THERMODYNAMICS AND STATISTICAL

MECHANICS热力学统计物理

School of Science, Jinan UniversitySchool of Science, Jinan University

Chapter 1 Equilibrium and State Quantities

I Thermodynamics

Thermo- thermometer （温度计）

Chapter 2 The Laws of Thermodynamics

Chapter 3 Phase Transitions

Chapter 4 Thermodynamic Potentials

Exercises-1

Ⅱ Quantum Statistics

Chapter 5 Number of Microstates Ω and Entropy S

Chapter 6 Applications of Boltzmann Statistics

Chapter 7 Bose and Fermi statistics

Chapter 8 Ensemble Theory and Microcanonical Ensemble

Exercises-2

Thermodynamics and statistical mechanics

I Thermodynamics

Chapter 1 Equilibrium and State Quantities

1 Introduction

As we know,many –particle systems can be found everywhere in nature.for example:the gases,fluids and solids.In the following we will see that all these completely different systems obey common and very

（平衡状态）

general physical laws.In particular, we will discuss the properties of such many-particle systems in thermodynamic equilibrium.

The task of thermodynamics is to define appropriate physical quantities(the state quantities).which characterize macroscopic properties of matter,the so-called macrostate,in a way which is as unambiguous as possible,and to relate these quantities by means of universally valid equations(the equations of state and the laws of thermodynamics).

Hence in the beginning,we have to define certain state quantities to formulate and substantiate （证实 substance- 物质，事实） the laws of thermodynamics,called state variables.

2 systems, phases and state quantities

The concept of a thermodynamics system can be uniquely and completely described by specifying certain macroscopic parameters

a.Isolated systems （孤立系）

These do not interact in any way with the surroundings.

b.Closed systems （闭系）

Here one allows only for the exchange of energy with the surroundings,but not for the exchange of matter.

c.Open systems （开系）

These systems can exchange energy and matter with their surroundings.Hence,neither the energy nor the particle number are conserved quantities.

It is obvious that at least the isolate system is an idealization,since in reality an exchange of energy with the surroundings cannot be prevented in the strict sense.

If the properties of a system are the same for any part of it,one calls such a system homogeneous. However,if the properties change discontinuously at certain marginal surfaces,the system is heterogeneous. One calls the homogeneous parts of a heterogeneous system phases and the separating surfaces phase boundaries. A typical example for such a system is a closed pot containing water,steam and air.

In general one distinguished two classes of state quantities:

a.extensive (additive )state quantities （广延量） These quantities are proportional to the amount of matter in a system.e.g.to the particle number or mass.Characteristic examples are the volume and the energy.

The macroscopic quantities which describe a system are called state quantities. The energy E,the volume V,the particle number N,the entropy S （熵） ,the temperature T,the pressure P and the chemical potential μ.

b.Intensive state quantities （强度量）

Intensive- 密集的；集中的 ；强烈的；增强的Extensive- 广阔的；广博的；大量的；【物】广延的；【逻】外延的

These quantities are independent of the amount of matter and are not additive for the particular phases of a system.Examples are:refractive index.density,pressure,temperature.etc.

3 Equilibrium and temperature-the zeroth law of thermodynamics

Temperature is a state quantity which is unknown in mechanics and electrodynamics. It is specially introduced for thermodynamics,and its definition is closely connected with the concept of (thermal) equilibrium. Thermodynamics state quantities are defined (and measurable)only in equilibrium.

Here the equilibrium state is defined as the one macroscopic state of a closed system which is automatically attained after a sufficiently long period of time such that the macroscopic state quantities no longer change with time.

One calls it also the zero law of thermodynamics.

As experience has shown, all systems which are in thermal equilibrium with a given system are in thermal equilibrium with each other. Equality of temperature of two bodies is the condition for thermal equilibrium between these bodies.

Hence systems which are in thermal equilibrium with each other have a common intensive property.which we denote as temperature.

