ESCI 341 Atmospheric Thermodynamics Lesson 9 Entropy

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1 ESCI 341 Atmosherc hermodynamcs Lesson 9 Entroy References: An Introducton to Atmosherc hermodynamcs, sons Physcal Chemstry (4 th edton), Levne hermodynamcs and an Introducton to hermostatstcs, Callen A REVIEW OF SAE VARIABLES he thermodynamc state of a one-comonent, closed system n equlbrum can be comletely descrbed by any two state varables (other than mass or moles). he state varables have two mortant roertes: ο he change n any of the state varables (say U) doesn t deend on the ath of the system on a thermodynamc dagram. It only deends on the endonts. b a du = U ( b) U ( a) ο he ntegral of a state varable around a closed ath s zero. du = 0 Mathematcally, ths means that dfferentals of state functons are exact dfferentals. HEA IS NO A SAE VARIABLE Imagne that an ar arcel starts out at Pont 1 on the skew- dagram shown below. he ntal ressure and temerature are 1 and 1. he arcel moves n a ersble, closed cycle around the ath shown. he three legs of the ath are: ο Leg A: Adabatc exanson to ressure. ο Leg B: Isothermal comresson back to ressure 1. ο Leg C: Isobarc heatng back to temerature 1. he ntegral of dq around the closed ath gves the total heat exchange between the arcel and ts surroundngs, and s just the sum of the heats from each leg, dq = q q q A + B + C. Note: We are very careful to wrte ths as q to ndcate that all the aths are ersble.

2 Leg A s ersbly adabatc, so q A = dq = 0 (1) Leg A ο he temerature at Pont s related to the temerature at Pont 1 va the Posson relaton Rd c = 1 1 Leg B s sothermal. he heat for ths leg s therefore. () Rd c q = R ln = R ln 1 1 B d d 1 1 Leg C s sobarc. he heat along ths ath s just. (3) qc = c ( 1 ) = c1 1 1 For the ath shown on the skew- dagram, 1 = 313K, 1 = 1000 mb, and = 700 mb. Usng these numbers n Eqs. (3) and (4) gves q A = 0 kj/kg q B = 8.95 kj/kg q C = kj/kg q = 1.5 kj/kg We ve demonstrated that the total heat around the closed, ersble ath s not zero. herefore, heat s not a state varable. R d c (4)

3 he work around the ath can be found from the frst law of thermodynamcs, du = dq + dw. Integratng around the closed ath gves du = dq + dw 0 = q + w w = q. Snce heat and work are not state varables (and are therefore not exact dfferentals), some texts gve ther dfferentals a horzontal slash symbol, dw and dq to show ths. ENROPY What f, nstead of evaluatng ( dq ) around the closed ath? For Leg A, dq was zero, so the ntegral s stll zero For Leg B (the sothermal leg) dq For Leg C (the sobarc leg) dq around the closed ath, we evaluate dq 0 = (5) Leg A αd d 1 = Rd Rd ln = =. (6) Leg B LegB LegB dq But, from Eq. (), we have c d d 1 = c c ln = =. LegC LegC LegC so = 1 1 Rd c, dq d 1 1 ln = c = Rd ln. (7) LegC Summng the results for the three legs [Eqs. (5), (6), and (7)] we have Leg A Leg B LegC R dq = dq dq dq Rd ln R ln + + =, c 1 1 d or 3

4 Eq. (8) mles that dq dfferental of a state varable. dq 0 = (8) s an exact dfferental, and therefore, must be the dq ds = (9) We call ths state varable the secfc entroy, and denote t by s. herefore, dq ds. (10) Multlyng Eq. (10) by mass, we get an extensve quantty called entroy, dq ds. (11) Entroy, S, s an extensve roerty, and has the unts of J K 1. he secfc entroy (s = S/m) s an ntensve roerty, and has unts of J K 1 kg 1. Note that entroy s only defned for equlbrum states, and that t s always defned n terms of the REVERSIBLE HEA! MAHEMAICAL EXCURSION A dfferental of the form dw = Mdx + Ndy (1) s an exact dfferental f M y N = x. (13) Sometmes an nexact dfferental can be made nto an exact dfferental by multlyng t by a sutable functon, called an ntegratng factor. Although the dfferental of heat, dq, s not an exact dfferental, t becomes an exact dfferental f we multly t by 1/. ο herefore, 1/ s an ntegratng factor for the dfferental, dq. 4

5 ENROPY AND HE FUNDAMENAL EQUAION he frst law of thermodynamcs can now be wrtten usng entroy nstead of heat, snce from Eq. (11) we have then become dq du = ds dv dh = ds + Vd = ds. he two forms of the frst law (14) or n ntensve form du = ds dα dh = ds + αd (15) ο Note that we are mlctly assumng only ersble rocesses when we substtute ds for dq. Recall from Lesson that the fundamental equaton relates nternal energy to entroy, volume, and the amounts of the consttuents, U = f ( S, V, n ). (16) akng the dfferental of Eqn. (16) for a system n materal equlbrum (n constant) we get Comarng Eqns. (14) and (17) we see that U U du = ds + dv. (17) S V V S and U = S V (18) U = V Equaton (18) s the thermodynamc defnton of temerature, whle Eqn. (19) s the thermodynamc defnton of ressure. S (19) ENROPY OF AN IDEAL GAS he frst of Eqn. set (14) can be rewrtten (for an deal gas) as 5

