页数：54页文件大小：2.99 MB来源： Atkins’PhysChem物理化学ppt课件
本课件进入热力学第二定律：由自发过程的不可逆性出发，引出熵的热力学定义，讨论熵作为状态函数的性质与常见过程的熵变计算，并用卡诺循环给出热力学温标与克劳修斯不等式。
In this chapter we will discuss: to predict the max. non - expansion work of a process to express a spo
The direction of spontaneous changes 4.1 The dispersal of energy 4.2 entropy 4.3 entropy changes accompanying specific processes 4.4 The third Law of thermodynamics Concentrating on the system 4.5 The Helmholtz and Gibbs energy 4.6 Standard Gibbs energy
Spontaneous changes, require no work to cool an object with a refrigerator to confine a gas to a smaller vo
The Second Law of thermodynamics Clausius expression
The Second Law of thermodynamics Kelvin’ s statement: No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Kelvin, Lord William Thomson (1824 - 1907)
The Second Law of thermodynamics Other expression The second kind of a permanent - moving machine cannot be made which could absorb heat from a single temperature reservoir and change it completely into work without any effect on the surroundings.
What determines the direction of spontaneous change?
能量的耗散
A ball bounce on a floor: the ball does not rise as high after each bounce. inelastic losses between ball and floor the kinetic energy converted into the energy of the thermal motion. The direction of spontaneous change is towards a the ball is at rest all its energy degraded into the thermal motion of the atoms 4.1 the dispersal of energy
能量的耗散
A ball resting on a warm floor has never been observed to start bouncing. In order to begin bouncing, some of the thermal motion of the atoms in the floor would have to accumulate in a single, small object, the ball. This accumulation requires a spontaneous localization of energy from the myriad of vibrations of the atoms of the floor into the much smaller number of atoms that constitute the ball.
4.1 the dispersal of energy
Whereas the thermal motion is disorderly, for the ball to move upwards its atoms must all move in the same direction. The localization of random motion as orderly motion is virtually impossible. 4.1 the dispersal of energy
The puzzling things are that something collapse into disorder can result the formation of such ordered substances as crystals or proteins. But in these cases, the organized structures can emerge as energy and matter disperse. 4.1 the dispersal of energy
The 1 stLaw allows uses the internal energy, Δ U, to assess a permissible change. The 2 ndLaw allows uses the entropy, S, to identify the spontaneous change among those permissible changes. Entropy is a state function. 4.2 Entropy
熵
The 2 ndLaw is expressed in terms of the entropy: The entropy of an isolated system increases in the process of a spontaneous change: Δ Stot > 0 Δ Stot is the total entropy of the system and its surroundings. Thermodynamically irreversible processes are all spontaneous changes, and hence must be accompanied by an increase in total entropy.
4.2 Entropy
熵
Thermodynamic definition of entropy: After a process, the extent of energy is transferred a she at (or in a disorderly manner). Heat stimulates disorderly motion in the surroundings, hence changes the degree of disorder, or entropy.
Work stimulates uniform motion of atoms in the surroundings, hence does not change the entropy. (a) The thermodynamic definition of entropy..
熵
Thermodynamic definition of entropy: To calculate Δ S between any two states of a system, we have to find a reversible path between them, and integrate the heat supplied a teach stage of the path divided by the temperature at which the heat is supplied.
But (a) The thermodynamic definition of entropy
For a isothermal expansion of a perfect gas ∵ Δ U = 0, q = - w, qrev = - wrev ∴ Ex: The calculation of Δ S
熵
If surroundings consist of a reservoir of const. V For a infinitesimal change of heat, dqsurdqsur = dUsur (dqsur = dHsurat constant pressure) That means dqsur is independent of whether the process is reversible or irreversible.
∴ Ex: The calculation of Δ S in surroundings
熵
For any adiabatic change, at const. Vqsur = 0, Δ Ssur = 0 Ex: H 2 (g) + ½ O 2 (g) H 2 O (l), Δ H = - 286 k J ∵ the process if at constant pressure, qsur = Δ Hsur = - Δ H ∴ Δ Ssur = - Δ H / T = 2.86 105 J / 298 K = 960 J K - 1 Ex: The calculation of Δ S in surroundings
In a thermodynamic cycle (b) The entropy as a state function
(b) The entropy as a state function … Hot source (combustion), Tsur, h Cold sink (cooling water) Tsur, cH 2
In a Carnot cycle (b) The entropy as a state function … 4. Reversible adiabatic compression from D to A. 3. Reversible isother
Carnot 循环
In a Carnot cycle 1. Reversible isothermal expansion from A to BatTh. 2. Reversible adiabatic expansion from B to C. Δ S = 0 (b) The entropy as a state function … 卡诺循环
In a Carnot cycle 3. Reversible isothermal compression from C to D atTc. 4. Reversible adiabatic compression from D to A. Δ S = 0 (b) The entropy as a state function … 卡诺循环
In a Carnot cycle The total Δ S around the cycle: For a adiabatic process (b) The entropy as a state function
熵
To apply the conclusion to any materials, not just a perfect gas. Consider the efficiency of a heat engine: The greater the work output for a given supply of heat from the hot reservoir, the greater is the efficiency of the engine.
