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CHAPTER II

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:

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CHAPTER II

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  1. 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)

  2. 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 • Subdivide the thermodynamic quantities into extensive (additive) and intensive. • 广延量 • 强度量

  3. 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： • 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: and We have further expression:

  4. Thermodynamic functions for variable-mass system • From above functions, we can define: • The Gibbs function of one mole is defined as: 定义式 • Thus, the differential (微分形式) of is expressed by: • 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。

  5. 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： • 单位Jocobian 行列式为：

  6. The differentials of free energy, enthalpy are treated in same way • Being an intensive quantity, the chemical potential is both independent of the number of moles and volume.

  7. 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.

  8. 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 ? • 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.

  9. 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?

  10. 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 • the variation in the total entropy is positive

  11. 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 performswork (P1-P2)dV = A=  Q • The increase in entropy is:

  12. Some examples: The Gay-Lussac Process • The Joule-Thomson process is at constant internal energy. • dU = TdS – P dV =0 Entropy increases during this irreversible free expansion.

  13. 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]： 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.

  14. 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)

  15. 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

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

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

  18. 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?

  19. 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-Lussac 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,

  20. 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？

  21. 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, Q1 = 0 second is not heat-insulated process: if heat is added dS2>0, Q2 = Q=TdS2 >0 if heat is removed dS2<0, Q2 = Q=TdS2 <0 • Results of inequality:

  22. 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) 用熵代替： 其他热力学函数

  23. 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.

  24. 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: Total entropy must increase: Inequality is So, at T=T0, and V=V0 T=T0, and P=P0

  25. 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: and

  26. Equalization condition • dG < 0, two phases: N1+N2 = const, dN1 = - dN2 • 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。

  27. 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。

  28. 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 • 由此平衡关系可以得到相变点的压强与温度的关系。 • 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？

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

  30. 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： • 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 化学势的全微分为： 当温度T和压强P发生变化时，两相化学势的变化相同：

  31. 将上式整理得： 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

  32. 化学势的变化图

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