5.5 Using Entropy to Calculate the Natural Direction of a Process in an Isolated System

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1 5.5 Using Entropy to Calculate the Natural Direction of a Process in an Isolated System 熵可以用來預測自發改變方向 我們現在回到 5.1 節引入兩個過程 第一個過程是關於金屬棒在溫度梯度下的自然變化方向 試問, 在系統達平衡狀態時, 梯度變大或更小? 為了模擬這過程, 考慮如圖 5.5 的模型, 一孤立的複合系統受 兩種金屬桿的系統, 在均勻但不同的溫度 T 1 >T 2 下被帶進熱接觸形式 熱量是從桿左邊 T 1 撤出, 同樣的推理可得桿溫度梯度變小 在下面的討論, 為了使用熱流, 計算 S 在此不可逆過程, 必須想像一個可逆過程是與不可逆過程一樣的最初的和最終狀態 在想像可逆過程中, 桿耦合到一個熱槽, 其溫度降低速度非常慢 那個桿和熱槽的溫度差異僅僅在整個進程中維持無窮小的量 棒的總溫度變化, T, 是與熱量 q p 相關 47

2 Figure 5.3 Figure 5.5 Two Systems at constant P, each consisting of a metal rod, are placed in thermal contact. The temperatures of the two rods differ by T. The composite system is contained in a rigid adiabatic enclosure (not shown) and is, therefore, an isolated system. 48

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4 5.5 Using Entropy Fig. When energy leaves a cold reservoir as heat, the entropy of the reservoir decreases. When the same quantity of energy enters a hotter reservoir, the entropy increases by a smaller amount. Hence, overall there is an decrease in entropy and the process is not spontaneous. Relative changes in entropy are indicated by the sizes of the arrows. 當能量離開低溫水庫 ( 冷儲熱槽 ), 水庫的熵下降 當同樣數量的能量進入一個高溫水庫 ( 較熱儲熱槽 ), 熵增加的量較低 總體是一個熵減少的過程, 因此不是自發的 熵的相對變化顯示在箭頭的大小 50 Chap 3 The Second Law

5 Fig. (a) The flow of energy as heat from a cold source to a hot sink is not spontaneous. As shown here, the entropy increase of the hot sink is smaller than the entropy DECREASE of the cold source, so there is a net decrease in entropy. 熱的能量流從冷源到熱水槽, 該流程不是自發的 這說明, 熱水槽的熵增加小於冷源的熵減小, 所以淨熵是減少的 51 Chap 3 The Second Law

6 5.5 Using Entropy When an energy is removed from a cool source at a temperature T c and then deposited in a warmer sink at a temperature T h, the process are not an spontaneous process, hence must be accompanied by an energy consumption. The change in entropy decreases in the course of this nonspontaneous change: ΔS < 0. dq dq ds T T c h dq Th Tc To know the power needed to cool objects in refrigerators, first we find out the work require to cool an object. 52 Chap 3 The Second Law

7 Figure 5.6 Figure 5.6 (a) An irreversible process is shown in which an ideal gas confined in a container with rigid adiabatic walls is spontaneouly reduced to half its initial volume. (b) A reversible isothermal compression is shown between the same initial and final states as for the irreversible process. 53

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10 5.8 Absolute Entropies and the Third Law of Thermodynamics The Third of Thermodynamics The entropy of a pure, perfectly crystalline substance (element of compound) is zero at zero kelvin. Under constant pressure conditions, the molar entropy of the gas can be expressed in terms of the molar heat capacities of the solid, liquid, and gaseous forms and then enthalpies of fusion and vaporization. Tf solid Tb liquid Tb gas p, m fusion p, m vaporization p, m C dt' H C dt' H C dt ' SmTSm0K T' T T ' T T' 0 f T f b T f 56

11 5.6 The Clausius Inequality 在上一節用兩個例子證明, 一個孤立系統中, S >0 提供了預測自然變動方向的要件 不考慮特定進程, 也可以獲得這一結果 考慮第一定律的微分可以形式的, 過程中, 只體積功是可能的 : d U q P external d (5.29) 公式 (5.29) 可逆性和不可逆過程都是有效的 如果這一進程是可逆的, 我們可以改寫格式如下 : du qreversible P dv T ds P dv (5.30) 由於 U 是一個狀態函數,dU 是獨立於路徑的, 公式 (5.30) 對為可逆和不可逆過程都成立, 只要相變或化學反應不存在, 在過程中, 只有體積功是可能的 V 57

12 為了推導克勞修斯不等式, 我們以 du 的表達方程式 (5.29) 等於 (5.30): dq q P P V reversible d external d (5.31) 如果 (P - P external )> 0 時, 系統會自發膨脹, 且 dv > 0 如果 (P - P external )< 0 時, 系統會自發收縮, 且 dv < 0, 在這兩種可能的情況下, 都是 (P - P external ) dv > 0 因此, 我們得到下式 dq reversible dq TdS dq 0 (5.32) 或 T ds dq. ( 此式只在可逆過程下成立 ) 58