Today one usually takes the melting point of ice as T=273.15K,where the unit is named in honor of Lord Kelvin,who made important contributions to the field of thermodynamics.

Historically,the unit of temperature was fixed by defining the temperature of the melting point of ice as 00C and that of boiling water as 1000C(at atmospheric pressure),which is Celsius’scale. The conversion to Fahrenheit’s scale is

329/5 00 FxCy

AB

酒精或水银

Example:1.1 the ideal gas

Such an ideal gas is characterized by the fact that the particles are considered (as in classical mechanics) to be noninteracting ,pointlike particles.

.0

00 constT

VP

T

PV

Since the expression PV/T is an extensive quantity ,

kTorPNkTPVNkT

VP

T

PV ,0

00 (1.2)

Boltzmann’s constant of proportionality k=1.380658×10-23Jk-1

We now show that the temperature of an ideal gas can be very simply understood as the mean kinetic energy of the particles. We write for the number of particales dN( ) in the velocity interval around

where is the velocity distribution , and of course it must hold that

v

v

vdvNfdN 3

vd

dN

Nvf

3

1

vf

13

vdvf

(1.3)

As mentioned above,the pressure of the gas originates from the momentum transfer of the particles when they are reflected at surface A.(e.g.,the wall of the box)

3 Kinetic theory of the ideal gas

Figure 1.4 scheme for calculating the pressure

A particle of velocity which hits that area transfers the momentum p=2mvz · Now the

v

v

question is how many such particles with velocity vector hit the surface element A during a time dt?As one infers from Figure 1.4,these are just all the particles inside a parallelepiped with basis area A and height vzdt.

the number of particles with velocity inside the parallelepiped is just

v

vdvfV

dVNdN

3 (1.4)

It holds that dV=Avzdt. Each particle transfers the momentum p=2mvz ， so that the impulse per area A is

V

AdtvdvfNmvdNmvdtdF zzA

3222

(1.5)

the total pressure then results by integrating over all possible velocities with a positive component vz (since otherwise the particles move in the opposite direction and do not hit the wall)

2

02

1zzyxA mvvfdvdvdv

V

NdF

Ap

(1.6)

Since the gas is at rest,the distribution cannot depend on the direction of ,but only on .Then,however, we can write the integral as and thus obtain

vf v

v

0 zdv

zdv21

23zvvfvdmNpV

(1.7)

The value ,however,has to be the same in all spatial directions because of the isotropy of the gas,i.e.,

22223yxzz vvvvvfvd

(1.8)

Or,since 2222zyx vvvv

22222

3

1

3

1yxzz vvvvv

(1.9 ）

So that we finally have

kinNvmNpV 3

2

3

1 2

Here is the mean kinetic

energy of a particle,I.e.,the quantity kT exactly measures this mean kinetic energy of a particle in an ideal gas.

kTvmkin 2

3

2

1 2

（ 1.10)

We will now determine the functional form of the velocity distribution in greater detail. Be- cause of the isotropy of the gas, can be a function only of ,or equivalently of .

Example 1.2 Maxwell's velocity distribution

vf

v 2v

2222222zyxzyx vfvfvfvvvfvf

三项同时确定 = 先确定一项，依次确定另外两项

(1.11)

The only mathematical function which fulfills the relationship(1.11)is the exponential function ,so that we can write 22 exp vaCvf

If we assume the function to be normalizable, it must obviously hold that a 〈0.The constant C can be determined from the normalization of the function for each component,

