Chapter 3 Homework Solutions

Transcription

1 Chapter Hmewrk Slutins. n = ml = 5 C = 98 K = 5 C = 98 K p = atm p = 5 atm. Cpm = (5/)R he entrpy changes fr the heating and cmpressin can e calculated separately and added. Heat at cnstant pressure S = nc ln( / ) = n(5 / ) Rln( / ) pm = (.0 ml)(5 / )(8. J / ml K)ln(98 / 98) =+ 8.0 J / K Cmpress at cnstant temperature. Because = cnstant pv = pv V/ V= p/ p= / 5 = 0.0 S = nr ln( V / V ) = ( ml)(8. J / ml K)ln(0.0) = 40. J / K S = S + S = +8.0 J/K -40. J/K = -. J/K In this particular example the negative entrpy change fr cmpressing the sample is larger than the psitive change fr heating the sample yielding a net negative S.. n = ml = 00 K = 50 K CVm = 7.5 J/ml-K Cpm = CVm + R = = 5.8 J/ml-K q = 0 [Adiaatic] dqrev S = = 0 [fr reversile Adiaatic prcess] w = U = nc ( ) = ( ml)(7.5 J / ml K)(50 K 00 K) = 45 J 4.kJ Vm H = nc ( ) = ( ml)(5.8 J / ml K)(50 K 00 K) = 570 J 5.4kJ pm. =98 K p = ar V = V ml n = 4 g = 0.50ml 8 g (a) Reversile Isthermal Expansin S = S = nr ln( V / V ) = (0.50 ml)(8. J / ml K)ln( V / V ) =+.88 J / K sys S = S =.88 J / K ecause it s reversile and Suniv = 0 surr sys

2 () Irreversile Isthermal Expansin against pex = 0 System: Cannt e calculated frm actual prcess which is irreversile. Hwever Ssys = +.88 J/K (part a) since S is a State Functin Surrundings: Usurr = 0 (since isthermal) wsurr = 0 (since pex = 0) qsurr = Usurr - wsurr = 0 herefre Ssurr = 0 Suniv = Ssys + Ssurr = = +.88 J/K > 0 (as expected fr irreversile prcess) (c) Reversile Adiaatic Expansin (dqrev)sys = (dqrev)surr = 0 herefre Ssys = 0 and Ssurr = 0.4 = 6 C = 5 K vaph = 9.4 kj/ml =.94x0 4 J/ml n = 40 g/9.4 g/ml =.0 ml qsys n vaph (.0 ml)(.94x0 J / ml) Ssys = = = = J / K 5 K 4 q qsys n vaph surr (.0 ml)(.94x0 J / ml) Ssurr = = = = = 76.4 J / K 5 K.5 = 78 C = 5 K vaph = 8.6 kj/ml p = ar = 00 kpa he prcess is a cndensatin. herefre we need: cndh = - vaph = -8.6 kj/ml = -.86x0 4 J/ml ml n = 50 g =.6 ml 46 g q = H = n H = (.6 ml)( 8.6 kj / ml) = 5.8kJ cnd ( liq gas ) gas w = p V V + pv =+ nr = (.6 ml)(8.x0 kj / ml K)(5 K) =+ 9.5kJ 9.5kJ U = q + w = = -6. kj qrev H 5.8kJ 0 J S = = = = 0.58 kj / K = 58 J / K 5K kj Nte that S is negative as expected fr a cndensatin ecause the entrpy f the liquid is lwer than the entrpy f the gas.

3 .6 m = -4 ml C = 59 K fush = 9.45 kj/ml p = ar = 00 kpa n = 50 g =.6 ml 46 g he prcess is a crystallizatin. herefre we need: crysh = - fush = kj/ml = J/ml q = H = n H = (.6 ml)( 9.45 kj / ml) = 0.8kJ cryst ( sl liq ) 0 w= pv V he vlumes f liquids and slids are very small and can e ignred. U = q + w = = -0.8 kj qrev H 0.8kJ 0 J S = = = = 0.94 kj / K = 94 J / K 59 K kj Nte that S is negative as expected fr a crystallizatin ecause the entrpy f the slid is lwer than the entrpy f the liquid..7 = 00 C = 7 K (actual iling pint) = 80 C = 5 K (nrmal iling pint) 7 K 5 K liquid gas Ssys We must calculate Ssys via a reversile path (see diagram ave). his path is: () Cl the liquid frm t (the nrmal iling pint) () Vaprize the liquid reversily at () Heat the gas ack t H ( = 5) S = S + S + S = C ( l) ln( / ) + + C ( g) ln( / ) vap pm pm 5.7x0 J / ml S = (8.7 J / ml K)ln(5 / 7) + + (5. J / ml K)ln(7 / 5) 5K = 7.64 J / ml K + 0. J / ml K +.9 J / ml K = J / ml K J / ml K

