Order (Boltzmann 1875) Information (Shannon 1948) Entropy (Clausius 1855) dq T Cycle. 2 nd LAW. 5. ENTROPY & 2 nd LAW OF THERMODYNAMICS.

Transcription

1 5. ENROPY & nd LAW OF HERMODYNAMIC. Order (Boltzmann 875) ln k B e nd LAW 0 Inormaton (hannon 948) Q bt k B ln e Entropy (Clausus 855) dq Cycle 0 4

2 5. he Clausus Inequalty. wo emprcal statements o the second law based on emprcal observaton have been ntroduced. ) he statement due to Kelvn and Planck s that t s mpossble to construct a devce whose only acton, when operatng n a cycle, s to extract heat rom a heat reservor and to delver an equvalent amount o work. and ) he statement due to Clausus that t was mpossble to construct a devce whose only acton, when operatng n a cycle, was to transer heat rom a cold to a hot reservor. It s now tme to look at ths law and try to get t onto a more mathematcal ootng. hs wll nvolve the dscovery o a new state uncton, entropy, and an examnaton o ts bass around the exchange o heat between bodes. o start to do ths requres a contnuaton o the study o the deal (reversble) Carnot engne operatng between two heat reservors. o ar engnes operatng on dealzed cycles have been consdered however real machnes wll nteract wth the envronment to a lesser or greater extent as they exchange heat wth many derent bodes at many derent temperatures. o nvestgate ths exchange o heat t s easest to begn wth reversble cycles beore movng on to rreversble cycles. Clausus heorem (855). Rudol Clausus (8 888) used the tool o the Carnot cycle, n a very elegant study, to ad the understandng o heat low around a general cycle by startng wth the reversble cycle as represented by the Carnot cycle. It has already been shown that or a Carnot cycle the heat lows and thermodynamc temperature are smply related as Q Q Q Q Clausus used these relatons n envsagng (n a thought experment) an arbtrary (unspeced) thermodynamc system undergong a cycle ncludng both heat low (nto and out o the system) and work processes. He constraned all o the heat lows to be rom a seres o reversble 5

3 Carnot engnes operatng rom a unversal hot reservor at temperature 0. hese multple Carnot engnes are used to drve ths arbtrary system rom one o ts equlbrum states to a neghbourng one, at temperature (that vares as the cycle s traversed) by supplyng (or extractng) ncremental heat dqys() at that temperature to/rom the system and causng t to do ncremental work dwys. hs wll nvolve the Carnot engne extractng (or gvng up) ncremental heat QC(0) rom/to the reservor, gvng up (or extractng) ncremental heat QC() to/rom the arbtrary system at temperature and dong ncremental reversble work WC. o drve the complete general cycle Carnot engnes wll also need to run n reverse at some ponts on the cycle and extract ncremental heat Q( / ) at temperature /. he dagram on the ollowng page llustrates ths conceptual process. Wth reerence to that dagram t s to be understood that; Insde the dashed lnes there s a composte system made up o the arbtrary system under consderaton and the Carnot engnes whose nputs and outputs are δqc(0), δqc(), δwc, dqys and dwys. NB. hs s generalzed to a system where the temperature vares around the cycle and multple Carnot engnes are used to supply/extract heat to/rom the system at the approprate temperature at any pont around the system cycle. he quanttes shown n the dagram are then;. δqc(0) s the heat receved rom the unversal reservor by the Carnot engne durng one complete cycle o the Carnot engne.. δqc() [=-dqys()] s the heat rejected to the general cycle n one complete cycle o the Carnot engne consstent wth constrants set by the requrements o the Carnot cycle,. δwc s the work perormed n one complete cycle by the Carnot engne consstent wth constrants set by the requrements o the Carnot cycle. 4. dwys s the ncrement o work perormed by the system at any partcular pont on the cycle at temperature. 5. dqys() [=-δqc()] s the ncrement o heat absorbed/gven out by the system at any partcular pont on the cycle at temperature. 6

4 Note that δqc(0), δqc() and δwc are cyclcal quanttes wrt each Carnot engne whle dqys and dwys are ncremental quanttes wrt the arbtrary system under consderaton. NB, o ad understandng a Carnot cycle run n reverse as a Carnot rergerator at the bottom o the dagram has been ncluded or completeness representng a part o the cycle where heat s extracted rom the general system. On ths dagram t appears as an upsde down verson o the rergerator dagrams prevously used. 7

5 Reservor at 0 QC(0) CARNO CYCLE (reversble heat engne) WC QC() = -dqys() dwys ARBIRARY CYCLE / dwys QC( / ) =-dqys( / ) CARNO CYCLE (reversble heat engne) WC Q(0) Reservor at 0 8

6 he analyss o the composte system (arbtrary cycle plus Carnot cycles nsde the dashed lnes) begns by ndng an expresson or the net work, the composte system. W Net, done by/on WNet W C dwys ystem ystem cycle cycle Usng the rst law o thermodynamcs the separate work elements, WC and dw ys, n the expresson or W Net may be examned. W C Q C Q U 0 0 C C WC QC 0 Q C And ystem cycle dw ys ystem cycle dqys du 0 ystem cycle dw ys dqys ystem cycle ystem cycle ystem cycle QC hereore; WNet QC 0 ystem cycle Q Q ystem cycle C ystem cycle C 9

7 WNet QC ystem Cycle 0 QRes Where Q Res s the heat low to/rom the unversal heat reservor at 0. here can be no net work done by the combned system and Carnot cycles as W Net Q Res would volate Kelvn-Planck statement Unversal Reservor at 0 Composte system I the net work s done by the composte arbtrary cycle plus Carnot cycles then W Net Q Res would volate the Kelvn-Planck statement as shown n above dagram. hereore net work must be ether zero or t must be work done on the system e postve; WNet 0 QC 0 ystem Cycle Usng the property o a Carnot cycle we can re-wrte ths nequalty δqc 0 δq 0 C 0

