tc/

D � �0 V

2 bC2T :

Solution 3.9.21

1. Maxwell relation for the free enthalpy (dG D �S dT C V dp)

�@S

@p

�T

D ��@V

@T

�p

:


Additionally:

�@S

@T

�p

D Cp

TH) �

dSp

D�Cp

dT

T

�p

:

Therewith one calculates:

Vˇ D�@V

@T

�p

D ��@S

@p

�T

D ��@

@p

Z T

0

�dSp

�T

D � @

@p

Z T

0

�Cp

dT 0

T 0

�p

!T

D �Z T

0

�@Cp

@p

�T0

dT 0

T 0

D �Z T

0

dT 0

T 0 .T0/x�a0 C b0T C c0T2 C : : :

D ��1

xTxa0 C b0

x C 1TxC1 C c0

x C 2TxC2 C : : :

�

D �Tx

�a0

xC b0

x C 1T C c0

x C 2T2 C : : :

�:

In the second line we have utilized that, according to the third law, the entropyvanishes at zero temperature. The lower limit of integration therefore does notcontribute. The subsequent T-integrations are to be performed along a path withp D const. Furthermore:

a0 D d

dpa I b0 D d

dpb I c0 D d

dpc : : :

That leads to:

Vˇ

CpD �

a0

x C b0

xC1T C c0

xC2T2 C : : :

a C bT C cT2 C : : :

The limiting value

limT!0

Vˇ

CpD � a0

ax

represents a finite constant.


2. From the T dS-equation (2.74)

T dS D Cp dT � TVˇ dp

it follows for an adiabatic process:

�dT

dp

�S

D TVˇ

Cp:

Because of 1.) it is then:

limT!0

�dT

dp

�S

D 0 :

Adiabatic relaxation does not lead in the limit T ! 0 to a lowering of thetemperature, which, in the last analysis, expresses nothing but the unattainabilityof the absolute zero as a consequence of the third law of thermodynamics.

Section 4.3

Solution 4.3.1

1. Clausius-Clapeyron equation:

dp

dTD QM

T.vg � vl/� QM

Tvg� p QM

RT2:

This means:

dp

pD d ln p � QM

R

dT

T2Õ ln p � �QM

RTC const.

Thus

p.T/ � ˛ exp

��QM

RT

�.˛ D const./

2. Thermal expansion coefficient:

ˇcoex D 1

V

�@V

@T

�coex

� 1

vg

�@vg

@T

�:


Along the line of coexistence we have

vg D vg.T; p.T//

Õ�@vg

@T

�coex

D�@vg

@T

�p

C�@vg

@p

�T

�@p

@T

�coex

:

The vapor can be seen as ideal gas:

�@vg

@T

�p

D R

pI�@vg

@p

�T

D �RT

p2I�@p

@T

�coex

� QM

T vg:

The last step is justified by the approximate Clausius-Clapeyron equation in part1). It remains:

ˇcoex � 1

vg

�R

p� RT

p2 QM

T vg

�D R

p vg

�1 � QM

p vg

�:

It follows eventually:

ˇcoex � 1

T

�1 � QM

RT

�:

The first term represents the contribution of the ideal gas. The second term, whicharises from the cohesive forces, which come into effect at the phase transition,however, in general dominates. Numeric example:

H2O W QM � 40 kJ=mole

RT � 3 kJ=mole at T D 373K Õ ˇcoex < 0 :

Altogether, a compression takes place with increasing temperature along thecurve of coexistence.

Solution 4.3.2

1. Maxwell relation for G.T; p/:

�@Si@p

�T

D ��@Vi

@T

�p

D �˛ip:

Integration over the pressure:

Si.T; p/ D �˛i ln p

p0C fi.T/ :


Heat capacities:

C.i/p .T/ D T

�@Si@T

�p

D T

�dfidT

�p

ŠD Cp.T/

Õ f 01.T/ D f 0

2.T/ Õ f1.T/ D f2.T/C � :

Third law of thermodynamics:

f1.T ! 0/ D f2.T ! 0/ D 0 Õ f1.T/ � f2.T/ :

This means:

Si.T; p/ D �˛i ln p

p0C f .T/ :

2. At the line of coexistence we obviously have S1 ¤ S2. Hence, it is about a phasetransition of first order. For that the Clausius-Clapeyron equation is valid:

d

dTpcoex D �QU

T.V1 � V2/:

The ‘transformation heat’“�QU is found by:

