CHAPTER II• 22 Systems with Variable Amount of Matter. The Chemic

al Potential.

• A system with constant quantity of matter is called

“ “ 闭 系 ”闭 系 ”• A system exchanged matter with external is called

“ “ 开 系 ”开 系 ”• Let us turn now to these thermodynamic systems.

• The examples of matter change:

• various chemical transformations of chemical compound.

• melting, crystalline, evaporation, and phase transition.

• ( 化合物的化学反应、溶化、结晶、蒸发、相变 )

The basic formulae of variable-mass system

• First is the internal energy U. U has not only its own natural variables, S and V, but also has a variable N, characterizing the number of moles. Therefore, it follows

dU = TdS – PdV + dN has the dimension of energy per mole and is referred to

as the “chemical potential” of a substance, which is

VSN

U

,

• Subdivide the thermodynamic quantities into extensive (additive) and intensive.

• 广延量• 强度量

广延量与强度量

• 体积 V 、熵 S 、内能 U 均是广延量。若系统内物质的摩尔数为 N ，定义单位摩尔的物理量：

SNSUNUVNVSUV~

,~

,~

:~

,~

,~

• 当系统的分子数发生变化时， V 、 S 、 U 将随之变化，而强度量 T 、 P 却不发生变化。 By virtue of formulae:

PVTSUGPVUHTSUF ,,

and

),(~

),,(),,(~

),,(

),(~

),,(),,(~

),,(

PTGNNPTGPN

SHNNPSH

N

VTFNNVTF

N

V

N

SUNNVSU

We have further expression:

Thermodynamic functions for variable-mass system

• From above functions, we can define:

,,

,,

dNVdPSdTdGdNVdPTdSdG

dNPdVSdTdFdNPdVTdSdU

PTPSVTVS N

G

N

H

N

F

N

U,,,, )()()()(

• The Gibbs function of one mole is defined as:

PTN

GPTPTG

N

G,)(),(),(

~

• Thus, the differential ( 微分形式 ) of is expressed by:dPVdTSd

~~

• 理解：单位摩尔物质，吉布斯函数是强度量的函数，物质分子数的增加，并不增加单位摩尔吉布斯函数，而是增加系统总的吉布斯函数。然而，强度量 T、 P的变化导致化学势的变化。

定义式

• 根据吉布斯的定义，摩尔数 N可以看做是一个特殊坐标，化学势是一个 conjugate generalized force(共轭广义力 ).

• 可以得到偏导数间的对应关系：

VNVSVNVSNVNS SN

P

SN

T

S

P

V

T

,,,,,,

,,

• 单位 Jocobian 行列式为：

,,1),(

),(

,,1),(

),(

,,1),(

),(

constSatN

VP

constVatN

ST

constNatVP

ST

The differentials of free energy, enthalpy are treated in same way

,,1),(

),(

,,1),(

),(

constTatN

VP

constPatN

ST

• Being an intensive quantity, the chemical potential is both independent of the number of moles and volume.

23 The Increase in entropy in equalization processes, The Gibbs paradox

• Previously, the equilibrium processes;

• Presently, concern non-equilibrium processes:

• 1) in a closed system, two or more parts, each equilibrium;

• 2) the initial and intermediate states are non-equilibrium;

• 3)come into a completely equilibrium;

• 4)an equilibrium adiabatic process, but entropy change.

What is Entropy

• We have known:• Entropy is a constant parameter in adiabatic process.• Entropy is a thermodynamic quantity like P、 V、 T.

• How we define Entropy ?,

T

QS

•The change in entropy in an isothermal process is the quantity of heat, the system absorbed, divides the temperature of heating.

•This definition is applicable to any thermodynamic process.

T

QC

Equalization of Temperature

• Only two equilibrium bodies are considered in a system.

• Temperatures are T1 and T2(T1 > T2), “contact”

• The system is heat-insulated, there exists heat transfer:• The left: Q1= - Q ;

• The right: Q2= Q .• The entropy also changed:

• The left: S1= - Q /T1 ;

• The right: S2= Q /T2.

• How about the total S?

The Total Entropy

• The change in entropy of each body depends 1) not on the manner of removed heat, in reversible or irreversible way;

• 2) on the quantity of heat.

• For a real irreversible process, the variation in the total entropy

0)11

(12

21 TT

QdSdSdS

•the variation in the total entropy is positive

Equalization of Pressure• 1)A piston,• 2) a heat-insulated cylinder between two

volumes, 3) at equal temperature, • 4) different pressure• Process is irreversible, isoenergetic

• Replace this by an imaginary process:

• The pressure is P1-P2- , the volume is dV.

is infinitely small.