8 W Net 0 QC 0 ystem Cycle W Net dqys 0 0 ystem Cycle nce 0 s postve ths gves HE CLAUIU INEQUALIY dscovered n 855 d ystem cycle Q ys 0 Whenever a system executes a complete cyclc process the ntegral o cycle s less than or equal to zero. dq around the For a reversble system cycle whch can be traversed n ether sense wth only the sgn o the heat low changng; dq 0 R dq and R 0 antclockw se clockwse hs must mean that or a REERIBLE cycle; dq 0 R Cycle REERIBLE hen dq 0 IRREERIBLE Note that the quantty dq R cycle sums to zero, c U, H, P and s behavng exactly as a state uncton where ts ntegral over a

9 5. Entropy he sgncance o the Clausus nequalty s stll not apparent so consder a reversble cycle; upper path P lower path For an arbtrary reversble cycle such as the one shown above, the contour ntegral may be wrtten n two parts one or the upper and the other or the lower path such that dqr upper upper dqr lower lower dqr 0 hereore, lower dq R lower lower dq R upper upper dq R lower But the lower and upper paths are any arbtrary paths connectng to and thereore the path ntegral o dq R s path ndependent (unlke the path ntegral o dq alone whch does depend on path) and ths means that dq R denes a new state uncton (Lke, P,, U etc). In 865 ater several years workng on ths new and engmatc quantty Clausus named ths new state uncton Entropy and gave t the symbol.

10 he change n the entropy o a system, ater undergong a process can be dscovered straghtorwardly; dqr Whle the value o the state uncton s a unque property o the equlbrum state t should be noted that wth ths denton only Entropy Derence may be dscovered. hat s to say, entropy s only dened up to an arbtrary constant whch could be taken to be the entropy o a reerence state (just lke potental energy or voltage). has unts o JK -. dq R re re Another thng worth consderng s that there s no such thng as an entropy meter!!! dqr Lookng at derental change o ; d and s a perect derental (t must be as t s a state uncton!). hs may be demonstrated by usng the prevously establshed rules or perect derentals. Example. Ideal Gas. Frst Law states du dqr Pd (assumng a reversble process) It ollows that and rom the equaton o state dq d R du P d U P P nr P nr U nr du nrd

11 o re-wrtng d d nr d nr a(, )d b(, ) d wth nr a nr b and are seen as the natural varables o ( and are both held constant the entropy s constant) and the test or perect derental status may be appled, thereore a nr 0 b a b nr hereore d s a perect derental. As has already been ound has and as natural varables and t s thereore possble to wrte ormally, d d d nr d 0 nr d and nr nr nrln ( ) nrln g( ) Both o these together gve (, ) nr ln nr ln const For mole o gas, n = 4

12 s the molar entropy, n C C n R cv the molar specc heat capacty at constant volume s( v, ) c ln R ln const I the gas s taken rom an ntal state wth temperature and volume temperature and volume, the entropy change s gven by;, to a nal state wth (, ) (, ) nrln nrln 5

13 Examples o calculatng Ex. ystem warmng Water, 0 o C Reservor, 00 o C Equlbrum state Irreversble Water, 00 o C Reservor, 00 o C Equlbrum state s unquely dened as both states are equlbrum states. o calculate t consder any reversble path that wll take the water rom to. Water, 0 o C d Water, 0. o C d Water, 0. o C Reservor, 0. o C Reservor, 0. o C Reservor, 0. o C Equlbrum state Equlbrum state Equlbrum state Make the change rom 0 o C to 00 o C n nntesmal steps, d dqr mcp d Where d QR mcp d has been used Note that the changes are occurrng at constant pressure. uppose the water has a mass m = kg 7.5 d mcp 9.5 mc ln P 6

14 NB. he temperature must be converted nto degrees Kelvn or the calculaton!! CP(HO) = 4. 0 JK - kg - HO kg 4.0 JK kg ln JK It s also nstructve to calculate the change n the entropy o the reservor where the temperature has remaned constant. dqres Re s dqres Heat released mcp (H0) Res mcp (H 0) kg 4. 0 JK kg JK 7.5 H O Res (04 899)JK 5JK 0 NB. Measurements o specc heats provde expermental methods or determnng. Ex.. ystem coolng plus constant temperature reservor. kg o water s cooled rom 40 o C to 0 o C, by placng a beaker outsde n the garden, calculate or the water and the surroundngs. H O H O dq mc d R P 9.5 HO d 9.5 R mcp kg 4.0 ln 77JK.5.5 For the surroundngs the temperature remaned constant at 9.5K 7

15 urr 9.5 QHO urr Qurr mcp d kg 4. 0 (9.5.5) JK Net 86 JK 77 JK 9 JK 0 NB the sgn o the net entropy change s ndependent o whether the water was heated or cooled. e. here s no conservaton o entropy urr 86 H O JK R 77 JK Ex. Adabatc Mxng. Mx kg o water at 0 o C wth 5kg o water at 0 o C n an adabatc enclosure () What s the nal state? he nal state s 6kg o water at. here s no heat nput so the total heat low must be zero. Q Net Qkg Q5kg 0 Convert temperatures nto Kelvn and use constant pressure heat capacty to nd heat lows; Q Net 4. Qkg kg 4. Jkg K Q5kg 5kg 4. Jkg K Jkg K 0 kg kg kg kg K he entropy change calculaton s as or the warmng ex. 8