�QU D T�S D �T.˛1 � ˛2/ ln pcoexp0

:

At the line of coexistence the volume, in addition, exhibits a jump:

.V1 � V2/coex D T.˛1 � ˛2/1

pcoex:

This immediately yields the slope of the line of coexistence

d

dTpcoex D �pcoex

Tln

pcoexp0

:

3. We put x D T and y D pkoex=p0 and have then to solve:

d

dxy D �y

xln y :

Rearranged that means:

dy

yD d ln y D �dx

xln y Õ d ln y

ln yD �dx

xD �d ln x :


That can also be written as follows:

d ln.ln y/ D �d ln x Õ d ln�x ln y D 0 Õ x ln y D x0 :

It is therefore:

y D exp�x0x

�:

If we eventually go back from the substitutions, we recognize that thecoexistence-pressure decreases exponentially with increasing temperature:

pcoex.T/ D p0 exp

�T0T

�:

The transformation heat is then constant along the line of coexistence:

�QU D �.˛1 � ˛2/ T0 :

Solution 4.3.3

1. The assignment is valid:

p ” B0 D �0H ;

V ” �m D �V M :

Clausius-Clapeyron equation (4.19):

dp

dTD �Q

T0 �V:

This means for the superconductor (Fig.A.16):

�Q D T0dB0CdT

.��m/ ;

�m D V .Mn � Ms/ � �V Ms D V HC :

Fig. A.16

×


The last step is due to the Meissner-Ochsenfeld effect:

dB0CdT

D �0dHC

dT

H) �Q D �T0 V �0

�HC

dHC

dT

�T D T0

:

2.

G.T;H/ D U � T S � �0 V HM ;

Mn very small H) Gn.T;H/ � Gn.T; 0/ ;

dG D �S dT � �0 V M dH :

Meissner-Ochsenfeld effect:

dGs D �Ss dT C �0 V H dH :

We are interested in the isothermal process:

.dGs/T D �0 V H dH

H) Gs.T;H/ D Gs.T; 0/C 1

2�0 V H2 :

Phase-equilibrium:

Gn .T;HC/ŠD Gs .T;HC/ � Gn.T; 0/ :

From that we get the ‘stabilization-energy’:

�G D Gs.T; 0/ � Gn.T; 0/ � Gs.T; 0/� Gs.T;HC/

H) �G D �12�0 V H2

C.T/ :

3.

Sn D ��@

@TGn.T;H/

�H

� ��@

@TGn.T;H D 0/

�HD0

;

Ss D ��@

@TGs.T;H/

�H

D � d

dTGs.T; 0/

H) Sn � Ss D � d

dT�G D ��0 V HC.T/

dHC.T/

dT:


This is in accordance with part 1. !Because of .dHC = dT/ < 0:

Sn.T/ > Ss.T/ :

The superconductor is therefore in a state of higher order. Because ofHC.Tc/ D 0

it holds at the critical point:

Sn .Tc/ D Ss .Tc/ :

4. Independently of the values of other parameters it must hold according to thethird law of thermodynamics:

Ss.T/ �!T ! 0

0 I Sn.T/ �!T ! 0

0 :

Since on the other hand it shall be

HC.T/ �!T ! 0

H0 ¤ 0 ;

according to part 3. it must be fulfilled

limT ! 0

dHC

dTD 0 ;

which is indeed guaranteed by our ansatz for HC.5.

Cs � Cn D T

@

@T.Ss � Sn/

�

D �0 V T

"�dHC

dT

�2C HC.T/

d2HC.T/

dT2

#;

dHC

dTD �2H0.1 � ˛/

T

T2c� 4 ˛H0

T3

T4c

D �2H0 TT2c

�1 � ˛ C 2 ˛

T2

T2c

�;

�dHC

dT

�2D 4H2

0

T2

T4c

�1 � ˛ C 2 ˛

T2

T2c

�2;

d2HC

dT2D �2H0

T2c

�1 � ˛ C 6 ˛

T2

T2c

�


H) Cs � Cn D �0 V T 2H20

T2c

˛ � 1C 3

T2

T2c

�1 � 4 ˛ C ˛2

C 15 ˛.1� ˛/T4

T4cC 14 ˛2

T6

T6c

�:

The critical point T D Tc is interesting:

.Cs � Cn/T DTc D 4�0 VH20

Tc.1C ˛/2 :

6. T < Tc

Sn.T/ ¤ Ss.T/

H) phase transition of first order.

