Fundamentals of thermodynamics

Study what rules must follow when the macroscopic parameters of thermodynamic system change.

§7-1 Internal Energy Heat &Work

its surroundings

1. Two basic concepts: is a system that can interact with its surroundings in at least two ways, one of which must be heat transfer.

Thermodynamic system:

: everything outside the system.

Surroundings(exterior)

Thermodynamic system

Thermodynamic systems in engineering:

Thermodynamic system

working substance

Gas: such as air. Vapor: such as steam. Liquid: such as freon( 氟利昂 ). Solid: such as semiconductor……..They are also called as working substance

Depends on T of the system only.

2. Internal energy

Internal energyIdeal gas

RTiTE2

M)(

ji

piji

kiE,

state quantity

dl

S

FdldW pSdl pdV

2

1

V

VpdVW p

VdV

pThe value of the work The value of the work equals the total area under equals the total area under the process curve of the process curve of pp~~VV diagram.diagram.

F

1V 2V

1

2

SF

SF

SF

SF

SF

SF

3.The work done by system during equilibrium 3.The work done by system during equilibrium processprocess

Work is process quantity

4. Heat4. Heat ---- macroscopic displaying ---- macroscopic displaying of thermal motionsof thermal motions

Transfer heatTransfer heat ：： one form of exchanging one form of exchanging energy between system and its environment energy between system and its environment as the temperature difference.as the temperature difference.

the molecules of the system and the molecules of the environment exchange their thermal kinetic energies.

Process quantity

5. The relationship between work and heat5. The relationship between work and heat

Heating Heating

stirringstirring

Work and heat have same effect for the system.Work and heat have same effect for the system. Joule proved that 1 cal of heat causes the same temperature increment as that of 4.18 joule of work does.i.e. 1cal= 4.186 J

Stipulation:Stipulation:

E:E: “+”“+”the internal energy of system increasesthe internal energy of system increases

WW:: “+” “+” thethe system does work on exterior.

Q:Q: “+” “+” the system absorbs heat from exterior the system absorbs heat from exterior “ “ －” －” the system gives out heat to exteriothe system gives out heat to exterior r

““ －” －” the internal energy of system decreathe internal energy of system decreasesses

““ －” －” exterior does work on the system.

1. The first law …1. The first law …WEEQ 12 WE

§7-2 The First Law of Thermodynamics

Note The FLT …can be used for any thermodynamic

process of any thermodynamic system.

---Conservation law of energy including the heat phenomena and thermal motion.

differential form ： for the equilibrium process of ideal gas ：

2

1

)(2 12

V

VpdVTTRiMQ

dWdEdQ

pdVRdTiMdQ 2

There is not any system(machine) that can do work forever without the supply of heat.

It is impossible to make out the first kind of perpetual motion machine.

Or

Other description of the FLT:

第一类第一类永动机：永动机：EE2 2 - - EE11= 0= 0 （循环）（循环）Q Q = 0= 0 （外界不供给能量）（外界不供给能量）W W > 0> 0 ( ( 对外界作功对外界作功 ))

Heat Worksystem

“Heat” and “Work” can not be transformed without thermodynamic system.

0dV pdVdW 0W

EQ

1. 1. Isochoric (Constant volume) process((VV=Const=Const..))

0

)(2 12 TTRiM

A

Bp

V0

0Q

§7-3 Application of the First Law to isochoric, isobaric & isothermal Processes

VPPi )(2 12

IfIf Q Q>0>0, then , then EE>0>0 ：：Absorb heat = increment of Absorb heat = increment of internal energy

　22.Isothermal (Constant temperature) process ((TT=Const.=Const.))

0dT 0E

WQ 2

1

V

VdV

VRTM

1

2lnVVRTM

Or 2

111 ln

ppVpQ

2

1

V

VpdV

DiscussionDiscussionWhenWhen Q Q>0>0 ，， we have we have VV22>>VV1 1 or or

pp1 1 >> pp22..The isothermal The isothermal expansion process.expansion process. Absorb Absorb heat from exterior heat from exterior == work for work for exterior.exterior.

A

B

p

V0 The isothermal expansion

process

0Q

WhenWhen Q Q<0<0 ，， we have we have VV22<<VV11. . The isothermal compression The isothermal compression process process . Exterior does work . Exterior does work to the system to the system == the system the system withdraws heat to the withdraws heat to the exterior.exterior.

3. 3. Isobaric ( Constant pressure) process((PP=Const.=Const.))

2

1

V

VdVpEQ

)()(2 1212 TTRMTTRiM

)(2

212 TTRiM

Or )(2

212 VVpiQ

)( 12 VVpE

DiscussionDiscussion WhenWhen Q Q>0>0 ，， TT22>>TT11((VV22>>VV11)) ，， EE>0 >0 , , WW>0>0. . IsoIso

baric expansionbaric expansion. The system absorbs heat. Par. The system absorbs heat. Part of heat is used for increasing the internal enet of heat is used for increasing the internal energy. Part of heat is used for doing work to extergy. Part of heat is used for doing work to exterior.rior.

WhenWhen Q Q<0<0 ，， TT22<<TT11 ，， EE< < ,,WW<0<0 ：： Isobaric Isobaric expression. expression. The work done by exterior for the The work done by exterior for the systemsystem++the decrease of the internal energy = tthe decrease of the internal energy = the heat withdrawning from the system.he heat withdrawning from the system.