16 5kg kg 5kg kg ln d d ln ot 6JK JK 9JK Note that the entropy s addtve e. o nd the net entropy change when there are two or more systems nvolved the entropy o each system may be added. In general the specc heats wll vary wth temperature whch complcates thngs the varaton s too large over the range o the temperature change. However ths eect has been gnored n these examples. NB. nce Entropy s a uncton o state t may change the state changes wth or wthout a low o heat Ex. 4 Joule Free Expanson. Joule ree expanson has been seen already n these notes; Where a gas s ntally conned to one hal o a rgd adabatc chamber and the partton s suddenly broken to allow unhndered expanson nto the whole chamber wth no lows o heat. hs s an rreversble process wth no heat lows and yet there wll be a change o entropy. Rln Rln = (no heat low and no work so no change n nternal energy). 9

17 Rln Rln 0 has ncreased wthout any net heat low. Ex. 5. Isothermal Mxng. Gas X x P, Gas Y (-x) P, Consderng the entropy o sothermally mxng wth ad o the dagram where two derent gases, X and Y, are occupyng ractons x and ( x) o a total volume respectvely and are physcally separated. he two gases are at the same temperature and pressure and by the openng o a valve are allowed to mx rreversbly. Gas A expands rom a volume x nto the volume and gas B expands rom a volume ( x) nto the volume. hs occurs wth no change n temperature. Because t s sothermal du 0and the rst law requres; dq d Pd d P d NkB d I the total number o molecules o both types o gas s N then N X xn and N Y xn he change n entropy or entropy o mxng may be ound xnkb da x A db B ( x) xnkb xln ( x) NkB xlnx has ncreased as both logarthms are negatve wth x and x both beng less than. 0

18 NkBln x Clearly, A and B had been the same gas then there would have been no change n equlbrum state and no entropy ncrease! he dea o dstngushablty s an mportant concept when dscussng entropy. Example 6. Change o state. As an example o calculatng entropy ater a phase change calculate the entropy change o kg o water at 0 o C placed n a reezer at -5 o C. here are three stages to consder. ) he coolng o the water, somethng we have already calculated; 7.5 d 7.5 mcp mcpln 4.0 ln 97 JK ) he phase change o water nto ce at 7.5 K. hs s a process that occurs at a constant temperature and reezng Q ml Note Q s negatve as heat lows out o the water as t reezes. ) Fnally there wll be entropy change as the ce cools to -5 o C = 68.5 K d 68.5 mcp.0 ln 8.8 JK JK

19 Entropy s addtve and the entropy change o the water s the sum o these separate entropy changes. HO JK 47. JK here wll be a change n entropy due to the acton o the reezer to be accounted or. he cold shel o the reezer wll reman constant throughout but heat, the cold shel to the water as t undergoes ths process. Q Freezer Q H O wll low rom QFreezer QH O water mc P water ce ce mc P ml reezng Q Freezer J Freezer Q Freezer Freezer he net change n entropy o water plus reezer s then H Freezer O JK JK 8.7 JK Agan the net entropy change s postve and a pattern s emergng. In all sx examples the overall entropy has ncreased.

20 5. he Prncple o Increasng Entropy & the nd Law o hermodynamcs. All o the vared examples seen so ar have nvolved an ncrease o rom ntal to nal state value when all potental sources o entropy change n the problem have been accounted or. An obvous queston presents tsel n the lght o ths observaton; s there a general prncple nvolved n the sgn o that can be uncovered. rreversble P reversble What can be sad about entropy change n rreversble processes? Consder the P ndcator dagram wth the ntal and nal states and. here s shown an reversble path rom to and an rreversble path rom to. ogether they make an rreversble cycle and thereore rom the Clausus nequalty t s possble to say that; dq 0 dqi d QR 0 Q I dqr dq I dqr d dq I In other words, n an rreversble process between ntal and nal states there s an actual change n entropy whch s that ound or a reversble process between the same two states (e. entropy s a state uncton). hs actual entropy change s larger than the change one would calculate on the bass o rreversble heat exchange.

21 In ths dervaton we used the act that or a reversble path we can wrte dqr dqr It s mportant to note that ths s not true or an rreversble path; dqi dqi I adabatc changes are consdered only then dq 0 and t ollows that; and or any nntesmal part o the process I 0 (Irreversble change) d 0 (Irreversble change) he above can be generalzed to non-adabatc processes as ollows; I there s heat exchange occurrng (eg an engne wth ts heat reservors) t s always possble to consder all o the nteractng elements beng contaned wthn an adabatc chamber and so there beng zero net heat low Adabatc enclosure Q ystem ystem I all the systems under consderaton are thermally solated then or rreversble processes Net 0 and or reversble processes 0. Net 4

22 In summary; ) he entropy o a thermally solated system can only ncrease or reman constant. ) pontaneous changes n solated systems lead to an ncrease n entropy. ) In a thermally solated system, wll tend to the maxmum value possble 4) In a totally solated system, wll tend to the maxmum possble value at xed U e. U wll reman constant. hese are all entropy statements o the second law o thermodynamcs. Clearly, the combnaton o the system (under study) plus the surroundngs plus everythng else must be consdered as a closed and thermally solated system and thereore, or any change that occurs, the grand statement can be made that Unverse 0 And the equalty part o ths statement only pertans the change s reversble. Ex. Water, Irreversble Water, R Reservor, R Reservor, R Equlbrum state A mass, m, o water at an ntal temperature,, s placed n thermal contact wth a reservor at temperature R and the water temperature approaches that o the reservor rreversbly untl t reaches a nal temperature, = R. Equlbrum state system mc ln R P and reservor dqres R dqres QRes R mc P R R 5