T D Tc

Sn .Tc/ D Ss .Tc/ ;

Cn .Tc/ ¤ Cs .Tc/ (finite jump)

H) phase transition of second order.

Solution 4.3.4

T D Tc."C 1/ :

f .T/ can be written as follows as function of ":

f ."/ D a Tc."C 1/ ln jTc "j C b T2c ."C 1/2 :

The critical exponent is then determined in the following manner:

' D lim"! 0

ln jf ."/jln j"j D lim

"! 0

ln ja Tc."C 1/ ln jTc"jjln j"j

D lim"! 0

ln ja Tc" ln jTc "j C a Tc ln jTc"jjln j"j D lim

"! 0

ln ja Tc ln jTc "jjln j"j

D lim"! 0

ln ja Tcj C ln j ln jTc "jjln j"j D lim

"! 0

ln j lnTc C ln j"jjln j"j

D lim"! 0

ln j ln j"jjln j"j D lim

"! 0

1j ln j"jj

1j"j

1j"j

D lim"!0

1

j ln j"jj D 0 :


Solution 4.3.5 According to the Ehrenfest-classification, phase transitions of sec-ond order are characterized by finite jumps of the second derivatives of the freeenthalpy or the free energy:

f ."/ �! A˙ .T ! T.˙/c / I AC ¤ A�

H) ' D lim"! 0

ln jf ."/jln j"j D lim

"! 0

ln jA˙jln j"j D 0 :

Solution 4.3.6

1.

T D Tc."C 1/ H) f ."/ D a T5c = 2."C 1/5 =2 � b

H) ' D lim"! 0

ln jf ."/jln j"j D 0 :

2.

f ."/ D a T2c ."C 1/2 C C

Tc

1

"

H) ' D lim"! 0

ln j CTc "

jln j"j D � lim

"! 0

ln j"jln j"j D �1 :

3.

f ."/ D apTcp

j"j C d

H) ' D lim"! 0

ln jdjln j"j D 0 :

Solution 4.3.7 We use (2.82):

�T .CH � Cm/ D �0 V T ˇ2H I ˇH D�@M

@T

�H

H) 1 � R D �0 V T ˇ2H ��1T

C�1H :

Critical behavior T ! T.�/c :

M � .�"/ˇ I ˇ2H � .�"/2ˇ� 2 I ��1T � .�"/� 0 I C�1

H � .�"/˛0

H) 1 � R � .�"/2ˇ� 2C � 0 C ˛0

:


From this we read off:

1. R ¤ 1 W The preceding equation is satisfied only if:

2 ˇ � 2C � 0 C ˛0 D 0 ” ˛0 C 2 ˇ C � 0 D 2 :

2. R D 1 W Then the left-hand side of the above equation is zero and can thereforebe fulfilled only if

2 ˇ � 2C � 0 C ˛0 > 0 ” ˛0 C 2 ˇ C � 0 > 2 :

Solution 4.3.8 A consequence of the scaling hypothesis (4.76) is equation (4.77).In that equation we use

� D .˙"/�.1 = a"/

and obtain with H instead of B0 D �0H:

M.";H/ D .˙"/.1� aB/ = a" M�˙1; .˙"/�.aB = a"/H :

We apply (4.78) and (4.80):

1 � aBa"

D ˇ I aBa"

D ˇı :

Therewith it follows immediately the assertion:

M.";H/

.˙"/ˇ D M�˙1; .˙"/�ˇı H :

One measures the magnetizationM for a multitude of external fields H as functionof the temperature (or "). If one then plots

M.";H/

j"jˇ versusH

j"jˇı ;

then this multitude will reduce to two curves, one for T < Tc and one forT > Tc,provided that the scaling hypothesis is valid.

Solution 4.3.9 We apply:

(4.78) W ˇ D 1 � aBa"

;

(4.79) W ı D aB1 � aB

;


(4.81) W � D � 0 D 2aB � 1a"

;

(4.82) W ˛ D ˛0 D 2˛" � 1a"

:

1. �.ı C 1/ D .2 � ˛/ .ı � 1/holds if and only if

2aB � 1a"

1

1 � aB

ŠD 1

a"

2aB � 11 � aB

is fulfilled. That is obviously the case!2. ı D .2 � ˛ C �/ = .2 � ˛ � �/

is valid if

aB1 � aB

ŠD 2 � 2a"�1a"

C 2aB�1a"

2 � 2a"�1a"

