PES 2130 Fall 2014, Spender Lecture 7/Page 1 Lecture today: Chapter 20 (ncluded n exam 1) 1) Entropy 2) Second Law o hermodynamcs 3) Statstcal Vew o Entropy Announcements: Next week Wednesday Exam 1! - and out: what wll not be covered on exam - Equaton Sheet ueston regardng W2: Monatomc vs datomc Indeed hydrogen s a datomc gas. hs s because monatomc hydrogen s very reactve and unstable. Monatomc hydrogen wll react wth almost any other chemcal. ydrogen gas = 2 Oxygen s smlar. he most stable s datomc oxygen O 2. Next s tratomc (ozone) O 3 and the least stable (very unstable) s monoatomc. Monoatomc oxygen s very shortly lvng speces combnng easly (due to very hgh electronegatvty) wth large number o elements ncludng another oxygen atom. Example o monatomc gases: helum, neon, argon, krypton, and radon Last lecture: Reversble and Irreversble Processes (Fundamentals o Physcs book) Any tme there s heat low through a nte temperature drop, t s an rreversble process. I you want a reversble process, heat only should low when the temperatures are nntesmally derent (whch would take orever). eat Engnes he ecency o a heat engne s dened as the rato o the work done to the amount o heat brought n rom the hot reservor: W L For a cyclc process, C 1 eat Pump (Rergerators) Devce that does work n order to move heat rom cold to hot. Rergerators are nothng more than engnes run n reverse: they take heat out o a cold reservor, and use work to put more heat nto the hot reservor. Coecent o Perormance: L K For a cyclc process, W he Carnot Cycle Carnot Engne Ecency L L C 1 1 K C L L
PES 2130 Fall 2014, Spender Lecture 7/Page 2 DEMO: Strlng engne he Strlng engne (heat cycle) wll run when t s placed on a dsh o crushed ce or on a mug o hot water. he workng volume may be lled wth helum whch makes the engne run aster. http://www.anmatedengnes.com/strlng.html Strlng Engne: Comparson wth the Carnot cycle shows that each engne has sothermal heat transers at temperatures and L. owever, the two sotherms o the Strlng engne cycle are connected, not by adabatc processes as or the Carnot engne but by constant-volume processes. o ncrease the temperature o a gas at constant volume reversbly rom L to (process da) requres a transer o energy as heat to the workng substance rom a thermal reservor whose temperature can be vared smoothly between those lmts. Also, a reverse transer s requred n process bc. hus, reversble heat transers (and correspondng entropy changes) occur n all our o the processes that orm the cycle o a Strlng engne, not just two processes as n a Carnot engne. Now we are ready to make the connecton to entropy! Entropy Any method nvolvng the noton o entropy, the very exstence o whch depends on the second law o thermodynamcs, wll doubtless seem too many ar-etched, and may repel begnners as obscure and dcult o comprehenson. - Wllard Gbbs, Graphcal Methods n the hermodynamcs o Fluds (1873) Let us start our dscusson o entropy by lookng at one o the last steps n our dervaton o the Carnot ecency: L L L L 0 Now, we dene a varable S, such that: d ds ( s constant)
PES 2130 Fall 2014, Spender Lecture 7/Page 3 and snce 1 ds d d S S S, where s the total energy transerred as heat durng the process hen, we can see that over the entre Carnot cycle, L 0 L the varable S, whch we now reer to as the entropy, returns to the same value. In other words the change n entropy adds to zero over the entre cycle. he net entropy change per cycle: S SL S 0 (In a Carnot engne there are two reversble energy transers as heat, and thus two changes n the entropy o the workng substance - one at temperature and one at L.) Entropy as a State Varable Now, perhaps that s just some artact o the Carnot cycle, but, usng calculus, we can show that over any cycle, the entropy returns to ts orgnal value (see proo n book). hereore, entropy s a state varable! It can descrbe the state o a system (lke the temperature, volume, pressure, E nt, etc.). A state uncton only depends on a state and not the path the system takes (unlke W and ). hs means that we want to calculate the entropy o an rreversble process we can use: Srr, gas S S Srev, gas d In other words, we use a correspondng reversble path to do the ntegraton. he 2nd Law o hermodynamcs All heat engnes have a maxmum ecency that s much less than 100% because o the second law o thermodynamcs. No engne can exceed Carnot ecency, because heat does not low spontaneously rom cold to hot. No process s possble whose sole result s the transer o heat rom a body o lower temperature to a body o hgher temperature. Rudol Clausus (1854) No process s possble n whch the sole result s the absorpton o heat rom a reservor and ts complete converson nto work. Lord Kelvn (1851) In other words, snce objects and ther envronment wll always reach a thermal equlbrum eventually, you can't keep gettng work out o the system! "here s no such thng as a perpetual moton machne!"
