Physcal Chestry I or Bochests Che34 Lecture 18 (2/23/11) Yoshtaka Ish Ch5.8-5.11 & HW6 Revew o Ch. 5 or Quz 2 Announceent Quz 2 has a slar orat wth Quz1. e s the sae. ~2 ns. Answer or HW5 wll be uploaded ths aternoon. Study t well or Quz 2 Questons ro HW6 to be covered n Quz 2: P5.2, 5.5, 5.6, 5.7, 5.8, 5.13, 5.14, 5.2, 5.22, 5.3 & Q1- Q2, where questons n red wll be studed today & green were nshed n Lecture 17. 1
w & q n varous process or deal gas ype o work w q U Expanson or P = U= C & H= C ext V P const sother -P ext V -w adabatc -P ext V -P ext V U/C V Reversble expanson/ copresson sother -nr ln(v n /V n ) -w adabatc C v C v n {(V n /V n ) a -1)} a=1-c P /C V =1- ype o work Irreversble P ext = const S or deal gas or constant w q S Isother -P ext V -w nrln(v /V ) Adabatc -P ext V nrln(v /V ) +nc v ln( / ) = + w/c V Reversble P =P ext sother -nr ln(v /V ) -w nrln(v /V ) adabatc C v n {(V n /V n ) a -1)} a=1-c P /C V =1-2
Calculated Change n Enthalpy (contnued) Case 5 (reversble) & 5 (rreversble) process: (, V ) (, V ) By calculatng S or (, V ) (, V ) (, V ) S nr [Q1 ln( ] V / V ncv, ln( / Case 6 & 6 (, P ) (, P ) By calculatng S or (, P ) (, P ) (, P ) (P/P = V /V ) S nrln( V / V ) nc P, ln( / ) [Q2 nrln( ] P / P ) nc P, ln( / ) P5.22) One ole o an deal gas wth C V, = 3/2 R s transored ro an ntal state = 6. and P = 1. bar to a nal state = 25. and P = 4.5 bar. Calculate U, H, and S or ths process. ΔU n C [Q1 ] ΔH n (C V, [Q2 ] V, (25-6) R)(25-6) S [Q3 nr ] ln( P / P ) nc ln( / ) ( P, 3
S 5.8 Absolute Entropes and he hrd Law o herodynacs he entropy o an eleent or a copound s experentally deterned ro Dq reversble =C p d Molar Entropy or Gas Sold Cp d ' H ( ) S( ) ' b C d " H " Lqud p, gas " vaporzaton, Cp, b b uson, d "' "' hrd Law o therodynacs - What s S ()? he entropy o a pure, perectly crystallne substance (eleent or copound) s zero at. C p, or O 2 P5.14) he standard entropy S at 298.15 o Pb(s) s 64.8 J 1 ol 1. Assue that the heat capacty o Pb(s) s gven by C P, Pb,s J ol 1 1 22.13.1172 1. 15 he eltng pont s 327.4 C and the heat o uson under these condtons s 477. J ol 1. Assue that the heat capacty o Pb(l) s gven by C P, Pb,l a. Calculate the standard entropy S o Pb(l) at 5 C. Sold 2 Lqud C[Q1] p [Q2] H, C[Q3] uson p, S( 2 ) S( 1 ) d ' d" ' " 1 b. Calculate H or transoraton Pb(s, 25 C) Pb(l, 5 C). Sold Lqud H ( 1 2 ) C d ' H, C, d" 1 p uson 2 2 J 1 ol 1 32.51.31 2 p 4
Dependence o S C p, / > and S vaporzaton, S uson > S ncreases as ncreases he orgn o C p, and S or Solds Nuber o degrees o reedo or olecule ade o n atos. N : ranslatonal: 3 N R : Rotatonal: Ato, Lnear 2, Non-Lnear 3 N V : Vbratonal: 3n N N R C p, & S hgher or a olecule wth ore atos. 5
5.9 Standard States n Entropy Calculaton (P12 For enthalpy, we dened H,A or the ost stable pure eleents at 298.15 ( ) and 1 bar (P ) as. We dene standard state o Entropy as S = S (P, ) What s the relatonshp between S (P) and S (1 bar)? S (1 bar P) = Rln(P/P ) S (P) = S (P ) Rln(P/P ) Molar Entropy S 5.1 Entropy Change n Checal Reacton At 298.15 and 1 bar A + 2B 2C + D S reacton = 2S C, + S D, S A, 2S B, In general SR ( ) XSX, X A() + 2B() 2C() + D() For a constant pressure Cpd ' S( ) S( 298 ) ' Q. How uch s C p or the above reacton? 298 C p 2C p, C Cp, D Cp, A 2C p, B Cp XCp, X 6
P5.2) Consder the oraton o glucose ro carbon doxde and water, that s, the reacton o the ollowng photosynthetc process: 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g). he ollowng table o noraton wll be useul n workng ths proble: Calculate the entropy and enthalpy changes or ths checal syste at (a) = 298 and (b) = 33.. Calculate also the entropy o the surroundng and the unverse at both teperatures. (a) S R = X S X (b) Cpd ' S( ) S( 298 ) ' 298 C p XC px 5.2 Heat Engnes and the Second Law o herodyacs Isother rev.: -PdV = -(nr/v)dv a b: q ab = -w ab =nr hot ln(v b /V a )> c d: q cd = -w cd = nr cold ln(v d /V c )< Adabatc: q= da & bc: q cd = q bc = Carnot Cycle For Cycle: U = w cycle + q ab +q cd = w cycle = -(q ab +q cd )< ( q ab > q cd ) Ecency = w cycle / q ab = q ab +q cd / q ab = 1 cold / hot < 1 7
