Lecture 3 Examles and Problems Mechancs & thermodynamcs Equartton Frst Law of Thermodynamcs Ideal gases Isothermal and adabatc rocesses Readng: Elements Ch. 1-3 Lecture 3, 1
Wllam Thomson (1824 1907) a.k.a. Lord Kelvn Frst wrote down Second Law of Thermodynamcs (1852) Became Professor at Unversty of Glasgow at age 22! (not age 1.1 x 10 21 ) Physcs 213: Lecture 3, Pg 2
Ideal Gas -, -T Dagrams vs at varous constant T s = NkT vs T at constant ncreasng T Isotherms Pressure Pressure olume For an deal gas at constant T, s nversely roortonal to the volume. 0 0 Temerature Pressure zero as T absolute zero, because the thermal knetc energy of the molecules vanshes. Lecture 3, 3
Last tme: The Frst Law of Thermodynamcs Energy s conserved!!! U = Q + W on change n total nternal energy heat added to system work done on the system alternatvely: U = Q - W by Note: For the rest of the course, unless exlctly stated, we wll gnore KE CM, and only consder nternal energy that does not contrbute to the moton of the system as a whole. Lecture 3, 4
Heat Caacty Look at Q = U + W by If we add heat to a system, there are two general destnatons for the energy: It wll heat u the system (.e., rase T). It can make the system do work on the surroundngs. Heat caacty s defned to be the heat requred to rase the temerature of a system by 1K (=1º C). Its SI unts are J/K. Q C (for small T) T The heat caacty wll deend on whether energy goes nto work, nstead of only ncreasng U. Therefore, we dstngush between: Heat caacty at constant volume (C ), for whch W = 0. Heat caacty at constant ressure (C ), for whch W > 0 (most systems exand when heated). Lecture 3, 5
Constant-olume Heat Caacty of an α-deal Gas Add heat to an deal gas at constant volume: W = 0 so U = Q = C v T U = αnkt = αnrt C = U/ T = αnk = αnr # avalable modes er atom (or molecule) Monatomc gas C = (3/2)Nk = (3/2)nR Datomc gas C = (5/2)Nk = (5/2)nR Non-lnear gas C = (6/2)Nk = (6/2)nR Hgh-T, non-metallc sold C = (6/2)Nk = (6/2)nR For an α-deal gas, C s ndeendent of T. Ths results from the fact that the number of avalable modes s constant. We wll see later n the course that: ths fals at low temerature, because there are fewer avalable modes. ths fals at hgh temerature, because there are more avalable modes. Lecture 3, 6
C ~ α Substances wth more nternal degrees of freedom requre more energy to roduce the same temerature ncrease: Why? Because some of the energy has to go nto heatng u those other degrees of freedom! The energy s arttoned equally equartton Lecture 3, 7
ACT 1 Consder the two systems shown to the rght. In Case I, the gas s heated at constant volume; n Case II, the gas s heated at constant ressure. Comare Q I, the amount of heat needed to rase the temerature 1ºC n system I to Q II, the amount of heat needed to rase the temerature 1ºC n system II. heat Q I heat Q II A) Q I < Q II B) Q I = Q II C) Q I > Q II Lecture 3, 8
ACT 1: Soluton Consder the two systems shown to the rght. In Case I, the gas s heated at constant volume; n Case II, the gas s heated at constant ressure. Comare Q I, the amount of heat needed to rase the temerature 1ºC n system I to Q II, the amount of heat needed to rase the temerature 1ºC n system II. heat Q I heat Q II A) Q I < Q II B) Q I = Q II C) Q I > Q II Aly the Frst Law: Q = U + W by In Case I, W by = 0, because the volume does not change. In Case II, W by > 0, because the gas s exandng. Both cases have the same U, because the temerature rse s the same. more heat s requred n Case II C > C v Lecture 3, 9
Work Done by a Gas When a gas exands, t does work on ts envronment. Consder a cylnder flled wth gas. For a small dslacement dx, the work done by the gas s dw by = F dx = A dx = (Adx)= d Ths s just the area under the - curve: Examles: W by = f d dx A W by W by W by The aths dffer because T vares dfferently along the aths. (Heat s added at dfferent tmes.) The amount of work erformed whle gong from one state to another s not unque! It deends on the ath taken,.e., at what stages heat s added or removed. That s why W s called a rocess varable. Lecture 3, 10
