06 Chapter 4 The Second Law：The concepts2.pptx

第1页：热力学第二定律：基本概念

第2页：能量的耗散

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

第3页：热力学第三定律

4.4 The Third Law of thermodynamics At T = 0 all thermal motion stoppe

第4页：能斯特

(a) The Nernst heat theorem … The entropy change accompanying any phys

第5页：能斯特

(a) The Nernst heat theorem … Ex: orthorhombic sulfur S (α) ↔ monoclin

第6页：热力学第三定律

The Third Law of thermodynamics The entropy of all perfect crystalline

第7页：热力学第三定律

(b) Third - Law entropies Third - Law entropies: entropies based on S

第8页：热力学第二定律：基本概念 例题

Ex 1. 1 mol of a perfect gas expanse isotherm ally at 27 ℃ from 50 dm

第9页：热力学第二定律：基本概念 例题

Ex 2.5 molofa perfect gas (Cp  m = 29  10 J K  1) colds down from 4

第10页：热力学第二定律：基本概念 例题

Ex 2.2 molofa perfect gas (C V  m = 3 R / 2) at 500 K, 405.2 kPa unde

第11页：热力学第三定律

(1) q 1 = 0,  U 1 = w 1 nCV  m (T 2 － T 1) (2) q 2 = 0,

第12页：热力学第三定律

(3)  V = 0, w 3 = 0, For the whole process:  U = 0,  H = 0, qtot =

第13页：热力学第三定律

热力学第三定律

1 mol, H 2 O (l) 101.325 kPaT 2 = 263.15 K 1 mol, H 2 O (s) 101.325 kPaT 2 = 263.15 KIrrev.  S (T 2) = ? 1 mol, H 2 O (l) 101.325 kPaT 1 = 273.15 K 1 mol, H 2 O (s) 101.325 kPaT 1 = 273.15 KRev.  S (T 1)  S 1  S 2  S (T 2) =  S 1 +  S (T 1) +  S 2 Ex: 1 molof over coldwater freezes to ice at 263.15 K and 101.325 kPa, calculate Δ S of the system.

The normal freezing heat of water is - 6020 Jmol  1. Cp  m (water) = 75.3 Jmol  1 K  1, Cp  m (ice) = 37.6 Jmol  1 K  1.

第14页：热力学第三定律

The heat of freezing at 263.15 K The result means that the freezing of the over cold water is spontaneous. Ex: 1 mol of over cold water freezes to ice at 263.15 K and 101.325 kPa, calculate Δ S of the system,  Ssurr and  S total.

第15页：热力学第三定律

The direction of spontaneous change? Δ S: both system and its surroundings Can we assess the direction just base on the data of the system? Concentrating on the system Δ A and Δ G Yes! Yes! Yes!

第16页：热力学第三定律

Suppose a system is in thermal eqmwithitssurrdgs, when a change occur in the system, the Clausius inequality is: dS - d q / T ≥ 0 We now develop this inequality in 2 ways according to the const. V or const. p 4.5 the Helmholtz and Gibbs energies

第17页：热力学第三定律

热力学第三定律

1 st consider heat transfer at const. V ∵ non on - expansion work, d q = dU, dS - dU / T ≥ 0 ∴ The expression means the criterion for spontaneous change solely in terms of the state functions of the system. orTdS ≥ dU (const.

V, no additional work) anddSU, V ≥ 0 (const. internal energy and const. V) d US, V ≤ 0 (const. entropy and const. V) (a) Criteria for spontaneity

第18页：热力学第三定律

dSU, V ≥ 0 (const. internal energy and const. V) The eqnmeans that the entropy of a system is increased in a spontaneous change at const. U and const. V such as in isolated system. (a) Criteria for spontaneity

第19页：热力学第三定律

热力学第三定律

d US, V ≤ 0 (const. entropy and const. V) The eqnmeans that at const. S and const. V, in a spontaneous change, the internal energy of a system is decreased as energy floesouttoitssurrdgsas heat. And hence the entropy of the surrdgs is increased.