QQ, , AA andand EE always have the always have the same “ same “ positive-negative signpositive-negative sign” ” in isobaric process. in isobaric process.

p

V0

A B

Isobaric expansion

0Q

2

i

iQE

22

iQ

W

Furthermore

Constant volume RdTidEdQ2

Constant temperatureConstant temperature pdVdQ

Constant pressureConstant pressure RdTidQ2

2

4.Differential forms of three processes above4.Differential forms of three processes above

[[ExampleExample] A monatomic ideal gas is contained ] A monatomic ideal gas is contained in a cylinder closed with a movable piston. The in a cylinder closed with a movable piston. The initial pressure is initial pressure is 33atm and the initial volume atm and the initial volume 11ll. The gas is heated first at constant pressure u. The gas is heated first at constant pressure until the volume become to ntil the volume become to 22ll. Then expands wit. Then expands with constant temperature until the volume becoh constant temperature until the volume become to me to 33ll. Finally cooled down at constant volu. Finally cooled down at constant volume until the pressure drops to me until the pressure drops to 11atm. Calculateatm. Calculate: The increment of internal energy, the work d: The increment of internal energy, the work done by the gas and the heat supplied to the gas one by the gas and the heat supplied to the gas in the three process.in the three process.

Pa)10013.1( 5p

)10( 33 mV

a b

cd1

1

2

2

3

3

SolutionSolution

ad EEE

)(2 ad TTRi

)(2 aadd VpVpi

0

Draw every process in Draw every process in p-Vp-V diagramdiagramInternal energy is the state Internal energy is the state quantity. It has nothing to quantity. It has nothing to do with process.do with process.

)( abap VVpW 35 10110013.13 J304

b

cbT V

VRTW lnb

cbb V

VVp ln J246

0VW

VTp WWWW J550WEQ J550

　§7-4 Heat Capacities of an Ideal Gas

1. Molar heat capacity1. Molar heat capacity

------The amount of heat added to cause unit rise The amount of heat added to cause unit rise of temperature (or withdrawn to cause unit of temperature (or withdrawn to cause unit fall of temperature) for fall of temperature) for 11molmol of a material. of a material.

i.e.i.e.

CC depends on process. depends on process.

dTdQC

dQdQ is a process quantity is a process quantity

2. The molar heat capacity at constant volume2. The molar heat capacity at constant volume CCvv

VV dT

dQC

dEdQ V )(

RiCV 2

RdTi2

Under constant volume process,

Absorb heat for increasing internal energy only

NoteNoteCCv v depends on depends on i i only.only.

Physical meaningPhysical meaning ：： The average kinetic energThe average kinetic energy of each freedom degree isy of each freedom degree is (1/2) (1/2)kTkT. The mor. The more numbers of freedom degree are, the more he numbers of freedom degree are, the more heat is needed.eat is needed.

For a random process:For a random process:

2

1

)(2 12

V

VpdVTTRiMQ

2

1

)( 12

V

VV pdVTTCM

　3. The molar heat capacity at constant pressure3. The molar heat capacity at constant pressure CCpp

pp dT

dQC

RdTidQ p 22)(

RCC Vp

Under constant pressure process,

NoteNote CCpp>>CCvv..

Physical meaning Physical meaning ： ： Absorb heat not only for increasing internal energy but for working for exterior.

)(2

212 TTRiMQ p

The ratio of molar capacity:The ratio of molar capacity:V

p

CC

i

i 2 1

)( 12 TTCMp

For Mkg gas

[[ExampleExample] Suppose the molar heat capacity of a] Suppose the molar heat capacity of an ideal gas is proportional with its temperature n ideal gas is proportional with its temperature CC==aT aT ((a a is a constant). Calculate the thermodynis a constant). Calculate the thermodynamic equation for amic equation for 1mol 1mol ofof this gas.this gas.SolutionSolution ：：

pdVdEdQ

molar heat capacity of molar heat capacity of 11molmol ideal gas ideal gas

dTdQC aT

According to the first law of According to the first law of thermodynamicsthermodynamics

Separate variablesSeparate variablesVdVR

TdTCadT V

bVRTCaT V lnlnIntegration constantIntegration constant

We getWe get

.consteVT RaT

RCV

dVVRTdTCaTdT V

Integration Integration

No heat transfer between a system and exterior.

§7-5 Application of the first Law of thermodynamics to Adiabatic Process

dQ = dQ = 001.character1.character

TT, , pp when adiabatic expandingwhen adiabatic expandingT T ,,pp when adiabatic compressingwhen adiabatic compressing.

0 WEAs As EW then

nkTp And And

i.e.i.e.

2. Equation of adiabatic process for ideal gas2. Equation of adiabatic process for ideal gas

pdVdEdQ

0

RTMpV

RdTMVdppdV

pdVdTCMV

And And

For ideal gas,For ideal gas,

Take differential ,Take differential ,

······

······

Eliminating Eliminating dTdT from from and and ：0)( RpdVVdppdVCV

i.e.i.e. 0)( VdpCpdVRC VV

0p

dpCVdVC Vp

Or Or 0p

dpVdV

Integration Integration

----The equation of adiabatic processThe equation of adiabatic process

.ConstpV

Using state equation of ideal gas, other forms of Using state equation of ideal gas, other forms of process equation can be gotten.process equation can be gotten.

.1 ConstTV

.1 ConstTp

Three constants Three constants are differentare different..A

BC

p

VO

(1)(1)Mathematical methodMathematical method ：： ccompare the two slopes of thompare the two slopes of the two curves.e two curves.

3. Adiabatic curve and is3. Adiabatic curve and isothermal curveothermal curve

isothermisotherm CpV 0 VdppdV

A

AT V

pdVdp

)(

adiabatadiabat CpV

01 dVpVdpV

A

AQ V

pdVdp )(

Iso-curveIso-curve

Adia-curveAdia-curve

i.e.i.e.

ABC

p

VOi.e.i.e.