23 Nothng may be sad about the respectve sgns o the entropy changes as t s not speced whether s greater or less than the reservor temperature although t s possble to say that they have opposte sgns. It s very mportant to stress at ths pont that they do not have equal magntudes (entropy s not a conserved quantty!!). eekng a more general statement. Unverse ystem reservo r Unverse mc P ln R mc P R R R Unverse mcp ln mcp mcp ln mcp R R R hs s n the orm Unverse mcp ln mcp R R mc ln Unverse P X mc X P where Unverse mcp ln X X X Note that; For coolng, R and X < 0 s negatve, For warmng, R and 0 < X. < Usng McClaurens seres expanson, x x ln( x ) x... R Unverse mcp X X X X... X mcp X... 6

24 5.4 Entropy as an Inequalty and the Kelvn-Planck and Clausus tatements We now have a mathematcal expresson o the second law o thermodynamcs prevously ormulated as statements grounded n emprcal observaton. How does the nequalty, 0 relate to those prevous statements?. ) Kelvn-Planck statement was that heat could not be converted nto work wth 00% ecency, or equvalently that some heat must always be ejected as waste rom the engne nto a cold reservor. hs can be related to the new orm o the second law by askng what happens to entropy heat can be converted to work wth no waste heat beng rejected to a cold reservor. he heat has come rom a hot body and thereore nvolves a decrease n the entropy o that body (the hot reservor) Res dq Q 0 (where the same notaton has been used as n the dscusson on engnes). For the system operatng n a cycle ys 0 d cycle Ie. there s only work then perormed whch o tsel has no eect on the entropy o the system or unverse the second law as an nequalty s volated, entropy has been reduced. Unverse Res ys dq Q 0 It can now be seen that the waste heat rejected nto the cold reservor n any real engne s absolutely necessary n order that the second law s compled wth e. It acts to rase that body s entropy by an amount dq Q such that Unverse Q Q 0 7

25 e. the entropy s ncreased and because the heat came rom a hot body and was rejected to a colder body, rom the measure o entropy where the temperature o the body appears n the denomnator, ths allows a smaller amount o heat Q Q to be rejected to the colder body but stll or there to be a net entropy ncrease to allow ull complance wth the second law and yet leave some energy, Q Q rom the hot body to be avalable to perorm useul work, W. ) Clausus statement that heat could not low spontaneously rom a colder to a hotter body can be related to the new entropc second law by magnng the scenaro, breakng Clausus statement, where a glass o water s placed n a warm oven and ce orms spontaneously as the colder body, the water, loses heat to the warmer body, the oven. he water as t cools loses heat and wll see a decrease n ts entropy dq Q H O HO HO whlst the warmer oven recevng the heat wll see an ncrease n ts entropy o Oven Q Oven It may be seen mmedately that the occurrence o the temperature n the denomnator o these expressons means that the same amount o heat has transerred rom the water and to the oven and yet the negatve change n entropy or the colder water s much larger than t s or the warmer oven and Unverse Q Oven Q HO H O Oven Q 0 HOOven he second law n ts entropc orm s agan volated. Heat cannot thereore leave a colder body and low to a hotter body as ths wll nevtable lead to a decrease n entropy n contraventon o the entropc orms o the second law. 8

26 As a thought experment consder the warmng o water reversbly by usng the heat output o a Carnot engne takng an nntesmal heat dq rom a heat reservor and delverng an nntesmal amount dq to the water whlst dong work dw and runnng many cycles, see dagram below. For a Carnot engne; thereore dq dq dq dq dq dq 0 d d 0 d Unverse 0 d 0 Unverse 5.5 he Carnot Engne Revsted. a he Carnot Cycle P Adabatc, d Q Isotherm, W Isotherm, Q b Adabatc, c 9

27 Fnally, wth the new state varable nsght may be ganed nto prevously studed systems. For example the Carnot cycle. Other state varables may be represented on dagrams equvalent to the P- dagram and the dagram oers an nterestng way to represent the Carnot cycle whose P- dagram s shown above. a Isotherm, Q b adabat adabat d Q Isotherm, c Q = a b On the - dagram the sotherms are represented by the horzontal lnes o constant temperature at and. In the process a b the volume ncreases and as has been seen ths has the consequence that entropy ncreases. he horzontal lne rom a b then goes rom let to rght towards ncreasng entropy. Durng the adabatc processes no heat s exchanged and s zero, thus the adabats are represented by the two vertcal lnes. he sotherm c d s at lower temperature and the adabat must be a downward process on the dagram. mlar reasonng may be appled to the other processes. Recallng that wth a P- dagram the ntegral Pd represents the area under the curve b a representng the process a b beng postve when gong rom small to large volumes 40

28 (expanson) and so the work b W Pd s mnus the area under the curve. mlarly, on a a - dagram the ntegral b a b d dq Q. a he area nsde the rectangle representng the Carnot cycle on a dagram s then; b a d d Q Q a b e. t s the total heat absorbed. By applcaton o the rst law and the act that over a cycle U = 0 the area s also the net work done. NB. he zero pont on the axs s arbtrary but the orgn o the temperature axs s absolute where the thermodynamc temperature scale may be used. Consderaton o the - dagram or a Carnot engne also allows the Carnot orm o the ecency to be obtaned straght away by smple calculaton o areas η W E Q As obtaned prevously or the Carnot engne. 4

29 5.6 he hermodynamc Identty. O great mportance n the study o thermodynamcs, entropy may be used to re-express the rst law, du dqr dwr dqr Pd by re-arrangng the denton o entropy and usng dq d ; du d Pd hs new expresson o the rst law contans only unctons o state or ther changes, du, d, d, and P! It s called; HE HERMODYNAMIC IDENIY hereore ths equaton s an dentty or any two equlbrum states nntesmally close. All derentals n ths dentty are now perect derentals, ndependent o path! d U d Pd hereore the natural varables o U are and, U = U(, ), and a ormal mathematcal statement may be wrtten usng these natural varables; hereore U U du d d U U P Or the hermodynamc Identty may be looked at another way. 4