(a) Criteria for spontaneity

第20页：热力学第三定律

2 nd consider heat transfer at const. p ∵ non on - expansion work, d q = d H, dS - d H / T ≥ 0 orTdS ≥ d H (const. p, no additional work) anddSH, p ≥ 0 (const. enthalpy and const. p) dHS, p ≤ 0 (const. entropy and const. p) (a) Criteria for spontaneity

第21页：热力学第三定律

dS - dU / T ≥ 0  dU - TdS ≤ 0 dS - d H / T ≥ 0  d H - TdS ≤ 0 Now introduce 2 thermodynamic quantities Helmholtz energy A = U - TSGibbs energy G = H - T S (a) Criteria for spontaneity

第22页：热力学第三定律

At const. T, after a state change of a system (a) Criteria for spontaneity … d AT, V ≤ 0 dGT, p ≤ 0 d A = dU - TdSdG = d H - TdS Then we obtain the

第23页：热力学第三定律

热力学第三定律

At const. T and V, a change of a system is spontaneous ifdAT, V ≤ 0 The criterion of eqm is d AT, V = 0 From d A = dU - TdS, ifdU < 0 and dS > 0, then d A < 0 That not means systems favour lower energy, for dSis the entropy change of the system, - dU / T is the entropy change of the surrdgs at const.

V, their total tends to a maximum. (b) Some remarks on the Helmholtz energy dS - dU / T ≥ 0

第24页：热力学第三定律

The great significant meaning of A is not solely the signpost of spontaneous change. d A is equal to the maximum work accompanying a process: dwmax = d A Hence, A is sometimes called the “ the maximum work function ”, or the “ work function ” (c) Maximum work

第25页：热力学第三定律

热力学第三定律

We have proved that a system does maximum work when it is working reversibly for a perfect gas. Now we turn to prove its universal validity. For Clausius inequality: dS - d q / T ≥ 0 and the 1 stLaw: dU = d w + dqFor a reversible change at constant T: dU = dwrev + dqrev = dwrev + TdSdwrev = dU - TdS = dAordwmax = d A (c) Maximum work

第26页：热力学第三定律

For a macroscopic isothermal change in the system: wmax = Δ A and Δ A = Δ U - T Δ S (c) Maximum work … The expression means that not all the change in internal energy may be

第27页：热力学第三定律

For a macroscopic isothermal change in the system: wmax = Δ A and Δ A = Δ U - T Δ S (c) Maximum work … If Δ S > 0, Δ U < 0, | wmax | > | Δ U | The explanation of this is tha

第28页：热力学第三定律

 A 的物理意义

• 系统从态1 恒温可逆变化到态2 引起的 A 等于过程的可逆功; (A 又称功函)；若是恒温恒容可逆变化则等于可逆非体积功.
• 如果经不可逆途径同样从态1 到态2,  A 的值是一样的, 但非体积功却变大了, 如电池放电(负功) 更少, 充电时耗电更多.
• 铜- 锌电池

第29页：热力学第三定律

热力学第三定律

Δ r U  = - 2808 k J mol - 1 Δ S = 182.4 J K - 1 mol - 1 How much of this energy change can be extracted as (a) heat at const. p, (b) work? Suppose all the gases involved are perfect. (a) Δ ν g = 0, Δ r H  = Δ r U  + Δ ν gRT = Δ r U  = - 2808 k J mol - 1 (b) Δ r A  = Δ r U  - T Δ S = - 2808 - 298.2  182.4 = - 2862 k J mol - 1 Ex: calculating the maximum available work

第30页：热力学第三定律

热力学第三定律

In the laboratory chemistry, we are usually more interested in const. p than const. V, so the Gibbs energy is more common than the Helmholtz energy. FromdGT, p ≤ 0 We can say that at constant T and p, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

(d) Some remarks on Gibbs energy

第31页：热力学第三定律

热力学第三定律

For a given chemical reaction at constant T and p, if G decreases as the reaction process, then the reaction has a spontaneous tendency to convert reactants into products. If G increases, then the reverse reaction is spontaneous.

(d) Some remarks on Gibbs energy

第32页：热力学第三定律

热力学第三定律

How to understand spontaneous endothermic reaction at constant T and p? ∵ Δ H > 0, and Δ G < 0, ∴ Δ S > > 0, then Δ G = Δ H - T Δ S < 0 Δ S > Δ H / T = - Δ Ssur The entropy change in the system overcomes the entropy changes in its surrdgs.

(d) Some remarks on Gibbs energy

第33页：热力学第三定律

热力学第三定律

The analogue of the maximum work interpretation of Δ A, the maximum additional (non - expansion) work w add, max, is given by change in Gibbs energy at const. T and p. dwadd, max = d G, orwadd, max = Δ G This expression is particularly useful for assess in t the electrical work that maybe produced by fuel cells and electrochemical cells.