(2)(2)Physical methodPhysical method ：：

--the adiabat is steeper than the isotherm--the adiabat is steeper than the isotherm

Isotherm:Isotherm:----VV increasing causes increasing causes nn decreasing decreasing p p decreasingdecreasing

Adiabat:Adiabat:----VV increasing causes increasing causes nn and and TT decreasing decreasing p p decreasingdecreasing

Find the reasons of Find the reasons of pp changingchanging

nkTp

[[Ex.Ex.]]II is adiabatic process for a gas as shown in is adiabatic process for a gas as shown in diagram. diagram. Judge:Judge: does does J J and and KK process process withdraw heat form outside or reject heat to withdraw heat form outside or reject heat to outside respectively?outside respectively?

p

V

IKJA

B

SolutionSolution ：： For For II process process 0IQ IWE For For J J processprocess

JJ WEQ

JI WW 0--Withdraw heat--Withdraw heat

For For KK processprocess

KK WEQ KI AA 0 --Reject heat--Reject heat

[[Ex.Ex.] A gas goes through three processes ] A gas goes through three processes respectively as shown in figure. respectively as shown in figure. ABAB is an isobaric is an isobaric process. process. AC AC is an isothermal process. is an isothermal process. AD AD is an is an adiabatic process. adiabatic process. Judge:Judge: ( (11) which process does ) which process does the maximum work? (the maximum work? (22) which process withdraw ) which process withdraw the maximum heat from outside?(the maximum heat from outside?(33)which )which process has the maximum change of internal process has the maximum change of internal energy?energy? p

VO

CD

A B

0V 02V

SolutionSolution ：：((11) Work = the area under the ) Work = the area under the

process curveprocess curve ABAB process does process does the maximum workthe maximum work

ACAC process process ：： 0E

ADAD processprocess ：： V V T T 0 E

321 WWW WEQ

321 QQQ

((33)) TCME V

TCompare Compare

A

A

B

B

VT

VT A

A

BB T

VVT AT2

((22) ) ABAB process process ：：V V T T 0 E

ABAB process process ：：

AAB TTTT

DDAA TVTV 11

AD

AD T

VVT 1)(

AT1)21(

AAD TTTT )2

11( 1 AT

ADAD process process ：：

I.e.I.e. AB AB process has the maximum change of process has the maximum change of internal energy. internal energy.

4. Adiabatic free expansion4. Adiabatic free expansion

A BRemoving baffle

Molecules AB quickly

Non-equilibrium process

Q = 0 W = 0 E= 0

When the system becomes equilibrium state,

T1 = T2

Use state equation 12 21 pp

Note(1) It is not adiabatic process though Q = 0

It is not isothermal process though T1 = T2

(2) For real gas, T T changes as the system gets changes as the system gets equilibrium state after adiabatic processequilibrium state after adiabatic process

ReasonsReasons ：：There are interactions between moleculesThere are interactions between molecules potential energies of moleculespotential energies of molecules

---Non-equilibrium process

If there are repulsive forces between moleculesIf there are repulsive forces between molecules

The repulsive forces do The repulsive forces do positive workpositive work during the gas expandsduring the gas expands Potential energiesPotential energies Kinetic energiesKinetic energies T T

If there are attractive forces between moleculesIf there are attractive forces between molecules

The attractive forces do The attractive forces do negative worknegative work during the gas expandsduring the gas expands

Potential energies Potential energies Kinetic energies Kinetic energies T T

ApplicationApplication——throttling (throttling ( 节流节流 ) process) process

Keep p1 > p2 when moving the pistons

The gas through 多孔塞 adiabatic free expansion

T T

--Joule-Thomson effect

F p1 p2

多孔塞piston piston

　1. Characters1. Characters

0E

When a system is carried When a system is carried through a cyclical process,through a cyclical process,

a

b

p

VOClassify: Classify: positive cyclepositive cycle(clockwise)—(clockwise)—heat engineheat engine

§7-6 Cyclical Processes

Inverse cycleInverse cycle(counterclockwise)--(counterclockwise)--refrigeratorrefrigerator

p 、 V 、 T return their initial values

2. positive cycle2. positive cycleWorking substanceWorking substance ：： for cyclefor cycleAssume working substance withdraws heat Assume working substance withdraws heat QQ11 from a from a high-high-temperaturetemperature source, and rejects heat source, and rejects heat QQ22 into into lower lower temperature sourcetemperature source, and does work , and does work WW to the exterior to the exterior meanwhile.meanwhile.

p

V

a

b

O

A1Q

2Q

the net heat absorbed during a the net heat absorbed during a complete cyclecomplete cycle

QA

21 QQQ

Net workNet work 21 QQ

= the area enclosed b= the area enclosed by the cyclical curvey the cyclical curve

Q1, Q2 are all positive

热水热水蒸汽蒸汽

冷水冷水

锅炉锅炉 进气阀进气阀

排气阀排气阀泵泵 冷凝器冷凝器

热机工作原理图热机工作原理图

A simplified heat engine:

1Q

W

Cold reservoirat T2

Heat engine

Hot reservoirat T1

2Q

The efficiency of heat engineThe efficiency of heat engine ：： the ratio of the the ratio of the useful work to the heat supplied.useful work to the heat supplied.

1QW

1

21QQ

i.e. i.e. what you getwhat you get divided by divided by what you pay forwhat you pay for

We hope that the work output from heat engine is as large as possible and the heat thrown away is as small as possible.

In 1794~1840, was about 3%~4%. So low!!!

Now, is about 30%~40%.

[[ExEx.].]1mol1mol oxygen is carried oxygen is carried out a cycle as shown if Fig. out a cycle as shown if Fig. AB AB is isothermal process.is isothermal process. BCBC is isobaric process. is isobaric process. CACA is is isochoric process. Find isochoric process. Find =?=?

p

VO

A

BC

1V 2V

1p

2p

Solution Solution

Find “Find “withdrawing heatwithdrawing heat” ” processprocess

CAAB QQQ 11

2lnVVRTA )( CAV TTC

--“--“AB,CA AB,CA “process“processThen Then

Find “Find “rejecting heatrejecting heat” process” process

)(2 BCp TTCQ )(27

1222 VpVp

)(25ln 1211

1

211 VpVp

VVVp

)(25ln

)(27

1211

1

211

1222

ppVVVVp

VpVp

1

21QQ

--“--“BCBC” process” process

Then Then

How can enhance the efficiency of heat engine?How can enhance the efficiency of heat engine?

abab, , cdcd----isothermal pisothermal processrocess

bc , da –adiabatic process

2T1T

a

b

cd

p

VO

§7-7 Carnot Cycle Reverse cycle§7-7 Carnot Cycle Reverse cycle

1.Carnot cycle:

Carnot answered this question.