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14 The Clausius Inequality 相同溫度下的等溫路徑可得 由熵值的熱力學定義 : dq rev dq T T dqrev dq ds T T 這個公式稱為克勞修斯不等式 (Clausius inequality) dq ds T 任一過程的熵值改變量必然大於或等於實際發生的熱量與溫度的比值. Chap 3 The Second Law 60 60

15 5.6 The Clausius Inequality Clausius inequality ds О dq T dq T 0 61

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17 5.7 The Change of Entropy in the Surroundings and S total = S + S surroundings 63

18 5.7 The Change of Entropy in the Surroundings For a system in thermal and mechanical contact with its surroundings at the same T, hot source T h and a cold sink at temperature T, then ds + ds sur 0 or ds -ds sur As ds sur = -dq/t where dq is the heat supplied to the system by surroundings during the process, ds dq/t This is the Clausius inequality, that is, in the isolated system the entropy cannot decrease when a spontaneous change take place. 64

19 5.7 The Change of Entropy in the Surroundings Cases I: A irreversible adiabatic expansion of ideal gas at T, since dq = 0 and constant T, so ds sur = -dq/t =0 but ds = dq rev /T dq/t for this spontaneous change, the system entropy increase, but surrounding entropy remains constant and total entropy obeys ds tot 0. Cases II: A spontaneous cooling of system in surroundings, When heat dq leaves hot source, entropy decrease by ds = - dq rev /T h = - dq /T h, when dq enter into cool sink, entropy increase by ds sur = dq /T c. Now ds tot = ds + ds sur = - dq /T h, + dq /T c = dq (1/T c -1/T h ) > 0. which is positive (because T h T c ) 65

20 5.7 The Change of Entropy in the Surroundings For any process, any change of state is accompanied by a change in entropy of the system ds, and of the surroundings, ds surr = -du /T surr Define the ds tot as ds tot = ds + ds surr (a) First, for a reversible process in system where ƌq rev is the heat supplied to the system by surroundings, as surroundings gains heat q rev, ds surr =-ƌq rev /T surr Since ds tot = ds + ds surr = 0 for a reversible process in an isolated system, (as total system is isolated) ds = -ds surr = ƌq rev /T surr = ƌq rev /T (T = T surr ) 66

21 5.7 The Change of Entropy in the Surroundings (b) Second, for a spontaneous process in a system in which the system gains heat ƌq irrev from its surrounding: ds tot = ds + ds surr > 0 (total system as isolated system) ds > - ds surr and ds surr = - ƌq irrev /T surr ds > ƌq irrev /T surr Combine both cases (a) and (b) : For a finite change, ΔS dq ds Tsurr dq T surr 67

22 5.7 The Change of Entropy in the Surroundings The Clausius theorem describe three types of processes: ΔS dq T surr ds > ƌq/t spontaneous and irreversible process ds = ƌq/t reversible process (T surr =T ) ds < ƌq/t impossible process The equality applied when a change is brought about reversibly. 68

23 5.7 The Change of Entropy in the Surroundings To an isolated system, because ƌq =0, the three possibilities are : dq ΔS ΔS > 0 ΔS = 0 ΔS < 0 spontaneous and irreversible process reversible process impossible process T surr That is, in the isolated system the entropy cannot decrease when a spontaneous change take place. 69

24 5.7 The Change of Entropy in the Surroundings The log dependence of entropy on volume ΔS = nr ln (V f /V i ) is path independence. The total entropy change does depend on how the expansion take place: Reversible expansion gives ΔS sur = -ΔS, and ΔS tot = 0. For free isothermal expansion gives q =0, ΔS sur = 0, and ΔS tot = ΔS. 70

25 5.7 The Change of Entropy in the Surroundings Summary Thermodynamic irreversible processes are spontaneous processes, hence must be accompanied by an increase in entropy. The total entropy of an isolated system increases in the course of a spontaneous change: ΔS tot > 0. ( 熱力學不可逆過程是自發過程, 因此必須伴隨熵的增量 自發變動期間, 一個孤立體系的總熵增加 : ΔS tot > 0) The direction of spontaneous change in any system is in the direction of increasing the entropy of the universe, thus indicates the time sequence of a spontaneous process. ( 在所有系統上的自發變化的方向是朝增加宇宙的熵的方向, 因而顯示一個自發過程的時間序列 ) 71

