S, S total, and S surroundings when the volume of 123 g of CO initially at 298 K

Transcription

1 hyscal Chestry II Che 40 Sprng 01 Chapter 5 ( c c 46) 5.9) Calculate S S total and S surroundngs when the volue o 13 g o CO ntally at 98 K and 1.00 bar ncreases by a actor o our n (a) an adabatc reversble expanson (b) an expanson aganst external = 0 and (c) an sotheral reversble expanson. ake C to be constant at the value 9.14 J ol 1 K 1 and assue deal gas behavor. State whether each process s spontaneous. he teperature o the surroundngs s 98 K. a) an adabatc reversble expanson S surroundngs = 0 because q = 0.S = 0 because the process s reversble. S total = S + S surroundngs = 0. he process s not spontaneous. b) an expanson aganst external = 0 and w = 0. hereore U = q = 0. V 13 g S nrln J ol K ln J K V 8.01 g ol S total = S + S surroundngs = 50.6 J K + 0 = 50.6 J K. he process s spontaneous. c) an sotheral reversble expanson = 0. hereore U = 0. V 13 g w q nr q reversble S 3 ln ol J ol K 98 K ln J V 8.01 g ol J = 50.6 J K 98 K 3 q J Ssurroundngs 50.6 J K 98 K S total = S + S surroundngs = 50.6 J K 50.6 J K equlbru. = 0. he syste and surroundngs are at

2 C 5.15) Usng the expresson ds d V d calculate the decrease n teperature that occurs one ole o water at 98 K and bar s brought to a nal pressure o 1.00 bar n a reversble adabatc process. Assue that = 0. Whle t s n the book you can go to the WEB to nd a value or the theral expanson coecent. I you do you wll nd a value ~.0.5 x 10-4 [K ]. C ds d V d S ds Cln V Because the process s reversble and adabatc S 0 Cln V ln V Cln C 998 kg kg [~ K ] 10 a 10 a 75.3 J K ol ln ~5.69 ~ 96.5 K ~ 1.5 K 75.3 J K ol ) Fro the data below derve the absolute entropy o crystallne glycne at = 300. K. You can peror the ntegraton nuercally usng ether a spread sheet progra or a curvettng routne and a graphng calculator (see Exaple roble 5.9)

3 he lne n the graph above o C (vertcal axs) aganst s the best t to the data o a polynoal o the or a + bx +cx +dx 3 gven by x x x K S (300 K)= J K ol 10.00K 1 1 We have ntegrated ro 10 K rather than 0 K because the ntegrand s not well dened at 0 K and because the contrbuton to S below 10 K s very sall. eperature (K) Heat Capacty C (J K ol )

4 ) One ole o H O(l) s supercooled to.5 C at 1 bar pressure. he equlbru reezng teperature o water at ths pressure s 0.00 C. he transoraton H O(l) H O(s) s suddenly observed to occur. By calculatng S S and S very that ths surroundngs transoraton s spontaneous at.5 C. he heat capactes are gven by C (H O(l)) = 75.3 J K 1 ol 1 and C (H O(s))= 37.7 J K 1 ol 1 and total H uson = kj ol 1 at 0.00 C. Assue that the surroundngs are at.5 C. [Hnt: Consder the two pathways at 1 bar: (a) H O(l.5 C) H O(s.5 C) and (b) H O(l.5 C) H O(l 0.00 C) H O(s 0.00 C) H O(s.5 C). Because S s a state uncton S ust be the sae or both pathways.] For pathway b) H O(l.5ºC) H O(l 0.00ºC) H O(s 0.00ºC) H O(s.5ºC) K H uson K S nc l ln n nc sln K K K K 6008 J ol 1 ol75.3 J ol K ln 1 ol K K J K.0 J K 0.31 J K 1.7 J K o calculate S surroundngs we rst calculate H = q K 1 ol 37.7 J ol K ln K

5 C 1 ol 6008 J ol 37.7 J ol K J ol K -.5 K H C H C C sold C lqud H 598 J q q 598 J Ssurroundngs 1.9 J K K surroundngs S total = 1.9 J K 1.7 J K = 0. J K > 0. he process s spontaneous. 5.3) One ole o an deal gas wth C V = 5/R undergoes the transoratons descrbed n the ollowng lst ro an ntal state descrbed by = 35 K and = 3.00 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. b. he gas undergoes an adabatc expanson aganst a constant external pressure o 1.50 bar untl the nal pressure s hal ts ntal value. c. he gas undergoes an expanson aganst a constant external pressure o zero bar untl the nal pressure s equal to hal o ts ntal value. a) he gas undergoes a reversble adabatc expanson untl the nal pressure s hal ts ntal value. q = 0 because the process s adabatc V ; ; V bar bar K 67 K J ol K U wncv 3 1 ol 67 K 35 K J J ol K H nc S = 0 because q reversble = ol 67 K 35 K J 1