Consists of two isothermal and two adiabatic processes

abab: : withdraw withdraw QQ11 from from TT11

Heat exchanging:Heat exchanging:

cdcd ：： reject reject QQ2 2 to to TT2 2

a

b

VVRTQ ln11

1Q

1T

a

b

cd

2T

p

VO

d

c

VVRTQ ln22 (absolute value)(absolute value)

2Q

The thermal efficiency The thermal efficiency of Carnot cycleof Carnot cycle

1

21QQ

c

ab

dc

VVTVVT

/ln/ln1

1

2

1Q

Cold reservoirat T2

Heat engine

Hot reservoirat T1

2Q

21 QQW

d

c

a

b

VV

VV

1

21TT

ab

dcc VVT

VVT/ln/ln1

1

2

1T

2Q

a

b

cd

p

VO

1Q

2T

According to adiabatic According to adiabatic equationequation

12

11

cb VTVT

12

11

da VTVT

2. Reverse cycle RefrigeratorA heat engine operating in reverse. It takes heat from a cold place and give it off to a warmer place.Performance coefficient（致冷系数）

21

22

QQQ

WQK

1Q

2Q

W

Hot reservoirat T1

Cold reservoirat T2

Refrigerator

For Carnot For Carnot refrigeratorrefrigerator

21

2

TTTK c

1T

2Q

a

b

cd

p

VO

1Q

2T冷凝器冷凝器 蒸蒸发发器器高压蒸气高压蒸气

高温液体高温液体

低压蒸气低压蒸气压缩机压缩机

节流阀节流阀传传感感器器

§§7-8 The second law of thermodynamics7-8 The second law of thermodynamics

question: question: Can man make a heat engine that its Can man make a heat engine that its ==100%100%??i.e. Q2=0 in ?

1

21QQ

Or: Can man make a refrigerator that its Or: Can man make a refrigerator that its KK ??i.e. W=0 in ?

WQK 2

It is impossible for any It is impossible for any system to undergo a process system to undergo a process in which it absorbs heat in which it absorbs heat from a reservoir at a single from a reservoir at a single temperature and converts temperature and converts the heat completely into the heat completely into mechanical work, with the mechanical work, with the system ending in the same system ending in the same state in which it began.state in which it began.

Hot reservoir T1

Cold reservoir T2

1Q

W

1. “ Engine” statement (Kelvin statement 开尔文说法）

Note:This statement lies in the difference between

the nature if internal energy and that of macroscopic mechanical energy.

Random motion of molecules

Coordinated macroscopic motion

The engine with The engine with ==100%100%----the second kind the second kind of perpetual motion machineof perpetual motion machine

So it’s impossible to make aSo it’s impossible to make a perpetual perpetual motion machinemotion machine

It is impossible for any It is impossible for any process to have as its sole process to have as its sole result the transfer of heat result the transfer of heat from a cooler to a hotter from a cooler to a hotter body.body.

Cold reservoirT2

Hot reservoir T1

2Q

1Q

2. “Refrigerator”statement 2. “Refrigerator”statement (Clausius statement (Clausius statement 克劳修斯说法克劳修斯说法 ))

Note :Note : The heat flows naturally The heat flows naturally

from a hotter to a cooler from a hotter to a cooler body. body. And the opposite And the opposite process can’t carry out process can’t carry out naturally.naturally.

Can realize by supplying some work to the system

A refrigerator can’t be operated without work A refrigerator can’t be operated without work input.input.There is no perfect refrigerator.There is no perfect refrigerator.

　3. Identity of the two statement of the second law3. Identity of the two statement of the second law A violation of either implies a violation of the other.A violation of either implies a violation of the other.Suppose the “refrigerator” statement is false. Suppose the “refrigerator” statement is false. i.e.i.e.

existing a perfect refrigerator that can transfer heat existing a perfect refrigerator that can transfer heat QQ22 naturally naturally TT22TT11(T(T22<<TT11)) without work input. without work input.

T2

T1

2Q

2QT1

T2

21 QQ

W

Then, we can make a heat engine and let it absorbs Then, we can make a heat engine and let it absorbs heat heat QQ11 from from TT11, rejects heat , rejects heat QQ22 to to TT22..

2Q

T1

T2

1QW

2Q

T1

T2

2Q

1QW

equivalent

effecti.e.

Violate the “engine” Statement

T1

T2

1Q

1QW

　

T2

T1

2Q

21 QQ

Inversely, suppose the “engine” statement is false. i.e. Inversely, suppose the “engine” statement is false. i.e. existing a perfect heat engine that can convert heat existing a perfect heat engine that can convert heat QQ11 from from TT11 completely into work. completely into work.Then, we can use the work to operate an ordinary Then, we can use the work to operate an ordinary refrigerator. refrigerator. The net effect : transfer heat The net effect : transfer heat QQ22 from from TT22 to to TT11..

equivalent

effect

T2

T1

2Q

2Q

Violate the “refrigerator” Statement

NoteNote ：：The first law of thermodynamics indicates that The first law of thermodynamics indicates that

all thermodynamic process obey the all thermodynamic process obey the conservation law of energy.conservation law of energy.

The second law of thermodynamics indicates The second law of thermodynamics indicates it is not sure whether a thermodynamic it is not sure whether a thermodynamic process can go or notprocess can go or not though it obeys the though it obeys the conservation law of energy.conservation law of energy.

i.e.there is a i.e.there is a direction direction for thermodynamic processfor thermodynamic process Work convert heatWork convert heat ：：

Regular kinetic energyRegular kinetic energyirregular onesirregular ones Heat conductHeat conduct ：：

Partly irregularityPartly irregularityentire irregularityentire irregularity

Free expansion of gasFree expansion of gas ：：irregularity in irregularity in small rangesmall range irregularity in irregularity in large rangelarge range

Any directional process can give a statement of the second law of thermodynamics

What is the essence of the second law of thermodynamics?