30 4 P U d d d = (U, ) wth U and the natural varables o. Formally; U U d U d d And comparng the two equatons or d; U P U Relatons such as these relatng partal derentals to physcal quanttes wll oten prove useul as wll be seen when lookng at the physcal meanng o entropy and other dened thermodynamc potentals later. Wth the explct recognton that d dq R another way s open to dene heat capacty prevously dened as; Q Lm C 0 hs was then moded n order to couch the denton n terms o state varables alone; U C or P P H C he new state uncton oers a new way o dong ths R Q C ln lmt lmt 0 0 P P P P R P Q C ln lmt lmt 0 0 Looked at another way ths denton o the heat capacty may be used to nd by measurng ether C or CP as a uncton o temperature. may be determned by ntegraton o C wrt and ts value may thereore be determned (up to a constant o ntegraton ) rom such measurements.

31 5.7 Extensve and Intensve varables. A very mportant and useul concept concernng thermodynamc varables, elds etc s that o extensty., U and all depend on the sze o the system (eg. number o moles) or a gven equlbrum state. I the sze s doubled then each o, U and s also doubled. arables that have ths property are called EXENIE thermodynamc varables. Quanttes such as P and are ndependent o system sze and are called INENIE thermodynamc varables. It s possble to show that A EQUILIBRIUM, or a OALLY IOLAED YEM, the temperature and pressure (any ntensve varables) must be UNIFORM throughout the system. Ex. U,, U,, Adabatc wall Heat Adathermal conductng wall wall Because s an extensve varable and U and may be used as ts natural varables we can wrte; ( U, ) ( U, ) (All are extensve varables) U U U constant e, U s extensve For equlbrum must be at a MAXIMUM and doesn t change urther wth tme even though energy exchange can occur through the adathermal wall and U U 0. du du du 0 du du It may then be stated that U U U U U 0 44

32 45 U U Or equvalently usng the relatonshp ound earler, U we can express ths as e. s ntensve. uppose and are at the same temperature but are separated by a moveable pston ), ( ), ( U U constant 0 d d d d d It may then be sad that 0 Or equvalently by usng the relatonshp, P U, ound earler we may express ths as; 0 P P P P (t s ntensve as also ound or.) Adabatc walls Adathermal Pston

33 5.8 he Mcroscopc Interpretaton o Entropy o ar the dea o entropy has been developed rom a macroscopc vewpont begnnng wth the propertes o the Carnot cycle and the Clausus nequalty that lead to a new state uncton beng dened. It s now necessary to ask whether there s a mcroscopc understandng o entropy analogous to the mcroscopc understandng o temperature and ts relatonshp to mcroscopc knetc energy. An ntutve understandng o such eatures as the act that or an solated system always ncreases or remans constant but never decreases would be useul!. Ludwg Boltzmann ( ) was the rst to gve such a mcroscopc explanaton o entropy (between 87 75). Boltzmann s conjecture may be expressed n the ollowng way; Entropy s a measure o the MICROCOPIC probablty o ndng the system under study n a gven MACROCOPIC equlbrum state. Probablty means here a number that s proportonal to the number o dstnct ways a system can arrange tsel mcroscopcally to acheve a partcular macroscopc equlbrum state. hs dea may be placed on a rmer ootng wth the help o an example. Imagne a smple lud or gas system wth xed volume, and nternal energy, U. and U are ndependent state unctons and as such are sucent to dene a macroscopc equlbrum state unquely. Introducng the quantty (U,) = number o dstnct mcroscopc arrangements o the system gvng the same U and. hen accordng to Boltzmann s conjecture the entropy s some uncton o (U,). ( U, ) ( ) o have a better understandng o what s sgned by, consder the gas n queston to be dlute. Mcroscopcally the state o the gas s descrbed by gvng the values o the poston and momentum o every molecule. I the poston or momentum o a sngle molecule s altered the mcroscopc state or mcrostate o the gas s changed. For current purposes all mcrostates are equally probable and ths s reasonable or a system o N dentcal partcles n equlbrum. 46

34 o begn to attack the problem o the value o one may ask how many ways there are o arrangng the postons o the N partcles o the gas n a volume? o begn to answer ths t s necessary to gve the partcles some choces. o do ths, magne dvdng up the volume nto lttle cubes or cells o sde length X. he number o cells s then; N Cell X hs s the number o places where a partcle can be put. he next partcle can also go nto NCell places and so on gvng us Poston... X X X X X Poston N X Next, the queston o how many ways may the partcle momenta can be arranged, needs to be answered. hs s somethng that requres a lttle more thought. Recall, U N mv N p m prms N m and U prms m N U s chosen o course because t s the other natural varable o (alongsde whch has already been consdered n the rst part). he momentum s to be spread over a regon o sde prms n momentum space. As wth the earler argument about poston and arrangements ths momentum space can be dvded up nto nntesmal cells or cubes o sde p and the number o choces s 47

35 N p prms p And the number o arrangements s N p rms Momentum p he total number o arrangements s the product o these two numbers ( U, ) pacemomentum N N p rms X p ( U, ) N p rms Xp Later on t s sometmes useul to relate the arbtrary X and p to quantum mechancs and the uncertanty prncple but or present purposes ths s unnecessary and these quanttes wll soon become rrelevant. Up untl now n ths argument t has been assumed that all o the partcles are dstngushable whch o course s not necessarly or even usually the case. Ie. arrangements lke the sx shown below have been treated as dstngushable, that s, as separate mcroscopc arrangements and ths s clear as the partcles P have been labelled wth subscrpts,, to dstngush them. P P P P P P P P P P P P P P P P P P I there are no numercal subscrpt labels then these! = 6 arrangements are equvalent and should be counted as. For N dentcal partcles the rst nave answer needs to be dvded by N! or pace and Momentum separately to avod over countng. 48