� 2aB�1a"

is fulfilled. That is indeed the case:

aB1 � aB

ŠD 2a" � 2a" C 1C 2aB � 1

2a" � 2a" C 1 � 2aB C 1

” aB1 � aB

ŠD 2aB2 � 2aB

Solution 4.3.10

1. We can start with the law of corresponding states (1.19):

�� C 3

v2

�.3 v � 1/ D 8 t ;

pr D � � 1 I Vr D v � 1 I " D t � 1

H)h.1C pr/C 3 .1C Vr/

�2i Œ3 .Vr C 1/� 1� D 8.1C "/

H) �4C 2Vr C V2r C pr

�1C 2Vr C V2r

.3Vr C 2/

D 8.1C "/�1C 2Vr C V2r

:

Arranging a bit the last equation leads to:

pr�2C 7Vr C 8V2r C 3V3r

D �3V3r C 8 "�1C 2Vr C V2r

:


2. In the critical region all the three quantities pr, Vr and " become very small. As afirst approximation we can therefore linearize the equation of state from part 1.:

pr � 4 " :

In the next step of approximation we insert this again into the equation of state:

4 "�2C 7Vr C 8V2r C 3V3r

D �3V3r C 8 "�1C 2Vr C V2r

H) 0 � Vr

�3V2r C 12 "C 24Vr "C 12 "V2r

H) 0 � Vr

�V2r C 8Vr "C 4 "

:

This equation has the solutions:

V.0/r D 0 I V.˙/r D �4 "˙ 2p�"p

1 � 4 " :

T>! Tc ” "

>! 0 WOnly Vr D 0 can be a solution, because V.˙/r are complex.

T<! Tc ” "

<! 0 WWe know that the solution Vr D 0 is unstable. The reduced volume of the van

der Waals gas is therefore:

V.˙/r D �4 "˙ 2p�"p

1 � 4 " � ˙2p�" :

3. ˇ determines the behavior of the order parameter (4.52):

��

2 �CD 1

2

�� C

�CD Vc

2

VC � V�

V� VC

D 1

2

�Vc

V�� Vc

VC

�D 1

2

�1

V.�/r C 1� 1

V.C/r C 1

�

� 1

2

�1 � V.�/r � �

1 � V.C/r

D 1

2

�V.C/r � V.�/r

H) ��

2 �C� 2

p�"

H) ˇ D 1

2I critical amplitude B D 2 :

4. T D Tc means " D 0. Then the equation of state from part 1. reads:

pr D �3V3r�2C 7Vr C 8V2r C 3V3r

�1:


Expanding for small Vr:

pr D �32V3r

�1 � 7

2Vr C O �

V2r�

:

5. The critical exponent ı is defined in equation (4.57):

p.0/c D nR TcVc

D 8

3pc :

We have utilized thereby (1.17). Hence it is:

p � pc

p.0/c

D 3

8

�p

pc� 1

�D 3

8pr :

Furthermore it holds:

�

�C� 1 D Vc

V� 1 D 1

Vr C 1� 1 D �Vr

Vr C 1

D �Vr�1 � Vr C 0

�V2r:

On the critical isotherm we therefore have, when we exploit part 4. and applyVr ! 0 for p ! pc:

p � pc

p.0/c

� 9

16

ˇ̌̌ˇ ��C � 1

ˇ̌̌ˇ3

:

The comparison with (4.57) yields:

ı D 3 I D D 9

16:

6. Compressibility:

�T D � 1V

�@V

@p

�T

D � 1VVc

�@Vr

@p

�T

;

dpr D d

�p

pc� 1

�D 1

pcdp ;

�T D � 1V

Vc

pc

�@Vr

@pr

�T

:


Normalization factor:

�.0/Tc

D 1

p.0/c

D Vc

n R TcD 3

8 pc:

In the last step we utilized again (1.17):

�T

�.0/Tc

D �83

1

Vr C 1

�@Vr

@pr

�T

:

According to part 1. we have:

�@pr@Vr

�T

D �9V2r C 16 " .1C Vr/

2C 7Vr C 8V2r C 3V3r

��3V3r C 8 "

�1C 2Vr C V2r

�7C 16Vr C 9V2r

�2C 7Vr C 8V2r C 3V3r

2 :

a) T>! Tc � D �C; i.e., Vr D 0

H)�@pr@Vr

�T

Vr D 0

D 8 "� 14 " D �6 " H) �T

�.0/Tc

D 4

9"�1 :

This holds not only for T>! Tc, but even everywhere along the critical

isochore .Vr D 0/, .