All natural process always proceeds spontaneously along the direction in which the irregularity of molecular thermal motion is increasing.

----Microscopic meaning of Second Law ----Microscopic meaning of Second Law

§7-9 Reversible process & Irreversible process Carnot Theorem

1. Reversible Process & Irreversible ProcessReversible Process: Process P(AB) if there is a way to make system back to the initial state without other effects.

AP(AB)

There is a way(other processes)

B A P(AB)

There is no way(other processes)

B

Irreversible process

Irreversible Processes: examples（ 1 ） The process in which work transfers into heat is irreversible

Do work by Do work by stirringstirring

work spontaneously heat

Can’t be spontaneously

(2) The process in which heat flows from a hotter body to a colder one is irreversible.

hot cold

High T spontaneously Low T

Can’t be spontaneously

(3)The free expansion of gas is irreversible;

vacuumvacuum

A B

(4)The rapid expansion of gas is irreversible;

A B

△V

………………….

Conclusion : All the real processes that relate with thermal phenomenon are irreversible

Reversible Processes must be satisfy:(1) must be equilibrium process

(2) No friction in each step of process.

(1) All Carnot engine operating between the same two temperatures T1 and T2 have the same efficiency, irrespective of the nature of the working substance.

2. Carnot Theorem

1

21TT

(2) No real engine operating between two given temperatures T1 and T2 can have a greater efficiency than that of Carnot engine operating between the same two temperatures

Carnot theorem points out the way to enhance the efficiency of heat engine.

1

21TT

Make the process near equilibrium process as possible.

Increase the temperature difference between two reservoirs

Theo. (1) provingTheo. (1) proving ：：设甲、乙两台可逆机工设甲、乙两台可逆机工作在相同的高低热源之间作在相同的高低热源之间

1

21QQ

T2

T1

1Q

2Q

A甲

'1Q

'2Q

'A乙'

'1'1

2

QQ

效率分别为效率分别为

甲乙组成复合机甲乙组成复合机 :: 从高温从高温热源吸热变为对外作功热源吸热变为对外作功

---- 违反热二律'

T2

T11Q

2Q

A甲

'1Q

'2Q

'A乙

11 ' QQ

AA '

T2

T1

' 不可能不可能即有即有

让甲逆向运行，并使让甲逆向运行，并使 Q2= Q2

'若若1

2

1

2 1''1

QQ

QQ

则则11' QQ

定理定理 22 ：：工作在相同高低温热源间的工作在相同高低温热源间的所有所有不可逆热机不可逆热机的效率不可能高于工的效率不可能高于工作在同样热源之间的可逆热机作在同样热源之间的可逆热机，，即即

可不 1

21TT不或或

同理若使乙机反向运行，可证有同理若使乙机反向运行，可证有 ' '

证明：证明：设乙机为不可逆机，甲机反向运行，则设乙机为不可逆机，甲机反向运行，则可证可证 不可

可不 因乙机不可反向，因乙机不可反向，即不可证即不可证

不可 T2

T1

1Q

2Q

A甲

'1Q

'2Q

'A乙

卡诺定理指出了提高热机效率的方向 卡诺定理指出了提高热机效率的方向 -------- 使实际的不可逆机尽量接近可逆机使实际的不可逆机尽量接近可逆机

there are N=3 molecules in A vessel at beginning. When the plank is removed, the three molecules may move in A+B vessel.

A B

§7-10 Statistical Meaning of the Second Law of Thermodynamics

Take the free expansion of gas as an example.

1. Macroscopic state and 1. Macroscopic state and

microscopic statemicroscopic state

微微观观态态

Ma.S.Ma.S. A3B0A3B0 A2B1A2B1 A1B2A1B2 A0B3A0B3

AA a b ca b c ab,ac,bcab,ac,bc a, b, ca, b, c 00

BB 00 c, b, ac, b, a bc,ac,abbc,ac,ab a b ca b c宏观态中包含宏观态中包含的微观态数的微观态数 11 3333 11

A B

is the number of microstates correspis the number of microstates corresponding to a given macrostate. onding to a given macrostate.

--It expresses the irregularity of molecular thermal motions

2.Thermodynamic Probability 热力学概率

(A3B0)=1

(A2B1)=3(A1B2)=3(A0B3)=1

Statistical mechanics assumes that the probability for each microstate is equal.

the position distribution for N=20 molecules

MacrostateL 20 R 0 1L 18 R 2 190L 15L 11L 10L 9L 5L 2L 0

R5R 9R 10R 11R 15R 18R 20

1550416796018476516796015504190

1 is different fo

r different macrostate.

The greater the number of molecules, the greater corresponding corresponding to a macrostateto a macrostate

Rules:

The macrostate that contains great number of microstates ( is great) has great appearing probability.

When N is much great, the macrostate corresponding to small nearly does not appear.

For a isolated system with huge N, its equilibrium state corresponds to the macrostate that gets maximum value.

If the system is not at equilibrium state, its does not get maximum value.

--non-equilibrium state

Then it must take place such process:

The state of the system changes spontaneously from the state with smaller to the one with biggest with the time extending.

gets maximum—equilibrium state

Stable state

The proceeding direction of natural process is always along the direction that increases.

Small corresponds non-equilibrium state

With time increasing

biggest . i.e. equilibrium state

3. Boltzmann entropy (3. Boltzmann entropy ( 玻尔兹曼熵 ))

DefinitionDefinition ：： lnkSto describe the irregularity(disorder) of the system quantitatively

NotesNotes ：：for an isolated system, the process always for an isolated system, the process always

proceeds spontaneously in the direction of proceeds spontaneously in the direction of increasing. i.e. increasing. i.e. initial > final

So So S = SS = Sff-S-Si i >0>0

The addition of entropyThe addition of entropyIf two subsystems have If two subsystems have 11 、、 2 2

respectivelyrespectivelyTotal Total of the system: of the system: = = 1122

21 lnlnln kkkS

21 SS

i.e. Natural process always proceeds spontaneously in the direction of increasing entropy.