2vf

ivf

James Clerk Maxwell 1831 ～ 1879

2exp1 iiii avdvCvfdv

aC

Now we are even able to calculate the constant for our ideal gas ,if we start from Equation (1.10)

kTvmkin 2

3

2

1 2

kTvm z 2

1

2

1 2

232zz vvfvdmvmkT

2zzzyyxx vvfdvvfdvvfdvm

2zzz vvfdvm

22exp zzz vavdva

m

because of the isotropy of the gas,i.e.,

vf vf 2vf

22

0exp2 zzz vavdv

amkT

If we substitute x= we find,with 2zav

x

dx

advz

2

1

xedxaa

amkT x

02

12

1

0

zx xdxez

2/1

11

2/2/3

zzz 1

kT

ma

2

kT

mv

kT

mav

avf i

ii 2exp

2exp

22

kT

vm

kT

mvf

2exp

2

22/3

(1.12)

（ 1.13)

Cmu 12

12

11 (1.14) 23100221367.6

1

1

u

gN A (1.15)

4 PRESSURE WORK AND CHEMICAL POTENTlAL

If a system consists of several kinds of particles,for instance NI,N2,...,Nn particles of n species,the SO-Called molar fraction X is a convenient quantity for measuring the chemical constitution,

n

ii NNN

NX

21

（ 1.16)

1i

iX

kinNNkTpV 3

2

kinkinV

N

V

NkTp

3

2

3

2

i.e.,the pressure is the (kinetic)energy density of the ideal gas.

As an example for the form of energy mentioned above we will use the concept of work from mechanics in thermodynamic problems.

(1.10)

We have

{1.18}dsFW i

Figure 1.5 Concerning the compressional work

As an example for work performed on a system we consider the compression of a gas against its internal pressure (Figure 1.5).

PAFi

If one pushes the piston a distance ds further into the volume against the force exerted by the system,the amount of work needed is just

0PAdsW

PdVW

Now A ds=-dV is just the decrease of the gas volume dV 〈 O in the container,and we have {1.20}

PdVW 2

1

v

v

We can illustrate this with further.

dqW (1.21)

elel DdEW

magmag DdBW

(1.22)

(1.23)

To complete our list of possible realizations of work we consider the work necessary to add another particle to a thermodynamic system.

dNW (1.24 ）

5 Heat and heat capacity

That heat is a special form of energy.

CdTQ (1.25 ）

To fix a unit for C we have to define a standard system;this is the amount of heat which warms 1g water from 14.50C to 15.50C. This corresponds to the definition =1cal/OC.

COgHC 0

2 15.1

1 cal of heat is equivalent to 4.184 Joule

The SI unit for the heat capacity is also J/K.

One can define an intensive quantity,the specific heat c,via

mcC (1.26)

Or molncC ANNn /

6 THE EQUATION OF STATE FOR A REAL GAS

NkTPV

We have the ideal gas law:

NAk=R=8.31451J K-l mol-l

The constant NAk=R is named the gas constant.

(1.29)

In thermodynamics one often assumes equations of state to be polynomials of a variable. If Equation (1.29)is correct for low pressures(p O),

As a first approximation terminates the virial expansion Equation (1.31) after the linear term;the coefficient B(T) can be determined experimentally.The quantity B(T) is called the first virial coefficient.

(1.31)pV=NkT+B(T)p+C(T)p2+ …

2' )('

V

NTC

V

NTBNkTPV

If one does not expand the equation of state of a real gas in terms of low pressure,but rather in terms of low density,one obtains an analogous equation,

We substitute for V the quantity V-Nb,where b is a measure for the proper volume of a particle.

On the other hand,the pressure of a real gas has to be smaller than that of an ideal gas. We can account for this in Equation （ 1.29 ） if we substitute the ideal gas pressure by ,were is the so-called inner pressure.

idP 0PPreal

0P

Another well-known equation of state for real gases is the equation of van der Waals (1873).

van der Waals1837~1923

0P is,however,not simply a constant,but depends on the mean distance between the particles and on how many particles are on the surface .

20 /VNaP

NkTNbVVNaP 2/

(1.33)

Here a and b are material constants,which are mostly cited per mole and not per particle.