H) � D 1 I C D 4

9:

b) T<! Tc

In the critical region we have now according to part 2.:

V2r � �4 " :

This means:�@pr@Vr

�"! 0

� 1

2.36 "C 16 "/� 1

456 " D 12 " ;

1

Vr C 1�!"!0

1 :


It remains:

�T

�.0/Tc

� �83

1

12 "D 2

9.�"/�1 :

so that the comparison with (4.55) yields:

� 0 D 1I C0 D 2

9D 1

2C :

Solution 4.3.11 Chain rule:�@V

@T

�p

�@T

@p

�V

�@p

@V

�T

D �1

” .V ˇ/

�@T

@p

�V

�� 1

V �T

�D �1

H) ˇ D �T

�@p

@T

�V

:

Especially for the van der Waals gas we get:

ˇ D �T

�n R

V � n b

�:

The expression in the bracket behaves analytically for T ! Tc, so that the criticalbehavior of ˇ corresponds to that of the compressibility �T .

Solution 4.3.12

1. According to (1.28) the equation of state of the Weiss ferromagnet reads:

M D M0 L

�mB0 C ��0M

kB T

�;

m��0M

kB TD M

M0

NV m

2��0

kB T(1.26)D bM 3 kB C �

kB T(1.30)D bM 3 Tc

T:

Therewith it follows immediately:

bM D L

b C 3 bM

"C 1

!:


2. L.x/ D .1 = 3/x � .1 = 45/x3 C 0.x5/

B0 D 0 H) b D 0 ;

T <! Tc H) bM very small.

Then:

bM � bM"C 1

� 3

5

bM3

."C 1/3

H) "

"C 1� �3

5

bM2

."C 1/3H) bM2 � �5

3"."C 1/2 :

Since ."C 1/2 ! 1 for T ! Tc, we have:

bM �r5

3.�"/1= 2 :

As for the van der Waals gas we get the critical exponent:

ˇ D 1

2:

3. Critical isotherm: T D Tc I B0 ! 0

H) " D 0 I bM and b very small.

This means:

bM � 1

3b C bM � 1

45

�b C 3 bM�3

H) 15 b ��b C 3 bM�3 ” b C 3 bM � .15 b/1= 3

H) 3 bM � .15 b/1= 3 � b � .15 b/1= 3 ; da b ! 0 :

That yields

b � 9

5bM3

and leads to the critical exponent

ı D 3 :


4.

�T D�@M

@H

�T

D M0 �0m

kB T

@bM@b

!T; bD 0

D 3

�."C 1/

@bM@b

!T; bD 0

:

In the critical region bM is very small:

@L

@b

ˇ̌̌ˇbD 0

D @x

@b

�1

3� 1

15x2�ˇ̌̌ˇbD 0

C : : :

@bM@b

ˇ̌ˇ̌ˇbD 0

D 1C 3

"C 1

@bM@b

ˇ̌ˇ̌ˇbD 0

! 1

3� 1

15

9 bM2

."C 1/2

!C : : :

H) @bM@b

ˇ̌ˇ̌ˇbD 0

1 � 1

"C 1C 9

5

bM2

."C 1/3

!D 1

3

1 � 9

5

bM2

."C 1/2

!:

T ! Tc means bM ! 0:

@bM@b

!T; bD 0

� 1

3

1

""C1 C 9

5bM2

."C1/2:

a) T>! Tc WAbove Tc we have bM � 0, so that with ."C1/ ! 1 for T ! Tc it follows:

@bM@b

!� 1

3"�1 :

This means for the susceptibility:

�T � 1

�"�1 H) � D 1 :

b) T<! Tc WAccording to part 2. we have to now insert bM2 � 5 = 3.�"/:

�T � 1

2�.�"/�1 H) � 0 D 1 :

The critical amplitude is equal to that of the van der Waals gas:

C0 D 1

2C :

Index

AAdiabatic compressibility, 57–61, 76Adiabatic demagnetization, 104, 110Adiabatic equations, 36, 37, 63, 70, 189, 194,

195, 212, 217, 218Adiabatic state changes, 35, 38, 107, 188, 190,

191, 194, 213Adiabatic susceptibility, 60, 110, 248Avogadro constant, 9

BBasic relation of thermodynamics, 53, 56, 75,

77, 95, 198Black body radiator, 32, 36–38Boltzmann constant, 9Boyle-Mariotte’s law, 8Boyle temperature, 25, 170

