Chapter 5 rd Law of Thermodynamics

Transcription

1 Entropy and the nd and 3 rd Chapter 5 rd Law o hermodynamcs homas Engel, hlp Red

2 Objectves Introduce entropy. Derve the condtons or spontanety. Show how S vares wth the macroscopc varables,, and. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

3 Outlne 1. he Unverse Has a Natural Drecton o Change. Heat Engnes and the Second Law o hermodynamcs 3. Introducng Entropy 4. Calculatng Changes n Entropy 5. Usng Entropy to Calculate the Natural Drecton o a rocess n an Isolated System 6. he Clausus Inequalty 7. he Change o Entropy n the Surroundngs and S S total S surroundngs Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

4 Outlne 8. Absolute Entropes and the hrd Law o hermodynamcs 9. Standard States n Entropy Calculatons 10. Entropy Changes n Chemcal Reactons 11. Rergerators, Heat umps, and Real Engnes 1. (Supplemental) Usng the Fact that S Is a State Functon to Determne the Dependence o S on and 13. (Supplemental) he Dependence o S on and 14. (Supplemental) he hermodynamc emperature Scale Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

5 5.1 he Unverse Has a Natural Drecton o Change Natural transormatons, also called spontaneous processes, s one that repeated agan and agan over the course o a human letme leads to the same outcome. Unnatural transormatons ndcate that n many human letmes, these transormatons do not occur. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

6 5. Heat Engnes and the Second Law o hermodynamcs Heat engnes produce work produced by release heat n a combuston process Contactng the cylnder wth hot or cold reservors changes n temperature whch generate mechancal to a rotary moton, whch s used to do work. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

7 5. Heat Engnes and the Second Law o hermodynamcs he dagram o reversble Carnot cycle conssts o two adabatc and two sothermal segments. he arrows ndcate the drecton n whch the cycle s traversed. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

8 5. Heat Engnes and the Second Law o hermodynamcs he ecency o the Carnot cycle can be determned by calculatng q, w, and or each segment o the cycle assumng that the workng substance s an deal gas. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

9 5. Heat Engnes and the Second Law o hermodynamcs 5.1 he Reversble Carnot Cycle he heat low n the two sothermal segments o Carnot cycle s: w < 0, so that q > cycle ab q cd Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

10 5. Heat Engnes and the Second Law o hermodynamcs he ecency, ε, o the reversble Carnot engne s dened as the rato o the work output to the heat wthdrawn rom the hot reservor. ε q q q qcd 1 < 1 because qab > qcd, qab > 0, and q < q ab cd cd ab ab 0 Ecency s always less than one shows that not all heat wthdrawn rom the hot reservor can be converted to work. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

11 5. Heat Engnes and the Second Law o hermodynamcs elvn lanck ormulaton second law o thermodynamcs: It s mpossble or a system to undergo a cyclc process whose sole eects are the low o heat nto the system rom a heat reservor and the perormance o an equvalent amount o work by the system on the surroundngs. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

12 5. Heat Engnes and the Second Law o hermodynamcs Equvalent statement o the second law by Clausus states that: It s mpossble or a system to undergo a cyclc process whose sole eects are the low o heat nto the system rom a cold reservor and the low o an equvalent amount o heat out o the system nto a hot reservor. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

13 5. Heat Engnes and the Second Law o hermodynamcs he ecency o the reversble Carnot heat engne wth an deal gas as the workng substance s ε w cycle q ab hot hot cold cold 1 <1 hot Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

14 Example 5.1 Calculate the maxmum work that can be done by a reversble heat engne operatng between 500. and J s absorbed at Soluton: ε 1 w εq ab cold hot J Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

15 5.3 Introducng Entropy he mathematcal statement o the second law s dq reversble 0 When the cyclc ntegral o s zero, the exact derental shows a state uncton called the entropy, S: ds dq reversble Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

16 Example 5. a. Show that the ollowng derental expresson s not an exact derental: R d Rd b. Show that, Rd Rd obtaned by multplyng the uncton n part (a) by, s an exact derental. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

