(1) One mole of an deal monatomc gas doubles ts temperature and doubles ts volume. What s the change n entropy of the gas? () 1 kg of ce at 0 0 C melts to become water at 0 0 C. What s the change n entropy of the ce? (3) Suppose you mx 1 kg of water at 0 0 C wth 1 kg of water at 100 0 C. What s the change n entropy for ths mxng process? (4) You have just made your frst mllon. It also happens that the world has a terrble energy crunch at ths tme, and effcent engne development s all the rage. An nventor approaches you and suggests that you nvest n hs engne whch works between 30 0 C and 100 0 C because t has an effcency of 70%. Do you nvest? Why or Why not? (5) Show that: (a) U s a functon of temperature only (b) C v =3/R and the other features of deal gases.
Entropy and the Carnot Cycle There are many ways to talk about entropy, rangng from nformaton theory, languages, computer scence to physcs. We ll here concentrate on the physcs formulaton, specfcally pertanng to thermodynamcs. We frst need to defne the state functon Entropy (whch s a measure of effcency of heat flow) s gven by: Q S= T Strctly sad, ths apples only to a small amount of heat added to a system at a constant temperature. In general, as temperature changes over a range, you would have to add up ndvdual changes as: S= Q T steps Snce entropy s a state functon, the change n entropy wll be zero for any closed reversble path. If the path s not reversble, then the change s greater than zero. I ll later show you examples of ths. f = dq T Calculus verson: S, where dq ndcates a small amount of Q. I ll later show you how to calculate some entropes. Now, t s mportant to realze that the area enclosed by a closed T-S plot represents work. Let s assume a small closed cyclc change occurs n a system at a temperature T. A small amount of heat wll be added to the system. From the frst law and the defnton of entropy, we then have = = = Q U+ W W S T T T U s zero because the cycle s closed. We thus have: T S = W ( ) It s not much harder to show ths n other ways but I thnk I ve shown my pont.
Calculaton of entropes for dfferent processes When we calculate the entropy changes for an deal gas, t s useful to use the frst law: U= Q W Q= U+ W Here, we have n general: U= nc T so long as we re talkng about an deal gas. ( ) We also have that the work s gven by: W= P = nrt Thus, f we wanted to calculate Q, we have: Q= nc T + nrt As t stands, ths can t be calculated n general for any process (because you don t know how T vares when t multples the last term). However, dvdng by T gves: Q T T= nc( T) + nr( ) You can now use the useful approxmaton: Tf f S= nc ln + nrln T Calculus verson: dq= nc dt + nrt d so, = nc + nr dq dt d T T and S = nc + nr dq dt d T T Thus Tf f S= nc ln + nrln T
completely ndependently of the process that the gas went though, so long as the process s reversble. Ths s nce although you probably want to work out specfc examples for the dfferent processes. Now, as promsed, I ll use entropy to show the Carnot cycle to be the most effcent cycle possble. How? Remember that for an adabatc process, Q=0 whch, n ths case, wll gve a vertcal lne on a TS plot. For an sothermal process, we end up needng to connect the two ends of these lnes whch then gves us a square for the Carnot cycle. It looks lke the one I ve shown before. Let s see where Q h and Q c become nvolved: What you want to do to make the most effcent engne possble s to maxmze the area (green stuff) for a gven length of red (and you must have Q c present). The shape that gves ths s the rectangle whch s the Carnot cycle. At least that s my smplfed verson of why the Carnot cycle s the most effcent cycle possble. Heat Engne dagram Heat engnes work lke the dagram shows. Another statement of the second law of thermo s you have got to reject some heat to a cold reservor (you can t convert heat drectly nto work wthout losng some heat n the process). It s perhaps less obvous why ths s true but I lke the analogy of a battery: the cold reservor s the termnal, the + termnal s the hot reservor and a lght bulb s connected between the two. Both