Act 2: Work along dfferent aths f 1) Consder the two aths, a, and af connectng onts and f on the dagram. Comare the work done by the system n gong from to a (W a ) to that done by the system n gong from a to f (W af ): a A) W a > W af B) W a = W af C) W a < W af 2) Consder the two aths, 1 and 2, connectng onts and f on the dagram. Comare the work W 2, done by the system along ath 2, wth the work W 1, along ath 1. A) W 2 > W 1 B) W 2 = W 1 C) W 2 < W 1 1 2 f Lecture 3, 11
Soluton f 1) Consder the two aths, a, and af connectng onts and f on the dagram. Comare the work done by the system n gong from to a (W a ) to that done by the system n gong from a to f (W af ): a W af s the area of the trangle W a and W af cancel here. A) W a > W af B) W a = W af C) W a < W af Not only s the area under a less than the area under af, but W a s negatve, because the volume s decreasng. The net work, W a +W af, s the (ostve) area of the trangle. 2) Consder the two aths, 1 and 2, connectng onts and f on the dagram. Comare the work W 2, done by the system along ath 2, wth the work W 1, along ath 1. A) W 2 > W 1 B) W 2 = W 1 C) W 2 < W 1 1 2 f Lecture 3, 12
Soluton f 1) Consder the two aths, a, and af connectng onts and f on the dagram. Comare the work done by the system n gong from to a (W a ) to that done by the system n gong from a to f (W af ): a A) W a > W af B) W a = W af C) W a < W af Not only s the area under a less than the area under ab, but W a s negatve, because the volume s decreasng. The net work, W a +W ab, s the area of the trangle. 2) Consder the two aths, 1 and 2, connectng onts and f on the dagram. Comare the work W 2, done by the system along ath 2, wth the work W 1, along ath 1. A) W 2 > W 1 B) W 2 = W 1 C) W 2 < W 1 1 2 f The area of the semcrcle s larger than the area of the trangle. Lecture 3, 13
Constant-Pressure Heat Caacty of an Ideal Gas Add heat to an deal gas at constant ressure, allowng t to exand. We saw n the Act that more heat s requred than n the constant volume case, because some of the energy goes nto work: Q = U + W by = U + For an deal gas at constant ressure, = Nk T work W by C P Q U = = + T T T = C + Nk = α + Nk ( 1) heat Q The rato of heat caacty at constant ressure to that at constant volume wll be useful: C C ( α +1) = γ α defnton Lecture 3, 14
Work Done by an Exandng Gas (1) Suose that 10 moles of O 2 gas are allowed to exand sothermally (T = 300 K) from an ntal volume of 10 lters to a fnal volume of 30 lters. How much work does the gas do on the ston? Lecture 3, 15
Soluton Suose that 10 moles of O 2 gas are allowed to exand sothermally (T = 300 K) from an ntal volume of 10 lters to a fnal volume of 30 lters. How much work does the gas do on the ston? f f d f Wby = d = nrt = nrt ln ( ) = = 4 10 8.314 300 ln 3 2.7 10 J Note: nrt = 10 8.314 300 = 0.01 = = 6 2.49 10 Pa 24.6 atm Lecture 3, 16
Adabatc (Q = 0) Process of an α-deal Gas How are and related when Q = 0? In ths case, U = -W by. U = W by NkT α Nk T = = α T = α dt = d T T ( T ) ( ) α ln = ln + constant α α ln( T ) + ln( ) = ln ( T ) = constant α T = constant Usng = NkT, we can also wrte ths n the form: γ = constant Note that s not constant. The temerature s changng. Lecture 3, 17
Comare Adabats and Isotherms Adabat: γ = constant γ = (α+1)/α Isotherms. = constant The adabat s steeer, because γ > 1. The temerature dros f the gas exands durng an adabatc rocess, because U s decreasng. Adabatc and sothermal (quas-statc) rocesses are reversble, because there s no heat flow from hot to cold. Ths s always true, not just for deal gases. Quas-statc means slow enough that the system s always near thermal equlbrum. We ll dscuss ths more later. Lecture 3, 18
Work Done by an Exandng Gas (2) Suose, nstead, that the gas exands adabatcally from 10 to 30 lters. How much work does the gas do? Lecture 3, 19