CCaloric equation of state, 77, 84, 107, 228,

229Carnot cycle, 41–45Celsius scale, 8–9Chemical potential, 31, 83, 85, 97, 98, 100,

108, 115, 120, 121, 123Classical theories, ix, 138, 141–145, 151Clausius-Clapeyron equation, 68, 123–124,

128, 134, 156, 157, 263–266Clausius inequality, 49, 51–52Clausius’ statement, 39, 40Closed system, 3, 31, 33, 55, 97–101, 120–122,

184Complete set, 4, 5Compressibility, 60, 76, 107, 130, 142, 159,

274, 276

Concave function, 129, 130, 148Continuous phase transition, 135, 136Convex function, 129Coopersmith-inequality, 146–147Corresponding states, 12, 169, 272Critical amplitude, 142, 154, 273, 278Critical exponent, 136–155, 158–160, 169,

269, 274, 277Critical fluctuations, 137Critical isotherm, 143–144, 149, 159, 274, 277Critical point, 11–12, 122, 132, 136, 138, 168,

169, 268, 269Critical temperature, 15, 136, 144Curie constant, 15, 62, 70, 110, 112, 143, 232Curie law, 15, 62, 70, 107, 177, 187, 222–223,

237, 249Curie temperature, 15–17, 141Curie-Weiss law, 17, 108, 238

DDiesel-process, 67Dieterici gas, 24, 25, 171, 172Differential Joule-Thomson coefficient, 92–94Differential work, 18Discontinuous phase transition, 135–136

EEfficiency, 40, 42–48, 66–70, 204, 208, 210,

213–215, 220, 221Ehrenfest classification, 132–136, 270Ehrenfest equations, 134–135Enthalpy, 79–81, 83, 85, 86, 92, 99, 104, 108,

111–112, 120, 123, 128–130, 135,151, 152, 155, 229, 232, 240, 241,258, 261, 270

© Springer International Publishing AG 2017W. Nolting, Theoretical Physics 5, DOI 10.1007/978-3-319-47910-1

279

280 Index

Entropy, 4, 30, 35, 48–55, 58, 63–66, 70, 77,84, 87–91, 94–96, 100–111, 118,131, 132, 135, 148, 150, 158, 182,192, 193, 199, 203–207, 217, 221,222, 228–234, 236–240, 244–247,250–251, 255, 260, 262

Entropy of mixing, 87–91, 108, 203, 204,236–237

Equation of state, 9–12, 14–17, 23, 24, 26, 36,45, 55, 56, 58, 61, 64, 65, 77, 84,86, 92–94, 101, 106, 107, 113, 124,148, 159, 167, 168, 172, 179, 180,184, 189, 194–197, 208, 212, 214,215, 227–229, 236, 273, 276

Equation of state of the ferromagnet, 16, 17,26, 148

Equation of state of the ideal gas, 9–10, 36, 45,84, 93, 184, 189, 201

Equation of state of the ideal paramagnet,14–15, 62, 109

Equilibrium, 4–8, 13, 20, 30, 36, 53, 54,71, 75–77, 91, 95–101, 112, 117,118, 120–123, 125, 126, 132, 157,189–190, 223–224, 256–257, 267

Equilibrium conditions, 77, 95–101, 117, 121,123, 125, 126

Equilibrium state, 5, 53, 75, 95, 118Evaporation heat, 131, 156Exchange constant, 16Experiment of Gay-Lussac, 32, 56, 85, 198,

249Exponent inequalities, 144–151Extensive state variables, 4

FFerromagnetism, 14, 17First law of thermodynamics, 1, 29–33, 35, 38,

42, 49, 50, 53, 91, 181, 184, 186,188, 197–198, 209, 216, 219, 220,258

Fluid system, 60, 136Free energy, 78, 86, 97, 103, 106–111, 126,

127, 129–131, 147, 148, 226,227, 229–231, 233, 234, 237, 240,246–248, 258, 260, 261, 270

Free enthalpy, 80, 81, 83, 85, 86, 92, 99, 104,108, 111, 120, 123, 128–130, 151,152, 155, 229, 232, 240, 241, 258,261

GGeneralized coordinates, 18, 33, 79, 82Generalized force components, 18

Generalized homogeneous function, 151, 152Gibbs-Duhem relation, 83, 123, 125, 229Gibbs free enthalpy, 80, 81Gibb’s paradox, 91, 237Gibbs phase rule, 117–122Gibb’s potential, 99Griffiths-inequality, 147–151