17 Soluton a. For the expresson, d g(, ) to be an exact derental, the condton (, )/ ] [ g(, )/ ] must be satsed. [ As R R and the condton s not ullled. ( ) d R 0 Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

18 Soluton b. Because [ ( R ) / ] R and [ ( R) / ] R, RYd Rd s an exact derental. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

19 5.4 Calculatng Changes n Entropy S must be calculated along a reversble path. As S s a state uncton, S s path ndependent. When U s zero or reversble sothermal expanson or compresson o an deal gas, S dq reversble 1 q reversble nr ln Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

20 5.4 Calculatng Changes n Entropy For deal gas that undergoes a reversble change n at constant or, S dq reversble nc v, m d nc v, m ln For a constant pressure process, S dq reversble nc p, m d nc p, m ln Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

21 Example 5.3 Usng the equaton o state and the relatonshp between C,m and C,m or an deal gas, show that can be transormed nto Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

22 Soluton We have S nrln nr ln nc n, m ln ( C R), m nrln ln nr nc, m ln nc, m ln Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

23 5.4 Calculatng Changes n Entropy When lqud s converted to a gas at a constant bolng temperature through heat nput at constant pressure S vaporzat on dq reversble q reversble vaporzat on H vaporzaton vaporzat on Smlarly, or the phase change sold lqud, S uson dq reversble q reversble uson H uson uson Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

24 5.4 Calculatng Changes n Entropy For the system undergong the change,,, S C d β d κ C ln β κ ( ) For the system undergong a change,,, S p d Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs C βd

25 Example 5.4 One mole o CO gas s transormed rom an ntal state characterzed by 30. and 80.0 L to a nal state characterzed by 650. and 10.0 L. Calculate S or ths process. Use the deal gas values or β and κ. For CO, Jmol C, m Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

26 Soluton Soluton For an deal gas, he entropy change or ths process s obtaned by [ ] [ ] nr nr κ β ln ln mol J L L mol J d S nr d C S

27 5.5 Usng Entropy to Calculate the Natural Drecton o a rocess s n an Isolated System When systems at constant are n thermal contact, the temperatures der by ; the two are solated rom outsde envronment. he process n whch S ncreases s the drecton o natural change n an solated system. he process wth S>0 s the drecton o natural change n ths solated system. he reverse process or whch S<0 s the unnatural drecton o change. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

28 5.6 he Clausus Inequalty Clausus nequalty or an rreversble process s ds > dq When dq reversble >dq rreversble, the equalty only holds or a reversble process. dq 0 Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

29 5.7 he Change o Entropy n the Surroundngs and As H and U are state unctons, the amount o heat enterng the surroundngs s ndependent o the path. It s essental to understand ths reasonng n order to carry out calculatons or S and S surroundngs. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

30 5.7 he Change o Entropy n the Surroundngs and I the system and the part o the surroundngs wth whch t nteracts are vewed as an solated composte system, the crteron or spontaneous change s S S total S surroundn gs > 0 Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

31 Example 5.7 One mole o an deal gas at 300. s reversbly and sothermally compressed rom a volume o 5.0 L to a volume o 10.0 L. Because t s very large, the temperature o the water bath thermal reservor n the surroundngs remans essentally constant at 300. durng the process. Calculate S, S, and S surroundngs total. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

32 Soluton As ths s an sothermal process, U 0, and q reversble w q reversble w nr ln d nr ln J Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

33 Soluton he entropy change o the system s gven by dq reversble qreversble S 7.6J 300. he entropy change o the surroundngs s gven by S surroundn gs q surroundn gs, reversble he total change n the entropy s gven by S S Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs total S surroundn gs q system J 0

34 5.8 Absolute Entropes and the 3 rd Law o hermodynamcs Under constant pressure the molar entropy o the gas can be expressed as S m ( ) S ( 0 ) m b C lqud, m ' 0 C d' m d' H ' sold, H vaporzaton b uson b C gas, m ' d' Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