(b) The entropy as a state function
To express the definition in terms of the heat transactions: The work supplied by the engine is the difference between supplied by the hot reservoir and returned the cold reservoir: (b) The entropy as a state function … To improve ε revTh orTc
The another statement of the 2 ndLaw of thermodynamics: All reversible engine have the same efficiency regardless of their construction. (b) The entropy as a state function
熵
Suppose two engines are coupled together and run between the same two reservoirs. For A: to acquire q h from hot source and to release qcin to cold sink. For B: to acquire q c from cold sink and to release a certain quantity of heat q’ h into hot source.
(b) The entropy as a state function
Suppose the engine A is more efficient than B. (b) The entropy as a state function … Not all the work that A produced is needed for the process, and differ
Now consider an arbitrary reversible cycle which can be approximated as a collection of Carnot cycle. (b) The entropy as a state function Interior cancel Perimeter survive Each of the infinitesimal cycle:
Suppose an engine works between two heat reservoir. (c) The thermodynamic temperature Temperature in cold sinkT = (1 - ε) Th The thermodynamic temperature s
Consider the work required to cool an object. The process is not spontaneous because not enough entropy is generated in the hot sink to overcome the entropy loss from the hot source. The thermodynamics of refrigeration
熵
To generate more entropy, energy must be added to the stream that enters the warm sink. To find the minimum energy that needs to be supplied. Coefficient of performance The less the work required, the greater the coeff. and more efficient the refrigerator.
The thermodynamics of refrigeration
熵
For q c is removed from the cold source, w is added to the energy stream, then the heat deposited into the hot sink is q h = q c + w The work to maintain a low temperature is also relevant to the design of refrigerators.
If the rate of heat leak is written A (Th - T c), the minimum power require to maintain the original temperature difference by pumping out that heat is: The thermodynamics of refrigeration
由－ w = q h + q c 可知向低温热源放热: q c = － w － q h = {－ (－ 0.4) － 1} k J = － 0.6 k J 例热源和冷却水温度分别为500 K 和300 K. 试问工作于此二温度间的热机, 从高温热源吸热1 k J 的热最多能作功若干? 最少需向冷却水放热若干?
q 1 为负表示向高温T 1 放热. 计算表明, 用电能驱动热泵, 可得到16 倍电能的热, 而通电予电炉却只能得到与电能等量的热. 例冬季利用热泵从室外0 ℃ 吸热, 向室内18 ℃ 放热. 若每分钟用100 k J 的功开动热泵, 试估算热泵每分钟最多能向室内供热若干?
熵
The entropy is a signpost of spontaneous change, Δ Stot ≥ 0 Consider a system in thermal and mechanical contact, but not necessary in equilibrium, with its surroundings at the same temperature. For an arbitrary change of the system, dS + dSsur ≥ 0, ordS ≥ - dSsur (d) The Clausius inequality … (The Clausius inequality)
熵
(d) The Clausius inequality … Suppose the system is isolated from its surroundings, d q = 0, and the ndS ≥ d q / T = 0 In an isolated system, the entropy cannot decrease when a spontaneous change takes place. For an irreversible adiabatic change, ∵ d q = 0, dS ≥ 0, dSsur = 0, ∴ dStot ≥ 0 For an irreversible isothermal free expansion of a perfect gas, ∵ d q = - d w = 0, dS ≥ 0, dSsur = 0, ∴ dStot ≥ 0
熵
For an irreversible change, spontaneous cooling, d q leaves from the hot source, the entropy changed S = - d q / Th. When d q enters the cold sink, the entropy changes bydS = d q / T c ∴ IfTc = Th, dStot = 0, at thermal eqm (d) The Clausius inequality
calculation of Δ S in a variety of basic processes 4.3 Δ S accompanying specific processes (a) The entropy of phase transition at transition temperature At norma
calculation of Δ S in a variety of basic processes 4.3 Δ S accompanying specific processes (a) The entropy of phase transition at transition temperature Trouton’
For an arbitrary path: (b) The expansion of a perfect gas But Δ Stot is depends on how the expansion takes place. For a reversib
For an arbitrary change from one T i to That const. p: (c) The variation of entropy with temperature
The entropy of a system at T is related to its entropy at T = 0 by measuring its Cpat different temperature, and evaluating the integral, and adding the transition entropy for all phase transition between T and 0. (d) The measurement of entropy
(d) The measurement of entropy The calculation of an entropy from heat capacity data.
(d) The measurement of entropy Debye extrapolation Cp = aT 3 Near T = 0
Homework Page 117, exercises 4.4, 4.6, 4.7, 4.8, 4.10 Page 118, problems 4.5, 4.6
Sad i Carnot Sad i Carnot (French, 1790 - 1832) 摄于1813 年, Ecole 理工专科学校一年级学生
Rudolf Clausius Born: 2 Jan 1822 in Kosl in, Prussia (now Koszal in, Poland) Died: 24
Kelvin, Lord (William Thomson) The Lord Kelvin Transcends Entropy
下载地址：{{downurl}}
{{{a}}}
{{#b}}{{{b}}}
{{/b}} {{/title}} {{#nolog}}登陆注册
{{/nolog}} {{#djsx}}点我刷新
{{/djsx}} {{#gocz}} {{/gocz}} {{#ffqr}} {{/ffqr}}