Soluton Suose, nstead, that the gas exands adabatcally from 10 to 30 lters. How much work does the gas do? We stll have: But now: So, W by = constant = γ f d f γ cons tant Wby = cons tant d = 1 γ 1 γ f But, what s the constant? It s constant, so just use and : 6 7 / 5 constant = γ = (2.49 10 )(0.01) = 3946 SI unts Therefore, W by = 2.2 10 4 J. It s smaller than the sothermal result. (why?) Lecture 3, 20
Four Thermodynamc Processes of Partcular Interest to Us Isochorc (constant volume) Isobarc (constant ressure) 2 1 2 1 Isothermal (constant temerature) Adabatc (Q = 0) 1 2 1 steeer lne 2 These rocesses wll llustrate most of the hyscs we re nterested n. Remember the FLT U = Q - W by Lecture 3, 21
Isochorc and Isobarc Isochorc (constant volume) W Isobarc (constant ressure) W = d = by by = d = U = αnk T * = α * Q = C T by 0 Q = U = C T = αnk T * = α * ( α 1) = U + W = + * Make sure you understand these equatons! Don t just memorze! 2 1 1 2 Q Temerature changes Q Temerature and volume change Beware!!! Many of these equatons (marked wth *) rely on the deal gas law. Lecture 3, 22
Isothermal and Adabatc Isothermal (constant temerature) U = 0* Q = W * by by d* W d NkT NkT 2 = = = ln 1 * 1 2 1 Q Thermal Reservor T olume and ressure change Adabatc (solated: no heat flow) 1 1 γ Q = 0 U = W = αnk T * = α * by ( ) 2 2 1 1 2 olume, ressure and temerature change Beware!!! Many of these equatons (marked wth *) rely on the deal gas law. Lecture 3, 23
Examle: Isothermal Comresson Suose we have 3 moles of an deal olyatomc gas ntally wth a volume of 2 m 3, and a temerature of 273 K. Ths gas s comressed sothermally to 1/2 ts ntal volume. How much heat must be added to the system durng ths comresson? Lecture 3, 24
Soluton Suose we have 3 moles of an deal olyatomc gas ntally wth a volume of 2 m 3, and a temerature of 273 K. Ths gas s comressed sothermally to 1/2 ts ntal volume. How much heat must be added to the system durng ths comresson? U = 0 Q = W on f f d = + d = nrt f = nrt ln = nrt ln2 = 4.7 kj Isothermal rocess - deal gas. FLT Defnton of work then use deal gas law Integral of d/ Note that the heat added s negatve - heat actually must be removed from the system durng the comresson to kee the temerature constant. Lecture 3, 25
Examle: Escae elocty How much knetc energy must a ntrogen molecule have n order to escae from the Earth s gravty, startng at the surface? Ignore collsons wth other ar molecules. How about a helum atom? At what temeratures wll the average molecule of each knd have enough energy to escae? Lecture 3, 26
Soluton How much knetc energy must a ntrogen molecule have n order to escae from the Earth s gravty, startng at the surface? Ignore collsons wth other ar molecules. How about a helum atom? At what temeratures wll the average molecule of each knd have enough energy to escae? KE = GM E m/r E = gmr E To escae, a molecule must overcome the = 2.9 10-18 J negatve otental energy. Smlfy usng GM E m/r E2 = g = 9.8 m/s 2. Use r E = 6.4 10 6 m (4000 m), and m N2 = 4.7 10-26 kg. T = 2<KE>/3k N Equartton tells us that <KE> = 3kT/2. 2 = 1.4 10 5 K That s hot! T He = 2 10 4 K. The mass of a helum atom s smaller by a factor of 4/28. KE and T needed for escae are reduced by the same factor. T s stll too low to let much He escae, but t does get hgh enough to get onzed by the Sun s U, and then other rocesses swee t away. Lecture 3, 27
Next Week Heat caacty of solds & lquds Thermal conductvty Irreversblty Lecture 3, 28
A Quck Probablty Problem We ll send a lot of tme calculatng robabltes. Here s a quck ntroducton. Ths lecture room s aroxmately a cube 15 m on a sde. Calculate the robablty that all the ar molecules wll be found n the left half of the room. The number of gas molecules n the room s: N = / kt ~ 10 29. Each molecule s equally lkely to be found n ether half, so the robablty that they are all n the same half s (1/2) 1029 ~ 10-3 1028. It s hard to conceve how small ths number s. All the molecules wll be n the left half of the room once every 10 +3 1028 years (.e., never). You can dvde ths tme by any concevable number you want (a bllon, a trllon, a google) wthout affectng ths result sgnfcantly. Lecture 3, 29