---the principle of the increase of entropy

In the interior of an isolated system, the entIn the interior of an isolated system, the entropy may increase, do not change or decrearopy may increase, do not change or decrease se for some individual subsystemfor some individual subsystem. .

The entropy of the system may decrease or The entropy of the system may decrease or do not change if the system is not isolateddo not change if the system is not isolated..

　§7-11 The formula of Clausius entropyFor Carnot heat engine, if For Carnot heat engine, if QQ22is algebraic value of is algebraic value of hear rejected -- i.e. hear rejected -- i.e. negativenegative 而而 instead of absolutinstead of absolute value. Then, e value. Then,

1

21

1

21

TTT

QQQ

02

2

1

1 TQ

TQ

　 p

VO

Popularize this conclusion to Popularize this conclusion to generalized (any) generalized (any) reversible cyclical processcyclical processdivide the cycle into madivide the cycle into ma

ny of small Carnot cycleny of small Carnot cycles.s.

0i

i

TQWe have We have

For any small Carnot cycle,For any small Carnot cycle,

0i

i

TQ

Then, for the sum of them Then, for the sum of them

when the numbers of small Carnot cycle when the numbers of small Carnot cycle ，， sawsaw

-toothed curve -toothed curve reversible cyclical curve.cyclical curve.

0TdQ

i.e.i.e. -- for reversible cycle-- for reversible cycle

i.e the calculation of i.e the calculation of does not depend ondoes not depend on the the

reversible cyclical process. cyclical process.TdQ

So we can introduce So we can introduce a state quantitya state quantity to express to express the calculation. the calculation.

entropyentropy

p

VO

A

Binitial

finalFor any reversible cycle, from initial state to final state,

iBfiAf T

dQTdQ

f

i TdQ For any reversible process

from initial state to final state

if SS

S – Clausius entropy, it’s a state function of the thermodynamic system. J/K

NotesNotes Only Only entropy changingentropy changing is significant for is significant for Cla

usius entropy.

If we want to find the If we want to find the Clausius entropy of the of the system at some equilibrium state, we must chosystem at some equilibrium state, we must choose the ose the reference point of zeroreference point of zero Clausius entropy.

From one equilibrium state to another From one equilibrium state to another

The integral path must be The integral path must be any reversible prany reversible processocess connecting initial and final state for ca connecting initial and final state for calculating the entropy changing. lculating the entropy changing.

If initial If initial →→ final state is final state is irreversible processirreversible process, , we can we can design any reversible integral path design any reversible integral path from initial to final statefrom initial to final state, and make a , and make a integration along this path to find the entropy integration along this path to find the entropy changing of the system. changing of the system.

i.e. the formula f

i TdQ

if SSS

can calculate S of the system form i state to f state no matter the process is reversible or not.

[[Exa.Exa.] ] 11molmol ideal gas from one ideal gas from one state state 11((TT11,,VV11) to another state ) to another state 22((TT22,,VV22) through some process. F) through some process. Find ind S =? =? SupposeSuppose CCv v is a constis a constant.ant.SolutionSolution

),( 12 VT

isochoricisochoric：： TdQS1

Such asSuch as a increasing temperature with a increasing temperature with constant volume process constant volume process + + aa increasing increasing volume with constant temperature processvolume with constant temperature process

Design a reversible process fromDesign a reversible process from 1 1 2. 2.

2

1

T

TV

TdTC

1

2lnTTCV

V

p

O),( 11 VT

),( 22 VT1 2

isothermalisothermal ：：

TdQS2 T

pdV 2

1

V

V VdVR

1

2lnVVR

1122

21 SSS 1

2

1

2 lnlnVVR

TTCV

说明：若此系统非孤立，S 可正、可负、可 =0

三三 .. 熵增加原理熵增加原理1.1. 不可逆循环不可逆循环

1

21

1

21

TTT

QQQ

不由卡诺定理由卡诺定理

即即 02

2

1

1 TQ

TQ

对任意对任意不可逆循环不可逆循环 0)( 不T

dQ-------- 克劳修斯不等式克劳修斯不等式

2.2. 非循环的不可逆过程非循环的不可逆过程 p

VO

1

2a

不可逆不可逆设系统由设系统由 11 经任一经任一不可逆过程不可逆过程

11aa22 22后又由后又由 22 经另一经另一可逆过程可逆过程 22bb11

回到回到 11 ，构成一不可逆循环，构成一不可逆循环 11aa22bb11

21 120)()(

a b TdQ

TdQ

可不

b可逆可逆

21)(

b TdQ

可 12 SS

2

112 )( 不TdQSS

　对任一微小不可逆过程：对任一微小不可逆过程：　　不)(

TdQdS

讨论：讨论：

-------- 在在 孤立系统孤立系统 中中 ，， 不可逆过程总不可逆过程总是 沿熵增加方向进行是 沿熵增加方向进行

对于一对于一孤立系统孤立系统：： dQdQ=0=0

0dS -------- 熵增加原理熵增加原理

与玻尔兹曼与玻尔兹曼熵增加原理一致熵增加原理一致

[[ 例例 ]] 求求 molmol 的某种理气进行绝热自由膨的某种理气进行绝热自由膨胀的熵变。设膨胀前后体积分别为胀的熵变。设膨胀前后体积分别为 VV11 和和 VV

