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1 Chapter 8 hermodynamics: he Second and hird Law 8-1 Entropy

2 Which is Spontaneous Change? 在室溫下 298K

3 he Second Law of hermodynamics No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. he entropy of an isolated system increases in the course of any spontaneous change.

4 he Second Law of hermodynamics 熱力學第二定律 ΔS tot = ΔS sys + ΔS surr = ΔS + Δs surr ΔS tot > 0:the forward process is spontaneous ΔS tot < 0:the reverse process is spontaneous ΔS tot = 0:the process has no tendency to proceed in either direction, i.e. the reversible process

5 什麼是熵? 跟亂度有何關係呢? 巨觀熵的定義 : 在恆溫狀態時,Δ=0 ΔS = q rev / 在變溫狀態時,Δ 0 ds = dq rev / 因為必須取 reverssible process 的 q, 所以 entropy S 為 state function 對理想氣體而言 : U E K m 3 R 2 U 越大表示理想氣體的平均動能越大, 系統的狀態應該就越亂內能 U 為何無法定義為系統亂度的指標呢?

6 What is q? 分子的觀點 : 動能的傳遞

7 何謂熵 ( 亂度 )? 如何計算?

8 Ex 8.1 A human body (36.5 ) generates heat at the rate of about 100W (1W=1 J/s) At what rate does your body heat generate entropy in your surroundings, taken to be at 20? How much entropy do you generate each day? Would the entropy generated be greater or less if you were in a room kept at 30?

9 S at Constant emperature 溶解過程時 ΔS fus = q rev / ΔS fus = ΔH fus / 蒸發過程時 ΔS vap = q rev / ΔS vap = ΔH vap /

10 ΔS vap ~ 85 J/(K mol) routon s rule

11 Example Use routon s rule to estimate the enthalpy of vaporization of liquid bromine, which boils at 59. Use routon s rule to estimate the boiling point of CH 3 NH 2, given ΔH vap = 25.6 kj/mol

12 熱傳現象的 S 1 2

13 Ex 8-14 Suppose that 100J of energy is taken from a hot source at 300 passes through a turbine that converts some of the energy into a cold sink at 20. What is the maximum of work that can be produced by the engine if overall it is to operate spontaneously? What is the efficiency of the engine, with the work divided by heat supplied expressed as a percentage? How could the efficiency be increased?

14 S at Different emperature ds dq rev Cd S i f Cd C ln f i

15 Example A sample of nitrogen gas of volume 20L at 5kPa is heated from 20 to 400 at constant volume. What is ΔS during this heating process? C V,m of N 2 = J/K

16 S at Different Volume At U S S S constant t 0 q 1 nr rev V V V q V i i f f rev emperature dv V w P w rev ex rev dv nr and 1 V ln V f i V V i f reversible nr V dv process

17 Example What is the change in entropy of the gas when 1 mol nitrogen gas expands isothermally from 22L to 44L?

18 理想氣體混合後的 S O 2 1mole 22 liter 298K N 2 1mole 22 liter 298K

19 S at Different Pressure At constant t emperature and reversible process, there is only an ideal gas in the system S nr ln V V f i nr ln P P i f

20 Example Calculate the change in entropy when the pressure of mol oxygen gas is increased from 0.3 atm to 12 atm at constant temperature.

21 Example In an experiment, 1 mol Ar(g) was compressed suddenly (and irreversibly) from 5 L to 1 L by driving in a piston (like a big cycle pump), and in the process its temperature increased from 20 to What is the change in entropy of the gas?

22 熱力學第三定律熱力學第三定律 (he 3rd Law of hermodynamics): he Entropies of all perfect crystals approach 0 as the absolute temperature approaches 0.

23 微觀 Entropy 的熵定義 S = k ln W k : Boltzmann s constant, J/K W : the number of ways that the atoms or molecules in the sample can be arranges 由 Boltzmann 的熵定義, 熵代表即時微觀狀態數的對數因此把熵作為能量散佈 (spread) 或彌散 (disperse) 到各個可能的微觀狀態的

24 微觀 Entropy 計算方式 S = k ln W 在 =0K 時,4 個 CO 分子堆積成奈米級的固體, 由於結晶製程不同, 使得此固體兩種可能的排法, 請分別計算其 residual entropy 四個分子排列呈同一方向, 如右藍色框內的排法隨意亂排 k : Boltzmann s constant, J/K

25 Example of Residual Entropy he molar entropy of FClO 3 (s) at = 0K is 10.1 J/K. Suggest an interpretation. k : Boltzmann s constant, J/K

26 Example On the basis of the structures of each of the following molecules, predict which ones would be most likely to have a residue entropy in their crystal forms at =0 (a) CO 2, (b)no, (c)n 2 O, (d)cl 2