4 Ssurr We will calculate the entrpy change f the surrundings as qsurr/(7). Hwever t determine qsurr we will have t knw vaph(=7 K). We can determine that frm the same prcess (ave). [ ] [ ] [ ] [ ] H ( ) = H+ H + H = C () l + H ( ) + C ( g) vap p m vap p m H ( ) = (8.7 J / ml K) 5K 7 K + H (5 K) + (5. J / ml K) 7K 5K vap = J ml + x J ml + J ml 774 / / 70 / = + 60 J / ml q qsys vaph ( ) surr 60 J / ml Ssurr = = = = 7K = 90.6 J / ml K 90. J / ml K vap Suniv Suniv = Ssys + Ssurr = = +5. J/ml-K Suniv > 0 as expected fr the spntaneus vaprizatn f a superheated liquid..8 n =. mles = 98 K Sm (98) = 9.45 J/ml-K S (98) = nsm (98) = ( ml)(9.45 J/ml-K) = J/K S () = S (98) + S c C pm = a + + a = 9.8 =.5x0 - c = -.6x0 5 c n a + + nc pm S = d = d = na d + n d + nc d [ ln( )] [ ] = na + n nc nc = na ln( / ) + n( ) (a) = 00 C = 7 K 5 ()(.6x0 ) ( ) S = ()(9.8) ln(7 / 98) + ()(.5x0 ) 7 98 (7) (98) =. J / K S () = S (98) + S = = J/ml-K

5 () = 500 C = 77 K 5 ()(.6x0 ) ( ) S = ()(9.8) ln(77 / 98) + ()(.5x0 ) (77) (98) = 8.5 J / ml K S () = S (98) + S = = J/ml-K.9 = 600 ml 00kPa C = 87 K n = 00 g =.6ml p = ar = 00kPa 8 g ar = a = 9.6 J/ml-K and =.5x0 J/ml Cpm a CVm = Cpm R= a R= ( a R) = a' a = a - R = =. J/ml-K nr (.6 ml) (8. kpa L / ml K)(87 K) V = = = 95.4 L p 00 kpa (a) Cl at cnstant pressure = 00 C = 57 K U = nc d = n a d = na n = na ' n ln / [ ] ( ) [ ] [ ] ' ' ln( ) Vm = ml J ml K K K ml x J ml (.6 )(. / )[57 87 ] (.6 )(.5 0 / ) ln(57 / 87) = 4700 J J = 90 J.9kJ H = nc d = n a d = na n = na n [ ] ln ( / ) [ ] [ ln( )] pm = ml J ml K K K ml x J ml (.6 )(9.6 / )[57 87 ] (.6 )(.5 0 / ) ln(57 / 87) = 40 J J = 8470 J 8.5kJ q = H = -8.5 kj (ecause p = cnstant) w = U - q = (-8.5) = +6.6 kj

6 nc pm a S = d = n d na [ ln ] n = = na ln( / ) + n = (.6 ml)(9.6 J / ml K)ln(57 / 87) + (.6 ml)(.5x0 J / ml) 57 K 87 K = 4.85 J / K +.94 J / K 9.9 J / K () Cl at cnstant vlume = 00 C = 57 K U = ncvm d.9 kj See part (a) H = nc pm d 8.5kJ See part (a) w = 0 (ecause V = cnstant) q = U = -.9 kj (ecause V = cnstant) ncvm a' S = d = n d na ' [ ln ] n = = na 'ln( / ) + n = (.6 ml)(. J / ml K)ln(57 / 87) + (.6 ml)(.5x0 J / ml) 57 K 87 K = 4.66 J / K +.94 J / K 0.7 J / K