Finally we investigate an equation of state corresponding to an ideal gas equation for solids.In this case,the temperature and pressure dependence of the volume is given in a range of values by

V(T,p)=V0{1+ α (T-T0)-K(p - po)}

7 Specific heat

(1.37 ）

i.e.,by a linear approximation. Here V( T0 ,po)=V0 is an arbitrary initial state.Theconstants α and k,  

00

1

PPT

V

V

00

1

TTP

V

V

（ 1.38)

(1.39)

are called the coefficient of expansion (at constant pressure)and compressibility (at constant temperature),respectively.

8 Changes of state-reversible and irreversible processes

It is a daily experience that a process in an isolated system proceeds by itself until an equilibrium state is reached.Since such processes do not reverse themselves,they are called irreversible.

It is characteristic for irreversible processes that they proceed over nonequilibrium states.

On the other hand,processes which proceed only over equilibrium states are called reversible.

Such changes of state are also called quasi reversible.

The importance of reversible changes of state is the following:for every small step of the process the system is in an equilibrium state with definite values of the state quantities,so that the total changes of the state variables can be obtained by integrating over the infinitesimal reversible steps.For irreversible processes this is not possible.

Reversible changes of state,however,can be simulated by small (infinitesimal) changes of the variables of state,where the equilibrium state is only slightly disturbed,if these changed happen sufficiently slowly compared to the relaxation time of the system.

Example1.3: Isothermal expansion

We consider the expansion of a gas at constant temperature (isothermal expansion,see Figure 1.12).

Figure 1.12 Isothermal system

We can simply accomplish the isothermal expansion of the gas from the volume V1 to V2 by removing the external force Fa which acts on the piston and maintains equilibrium.

This process happens by itself and would never reverse itself.Therefore it is irreversible.

However,we can also perform this isothermal expansion reversibly,or at least quasi reversibly,if we decrease the force at each step only by an infinitesimal amount and wait for the establishment of equilibrium in the new situation.

If we consider an ideal gas in our case,we have p=NkT/V,and we can calculate the total amountof work performed in the expansion of the system,

1

22

1ln

2

1

2

1 V

VNkT

V

dVNkTPdVdW

V

V

V

V

Real expansions,of course,lie between the extreme cases of the completely irreversible expansion ( W =0) and the completely reversible expansion ( W= -NkTln(V2/V1)).

By the way, this is also the case if we consider isothermal compression.For a reversible process we need in this case the work

Here it is assumed that in each step the force exerted on the piston is only infinitesimally increased.If we instead push the piston spontaneously with large effort,we have to spend more work,which is consumed in turbulences and finally transferred to the heat bath in the form of heat.

0lnln1

2

2

11

2

1

2

1

2

V

VNkT

V

VNkT

V

dVNkTPdVdW

V

V

V

V

Exercises:

1. Ask for the ideal gas coefficient of expansion (at constant pressure)αand the compressibility(at constant temperature)κ.

(1) If at constant volume, how much does the pressure increase.

(2) If the pressure increase to 100Pn, how much does the volume change.

2.At 1pn and in 00C,the coefficient of expansion and the compressibility of a cuprum block are 4.85×10-

5K-1 and 7.8×10-7 Pn-1 respectively,we assume that they are constants,now,in order to warm the cuprum block into 100C

3. 1mol of ideal gas proceed an isothermal expansion.And their pressure decrease from 20Pn to 1Pn,consider the gas performed work and the heat which is absorbed by the gas during the process.

4. At 1pn and in 00C,the density of the atmosphere is 1.29kgm-3,and its specific heat ,here is an ideal gas in a container whose volume is 27m3 , (1)if at constant volume,please calculate the amount of heat needed to warm the gas from 00C to 200C.

11996.0 KkgJc p11706.0 KkgJcV

(2)if at constant pressure,please calculate the amount of heat needed to warm the gas from 00C to 200C.

(3)if the container has a split, and the external pressure is always 1pn,please calculate the amount of heat needed to warm the gas from 00C to 200C.