27 微觀 Entropy 與體積的關係盒中粒子的能階公式 E n E 2 n h 8 ml E n E n 2 n 1 8 ml 2 h 2

28 微觀的與巨觀的 Entropy 之相同性 f f f f N f f N i N f i f i f i f N V V nr V V Nk V V k V V k S W W k W k W k S S S V W N V W V W ln ln ln ln ln ln ln = 個分子的排列方式對於 = 式對於單一分子的排列方

29 微觀的 Entropy 與溫度的關係

30 微觀的 Entropy 與分子量的關係 E n E 2 n h 8 ml E n E n 2 n 1 8 ml 2 h 2

31 Example 在 298K 下何者的 S m 比較大 CO 2 (g) Ar (g) Ne (g) H 2 O (l) 在 298K 下何者的 S m 比較大 C (s, diamond) H 2 O (g) H 2 O (l) H 2 O (s)

32 何者熵比較大?

33 微觀 Entropy 的物理意義 : Dispersal of Energy 微觀 Entropy 應解釋為 : 將能量散佈 (spread) 或彌散 (disperse) 到各個可能的微觀狀態而亂度 (Disorder) 並不是非常正確的敘述!

34 HE QUES FOR ABSOLUE ZERO One method to approach = 0 isothermal magnetization adiabatic demagnetization

35 Standard Molar Entropies S m S m : 純物質在 1 bar 下的莫爾熵 S( ) S(0) S( ) S( ) S(0) S(heating from 0 to ) S(heating from 0 to ) 0 (he 3rd Law of At constant pressure, 0 CPd hermodynamics) 0 Cd

36 圖示標準莫爾熵的計算方式 S H fus fus or H vap vap S 1 2 C P d

37 Standard Reaction Entropies S r S 0 r 0 product ns reactant 0 ns m m Consider one equation : N 2 (g) + 3H 2 (g) 2NH 3 (g) Without doing and calculation, estimate the sign of ΔS r Calculate the standard reaction entropy ΔS r of the above reaction equation at 25

38 標準莫爾熵 S m 資料表

39 回顧熱力學第二定律熱力學第二定律 S S S S tot tot tot tot S sys S surr S S 0: the spontaneous process 0: the reverse process is spontaneous 0: the process has no tendency to proceed in either direction; the reversible process surr

40 如何計算 S surr?

41 Example H 2 O (s) H 2 O (l) ΔH fus =6.01kJ/mol 為什麼吸熱反應在室溫 298K 底下會自然進行呢?

42 Entropy Change of Isolated Systems 熱力學第二定律 S S S S S tot tot surr tot 0, for a isolated system S S sys sys S S surr qsurr q H Sr r S S 對於恆溫與恆壓的化學反應 H surr

43 Example 已知在 298K 時 N 2 (g) + 3H 2 (g) 2NH 3 (g) 的反應焓 ΔH r = kJ, 請問此放熱反應是否會在室溫下會自然反應呢?

44 化學反應的 H r 如何影響 S surr

45 吸熱或放熱反應影響化學反應的自發性

46 S of Reversible and Irreversible Process 理想氣體在等溫膨脹過程, 其內能的變化 : ΔU = q rev + w rev = q irrev + w irrev = 0 因為對外做功 w rev > w irrev q rev > q irrev 因此系統的 ΔS 為 qrev S in general S q q irrev for reversible, for irreversible Clausius inequality

47 證明 S tot 0 S S S S q (Clausius inequality) q 0 tot surr 0 q, S S surr 0 S tot S tot Equilibrium S tot =0 0 反應路徑 S? tot 0 : 自發反應路徑

48 Equilibrium 熱力學第二定律 S S tot tot S sys S surr S 0: Equilibrium S he process has no tendency to proceed in either direction; which is known as the reversible process surr

49 Example Calculate ΔS, ΔS surr, and ΔS tot for the two following expansion processes, where 1 mol of ideal gas molecules expands from 8 L to 20 L at 292K. he isothermal, reversible expansion process. he isothermal, free expansion process.

50 Example Calculate the standard entropy of vaporization of water at 85, given that its standard entropy of vaporization at 100 is 109 J/K/mol and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J/k/mol and 33.6 J/K/mol, respectively, in this range.

51 Review: Definition of Entropy 巨觀的定義 At constant temperature ΔS = q rev / When the temperature is changed ds = dq rev / Entropy is the state function. 微觀的定義 S = k ln W k : Boltzmann s constant, J/K W : the number of ways that the atoms or molecules in the sample can be arranges

hap First Law of hermodynamics Isochoric process: Isochoric rocess: hange in the state at constant volume,, ƌq /d (U / ) ; m mol J K - mol - kg J K -

hap First Law of hermodynamics.6 hange in State at onstant olume : : heat capacity : ƌq / d (ƌ: inexact differentials, ƌq ) U,, du,, :,, U f (, ) : du (U/ ) d + (U/) d, du ƌq ex d (for -work only, ƌw -