6 du d dv ds = + dv = Cv + nr. V If ths s ntegrated from some ntal state (wth subscrt 0) we get V S(, V ) = S Cv ln nr ln, (0) 0 V0 where S 0 = S( 0,V 0 ). Usng the second form of the frst law of thermodynamcs the entroy of an deal gas can be shown to also be S(, ) = S + 0 C ln nr ln, (1) 0 0 where S 0 = S( 0, 0 ). he change n entroy between two states (State 1 and State ) for an deal gas s S = S V S1 1 V1 = Cv + nr f the state varables are and V, or V V (, ) (, ) ln ln 1 1 S = S S 1 1 = C nr f the state varables are and. (, ) (, ) ln ln 1 1 () (3) ENROPY CHANGE IN AN ADIABAICALLY EXPANDING IDEAL GAS Consder two scenaros: ο An deal gas ersbly and adabatcally exands from volumes V to V f he change n entroy s zero, because dq = 0. ο An deal gas undergoes an adabatc free exanson from volumes V to V f No work s done n a free exanson. Snce no heat s added ether, the nternal energy does not change, so the temerature remans constant. Snce the exanson s sothermal, the change n entroy s equal to nrln(v f /V ) ο he two cases do not connect the same states on a thermodynamc dagram. hough they end u at the same fnal volume, the fnal ressures wll dffer (see dagram). 6

7 In both cases the rocess s adabatc, yet the free exanson has a non-zero entroy change. Why? ο Because the free exanson s not a ersble rocess, and entroy s defned n terms of ersble rocesses. ο You could connect onts and f wth a ersble sothermal rocess, but ths would requre dq 0. ο here s no ersble, adabatc rocess connectng onts and f, and so the entroy change s non-zero. ENROPY OF A MIXURE OF IDEAL GASES o fnd the entroy of a mxture of deal gases, start wth the frst law for a mxture of gases dq = ( Cv ) d + ( )dv. Dvdng through by, and substtutng for artal ressure from the deal gas law, we get 7

8 Integratng ths gves ds d dv ( Cv ) + R( n ) V =. = ( ) ( ) + + V S(, V ) S 0 Cv ln R n ln. (4) 0 V0 ο he entroy of the th comonent of the mxture s V S + (, V ) = S0 + Cv ln Rn ln, (5) 0 V0 where S 0 = S ( 0,V 0 ), whch s just the entroy of the th comonent f t occued the whole volume at temerature. Comarng Eqs. (4) and (5) we see that S(, V ) = S (, V ) (6) Eq. (6) s known as Gbbs heorem, whch n words states: he entroy of a mxture of deal gases s the sum of the entroes that each gas would have f t alone were to occuy the volume V at temerature. In terms of temerature and ressure, the entroy of a mxture of deal gases can be derved from the other form of the frst law of thermodynamcs, so that S(, ) = where S 0 = S ( 0, 0 ). ds = C d dq, = Vd C d R n d S ( ) 0 + C ln R n ln, (7) 0 0 ο he entroy of the th comonent of the mxture s S (, ) = S0 + C ln Rn ln. (8) 0 0 8

9 Comarng Eqs. (7) and (8) we get S(, ) = S (, ). (9) hs form s stll comletely consstent wth Gbbs heorem. In ths form, we state that he entroy of a mxture of deal gasses s the sum of the entroes that each gas would have f t were by tself at a ressure equal to ts artal ressure n the mxture. ENROPY OF MIXING Imagne several gasses searated by rgd, mermeable arttons. Each gas s at the same temerature and ressure. ο he total entroy of the gasses n ther contaners ror to mxng s gven by S(, ) = S + 0 ln. (30) ( C ) ln ( ) R n 0 0 If the arttons are removed and the gases allowed to mx, the ressure and temerature of the mxture wll not change, and the entroy of the mxture s gven by Eq. (7). he entroy of mxng, S mxng, the dfference n entroy between the mxture and the searate gasses, and s found by subtractng Eq. (7) from Eq. (30). he result s S + mxng = R n ln R n ln. (31) 0 0 Snce the artal ressure can be wrtten as n =, n we can substtute ths nto the frst exresson on the RHS of Eq. (31) to get 9

10 n S = + mxng R n ln R n ln. n 0 0 Exandng the frst logarthmc term, whch allows the ressure terms to cancel each other, we are left wth n Smxng = R n ln n. (3) hs can be wrtten n terms of mole fracton, so that the entroy of mxng s n n χ = (33) S = nr χ ln χ. Entroy of mxng (34) mxng he entroy of mxng s always a ostve value, so f the gases are mxed, the entroy ncreases. EXERCISES 1. Startng wth the frst law of thermodynamcs, show that the secfc entroy for an deal gas s s(, ) = s + 0 c ln R ln What are the unts for entroy? For secfc entroy? 3. a. 640 J of heat are suled to moles of helum (He, molecular weght 4 g/mol) at constant volume. he ntal temerature and ressure are 15 C and 1000 mb. What are the fnal temerature and ressure? b. he helum s now allowed to exand sothermally to twce ts ntal volume. What s the new ressure? c. What s the total change n entroy and secfc entroy for the helum? 4. a. 100 grams of helum (He) s mxed wth 100 grams of ntrogen (N, molecular weght 8 g/mol) at a ressure of 1000 mb and temerature of 0 C. What s the entroy of mxng? 10

11 b. he mxng rocess s adabatc, yet your answer from art a. s not zero. Why? 5. he Frst Law of hermodynamcs for an deal gas can be wrtten as dq = c d αd (35) a. Show that Eq. (35) can also be wrtten as R dq = cd d (36) b. Eq. (36) has the form of Eq. (1) where x ; y ; M = c ; N = R. Show that Eq. (36) s not an exact dfferental by showng that the condton of Eq. (13) s not met. c. Show that dq s an exact dfferental, by multlyng Eq. (36) by 1 and then showng that the condton of Eq. (13) s met. 11