7 (c) Isthermal Cmpressin t p = 5 ar = = 600 C = 87 K U = nc d = Vm 0 H = nc d = pm 0 V p ar Isthermal: pv = pv = = = 0.40 V p 5ar w = nr ln( V / V ) = (.6 ml)(8. J / ml K)(87 K)ln(0.40) = J = 7.5kJ q = U - w = 0 - w = -7.5 kj ( ) S = nr ln V / V = (.6 ml)(8. J / ml K) ln(0.40) = 0.0 J / K.0 (a) CHCHO(g) + O(g) CHCOOH(l) S = [ S ( CH COOH )] [ S ( CH CHO) + S ( O )] r m m m = [( ml)(59.8 J / ml K)] [( ml)(50. J / ml K) + ( ml)(05. J / ml K)] = 86. J / K () Hg(l) + Cl(g) HgCl(s) S = [ S ( HgCl )] [ S ( Hg) + S ( Cl )] r m m m = [( ml)(46.0 J / ml K)] [( ml)(76.0 J / ml K) + ( ml)(. J / ml K)] = 5. J / K. (a) CHCHO(g) + O(g) CHCOOH(l) G = [ G ( CH COOH )] [ G ( CH CHO) + G ( O )] r f m f m f m = [( ml)( 89.9 kj / ml] [( ml)( 8.9 kj / ml) + ( ml)(0 kj / ml)] = 5.0kJ () Hg(l) + Cl(g) HgCl(s) G = [ G ( HgCl )] [ G ( Hg) + G ( Cl )] r f m f m f m = [( ml)( 78.6 kj / ml] [( ml)(0 kj / ml) + ( ml)(0 kj / ml)] = 78.6 kj

8 . 4 HCl(g) + O(g) Cl(g) + HO(l) at = 98 K S = [ S ( Cl ) + S ( H O] [4 S ( HCl) + S ( O )] r m m m m = [( ml)(69.9 J / ml K) + ( ml)(. J / ml K)] [(4 ml)(86.9 J / ml K) + ( ml)(05. J / ml K)] = 66.7 J / K = kj / K H = [ H ( Cl ) + H ( H O] [4 H ( HCl) + H ( O )] r f f f f = [( ml)(0 kj / ml) + ( ml)( 85.8 kj / ml)] [(4 ml)( 9. kj / ml) + ( ml)(0 kj / ml)] = 0.4 kj G = H S = 0.4 kj (98 K)( kj / K) = 9.kJ r r r. = 40 C = K (isthermal) n = 70 g/8 g/ml =.5 ml V = 5 L V = 500 ml = 0.50 L Because = cnstant pv = pv p/ p= V/ V = 5 / 0.50 = 50. G = nr ln( p / p ) = (.5 ml)(8. J / ml K)( K)ln(50) = 5440 J 5.4 kj.4 G = a + a = +560 J = 8.0x0 - J/K = 0 C = 0 K G G dg = Sd + Vdp = d + dp p G herefre: S = p Fr a reactin (prcess) this can e written as: herefre p G S = + = = = = [ a ] S 0 (8.0x0 J / K )(0 K) 48.5 J / K p p.5 p =. atm p = 00 atm m = 0 ml x 78 g/ml = 560 g V = 560 g x ml/0.88 g = 77 ml =.77 L p G = Vdp = V ( p p) [Fr slid r liquid V cnstant] p 0J = = = = L atm 4 G (.77 L)(00 atm atm) 75. L atm.77x0 J 7.7 kj

9 .6 n = 0 ml = 98 K p = atm p = 00 atm nr G Vdp dp nr ln( p / p) p p p = = = p p 5 = (0. ml)(8. J / ml K)(98 K) ln(00 /) =.8x0 J = 8kJ Nte: It is nt surprising that G fr the pressure increase in H(g) is much greater than fr C6H6(l) ecause the vlumes f gases are much greater than the vlumes f liquids (and slids)..7 G = 6.0x0 J/ml = 6.0x0 kpa-l/ml Vm = -5 ml/ml = -.5x0 L/ml At equilirium: G = G + V ( p p ) = 0 G x kpa L ml ( ) / 5 p p = = = x kpa Vm.5x0 L / ml m p = 4.0x0 5 kpa + p = 4.0x0 5 kpa + 00 kpa = 4.0x0 5 kpa x ar/00 kpa = 4000 ar.