36 U Identcal (, ) N! N p rms Xp he next problem to be dealt wth s how to nd what the uncton () should be lke n order that t gves the entropy or somethng wth smlar propertes. One mportant property that was touched upon earler s that s extensve e. addtve. U U U U U ( U, ) ( U, ) ( U, ) But, also n the above dagram t s the case that each o the two subsystems has ts own number o arrangements, U and U,, system the number o arrangements s gven by the product.. It s also known that or the combned ogether wth the Boltzmann conjecture that () ths mples; ( ) ( ) ( ) ( ) he uncton sought then has the property ( ) ( ) ( ) here s a uncton that assgns ths addtve property to a product that mmedately suggests tsel. ln( ab) lna lnb More generally t may be concluded that; 49

37 = Cln + C where the constant C s ncluded to gve the most general orm. Checkng the propertes are those o an extensve quantty, ln ln C C ln C C s a constant (or a gven macrostate) so t makes sense to dene = 0 when = and thus C 0 e. he arbtrary choce may be made that s lowest when there s only one arrangement that wll gve rse to the macrostate! hs stll leaves C to be ound. he argument or the orm o can be made n a mathematcally more ormal way and ths s presented, or those nterested, n an appendx to ths set o notes. o nd the constant C t s necessary to return to the dlute gas. ( U, ) C ln Identcal ln Identcal ln N! prms ( Xp) N prms N ln ( Xp) lnn! ln Identcal Nln N lnprms N lnx N lnp lnn! he terms n curly brackets are constants whch wll cancel and be lost when an entropy change s calculated ( s only known to wthn a constant). 50

38 Usng the act that to obtan ln ln Identcal N ln N lnp constants Identcal p rms N ln N lnu mu N rms m N ln N ln Identcal N ln N lnu constants constants ( U, ) CN ln CN lnu constants It wll be noted that all o those arbtrary seemng X and p used to nd relegated to rrelevance n the term constants. Identcal have been Now or a neat trck! It was ound prevously that U whch can be used to complete ths dervaton. U C N U U CN U CN nr N N R A Gves us C R C kb NA (U,)=kBln(U,) 5

39 I we accept Boltzmann s conjecture t can now be seen why may ncrease even n the absence o heat low when the system can spontaneously pass rom a LE PROBABLE (small ) to a MORE PROBABLE (large ) state. It may never nd ts way back to the less probable orgnal state and contnue evolvng to states whch are avalable wth more ways o achevng them. At the tme that Boltzmann proposed ths, all other mportant and well known quanttes such as energy, lnear momentum, mass, angular momentum were conserved. In what was a relgous socety ths was vewed as a result o God s benecence whereby He had gven us exactly what we needed, no more and no less! hs new quantty, entropy, was not conserved!! Consequently ths mcroscopc explanaton o entropy due to Ludwg Boltzmann was never accepted n hs letme and ths was understood to be one o the man reasons behnd hs sucde. oday on hs tombstone are nscrbed the letters = klogw 5.9 he Equvalence between Mcroscopc Entropy and Macroscopc Entropy here are now two descrptons o entropy change; ) he mcroscopc nterpretaton o entropy due to Boltzmann has gven us; where, as an example, or a gas; and k B ln U Identcal (, ) N! k B ln Identcal N p rms Xp ( U, ) k B ln N! p rms N Xp ) From macroscopc thermodynamcs usng the hermodynamc Identty or a lud; du d Pd 5

40 d du P d R d R d Rln Rln I the Boltzmann conjecture s ndeed correct then t must be the case that the Boltzmann mcroscopc theory o entropy should oer predctons that all nto agreement wth the Clausus macroscopc theory ound earler. hs can be tested on specc examples; Consder as an example JOULE EXPANION ake mole o an deal gas (N = NA) n a box dvded nto two dentcal halves each o value. tart wth all o the gas n the let hand hal o the box conned by a membrane wth a vacuum to the rght. Break the membrane to produce a ree expanson nto a volume Beore Ater Ater the expanson the number o arrangements o the atoms s changed rom () to (). he NA atoms each now have twce as many choces than ntally as they have twce the volume they may occupy ( )... N ( ) A ( ) N ( ) k ln ( ) ln A B kb ( ) N ( ) kb ln A kb ln( ) ( ) kbna ln And nally the change n entropy s ( ) ( ) ( ) kbna ln Rln But rom the earler macroscopc argument or mole o a monatomc gas; 5

41 ( ) ( ) Rln Rln hus, the mcroscopc argument reproduces the same result as the macroscopc argument as requred. he equvalence between thermodynamc entropy (Clausus) and statstcal entropy (Boltzmann-Gbbs) may be demonstrated more generally as ollows; Boltzmann s entropy gves; usng ( U, ) k Nln k Nlnp constants B B rms U N mv N mvrms prms N m Usng ths n the Boltzmann entropy; mu p rms N (, ) ln ln mu U k B N kbn constants kbn ln kbn lnu constants N ( U, ) kb N ln kbn ln nr constants ( U, ) kb N ln kbn ln constants or wth urther smplcaton usng R k B and N A N n ; N A R R ( U, ) N ln N ln constants nr ln nr ln constants NA NA 54