• The real irreversible and imaginary process have the same initial and final states.The system performs work (P1-P2)dV = A= Q

•The increase in entropy is: 021

dVT

PP

T

QdS

Some examples: The Gay-Lussac Process

• The Joule-Thomson process is at constant internal energy.

• dU = TdS – P dV =0

0

0)()()(

)()()(

0)()(

dVT

PdS

T

P

U

S

V

U

V

S

dVV

SdV

U

S

V

UdS

PdVTdSdVV

UdS

S

UdU

VSU

UVS

SV

Entropy increases during

this irreversible free expansion.

The Joule-Thomson Process

• An equivalent reversible process occurs at a constant heat content (一个等价的可逆过程以等焓发生 )。

• dH = T dS + V dP,• Whence(由此 )， [hence]： 0

T

V

P

S

H

Therefore, the JT process is an entropy increase process.

.0So,0As SP

For examining the increase in entropy, the molar entropy of a perfect gas is considered from famous Eq.4.6.

)~

~ln(

1

~~1

00

1

0

VT

VTRSS

1000

1 ~ln

1

~)

~ln(

1

~

VT

RSVT

RS

ATVNR

VTR

SNNNRTVNR

VTR

SNVTNR

S

)ln(1

]~

ln1

~[ln)ln(

1

]~

ln1

~[)

~ln(

1

1

1000

1

1000

1

The entropy of a mixture of two different perfect gases

• Two cylinders have equal volumes V. Vessel 1 and 2 contains N1 a

nd N2 moles of the first gas A and second gas B, respectively. • C is a heat-insulating enclosure.• Two cylinders can freely entering each other without friction( 磨

擦 )

• Gibbs: Carry out mentally an experiment. ( 在大脑中进行实验 )

• Vessel 1: a is the wall, permeable to the molecules B, not A.

• Vessel 2: b is the wall, permeable to the molecules A, not B.

• Let vessel 2 insert slowly into vessel 1, then the space ab contains the mixture of gases.

• As freely permeating of two molecules, A no pressure on b. Two walls of V2 are exposed to the same pressure.

• T is not changed on the condition of A = 0,Q = 0.

• The Entropy in this reversible mixture process is

),(),(),( 21 VTSVTSVTS

Interdiffusion of Two Gases

• A heat-insulated vessel separated by a partition( 隔离物 ).

• Two different perfect gases, same T, P.

•The entropy satisfied

2,12,12,12,11

2,12,1

2,12,1 ln)ln(

12,1 NRNNBTV

RNS

•Partition is removed, and A=0, Q=0, U=0, the entropy is

0)ln()ln(2

212

1

211

V

VVRN

V

VVRNS

Discussion• Last formula is so-called Gibbs paradox( 吉布斯谬论 ) ，因

为等号右边没有两种气体性质的参量。因此，当两种气体完全相同时，“熵增加”。

• 解释：当两种气体完全相同时，不发生“相互渗透” . “ 熵不增加”。• Impermeable

• 广延量熵的表述为：

BTN

VRNS 1)ln[(

1

结论： moles N and volume V are increased n times, the entropy will increases in the same proportion.

The Entropy increase principle• Several examples demonstrate that the en

tropy of an isolated system undergoing equalization processes increases, dS > 0. Prove? No, but in statistics.

• In any isolated system, the entropy reaches its maximum value in the state of equilibrium.

• How about universe? Silence finally?

Reversible and Irreversible?• 热力学第一定律 ( 能量定律 ) 允许的过程• 1 。热量从高温物体流向低温物体，及其逆过程。• 2 。物体因磨擦力而减速运动的过程；• 3 。气体向真空自发 spontaneous 膨胀的过程 (the Gay-L

ussac process) ，其逆过程为自发压缩；• 4. 两种不同的气体相互扩散和混合气体自发分离的过程。• 还可以找出很多这样的过程• 上述四个过程的逆过程实际是被禁止的，熵增加原理允

许自发的熵增加的过程，禁止熵减小的自发过程 ( 第二定律 ) ，除非熵减小的过程是非自发的。

作业： P121 problem,

24.Eetrema of Thermodynamic Functions 极限

• 自发的绝热过程，熵会增加，下面考虑两个问题：

• 1) 非绝热的自发过程或可逆过程，熵如何变化；

• 2) 其他热力学函数是否存在类似的极限规律？

第一个问题的分析• Consider a not heat-insulated irreversible process. Heat

Q is added to or removed from the system. • This process is mentally visualized as two processes: one is spontaneous process:

dS1>0, Q 1 = 0 second is not heat-insulated process:

if heat is added dS2>0, Q 2 = Q=T dS2 >0

if heat is removed dS2<0, Q 2 = Q=T dS2 <0• Results of inequality:

TdSQT

QdSdSdS

,21

第二个问题的分析• 如果熵的不等式成立，则热力学函数的不等式将很容易解决。

• 考虑能量定律 (第一定律 )

PdVQdU 用熵代替：

PdVTdSdU

其他热力学函数

PdVSdTdG

VdPTdSdH

PdVSdTdF

Corollaries stemming from inequalities( 源于不等式的推论 )

• V = const, and T = const : dF < 0;• P = const, and T = const : dG < 0;• P = const, and S = const : dH < 0.• Inequalities stem from irreversible process.• dS>0; dF < 0; dG < 0; dH < 0.• If we put the material in “V = const, and T = const “, and s

uddenly a disturb make the materials in inequality.