P5.2 Consder the reversble Carnot cycle shown n Fgure 5.2 wth 1 ol o an deal gas wth C V = 3/2R as the workng substance. he ntal sotheral expanson occurs at the hot reservor teperature o hot = 6 C ro an ntal volue o 3.5 L (V a ) to a volue o 1. L (V b ). he syste then undergoes an adabatc expanson untl the teperature alls to cold = 15. C. he syste then undergoes an sotheral copresson and a subsequent adabatc copresson untl the ntal state descrbed by a = 6. C and V a = 3.5 L s reached. a. Calculate V c and V d. (V c /V b ) 1- =( [Q1 c / b ]) V c = V b (x c [Q2 / b ) (1/1-) ] (V a /V d ) 1- = ( a / d ) b. Calculate w or each step n the cycle and or the total cycle. w ab = -nr [Q3 a ln(v ] b /V a ) w bc = U bc = C v ([Q4 c - b )] c. Calculate and the aount o heat that s extracted ro the hot reservor to do 1. kj o work n the surroundngs. = w / q ab 5.11 Rergerator, Heat Pup and Real Engnes Carnot Cycle: w cycle < & q cycle > Reversed Carnot cycle: w cycle >&q cycle < q ab > q ab < Hot snk heated q cd < q cd > Cold snk cooled 8
Carnot Cycle Rergerator q ab > q ab < q cd < q cd > Rergeraton ecency or a reversble Carnot rergerator r = q cold /w = cold /( hot cold ) Ex. cold =.9 hot =9 (1 J o work 9 J o coolng) cold =.8 hot = 4 Rergeraton ecency or a reversble Carnot heat pup hp = q hot /w = hot /( hot cold ) Ipossble Heat Pup Calculated Change n Enthalpy (contnued) Case3 (3 ). Reversble change n or a xed V Dq ncv, d reversble S ncv, ln( / Case4 (4 ) ). Reversble change n or a xed P Dq ncp, d reversble S ncp, ln( / Case 5 (reversble) & 5 (rreversble) process: (, V ) (, V ) By calculatng S or (, V ) (, V ) (, V ) S nr ln( V / V ncv, ln( / Case 6 & 6 (, P ) (, P ) By calculatng S or (, P ) (, P ) (, P ) (P/P = V /V ) S nrln( V / V ncp ln( / nrln( P / P nc, ln( / ), P 9
When Cp or Cv s not constant P5.6) One ole o N 2 at 2.5 C and 6. bar undergoes a transoraton to the state descrbed by 145 C and 2.75 bar. Calculate S C P, J ol 1 1 3.81 11.87 13 2.3968 2 15 2 1.176 1 8 3 ΔS n p p n 3 C p, 2.75 bar d n ln [Q2] 6 6. bar nr ln [Q1] R 418.15 293.65 a b c 2 / d 3 d J 1 Equatons to be eorzed or deal gas () For constant C v, or C p, U = nc v, [1] H = nc P P, = n(c v v, + R) [2] (1) For a reversble sotheral process, U = H = S = nrln(v /V ) w = -q = -nrln(v /V ) (2) For a reversble adabatc process (or C p,, C v, const) V V 1 p p 1 Useul or H &U calc. n [1] & [2] S [Q1] S nr ln( V / V ncv, ln( / S nrln( P / P ncp, ln( / 1
P5.13) One ole o an deal gas wth C V, = 5/2 R undergoes the transoratons descrbed n the ollowng lst ro an ntal state descrbed by = 25. and P = 1. bar. Calculate q, w, U, H, and S or each process. a. he gas undergoes a reversble adabatc expanson untl the nal pressure s hal ts ntal value. (a) q rev =. ds = Dq rev / = [Q1] V V 1 1 p p 1 1 p p p p Use X a =Y b X =Y b/a 1 Obtan w = U = [Q2] nc v, ; H = nc [Q3] P, P5.3) Calculate S surroundngs and S total or the processes descrbed n parts (a) and (b) o Proble P5.13. Whch o the processes s a spontaneous process? he state o the surroundngs or each part s as ollows: a. 25.,.5 bar (a) Reversble adabatc expanson b. 3.,.5 bar (b) Adabatc expanson at constant P ext =.5 bar (a, b) q surroundngs = S surroundngs = (a) S syste = [Q1] 11
(3) For an deal gas n a reversble adabatc process or -dependent C v () or C p () (Meorze) U nc, ( d H ncp, ( ) d V ) ncc P, ( ) d S nr ln( P / P ) S nr ln( V nc / V ) V, ( ) d (4) For an deal gas n a rreversble adabatc process at a constant external pressure (Cv, Cp const) U n C V, n R pexternal w Solve ths or p p H nc P, S nrln( P / P ncp, ln( / Q. What are U, H, S or P ext =? S nr ln( V / V ncv, ln( / P5.6) One ole o N 2 at 2.5 C and 6. bar undergoes a transoraton to the state descrbed by 145 C and 2.75 bar. Calculate S C P, J ol 1 1 3.81 11.87 13 2.3968 2 15 2 1.176 1 8 3 ΔS n 3 p C p, 2.75 bar nr ln n d n ln R p 6 6. bar 418. 15 293. 65 a b c 2 d 3 d J 1 12
P5.13) One ole o an deal gas wth C V, = 5/2 R undergoes the transoratons descrbed n the ollowng lst ro an ntal state descrbed by = 25. and P = 1. bar. Calculate q, w, U, H, and S or each process. b. he gas undergoes an adabatc expanson aganst a constant external pressure o.5 bar untl the nal pressure s hal ts ntal value. U n C V, H ncp, p p n R p w external S nrln( P / P ncp, ln( / Solve ths or Q. Whch equaton should we use or S? 13