42 5.0 Entropy and Dsorder I, or a gven equlbrum state, there are only a small number o mcroscopc arrangements possble that wll reproduce the macroscopc state then t s extremely unlkely that the macroscopc state wll be ound as compared wth a macroscopc state that can be reproduced by many alternatve arrangements o mcroscopc states. he spontaneous ncrease n entropy o an solated system, 0, s a result o the change o a system s macrostate rom a less probable, low, arrangement to a more probable, hgh, arrangement. Once ths has happened the system wll nd t extremely mprobable to go backwards to the low macrostate. hs s ntmately lnked to the concept o dsorder. o see ths some examples wll help; nce, as the system evolves to a more dsordered arrangement the greater the potental number o equvalent mcroscopc arrangements becomes, t s seen that entropy s greater as the dsorder ncreases and that thereore entropy can be vewed as a measure o dsorder. In other words the entropc declaraton o the second law or prncple o ncreasng entropy, 0, s a statement that a system wll spontaneously evolve to greater dsorder. Example. Polymer chans, random col and rgd rod. In the above pcture two lasks o an dentcal polymer n soluton ndcate the transton rom order to dsorder. Recallng that a polymer chan s a very large number o monomer unts bonded together lke a strng o pearls there are two extreme possbltes that may be dented (and many n between) In the yellow lask s a polymer (polydacetylene 4BCMU) n soluton where the long polymer chan s allowed to orm a random col where the drecton one segment (monomer) o the chan s pontng n s (almost) ndependent o the drecton o the precedng segment. I condtons are sutable (temperature reduced) n a ew specal polymers the chan may be constraned to straghten out nto a rgd rod where each segment s pontng n the 55

43 same drecton as the prevous segment, the red soluton. Clearly the yellow soluton s much more dsordered than the red soluton and thereore o hgher entropy. I we are told the drecton o any segment n the random col we know nothng about the drecton n whch any other segment may be pontng whereas just knowng the drecton o one segment n the rgd rod conormaton wll allow us to know (n prncple) n whch drecton any other segment s pontng. We may equvalently say that the normaton content s much hgher n the latter case. Example. Dssolvng sugar n tea. ugar added to a cup o (hot) tea s another example o ncreasng entropy as the sugar dssolves. he sugar represents a large number o small crystals each crystallte representng an outstandng degree o order where knowng the poston o one sugar molecule n the crystallte allows ull knowledge o the postons o all the others n that partcular crystallte (n prncple). he dssoluton and mxng o the sugar molecules throughout the volume represents a very large ncrease n dsorder and a loss o normaton as we now know very lttle about the poston o each sugar molecule except that t s somewhere n the tea! he heat requred or ths ncrease n entropy s the latent heat o meltng and comng rom the surroundng tea causes the drnk to cool rapdly. All phase changes represent an abrupt change n the degree o order o a system whether magnetzng a pece o ron and algnng the electron spns thus ncreasng the order and concurrently reducng entropy or algnng the molecules o a lqud crystal by applyng some external eld. Heat wll be requred to drve the process o ncreasng entropy/dsorder. 5. he arrow o tme. hat dsorder always ncreases wth tme has come to be known as the arrow o tme as ths represents the only law o physcs whch does not appear the same under tme reversal. All o Newtons laws are symmetrc under tme reversal or example and a collson o snooker balls s lmed on a rcton ree table t s not possble by observng the acton o the balls to say whether the lm s beng run n reverse or not. However, at the start o the game when teen perectly ordered red balls n a trangle (and all other colours n a desgnated poston) are broken up by the cue ball t s perectly straghtorward to say whether the lm s run n reverse or not as the probablty o a seres o randomly dstrbuted balls all colldng and comng to rest wth a trangle o reds and all colours on ther spots s hghly unlkely to occur n the real world! 56

44 5. Loss o avalable work n rreversble processes. he Heat Death. here s oten talk about an energy crss nowadays but energy s never lost or created! What we have s n act an entropy crss! he amount o energy avalable to do useul work s orever decreasng. hs dea was orgnally realzed and explored by homson (Lord Kelvn) and termed the Heat Death o the unverse. It s possble to show that the acton o an rreversble process (most processes are) and the second law wll result n the nevtable loss o avalable work. hat s potental work that could be accessed by the unverse. o see ths consder the ollowng stuaton; Heat ource. Heat ource. Q Q Engne WIrr Q Carnot Engne WC Intermedate Heat nk. Q WResdual Lowest avalable heat snk. 0 here s a heat source at temperature to supply heat Q whch wll do work. here are two heat snks, wth one at 0 beng the at lowest avalable temperature (n the unverse eg. cosmc background temperature) to accept rejected heat rom a Carnot engne. here s also an ntermedate heat snk at temperature avalable to accept rejected heat rom an rreversble (standard) engne. he maxmum avalable work s that carred out by a reversble Carnot engne operatng between the two temperatures, and 0; 57

45 WC C Q Q 0 I an rreversble engne s run between the source at temperature and the ntermedate heat snk at temperature ths wll gve an amount o work; W Irr Q Q here s stll an amount o resdual work to be extracted n the second scenaro by runnng a Carnot engne between the ntermedate heat snk at temperature and the heat snk at 0; 0 WResdual C Q Q he derence between the maxmum avalable work and that extracted usng an rreversble engne s; WLost WC 0 0 WIrr WResdual Q Q Q Q W Lost 0 Q he entropy change n the reversble process s zero and or the rreversble process the entropy change s Q Q Q source nter hereore the derence between the maxmum avalable work and that extracted usng an rreversble engne s; Q Q W Lost 0 0 he work lost s the product o the entropy ncrease related to the use o an rreversble engne and the lowest temperature avalable to act as a snk or the engne dong work. As entropy ncreases due to rreversble processes that ncrease renders more and more o the heat energy avalable or dong work unusable. e. as the unverse ages and all ts parts come to 58