26 5.7 The Change of Entropy in the Surroundings 由於內能是狀態函數, 在相同的起始和最終狀態間的可逆與不可逆兩種不同條件的路徑中, 內能的改變值應相同 因此可表示為 du = ƌq + ƌw = ƌq rev + ƌw rev 又因為在可逆條件下的輸出功大於不可逆條件下的輸出功, 亦即 : dw rev dw, 或 dw dw rev 0 內能的改變量 du 表示式經移項後可得 Ƌq rev - ƌq = ƌw - ƌw rev 0 或 ƌq rev ƌq Chap 3 The Second Law 72 72

27 Figure 5.7 Figure 5.7 A sample of an ideal gas (the system) is confined in a piston and cylinder assembly with diathermal walls. The assembly is in contact with a thermal reservoir that holds the temperature at a value of 300 K. Sand falling on the outer surface of the piston increases the external pressure slowly enough to ensure a reversible compression. The directions of work and heat flow are indicated. 73

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29 Example Problem 5.7 Example Problem 5.7 One mole of an ideal gas at 300 K is reversible and isothermally compressed from a volume of 25.0L to a volume of 10.0 L. Because it is very large, the temperature of the water bath thermal reservoir in the surroundings remains essentially constant at 300 K during the process. Calculate S, S surroundin S. gs, total 75

30 Example Problem 5.7 Solution Because this is an isothermal process, U 0, and q w. From Section 2.7, f dv qreversible wnrt nrt V V V mol 8.314Jmol K 300K ln i 3 J V ln V The entropy change of the system is given by f i reversible 10.0L L 3 dqreversible qreversible J Ssystem 7.62JK T T 300K The entropy change of the surroundings is given by 1 3 qsurroundings qsystem J Ssurroundings T T 300K The total change in the entropy is given by 7.62JK 1 S S S JK JK total system surroundings

31 Example Problem 5.8 Example Problem 5.8 One mole of an ideal gas at 300 K is isothermally compressed by a constant external pressure equal to the final pressure in Example Problem 5.7. At the end of the process,. Because at all but the final state, this process is irreversible. The initial volume is 25.0 L and the final volume is 10.0L. The temperature of the surroundings is 300 K. Calculate S, S surroundin S. gs, total 77

32 Example Problem 5.8 Solution We first calculate the external pressure and the initial pressure in the system: nrt V 1 1mol J mol K 3 1m 10.0L 3 10 L 300K 1 P external nrt V 1 1mol Jmol K 3 1m 25.0 L 3 10 L 300K 1 P i Pa Pa Because P, we expect that the direction of external P i spontaneous change will be the compression of the gas to a smaller volume. Because U 0. 78

33 Example Problem 5.8 qwp V V Pa m m external ( f i ) ( ) 3 J The entropy change of the surroundings is given by 3 qsurroundings q J Ssurroundings 12.45JK T T 300K 1 The entropy change of the system must be calculate on a recersible path and has the value ovtained in Example Problem 5.7: 3 dqreversible qreversible J S 7.62JK T T 300K 1 It is seen that S 0, and S 0. The total change in the entropy is given by surroundings S S S 7.62 JK JK JK total surroundings

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36 Figure 5.8 Figure 5.8 The experimentally determined heat capacity for O 2 is shown as function of temperature below 125K. The dots are data from Giauque and Johnston [J. American Chemical Society 51 (1929), 2003]. The red solid lines below 90K are polynomial fits to these data. The red line above 90K is a fit to data from the NIST Chemistry Webbook. The yellow line is an extrapolation from to 0 K as described in the text. The vertical dashed lines indicate constant temperature-phase transitions, and the most stable phase at a given temperature is indicated in the figure. 82

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38 Figure 5.9 Figure 5.9 C P /T as a function of temperature for O 2. The vertical dashed lines indicate constant temperature-phase transitions, and the most stable phase at a given temperature is indicated in the figure. 84

39 Figure 5.10 Figure 5.10 The molar entropy for O 2 is shown as a function of temperature. The vertical dashed lines indicate constant temperature-phase transitions, and the most stable phase at a given temperature is indicated in the figure. 85

40 Figure 5.11 Figure 5.11 A useful model of a solid is a three-dimensional array of coupled harmonic oscillators. In solid with a high binding energy, the atoms are coupled by stiff springs. 86

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42 Example Problem 5.9 Example Problem 5.9 The heat capacity of O 2 has been measured at 1 atm pressure over the interval K<T< K. The data have benefit to the following polynomial series in T/K, in order to have a unstressed variable: , , 1 1, , , , ) ( ) ( ) ( ) ( ) ( ) ( K T K T K T K mol J T C K T K K T K T K T K mol J T C K T K K mol J T C K T K K T K T K T K mol J T C K T K K T K T K T K mol J T C K T K K T K mol J T C K T m P m P m P m P m P m P : : : : : : 88