HHeat, ix, 1, 29, 124, 184Heat bath, 3, 8, 9, 36, 39–41, 43–46, 48, 49,

53, 54, 65, 66, 89, 97–100, 110, 204,213, 215, 219, 222

Heat capacity, 33–35, 37, 55, 56, 61–66, 70,71, 83, 102–103, 107–113, 133,141, 150, 155, 157, 158, 184, 195,196, 204, 216, 225, 231, 233, 234,238–239, 248, 261, 265

Heat engine, 39–42, 44, 45, 49, 67, 70, 219Heat-exchange contact, 3Heat pump, 42, 44, 49Heat radiation, 36Heisenberg model, 138, 142–144Homogeneity relations, 81–83, 245Homogeneous function, 151, 152Homogeneous phase, 13, 81, 125H2O-phase diagram, 132

IIdeal gas, 8–10, 13, 14, 25, 32, 34, 36, 38,

42, 45, 48, 53, 54, 56, 61–67, 69,83–89, 93, 102, 108, 112, 142, 143,156, 171, 175, 181–185, 189–191,194–196, 198, 199, 204, 206, 207,209, 214, 236, 249, 257, 264

Ideal paramagnet, 14–15, 25, 62, 109, 110,112, 143, 244

Independent state variables, 4, 7, 19, 30, 55,58, 76, 97

Isolated system, 3, 5, 6, 30, 31, 53, 73, 95, 96,97, 98, 115, 17, 120, 160

Integrability conditions, 20, 55, 76, 108, 109,181, 183, 192, 193, 216, 222, 232,238–241, 243

Integrating factor, 19, 51, 61, 182Intensive state variables, 4–7, 81Internal energy, 29–33, 36, 37, 56, 64, 65, 70,

75–78, 81, 85, 89, 95, 106–112, 194,195, 199–200, 207, 209, 214, 218,229–232, 234, 236–239, 245, 246,248, 251

Intrinsic pressure, 11, 171

Index 281

Inversion curve, 94Irreversible intermixing, 87, 115, 203Ising model, 138, 141–145, 151, 152Isobaric expansion coefficient, 113Isobaric thermal expansion coefficient, 23, 57,

159Isothermal compressibility, 24, 25, 32, 57,

107Isothermal expansion of the ideal gas, 53Isothermal magnetic susceptibility, 108Isothermal state changes, 37–38, 214, 222,

227Isothermal susceptibility, 60, 109, 147

JJoule-Thomson coefficient, 92–94Joule-Thomson effect, 91–95Joule-Thomson throttling experiment, 92

KKadanoff construction, 152Kelvin scale, 9Kelvin’s statement, 39, 40, 68

LLagrange multipliers, 118, 119Lagrange parameters, 119, 120Langevin function, 15, 160Latent heat, 128, 131, 135, 157Law of corresponding states, 12, 169, 272Legendre transformation, 77–81, 241Lever rule, 127Line of coexistence, 123, 128, 132, 134, 156,

157, 264–266

MMagnet, 4, 14–18, 21, 22, 25, 26, 35, 60,

70, 71, 104, 108–110, 112, 113,130, 135, 136, 138, 141–144, 151,157–160, 176, 179, 261

Magnetic moment, 4, 14, 18, 21, 26, 109, 112,143, 160, 176, 179, 222

Magnetization, 4, 14–17, 21–22, 70, 104,108–110, 112, 142, 147, 148, 150,157, 159, 160, 176, 271

Magnetization work, 21–22Mathematical formulation of the second law,

52–53Maxwell construction, 124–127, 174,

181

Maxwell relation, 76, 79–81, 92, 103, 104, 134,147, 148, 197, 229–231, 240, 241,253, 255, 258, 261–262, 264–265

Meissner-Ochsenfeld effect, 157, 267Method of Lagrange multipliers, 118Molar evaporation heat, 156Molecular heat, 33–35Mole volume, 124

NNatural state variables, 75–77Natural variables, 76–79, 81, 84–86, 95, 229,

233Nernst’s heat theorem, 101–105Nernst theorem, 103

OOpen system, 3, 31–32Ordering axiom, 6Order of the phase transition, 132, 133Order parameter, 141–142, 150, 273Otto motor, 213