22A B

1V 2V

A B

21 VV

解：解：系统绝热系统绝热 0Q

对外不做功对外不做功 0A

0 E

即 即 T T 不变不变

设想气体经一个可逆等温膨胀过程设想气体经一个可逆等温膨胀过程由由 VV11VV22

TdQSS 12 dQ

T1

1

21ln1V

VVRTT

1

21lnV

VVR 0 熵增加

可劳修斯可劳修斯熵熵与玻尔兹曼与玻尔兹曼熵的区别：熵的区别：可劳修斯可劳修斯熵：熵： 玻尔兹曼玻尔兹曼熵：熵：

2

1)( 可T

dQS lnkS

只对平衡态有意义只对平衡态有意义为平衡态的函数为平衡态的函数

熵变：熵变：平衡态平衡态 1 1 22 的熵变的熵变

对 平 衡 态 、 非 平 衡 态对 平 衡 态 、 非 平 衡 态均有意义均有意义非平衡态时也有非平衡态时也有 SS ，平，平衡态时的衡态时的 SS 最大最大

可氏熵是玻氏熵的最大值可氏熵是玻氏熵的最大值非平衡态 非平衡态 平衡态的熵变平衡态的熵变

熵具有可加性熵具有可加性11 、、 22-------- 两子系统的热力学概率两子系统的热力学概率则总系统的则总系统的 = = 1122

21 lnlnln kkkS

21 SS

在孤立系统内部在孤立系统内部，，个别物体的熵值可增加个别物体的熵值可增加、 不变或减少、 不变或减少

非孤立系统熵值可不增加非孤立系统熵值可不增加，，也可减少也可减少。。

即： 即： SS>0>0 (( 孤立系，自然过程孤立系，自然过程)) -------- 熵增加原理熵增加原理

四四 .. 玻尔兹曼熵的宏观表达式玻尔兹曼熵的宏观表达式用宏观状态量表示熵

只适用于平衡态对 mol 单原子理气：

平衡态可用 V 、 T 确定，设此时 ),( TVSS = ？先求 ),( TV = ？

由分子的位置和速度确定vp

分子按位置分布的可能微观状态数分子按速度分布的可能微观状态数

一个分子按位置分布的可能微观状态数 x,y,z = V

一个分子按速度分布的可能微观状态数 Mol 气体： 0N

p V

速度立方盒的体积

很大 可取 ppi vorvv 1000100~

zyxi ,, 而 21

Tv p

23

T

23

T

Mol 气体： 0)( 23

Nv T

0)( 23

Nvp VT

令 0)( 23

NVTC 比例常数

则 lnkSCkTkNVkN lnln

23ln 00

R

0lnln STCVRS v

R23 单原子理气的 vC

以相应单位表示的相应物理量的数值---S 的宏观表达式宏观表达式

对双原子分子 0)( 25

Nv T

转动速度与平动速度对 v 的贡献等价对多原子分子 0)( 2

6N

v T

上述宏观表达式仍成立！上述宏观表达式仍成立！五五 .. 热力学过程的熵变热力学过程的熵变

对 S 的宏观表达式微分：宏观表达式微分：TdTC

VdVRdS v

dTCVdVRTTdS v

对理气 =pV

dTCpdV v

对可逆过程 =dA

dQ

TdQdS

可逆)(2

112 TdQSS

克劳修斯熵公式

对于孤立系统、可逆过程：0dQ

0dS

适用于任意系统的可逆过程

说明说明 对可劳修斯对可劳修斯熵，只有熵变才有意义。熵，只有熵变才有意义。

一平衡态一平衡态另一平衡态的熵变另一平衡态的熵变

计算熵变时的积分路径必须是连接始末计算熵变时的积分路径必须是连接始末状态的状态的任一可逆过程。任一可逆过程。如如 11→→22 是一是一不可不可逆过程逆过程，可，可在在 11→→22 间想象一间想象一可逆过程可逆过程，，再计算再计算

某状态的某状态的熵，用 计算熵，用 计算 lnkS

由由

2

112 )( 可TdQSSS

与与1

212 ln

kSSS

计算的熵变相同计算的熵变相同21 、 ---1---1 、、 22 两平衡态的热力学概两平衡态的热力学概

率率

[[ 例例 ]]11kg kg 0000CC 的冰，在的冰，在 0000CC 时完全化成水。已知时完全化成水。已知冰在冰在 0000CC 时的熔化热时的熔化热 =334J/g=334J/g 。求冰。求冰水的熵水的熵变。并计算冰变。并计算冰水时微观状态数的变化。水时微观状态数的变化。解：冰冰水的过程，水的过程，

可设想为是等温吸热的可逆过程可设想为是等温吸热的可逆过程则则

水

冰 TdQS

TQ

KJT

m /1022.1 3

又又1

2ln

kS

KS

e

1

2 251084.8 e

[[ 例例 ]] 作业作业 P206 4.8 P206 4.8 （（ 11 ）求）求 11kg kg 0000CC 时的水，放到时的水，放到 10010000CC 的恒温的恒温热库达平衡后，水、热库的熵变。热库达平衡后，水、热库的熵变。（（ 22 ）若先放到）若先放到 505000CC 恒温热库达平衡，再放到恒温热库达平衡，再放到10010000CC恒温热库达平衡后，求水、两热库的熵恒温热库达平衡后，求水、两热库的熵变变解：（（ 11 ）） 2