6 b) he gas undergoes an adabatc expanson aganst a constant external pressure o bar untl the nal pressure s hal ts ntal value. q = 0 because the process s adabatc. ncv nrexternal nr external nr external ncv ncv R external J ol K 0.500bar Cv J ol K 35 K 3.00bar Rexternal J ol K 0.500bar Cv J ol K bar 79 K J ol K U wncv 1 ol 79 K 35 K965 J J ol K H nc S nrln nc 3 1 ol 79 K 35 K J ln 1.50 bar J ol K 79 K 1 ol J ol K ln + 1 ol ln 3.00 bar 35 K J K 4.49 J K 1.8 J K c) he gas undergoes an expanson aganst a constant external pressure o zero bar untl the nal pressure s equal to hal o ts ntal value. and w = 0. hereore U =H = 0 and q = U w = bar S nrln J ol K ln = 5.76 J K 3.00 bar 5.4) he ollowng heat capacty data have been reported or L-alanne: (K) C (J K ol ) By a graphcal treatent obtan the olar entropy o L-alanne at = 300. K. You can peror

7 the ntegraton nuercally usng ether a spread sheet progra or a curve-ttng routne and a graphng calculator (see Exaple roble 5.9) he above graph shows C as a uncton o. he lne s the best t to the data o a polynoal o the or a + bx +cx +dx 3.gven by x x x K S (300 K)= d 135. J K ol 10.K 1 1 We have ntegrated ro 10 K rather than 0 K because the ntegrand s not well dened at 0 K and because the contrbuton to S below 10 K s very sall. 5.45) For proten denaturaton the excess entropy o denaturaton s dened as trs C Sden d where 1 trs C trs s the transton excess heat capacty. he way n whch C can be extracted ro derental scannng caloretry (DSC) data s dscussed n secton 4.10 and shown n Fgure 4.7. he DSC data below are or a proten utant whch denatures between 1 = 88 K and = 318 K. Usng the equaton or Sden gven above calculate the excess entropy o denaturaton. In your calculatons trs use the dashed curve as the heat capacty base lne whch denes C as shown n Fgure 4.7. Assue the olecular weght o the proten s g ol.

8 You can peror the ntegraton nuercally usng ether a spread sheet progra or a curvettng routne and a graphng calculator (see Exaple roble 5.10) or by tracng the curve on a pece o paper and weghng the paper. Below we break up the area under the curve nto rectangles and evaluate the area whch s C. We then dvde by or each rectangle and su the contrbutons and nally ultply by the olar ass. hs estate approaches the value o the ntegral as the value approaches zero. he result s 49 J ol K. trs trs C C Sden d J K g 0.06 J K g J K g 0.15 J K g J K g 91 K 9 K 93 K 94 K 95 K 0.67 J K g J K g 96 K 97 K 98 K 99 K 300 K 1.04 J K g 1.3 J K g 1.38 J K g 1.40 J K g 1.33 J K g M 301 K 30 K 303 K 304 K 305 K 1.17 J K g J K g J K g 0.5 J K g J K g 306 K 307 K 308 K 309 K 310 K 0.51 J K g J K g J K g 311 K 31 K 313 K 67 J ol K 0.9 J K g J K g J K g 1 1

9 he graph above shows C as a uncton o. 5.46) he standard entropy o b(s) at K s J K 1 ol 1. Assue that the heat capacty o b(s) s gven by C (b s) J ol K K K he eltng pont s 37.4 C and the heat o uson under these condtons s J ol 1. Assue that the heat capacty o b(l) s gven by C (b l) J K ol K a. Calculate the standard entropy o b(l) at 650. C.

10 b. Calculate H or the transoraton b(s 5.0 C) b(l 650. C). 0 a) S b l K S b s98.15 K C H uson C d / K d / K /K /K uson J ol K /K / K / K 4770 J ol K / K d / K / K d/ K J ol K 0.40 J ol K 7.94 J ol K 13.0 J ol K J ol K sold total uson lqud b) H C d /K + H + C d /K 8918 J ol 4770 J ol 9748 J ol J ol 3