Transcript CHAPTER II • 22 Systems with Variable Amount of Matter. The Chemical Potential. • A system with constant quantity of matter is called “闭系” •

CHAPTER II
• 22 Systems with Variable Amount of Matter. The
Chemical Potential.
• A system with constant quantity of matter is called
“闭系”
• A system exchanged matter with external is called
“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
U 


N

 S ,V
 

• 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：
~ ~ ~
~
~
V ,U , S : V  NV , U  NU ,
~
S  NS
• 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:
F  U  TS , H  U  PV , G  U  TS  PV
and
V
~ S V
~
U ( S , V , N )  NU ( , ), F (T , V , N )  NF (T , )
N N
N
~
~ S
H ( S , P, N )  NH ( , P ), G (T , P, N )  NG (T , P )
N
We have further expression:
Thermodynamic functions for variablemass system
dU  TdS  PdV  dN, dF  SdT  PdV  dN,
dG  TdS  VdP  dN, dG  SdT  VdP  dN,
• From above functions, we can define:
 (
U
F
H
G
) S ,V  ( )T ,V  (
) S , P  ( )T , P
N
N
N
N
• The Gibbs function of one mole is defined as:
G
G
~
 G (T , P )   (T , P )  (
)T , P
N
N
定义式
• Thus, the differential (微分形式) of is expressed by:
~
~
d  S dT  VdP
• 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。
• 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：
 T 
 P 
 T 
  
 P 
  
   , 


    ,
  
 V  S , N
 S V , N  N  S ,V  S  N ,V  N  S ,V  S  N ,V
• 单位Jocobian 行列式为：
 (T , S )
 1, at N  const,
 ( P, V )
 (T , S )
 1, at V  const,
( , N )
 ( P, V )
 1, at S  const,
( , N )
The differentials of free energy,
enthalpy are treated in same way
(T , S )
 1, at P  const,
( , N )
 ( P, V )
 1, at T  const,
( , N )
• 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 ?
S 
Q
,
T
C 
Q
T
•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.
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
1 1
dS  dS1  dS2  Q(  )  0
T2 T1
•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
Q
P1  P2

dV  0
•The increase in entropy is: dS 
T
T
Some examples: The Gay-Lussac Process
• The Joule-Thomson process is at constant internal energy.
•
dU = TdS – P dV =0
U
U
)V dS  (
) S dV  TdS  PdV  0
S
V
U
S
S
dS  (
)S (
)V dV  (
)U dV
V
U
V
S
U
S
P
(
)U  (
)S (
)V   0
V
V
U
T
P
dS  dV  0
T
dU  (
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,
S
V
• Whence(由此)，[hence]：  P    T  0

As P  0,
H
So
S  0.
Therefore, the JT process is an entropy increase process.
For examining the increase in entropy, the molar entropy
of a perfect gas is considered from famous Eq.4.6.
~
R
TV  1
~ ~
S  S0 
ln( ~  1 )
  1 T0V0
R
R
~
~
~ 1
~  1
S 
ln(TV )  S0 
ln T0V0
 1
 1
NR
R
~
~  1
~  1
S
ln(TV )  N [ S 0 
ln T0V0 ]
 1
 1
NR
R
~
~  1
 1

ln(TV )  NR ln N  N [ S 0 
ln T0V0 ]
 1
 1
NR

ln(TV  1 )  A
 1
The entropy of a mixture of two
different perfect gases
• Two cylinders have equal volumes V. Vessel 1 and 2 contains N1
and 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
S (T ,V )  S1 (T ,V )  S2 (T ,V )
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
S1, 2 
N1, 2 R
 1, 2  1
 1 , 2 1
ln(TV1, 2
)  B1, 2 N1, 2  RN1, 2 ln N1, 2
•Partition is removed, and A=0, Q=0, U=0, the entropy is
V1  V2
V1  V2
S  N1 R ln(
)  N 2 R ln(
)0
V1
V2
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：
 R

V  1
S  N
ln[( ) T  B
N
  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
entropy 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 GayLussac 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:
dS  dS1  dS 2 
Q
T
, Q  TdS
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)
dU  Q  PdV
用熵代替：
其他热力学函数
dU  TdS  PdV
dF   SdT  PdV
dH  TdS  VdP
dG   SdT  PdV
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
suddenly 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:
U  U0  T0S0  P0V0  T0S0  P0V
Total entropy must increase:
Inequality is
S0  S  0
(U  T0 S  P0V )  0
So, at T=T0, and V=V0
T=T0, and P=P0
dF  0, F  Fmin
dG  0, G  Gmin
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:
dG  SdT  VdP  1dN1  2 dN2
dGT , P  1dN1  2dN2 and 1  ( G )T , P, N ,  2  ( G )T , P, N
N1
2
N 2
1
Equalization condition
• dG < 0, two phases: N1+N2 = const, dN1 = - dN2
(1  2 )dN1  0
If
dN1  0, 1  2 , If
dN1  0, 1  2
• 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
potentials 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
1 (T , P)  2 (T , P)
• 由此平衡关系可以得到相变点的压强与温度的关系。
P  P(T ), or T  T ( P)
• 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。
d (U1  PdV1  1dN1 )  d (U 2  PdV 2   2 dN 2 )
dS  dS1  dS 2 
T
因 dU1 +dU2 = 0， dV1 +dV2 = 0， dN1 +dN2 = 0。
( 1   2 )dN 2
dS 
T
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：


~
~
( ) P   S , ( )T  V
T
P
• 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
~ ~
  T (S2  S1 )
~
~
化学势的全微分为： d  S dT  VdP
当温度T和压强P发生变化时，两相化学势的变化相同：
~
~
~
~
d1  S1dT  V1dP  d2  S2dT  V2 dP
• 将上式整理得：
~
~
dP S 2  S1

 ~ ~ 
~ ~
dT V2  V1 T (V2  V1 )
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