(e) Maximum non - expansion work

第34页：热力学第三定律

To prove it

(e) Maximum non - expansion work

第35页：热力学第三定律

(e) Maximum non - expansion work

第36页：热力学第三定律

系统从态1 恒温恒压可逆变化到态2 引起的 G 等于过程的可逆非体积功.

• 如果经不可逆途径同样从态1 到态2,  G 的值是一样的, 但非体积功却变大了, 如电池放电(负功) 更少, 充电时耗电更多.  G 的物理意义

第37页：热力学第三定律

热力学第三定律

How much energy is available for sustaining muscular and nervous activity from the combustion of 1.00 molof glucose molecules under standard condition at 37 º C? Δ rS  = 182.4 J K - 1 mol - 1 Answer from Table 2.5, Δ r H  = - 2808 kJmol - 1 Δ r G  = Δ r H  - T Δ rS  = - 2808 kJmol - 1 – (310.2 K)  (182.4 J K - 1 mol - 1) = - 2865 kJmol - 1  w add, max = - 2865 kJmol - 1 Ex: calculating the maximum non - expansion work of a reaction

第38页：热力学第三定律

· A 和G 是两个人为组合的状态函数, 绝对值未知.

• A 和G 本身无明确的物理意义, 之所以定义它们是为了得到两个新的判据. (使用方便)
• A 和G 都是系统的容量性质, 具有能量单位， J (k J).
• 任何过程都会使状态函数A 和G 发生变化:  A =  U －  (T S);  G =  H －  (T S)  (T S) = (T S) 2 － (T S) 1 但只有符合判据条件时,  A 和 G 的正负才可以用来判别过程变化的方向.
• 与熵判据相比, 这两个判据适用于封闭体系, 无须计算环境的 A 或 G, 但增加了过程条件的限制. 对Helmholtz energy 判据和Gibbs energy 判据的几点说明

第39页：热力学第三定律

可逆与“ 平衡” 以及不可逆与“ 自发” 的关系适用条件: 隔离系统的任何过程封闭体系的恒温恒容不作非体积功的过程封闭体系的恒温恒压不作非体积功的过程

• 不作非体积功条件下可能进行的过程才是自发过程. 电解水时, 整个电解装置(包括电源) 是一个隔离体系, 电解过程是可能进行的(体系的熵增大), 但不能说水自发地分解为氢气和氧气.
• 不作非体积功条件下的可逆过程, 系统内部才无限趋于平衡. 若将电池串以相差无限小的反电动势, 电池将可逆地工作, 但电池内的反应物质并未达到化学平衡(平衡时电动势为零).

第40页：标准摩尔吉布斯自由能

Standard Gibbs energy of reaction can be obtain from standard entropies, Δ rS , and standard enthalpies, Δ r H  of the reaction 4.6 Standard molar Gibbs energy … Δ r G  = Δ r H  - T Δ rS  Δ f G : Standard Gibbs energy of formati

第41页：标准摩尔吉布斯自由能

标准摩尔吉布斯自由能

Calorimetry (for Δ H directly, and for Δ Sviaheat capacities) is only one of the ways of determining the values of Gibbs energy. They may also be obtained from equilibrium constants (Chapter 9) and electrochemical measurements (Chapter 10).

They maybe calculated using data from spectroscopic observations (Chapter 20). 4.6 Standard molar Gibbs energy

第42页：热力学第二定律：基本概念 例题

热力学第二定律：基本概念 例题

Ex: 1 mol water freezes at 273.15 K and 101.325 kPa. Calculate  T, pGn 1 = 1 mol, H 2 O (l) p 1 = 101.325 kPaT 1 = 273.15 K n 2 = 1 mol, H 2 O (s) p 2 = 101.325 kPaT 2 = 273.15 K reversible  T, pGFromG = H － TSat const.

T  G =  H － T  S For reversible phase transition   G = 0 表明恒温恒压固- 液平衡条件下的凝固过程(可逆相变) 的 T, pG = 0.

第43页：热力学第二定律：基本概念 例题

Ex

1 mol, 263.15 K 的过冷水于恒压101.325 kPa 下凝固为同温的冰, 求 G. 已知Cp, m (水) = 75.3 J  mol － 1  K － 1， Cp, m (冰) = 37.6 J  mol － 1  K － 1 恒温下 G =  H － T  S  G =  H － T  S = {－ 5643 － 263.15  (－ 20.63)} J = － 214.2 J  G < 0, 表明在恒温恒压下过冷水的凝固是能够自发进行的.