46 a unormty o dsorder equlbrum s reached and entropy has been maxmzed there wll be less and less capacty or dong work and thereore or le. No comment s made here about urther and more recent cosmologcal speculatons such as the bg crunch. 5. Inormaton and Entropy here s an ntmate lnk between normaton and entropy that wll be brely consdered here. Claude hannon (96-00) was one o the rst scentsts to poneer normaton theory and has a number o mportant dscoveres to hs name ncludng the lnk between thermodynamc entropy and normaton (948). o begn; a method o quantyng normaton s requred and to see how ths may be done a concrete example helps. Consder the three true statements concernng the brthday o physcst Albert Ensten; ) Albert Ensten s brthday alls on a partcular day o the year. ) Albert Ensten s brthday alls n the rst hal o the year. ) Albert Ensten s brthday alls on the 4 th day o a month. It may be commonly agreed that qualtatvely, the rst statement contans no normaton whatever thereore t s normaton content = 0. he second statement has lmted the month n whch the brthday alls to one o 6 months or one o 8 days, clearly a greater normaton content than the rst statement. he nal statement has even more normaton as t now lmts the brthday to one o days and thereore has the greatest normaton content. hs normaton content now needs to be quanted by realzng that n ths example the greater the probablty o the statement beng true the lower the normaton content n the absence o any other normaton. By tsel; 65 tatement has a probablty (o beng true) P tatement has a probablty (o beng true) P tatement has a probablty (o beng true) P

47 More mportantly the statements are ndependent peces o normaton, as n these examples, then t s desrable that however t s dened ther normaton content should add. 6 From probablty theory the probablty o both and beng true s P P P o ensure that normaton content, I, s addtve and I I I I then ollowng hannon t s useul to dene the normaton content as; I k lnp Where k s a postve constant. Choosng k = and usng log the normaton content wll be gven n bts. Wth ths denton then; For statement P and the normaton content s I k ln 0 For statement P 0. 5 and the normaton content s I k ln 0.5 For statement P 0. 0 and the normaton content s I k ln For statements and together; P gven the denton o I then I I I k(ln 0.5 ln 0.0) k ln I k k B and log e were chosen then the denton o normaton content and entropy are dentcal where P s recognzed as the probablty o ndng a partcular macrostate gven the act that there arrangements o mcrostates that wll make up ths macrostate. I k lnp kb ln kb ln Gven a set o statements wth probablty hannon s gven by; s known as the hannon entropy. I P then the average normaton,, as dened by I P k P log P 60

48 Usng the Ensten brthday example. And takng all three statements together, clearly statement adds nothng but statements and taken together cut the possbltes to o 6 days. kp log P k he connecton between ths and physcal entropy s drect. In act normaton tsel s now to be understood as a physcal quantty. hs s most easly seen by magnng some physcal data storage devce contanng N bts o normaton n the orm o s and 0s. hs memory s to be our thermodynamc system and has N possble arrangements or states (equvalent to n prevous dscusson). o completely erase ths normaton all the bts need to be made nto 0s (or s) and erasure must be rreversble. Imagnng ths memory to be n contact wth a heat reservor at temperature, all N bts are made 0s (the macrostate) then the number o ways the states o the memory may be arranged to acheve ths s now and the erasure process has reduced the number o states o the memory by N and the physcal entropy o the memory s reduced by; N k B log e Nk B log e o avod a net decrease n the entropy o the unverse requres that the entropy o the reservor ncrease by at least the same amount; Res Nk B log e For ths to happen an amount o heat needs to low rom the memory to the reservor Res Q Nk B loge Each bt erased requres a heat Q bt dq bt Q k B log e N In other words the erasure o a bt o normaton requres that there s a mnmum heat added to the surroundngs (reservor), dq Bt k B log e. 6

49 o do the nverse and add a bt o normaton to the memory devce an equvalent heat would be taken up by the memory (ts entropy has ncreased). hs s the thermodynamc mnmum and represents a undamental lmt that s not to be crcumvented by mproved technology! hought or the day. Is ths gong to cause global warmng? ) Next tme a ggabyte o no s downloaded to your phone that s bts.80 JK 00K loge Joules o - heat. 0 J It seems global temperatures are sae rom YOUR phone! 50 ) Google processed 0 petabytes (or 8 0 bts ) per day n 04 and Watts was constantly beng consumed by googlers as a result. hs wll have greatly ncreased by the tme you read ths. ) In 05 there were Bytes o new data created every day or Joules worth equvalent to Watts o power creatng new data. he concluson mght be that the thermodynamc prce we pay or normaton wll not cost the Earth! But remember ths calculaton only ncludes the creaton o new data not ts downloadng by users onto multple machnes. he cost o that normaton upload gong vral wll be substantally greater. Fnally, just as wth all o those rctonless pstons whch I have convenently ntroduced rom tme to tme, there are hdden power costs due to dsspaton nvolved n swtchng a CMO nverter (Joule heatng n the swtchng devces), or each new bt, that are way above ths thermodynamc lmt! Use your devces responsbly. 6

50 Appendx he argument or the orm o () can be made more ormal as ollows; ( ) ( ) ( ) and thereore holdng constant and take the partal derental o () wrt reatng ths as a uncton o a uncton wth a constant the derentaton yelds / / ( ) ( ) / And holdng and derentatng wrt usng the product rule on ( ) wrtng / ( ) ( ) 0 // y // ( ) / y y ( y ) 0 And thus // ( y) / ( y) y Now wrtng / g / g ( y) d lng( y) g( y) dy y hereore by ntegraton o the above wrt y lng( y) lny C akng exponentals o both sdes / exp( C C y g y ) ( ) ( ) y y 6

51 thereore ( y) C lny C 64