1

T

T TdQS水

2

1

T

T TcmdT

1

2lnTTcm

)/(1018.4 3 KkgJc

2

2

T

T TdQS库

2TQ

2

12

TTTcm

库水 SSS )/(184 KJ

（（ 22 ））

2

1

T

T

T

T TdQ

TdQS水 水S

2

2

TQ

TQS

库

2

21

TTTcm

TTTcm

库S

库水 SSS )/(97 KJ

结论：中间热库越多， S越小，中间热库多， S 0 ， ---- 可逆过程

[[ 例例 ]]11molmol 理气由初态理气由初态 11((TT11,,

VV11)) 经某一过程到达末态经某一过程到达末态 22((

TT22,,VV22)) ，求熵变。设，求熵变。设 CCVV 为为常量常量解：解：

),( 12 VT

等体：等体：

TdQS1

过程过程 ++ 一可逆等温膨胀过程一可逆等温膨胀过程 ((如图如图 ))设计一可逆等体升温设计一可逆等体升温

2

1

T

TV

TdTC

1

2lnTTCV

V

p

O),( 11 VT

),( 22 VT1 2

等温：等温：

TdQS2 T

pdV 2

1

V

V VdVR

1

2lnVVR

1122 过程过程21 SSS

1

2

1

2 lnlnVVR

TTCV

说明：若此系统非孤立，S 可正、可负、可 =0

三三 .. 熵增加原理熵增加原理1.1. 不可逆循环不可逆循环

1

21

1

21

TTT

QQQ

不由卡诺定理由卡诺定理

即即 02

2

1

1 TQ

TQ

对任意对任意不可逆循环不可逆循环 0)( 不T

dQ-------- 克劳修斯不等式克劳修斯不等式

2.2. 非循环的不可逆过程非循环的不可逆过程 p

VO

1

2a

不可逆不可逆 设系统由设系统由 11 经任一经任一不可逆过程不可逆过程

11aa22 22

后又由后又由 22 经另一经另一可逆过程可逆过程 22bb11回到回到 11 ，构成一不可逆循环，构成一不可逆循环 11aa22bb11

21 120)()(

a b TdQ

TdQ

可不

b可逆可逆

21)(

b TdQ

可 12 SS

2

112 )( 不TdQSS

　 对任一微小不可逆过程：对任一微小不可逆过程：　　

不)(TdQdS

讨论：讨论：

-------- 在在 孤立系统孤立系统 中中 ，， 不可逆过程总不可逆过程总是 沿熵增加方向进行是 沿熵增加方向进行

对于一对于一孤立系统孤立系统：： dQdQ=0=0

0dS -------- 熵增加原理熵增加原理

与玻尔兹曼与玻尔兹曼熵增加原理一致熵增加原理一致

[[ 例例 ]] 求求 molmol 的某种理气进行绝热自由膨的某种理气进行绝热自由膨胀的熵变。设膨胀前后体积分别为胀的熵变。设膨胀前后体积分别为 VV11 和和 VV

22A B

1V 2V

A B

21 VV

解：解：系统绝热系统绝热 0Q

对外不做功对外不做功 0A

0 E

即 即 T T 不变不变

设想气体经一个可逆等温膨胀过程设想气体经一个可逆等温膨胀过程由由 VV11VV22

TdQSS 12 dQ

T1

1

21ln1V

VVRTT

1

21lnV

VVR 0 熵增加

[[ 例例 ] ] 理气热传导过程的熵变理气热传导过程的熵变。。为简单起，设为简单起，设 AA 、、 BB 中理气种中理气种类、质量完全相同，开始温度类、质量完全相同，开始温度分别为分别为 TTAA 、、 TTBB

TA TB

BA TT 21平衡后温度 BA TTT

21

熵变 BA SSS

T

T

T

T BA TdQ

TdQ

T

TvT

Tv

BA TdTC

TdTC

BAv TT

TC2

ln 0 熵增加

可劳修斯可劳修斯熵熵与玻尔兹曼与玻尔兹曼熵的区别：熵的区别：可劳修斯可劳修斯熵：熵： 玻尔兹曼玻尔兹曼熵：熵：

2

1)( 可T

dQS lnkS

只对平衡态有意义只对平衡态有意义为平衡态的函数为平衡态的函数

熵变：熵变：平衡态平衡态 1 1 22 的熵变的熵变

对 平 衡 态 、 非 平 衡 态对 平 衡 态 、 非 平 衡 态均有意义均有意义非平衡态时也有非平衡态时也有 SS ，平，平衡态时的衡态时的 SS 最大最大

可氏熵是玻氏熵的最大值可氏熵是玻氏熵的最大值非平衡态 非平衡态 平衡态的熵变平衡态的熵变

§§4-9 4-9 温 熵 图温 熵 图由 可)(

TdQdS 或或 TdSdQ

得可逆过程系统吸热 TdSQ

S

TA

B

温熵图Q

绝热线

循环过程：

S

TA

B

CD

M N

AC 过程：B dS>0吸热过程吸热 Q1 = SABCNMA

CA 过程：D dS<0放热过程 放热 Q2 = SCDAMNC功 A=SABCDA

ABCNMA

ABCDA

SS

卡诺循环：

S

Ta b

cd

m n

T1

T2abcnma

abcda

SS

卡

amad

1

21TT

由图可知：T1 、 T2 不变，改变等温过程长度时，

卡 不变可逆 卡 只与 T1 、 T2 有关

不可逆 < 可逆 卡

§§4-10 4-10 熵和能量退降熵和能量退降不可逆过程系统熵增加，导致部分能量 Ed 从能做功形式 不能做功形式

退降的能量 不可)( S

系统熵增加是能量退降的量度1.焦耳实验

设 M下降 dh 时，带动叶片转动使水温 dTTT Mgdh 水的内能 E

有用功 只能利用热机将部分变为有用功

E 可转换为有用功的最大值：EA 卡 Mgdh卡 )1( 0

dTTTMgdh

0T --- 低温热库的温度降退的能量： AMgdhEd

dTTMgdhT

0

TMgdhT0

这一过程的熵变：TdQdS

TcmdT

由能量守恒： MgdhdQ

TcmdTTEd

0 dST0

2. 有限温差的热传导设 A 、 B 两物体有 TA > TB ，A B ，dQ

dQ 在 A 中时，可做功的最大值：dQA 卡1 dQ

TT

A

)1( 0dQ 传到 B 中后，可做功的最大值：

dQA 卡2 dQTT

B

)1( 0热传导前后：

21 AAEd dQTTT AB

0)11(

熵变：BA dSdSdS

BA TdQ

TdQ

dSTEd 0

3. 理气绝热自由膨胀 mol 理气，温度 T ，体积 21 VV 对外做功 =0若使其回到 1V

则需外界对其做功1

21 ln

VVRTA

变为热 Q

Q再转变为有用功 QA 卡2 10 )1 A

TT（

21 AAEd 1

20 ln

VVRT

而TQS

（设为等温压缩）

1

2lnVVR

STEd 0

不可逆过程中退降的能量 Ed 与系统熵的增量成正比结论：