PES 2130 Fall 2014, Spender Lecture 7/Page 4 Statements above are concerned wth work and are consdered the rst statements o the second law beore entropy was dened. Entropy - Measures the amount o dsorder n a system. Let s examne now an nntesmal sothermal expanson: snce d E nt = 0 ( = 0, b/c o constant temperature) d = dw = pdv = nr dv V dv d ds V nr Now, t turns out that the ncremental ractonal volume ncrease (or added heat) s drectly related to the ncrease n dsorder o the gas. Snce there s more space to occupy, there are more ways whch the molecules can exst where the gas has the same state. he second law o thermodynamcs can be stated n terms o entropy: No process s possble n whch the total entropy o an solated system decreases. S 0 he term solated system here s crucal. he entropy o some object can decrease, t s n contact wth another object whose entropy ncreases just as much or more. It s certanly possble or somethng to have more order than t dd at some pont n the past, but t just means that somethng else, n contact wth t now has less order. rreversble process: S rr 0 reversble process: S rev 0 We just talked about entropy beng a state varable and that we want to calculate the entropy o an rreversble process we can use a correspondng reversble path to do the ntegraton: Srr, gas S S Srev, gas d But doesn't ths volate the 2nd law o thermodynamcs? No, snce we have not consdered the whole system, only the gas. he second law apples to the system as a whole, not just one component. We would also need to calculate the changes n entropy o the surroundngs whch can be dcult.
PES 2130 Fall 2014, Spender Lecture 7/Page 5 Example: A 250g potato at 293K s thrown nto a 2.00 L pot o water at 373K. a) What s the nal equlbrum temperature o the potato-water system? b) What s the change n entropy o ths system, assumng t s solated rom the surroundngs? Specc heat o potato: c p = 3430 J kg -1 K -1 Specc heat o water: c w = 4187 J kg -1 K -1
PES 2130 Fall 2014, Spender Lecture 7/Page 6 A statstcal Vew o Entropy and the Arrow o me You can slde a book across a table. It stops due to energy lost rom rcton. But, have you ever seen a book start sldng takng energy rom the thermal energy o the table? You could start out wth all o the ar n ths room n one hal and vacuum n the other hal, separated by a partton. hen remove the partton and the ar would ll the room. But, you wouldn t expect or a room lled wth ar to spontaneously separate nto hal lled and hal empty. Nether o these scenaros s prohbted by the conservaton laws or by the rst law o thermodynamcs. Why don t many thngs that happen spontaneously n nature also happen n reverse? Macroscopc and Mcroscopc States In order to understand ths apparent drecton n tme we need to understand how to nd a partcular macro-state gven the statstcs o the mcro-states whch make t up. As example, let s look at a smpler problem: a con toss. Let s say we have one con. For smplcty, t s motonless and can only be n one place. Its state, whch we now specy as ts macro-state s dened as whether t s a heads or a tals. ow many ways can you make each macro-state?.e. what s the number o possble mcro-states (called "multplcty" n your textbook). Only one way or each state - one mcro-state or each way. So, we toss a sngle con n the ar, what s the probablty that t wll land n each macro-state? 1 or 1? P 1 # mcrostates 1 1 1 mcrostates 11 2 1 1 mcrostates mcrostates mcrostates
PES 2130 Fall 2014, Spender Lecture 7/Page 7 P 1 # mcrostates 1 1 1 mcrostates 11 2 Now, lets do the same wth two cons. here are 3 derent macro-states, but or each macro-state, there s a derent number o ways to make t. For one tal and one head, there are 2 ways to make t. So, we toss 2 cons n the ar, what s the probablty that they wll come up all heads? ( ) ( ) What s the probablty that they wll come up 1 head and 1 tal? ( ) ( ) Notce that the total number o mcro-states = 2 N = where N s the number o cons. mcrostates So, or 100 cons, there are 2 100 mcro-states and the possblty o throwng 100 heads s 1 n 1,267,650,600,228,229,401,496,703,205,376! Now, the number o molecules n ths room s approxmately 1 x 10 29. Consder the macro-states as beng (heads) a molecule s n one hal o the room and (tals) t s n the other hal. What s the probablty that on a random samplng o the molecules poston, we nd them all n one hal o the room? So, ths tendency towards dsorder s just a consequence o the probablty o ndng a partcular macro-state gven the statstcs o the mcro-states whch make t up! hs s the underlyng nature o entropy, and the underlyng nature o the drecton o tme. Calculatng Entropy
PES 2130 Fall 2014, Spender Lecture 7/Page 8 One can calculate the entropy, rst expressed by Ludwg Boltzmann (1844 1906) an Austran physcst and phlosopher), as: S k w, B ln where k B =1.38 x 10-23 JK -1, s the Boltzmann constant and w s the number o all mcrostates avalable to the system. Or, snce only the derence n entropes are ever used n calculatons: S S S k ln w k ln w w SkB ln w B B