PParamagnetism, 14, 16, 17Particle density, 13, 25, 117Particle-exchange contact, 41Perpetuum mobile of the first kind, 30–31, 157Perpetuum mobile of the second kind, 39Phase, 10, 101, 117, 169Phase transition, 10, 13, 117–160, 169, 174,

264, 269, 270Phase transition of first order, 124, 128,

131–133, 135, 136, 265, 269Phase transition of second order, 133–136,

169, 269Photon, 2, 32, 37, 38, 56, 65, 106, 228, 229Photon gas, 32, 37, 38, 56, 65, 106, 228, 229Piezoelectricity, 108Piston pressure, 20, 21Postulate of homogeneity, 152–156Potentials of the ideal gas, 83–86Power-law behavior, 137Process, 2, 5–6, 35, 38–40, 42, 43, 51–55, 58,

59, 61, 66–71, 75, 77, 88–92, 94,96, 97, 99–101, 104, 105, 111, 128,145, 189, 191, 197, 198, 203, 205,209, 210, 214, 219–221, 225, 226,263, 267

Proper volume, 10, 25, 171, 173

282 Index

QQuasi-static, 5, 18, 21, 48, 62, 63, 66, 75, 88,

222

RRadiation pressure, 37Rüchhardt experiment, 63Relaxation time, 5Reversible auxiliary process, 52, 88, 89, 115,

203, 221Response function, 57, 60, 76, 103, 114, 133Reversible, 5, 7, 39, 41, 42, 44–46, 48, 51–55,

59, 63, 64, 66–68, 75, 88–91, 101,104, 105, 107, 111, 178, 192, 194,197, 201, 203, 205, 221, 225, 226,235, 241, 250

Rushbrooke-inequality, 145

SSaturation magnetization, 15, 147, 160Scaling hypothesis, 145, 151–156, 159, 271Scaling laws, 139, 155Second law of thermodynamics, 1, 2, 10,

38–40, 43, 44, 48, 50, 52, 53, 95,101, 105, 197

Semipermeable wall, 88, 89, 91Sign convention, 18, 231Specific heat, 33, 72Spin dimensionality, 137Spontaneous magnetization, 16, 142, 147, 148,

150Stability conditions, 129State, 4, 30, 75, 117, 167State functions, 4, 19, 32, 35, 58State space, 4, 5, 19, 51, 101, 244State variables, 4–7, 12, 19, 30, 33, 55, 58, 60,

75–77, 79, 81, 97, 102Stefan-Boltzmann law, 56, 73Stirling cycle, 69Superconductor, 133, 134, 157, 158, 266, 268Superconductor of the first kind, 157

TT dS-equations, 73Term of temperature, ix, x, 6–7Theorem of Nernst, 101–105Thermal contact, 3, 7, 65, 98Thermal equation of state, 24–25, 58, 61, 107,

195, 196, 236Thermal expansion coefficient, 57, 107, 111,

156, 234, 263

Thermodynamic cycle, 6, 31, 41, 45, 49, 50,66–70, 209, 212

Thermodynamic potential, 75–113, 230, 241Thermodynamic potentials of the ideal gas,

83–86Thermodynamics, v, ix, 1, 29, 75, 117, 181Thermodynamic system, 2–4, 6, 38, 39, 41,

48–50, 65, 77, 81, 117Thermodynamic temperature, 10, 45–48, 53Thermodynamic temperature scale, 45–48Thermometer, 7Third law of thermodynamics, 2, 101–105,

113, 158, 229, 249, 263, 265, 268Throttled adiabatic relaxation, 91Total differential, 19, 22, 31, 51, 55, 75, 76,

78–81, 163, 164, 166, 167, 181, 182,186, 192, 197, 216, 222, 241

Transformation heat, 128, 131, 135, 157, 265,266

Transitivity, 6Triple point, 48, 122

UUnattainability of the absolute zero, 2, 104–105Universal gas constant, 9, 24Universality hypothesis, 138, 169

VVan der Waals equation of state, 11, 12, 94,

168Van der Waals gas, 10–14, 56–57, 63, 64,

93–95, 124, 134, 141, 159, 160,193–196, 276–278

Van der Waals model, 12, 171Vapor pressure, 123–124, 128, 156, 258Vapor pressure curve, 123–124Virial coefficient, 13, 25Virial expansion, 13–14, 25Volume work, 20–21, 58

WWidom-inequality, 156Work-exchange contact, 3

XXY model, 138, 142–144

ZZeroth law of thermodynamics, 1, 6

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