平衡条件下的物理意义• 一个系统是否达到了平衡，其判断的依据是什么？• 考虑一个物体与外界接触，且不为热平衡。• 外界： T0、 P0；物体： T、 P。外界的条件始终不变。• For a closed system:

VPSTVPSTUU 00000000

Total entropy must increase: 00 SS

Inequality is 0)( 00 VPSTU

So, at T=T0, and V=V0

T=T0, and P=P0 min

min

,0

,0

GGdG

FFdF

26. Phase Equilibrium. First-Order Phase Transitions( 一阶相

变 )• We have considered homogeneous systems.( 均匀系统 )• Turn to system comprising several phases in equilibrium.• a “phase( 相 )” : a homogeneous part of a system.• Several phases are separated by a well-defined boundary.• Example: two phases of a liquid and a vapour at constant

temperatures T and pressures P. • The number of moles in each phase, N1 and N2.• We use G to describe the transition:

2211 dNdNVdPSdTdG

2211, dNdNdG PT and12 ,,

22,,

11 )(,)( NPTNPT N

G

N

G

Equalization condition

• dG < 0, two phases: N1+N2 = const, dN1 = - dN2

0)( 121 dN

211211 ,0If,,0If dNdN• Discussion:

• dN1< 0 表示 N1 减少， N2 增加。即化学势大的物质减少，化学势小的物质增加。将化学势对应某种能量或能级，则可认为粒子从高能级自动流向低能级，如果粒子的流动会引起化学势的变化，可以预见，平衡时化学势相等。

等化学势的物理意义• 不平衡的系统中，• “ 温度差导致了 heat transfer”• “ 压强差导致了 gas flow ”• “ 化学势差导致了 mass flow”• Mass transfer terminates when the chemical po

tentials become equal.• ( 等化学势终止了质量流 )• 化学势也是一个热力学参量，如温度和压强，

也是广延量。

由化学势所得到的：• 化学势虽然与物质的性质有关 ( 如水与汽 ) ，但同时也

与热力学参量 P 和 T 密切相关，平衡时可以表达为

),(),( 21 PTPT

• 由此平衡关系可以得到相变点的压强与温度的关系。)(),( PTTorTPP

• 例如水与汽平衡，一个大气压时的温度为固定值。压强变化则温度相应变化。其关系称之为“相图”。

相变发生时，热力学量是如何变化的？

融化吸热相变• 相变时，如未饱和水、汽共存时，水会继续蒸发。设

水为物质 1 ，汽为物质 2 。共存温度为 T ，压强为 P ，外界提供热量 dQ 。熵变为 dS = dS1 +dS2 。

T

dNPdVUddNPdVUddSdSdS

)()( 2222111121

因 dU1 +dU2 = 0 ， dV1 +dV2 = 0 ， dN1 +dN2 = 0 。

T

dNdS 221 )(

因此，提供的热量 TdS 使 dN2 的水化为汽，其比例系数为化学势之差。为了讨论的方便，引入摩尔熵 $ 。

一阶相变• 将化学势等效于摩尔吉布斯函数，得到如下关系式：

VP

ST TP

~)(,

~)(

• 摩尔熵和摩尔体积为吉布斯函数的一阶导数。将摩尔熵和摩尔体积发生突变的相变称之为一阶相变。水变成汽的相变为一阶相变。定义相变的摩尔热量为，则

)~~

( 12 SST 化学势的全微分为： dPVdTSd

~~

当温度 T 和压强 P 发生变化时，两相化学势的变化相同：

dPVdTSddPVdTSd 222111

~~~~

• 将上式整理得：

)~~

(~~

~~

1212

12

VVTVV

SS

dT

dP

上式称之为 Clapeyron-Clausius equation 。 “克拉珀龙 -克劳修斯方程 ”

• 例：冰在 1pn 下的熔点为 273.15K ，熔解热为 =3.35Jkg-1 ，冰与水的比容分别为 1.0907*10-3 和 1.00013*10-3m3kg-1 ，

• 由此计算得到： dT/dP = -0.00752 ，• 实验： dT/dP = -0.0075 。

作业， ( 中文 )P144 ： 3.4(1) 、 3.7 、 3.9 、 3.16

化学势的变化图