35 5.8 Absolute Entropes and the 3 rd hermodynamcs Law o he thrd law o thermodynamcs can be stated n the ollowng orm, due to Max lanck: he entropy o a pure, perectly crystallne substance s zero at zero kelvn. S k k B w B ln w No. o (statstc al denton 3 J / o (Boltzman constant) mcrostates n the system entropy) Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

36 5.9 Standard States n Entropy Calculatons he thrd law provdes S wth a natural denton o zero, namely, the crystallne state at zero kelvn. he molar entropy o an deal gas decreases exponentally wth pressure at constant. S Choosng nrln S( ) ns Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs m nrln 1bar, nrln ( bar)

37 5.9 Standard States n Entropy Calculatons For an deal gas at constant where 1 bar, the molar entropy S m s S m ( ) S m R ln ( ) bar Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

38 5.10 Entropy Changes n Chemcal Reactons he entropy change S R n a chemcal reacton s a major actor n determnng the equlbrum concentraton n a reacton mxture. S R v S However, t s oten necessary to calculate at other temperatures as ollow: S S C p d ' ' Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

39 Example 5.10 he standard entropes o CO, CO, and O at are he temperature dependence o constant pressure heat capacty or o CO, CO, and O s gven by C J C J C J ( CO, g ) 1 ( CO, g ) 1 ( O, g ) 1 mol mol mol 1 Calculate S or the reacton CO(g) 1/ O (g) CO (g) at Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs 1 1 S S S ( CO, g) J mol 1 ( CO, g) 13.74J mol 1 1 ( O, g) J mol

40 Soluton Soluton We have ( ) ( ) ( ) ( ) mol J C ( ) ( ) ( ) ' ' 86.50, 1,, mol J mol J d d C S S mol J g O S CO g S g CO S S

41 5.11 Energy Ecency: Rergerators, Heat umps, and Real Engnes he nteror o a rergerator acts as the system, and the room n whch the rergerator s stuated s the hot reservor. he coecent o perormance, n r, s dened as the rato o the heat wthdrawn rom the cold reservor to the work suppled to the devce: η r q w cold q hot q cold q cold hot cold cold Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

42 5.11 Energy Ecency: Rergerators, Heat umps, and Real Engnes A heat pump s used to heat a buldng by extractng heat rom a colder thermal reservor. he maxmum coecent o perormance, n hp, s dened as the rato o the heat pumped nto the hot reservor to the work nput to the heat pump: η hp q hot w q hot q hot q cold hot hot cold Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

43 5.11 Energy Ecency: Rergerators, Heat umps, and Real Engnes Otto engne o our strokes o pston. Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

44 5.11 Energy Ecency: Rergerators, Heat umps, and Real Engnes Otto engne process: e c: Constant ressure c d: Adabatc Compresson d a: Heat n a b: Adabatc Expanson b c: Heat Out Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

45 5.11 Energy Ecency: Rergerators, Heat umps, and Real Engnes Desel engne process: e d: Constant ressure d a: Adabatc a b: Heat In b c: Adabatc c d: Heat Out Constant olume Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

46 5.1 Usng the Fact that S Is a State Functon to Determne the Dependence o S on and he total derental ds n terms o the partal dervatves wth respect to and s: ds S Equatng the coecents o d and d S d d S C S 1 and U Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

47 5.13 he Dependence o S on and As chemcal transormatons are normally carred out at constant pressure we need to know how S vares wth and. he total derental ds s ds S d S he pressure dependence o the entropy at constant temperature can be wrtten as S d Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

48 5.13 he Dependence o S on and Integratng along a reversble path gves S C d - βd Chapter 5: Entropy and the nd and 3 rd Law o hermodynamcs

49 5.14 he hermodynamc emperature Scale he reversble Carnot cycle provdes a bass or the thermodynamc temperature scale, a scale that s ndependent o the choce o a partcular thermometrc substance. All reversble Carnot engnes have the same ecency, regardless o the workng substance. he bass or the thermodynamc temperature scale s the act that the heat wthdrawn rom a reservor s a thermometrc property.