第44页：标准摩尔吉布斯自由能

The basic thermodynamic equation sHUpVpVATSTSGU、H、S、A、G、p、V、TH = U + p V， A = U - T S， G = H - T S Definition 以线段相对长度表示：

第45页：标准摩尔吉布斯自由能

标准摩尔吉布斯自由能

d H = d (U + p V) = dU + pdV + Vdp = TdS + VdpdA = d (U  T S) = dU  TdS  SdT =  SdT  pdVdG = d (H  T S) = d H  TdS  SdT =  SdT + Vdp 对不作非体积功的可逆过程dqrev = TdS, dwrev =  pdVdU = dqrev + dwrev = TdS  pdV 以上4 式称为热力学基本方程, 是封闭体系状态在不作非体积功且可逆条件下发生微变时各状态函数间的定量关系.

dU = TdS  pdVdH = TdS + VdpdA =  SdT  pdVdG =  SdT + Vdp 以上公式的应用条件是： (i) closed system: (i i) no additional work; (i i i) reversible The basic thermodynamic equations

第46页：标准摩尔吉布斯自由能

由热力学基本方程计算纯物质pVT 变化过程的 A 和 G (1) 恒温过程恒温下： d AT = - pdVdGT = Vdp 对理想气体，将p V = nRT 代入，积分得对凝聚态物质：压力改变不大时：  G ≈ 0；压力改变较大时：  G 不容忽略。

第47页：标准摩尔吉布斯自由能

(2) 恒容、恒压变温过程恒容变温

dAV = - SdT 恒压变温： dGp = - SdT 式中： S 是系统的熵值。由于熵的绝对值不知，故不能用规定熵代入积分。但恒压下，温度对于相变化、化学变化过程Gibbs energy 变的影响时，对相变化、化学变化前后的各物质均应用上式，可得d  G = -  SdT 式中 G，  S 分别为相变化、化学变化过程的Gibbs energy 变和熵变。  S 为T 的函数。

第48页：标准摩尔吉布斯自由能

• 纯物质相变过程 G 的计算(1) 可逆相变化 GT, p = 0;  AT, p =  G   (p V) = p  (V) (2) 不可逆相变化可设计可逆途径计算. 恒温恒压, T, p , T, p , T 1, p , T 1, p , T, p 1 , T, p 1  已知体积或密度数据时, 设计变压途径.  已知热容数据时, 设计变温途径.
• (T 1, p) 和(T, p 1) 是两对相平衡温度和压力.
• 当一相是气体时, 相平衡压力亦即饱和蒸气压.

第49页：热力学第二定律：基本概念 例题

热力学第二定律：基本概念 例题

Ex: 已知在100 kPa 下水的凝固点为0 ℃，在－ 5 ℃，过冷水的比凝固焓Δ lsh = － 322.4 J g  1。过冷水和冰的饱和蒸气压分别为p (H 2 O, l) = 0.422 kPa， p (H 2 O, s) = 0.414 kPa。今在100 kPa 下，有－ 5 ℃, 1 kg 的过冷水变为同样温度、压力下的冰，设计可逆途径，计算过程的Δ S 及Δ G。  G H 2 O (l), － 5 ℃, 100 kPaH 2 O (s), － 5 ℃, 100 kPaH 2 O (l), － 5 ℃, 0.422 kPaH 2 O (g), － 5 ℃, 0.422 kPaH 2 O (g), － 5 ℃, 0.414 kPaH 2 O (s), － 5 ℃, 0.414 kPa  G 1  G 2 = 0  G 3  G 4 = 0  G 5  S 1 = 0  S 2  S 3  S 4  S 5 = 0 设计如下可逆途径： 1 kg 水；恒温变化：  S =  S 1 +  S 2 +  S 3 +  S 4 +  S 5

第50页：标准摩尔吉布斯自由能

由于压力变化不大，对液、固体恒温变化

 S 1 ≈ 0;  S 5 ≈ 0  G =  G 1 +  G 2 +  G 3 +  G 4 +  G 5   G 3

第51页：热力学第二定律：基本概念 习题

Homework p 3.8

第52页：热力学第二定律：基本概念 习题

Homework p 3.8

第53页：热力学第二定律：基本概念 习题

Homework p 3.8

第54页：热力学第二定律：基本概念 习题

Exercises (a) 4.11, 4.15, 4.16, 4.23, 4.25 Problems: 4.7 Homework

第55页：能斯特

Walther Hermann Nernst Walther Hermann Nernst was born in Briesen, West Prussia, on June 25,