termnals must be connected n order for the bulb to lght. Here, f you do work you have extracted some heat from the gas that dd work and thus, you automatcally have a colder gas left or somethng lke that. It s also mpossble to have energy go from cold to hot wthout dong work. (ths s the mpossble refrgerator). Interestngly enough, although we don t have the tme to explore ths completely, another statement of the second law of thermodynamcs s that the entropy of the unverse ncreases n all processes. I ll show you ths n the context of mxng two fluds. (1) One mole of an deal monatomc gas doubles ts temperature and doubles ts volume. What s the change n entropy of the gas? Soluton: For a monatomc gas, C v =3/R. Thus, Tf f S= nc ln + nrln T ( ) S= 1 Rln + 1Rln = Rln = 14. 5 3 T 5 J T K Notce that I don t need to talk about the path (so long as t s reversble) to answer ths queston. () 1 kg of ce at 0 0 C melts to become water at 0 0 C. What s the change n entropy of the ce? Soluton: Start wth: Q S= But f you use T=0 0 C, you re just about as wrong as you can hope to be. You must work n K here. The soluton? ml f 333. x10 J S = = 5 73 73 = 119. 9 K (3) Suppose you mx 1 kg of water at 0 0 C wth 1 kg of water at 100 0 C. What s the change n entropy for ths mxng process? Tm ( ) ( ) 4186 S= + = mc + = ln + ln = Q1 Q T1 T T T T T T T T T 1 1 33 33 ( ) ( ) 4186 ln + ln = 4186 0. 16818 0. 14393 = 101. 5 73 373 (4) You have just made your frst mllon. It also happens that the world has a terrble energy crunch at ths tme, and effcent engne development s all the rage. An nventor approaches you and suggests that you nvest n hs engne whch works between 30 0 C and 100 0 C because t has an effcency of 70%. Do you nvest? Why or Why not? c H J K
Soluton: If you calculate ε = = = Tc 30 c 1 1. 7 70 % T H 100 and then nvest, you ve just lost your frst mllon to a scam artst. If you calculate: 303 ε c = 1 373= 1 0. 81= 0. 188 18. 8 % and then don t nvest, you re just seen the value of my physcs course. You wll probably want to be sure to recommend that ths guy vst your competton. Note: each year when I use ths problem, t gets closer and closer to beng true!
(5) Show that: (a) U s a functon of temperature only (b) C v =3/R and the other features of deal gases. The knetc theory of an deal gas envsons pont partcles bouncng off of a wall and not nteractng wth each other. The force exerted by one molecule on the wall s gven from Newton s law as: p F= t The tme between successve collsons for a box of length L s: L t= for a one dmensonal box. The change n momentum s gven by: p= mv x If we permt what I refer to as the great le to preval, we ll use that t and p n the force equaton to get: mvx F= 1 L / = L mv x x In general, the partcle has equal amounts of veloctes n the x, y and z drectons. Thus, 1 v = v = v = v x y z where v s the magntude of the velocty of the molecule. We can thus fnd that for a 3-d box, 1 F= 3 L mv Snce we re nterested n pressure, and snce the wall of nterest has an area LxL, we fnd: P F 1 mv 1 = = = mv 3 L 3 L 3 Multply by the volume to get: 1 P= mv 1 = mv = K.E. x 3 ( ) 3 3 3 We n general have N atoms present to gve addtons as: P N K.E. Compare ths to the IDG: = 3 N K.E. = NkT T= 3 3 Ths means that the temperature and the average knetc energy of an atom are connected as shown. But, the knetc energy s ndeed the nternal energy of the deal K.E. k
gas so thus, the nternal energy of an deal gas depends only upon temperature. (N s constant). We can wrte ths as: U= N K.E. and thus, we fnd the exact functonal dependence: U NkT How does U change wth temperature? In terms of moles: so, Thus, we have the result: = 3 U T = 3 Nk U Nk = nr = 3 T nr = nc v C R = 3 Notce the 3.. t s very suggestve. Indeed, there are 3 degrees of freedom and we have 3 (1/R) contrbutons to C v. Suppose that we have a datomc molecule. The rotatonal modes have 3 degrees of freedom but only of these modes are capable of storng energy for a true pont-partcle system. Ths would add: / R to C v. If the molecule can also vbrate, we have addtonal consderatons. There are ways the molecule can vbrate and ths adds an addtonal / R to C v. Thus, we have potentally: 3 C = R+ R+ R+ v translatonal rotatonal vbratonal But some of the modes wll freeze out at low temperatures. Thus, we have the concluson: ½ R s added to C v for each degree of freedom whch s realzed n a system. Here s an mportant quote regardng the vbratonal contrbuton (From Modern Physcs for Scentsts and Engneers, 3 rd. Edton (006) (Brooks/Cole Publshers), page 301. How many degrees of freedom does ths add? One may be tempted to say just 1 one, because of the potental energy κ( r r) 0, where κ s the sprng s force constant, r the separaton between atoms, and r 0 the equlbrum separaton between the atoms. But another degree of freedom s assocated wth the dr vbratonal velocty ( ) m. 1 dr dt, because the vbratonal knetc energy s ( dt)