43 Example Problem 5.9 The transition temperatures and the enthalpies for the transitions indicated in Figure 5.8 are as follow: Solid III Solid II K 93.8 J mol -1 Solid II Solid I 43.76K 743 Jmol -1 Solid I liquid 54.39K 445 Jmol -1 Liquid gas 90.20K 6815 Jmol -1 a. Using these data, calculate for O 2 at K. b. What are the three largest contributions to? 89

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46 Solution Example Problem 5.9 a. S o m (298.15K) C 445J 54.39K soild,iii P, m JK 8.181JK T JK dt J 23.66K C liquid P, m T dt JK JK C 6815J 90.20K 19.61JK soild,ii P, m JK T dt C 743J 43.76K lg as P, m T dt JK JK C JK soild,i P, m This is an additional small correction for nonideality of gas at 1 bar. The currently 1 1 accepted value is (298.15K) Jmol K S o m (Linstrom, P. J., and Mallard, W. G.., NIST Chemistry Webbook:NIST Standard Reference Database Number 69, National Institute of Standards and Technology, Gaithersburg, MD, retrieved from b. The three largest contributions to S m are S for the vaporization transition, Sfor the heating of the gas from the boiling temperature to K, and S for heating of the liquid from the melting temperature to the boiling point. T 1 dt 92

47 5.9 Standard States in entropy Calculation Standard State in entropy Calculations The values of S varies most strongly with P for a gas. Vf Sm Rln R V Choosing P S P S Rln m P i m i P P 1 bar f P i bar P 93

48 5.10 Entropy Changes in Chemical Reactions Entropy Changes in Chemical Reactions S S R i i i T CP T ' T S S dt ' 94

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50 Figure 5.12 Figure 5.12 The molar entropy of an ideal gas is shown as a function the gas pressure. By definition, at o 1 bar, S m S m, the standard state molar entropy. 96

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55 Solution Example Problem 5.10 o CP T JK mol 2 2 K 1 T K T T T K K K o o o 1 o S S298.15( CO2, g) S298.15( CO, g) S298.15( O2, g) JK mol JK mol JK mol JK mol T K o o o CP T T S S dt 86.50JK mol JK mol 1.757JK mol 88.26JK mol T 4 T 8 T K K K T d JK mol T K K JK mol JK mol o The value of S is negative at both temperatures because the number of moles is reduced in the reaction. T K

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57 Refrigeration The process is not spontaneous because not enough entropy is generated in the warm sink to overcome the entropy loss from the hot source. To generate more entropy, energy must be added to the stream that enters the warm sink. The outcome is expressed as the coefficient of performance, η r, r energy transfered as heat energy transfered as work qc w The less the work that is required to achieve a given transfer, the greater the coefficient of performance η r, and the more efficient the refrigerator. Our task is to find the minimum energy that needs to be supplied. 103 Chap 3 The Second Law

58 5.11 Refrigerators, Heat Pumps and Real Engines 104

59 Figure 5.13 Figure 5.13 Reverse Carnot heat engines can be used to induce heat flow a cold reservoir to hot reservoir with the input of work. The hot reservoir in both (a) and (b) is a room. The cold reservoir can be configured as the inside of a refrigerator, or outside ambient air. 105

60 Refrigeration Because q c is removed from the cold source, and the work w is added to the energy stream, the energy deposited as heat in the hot sink is q h = q c + w. Therefore, 1 r w q c q h q c q c q q h c 1 To express this result in terms of the temperature alone, we have the thermodynamically optimum coefficient of performance, η r r Tc T T h c 106 Chap 3 The Second Law

61 Refrigerators, Heat Pumps and Real Engines 294 K r Tc T T h c K

62 The process becomes feasible if work is provided to add to the energy stream. Then the increase in entropy of the hot sink can be made to cancel the entropy decrease of the COLD source. 108 Chap 3 The Second Law

63 Example For a refrigerator withdrawing heat from ice-cold water (T c = 273 K) in a typical environment (T h = 293 K), c = 14, so, to remove 10 kj (enough to freeze 30 g of water), requires transfer of at least 0.71 kj as work. Practical refrigerators, of course, have a lower coefficient of performance. 109 Chap 3 The Second Law

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65 5.11 Refrigerators, Heat Pumps, and Real engines The coefficient of performance of a reversible Carnot refrigerator qcold qcold Tcold r w q q T T hot cold hot cold The maximum coefficient of performance of a heat pump qhot qhot Thot hp w q q T T hot cold hot cold 111

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67 Figure 5.14 Figure 5.14 Illustration of the four-stroke cycle of an Otto engine, as explained in the text. The left valve is the intake vale, and the right valve is the exhaust valve. 113

68 Figure 5.15 Figure 5.15 Idealized reversible cycles of the (a) Otto and (b) diesel engines. 114

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