Chapter 18: The Laws of Thermodynamics

Transcription

1 Capter 18: e Laws o ermodynams Answers to Even-Numbered Coneptual uestons. (a) Yes. Heat an low nto te system at te same tme te system expands, as n an sotermal expanson o a gas. (b) Yes. Heat an low out o te system at te same tme te system s ompressed, as n an sotermal ompresson o a gas. 4. No. e eat mgt be added to a gas undergong an sotermal expanson. In ts ase, tere s no ange n te temperature. 6. Yes. In an sotermal expanson, all te eat added to te system to keep ts temperature onstant appears as work done by te system. 8. e nal temperature o an deal gas n ts stuaton s ; tat s, tere s no ange n temperature. e reason s tat as te gas expands nto te vauum, t does no work beause t as notng to pus aganst. e gas s also nsulated, so no eat an low nto or out o te system. It ollows tat te nternal energy o te gas s unanged, w means tat ts temperature s unanged as well. 10. s would be a volaton o te seond law o termodynams, w states tat eat always lows rom a g-temperature objet to a low-temperature objet. I eat were to low spontaneously between objets o equal temperature, te result would be objets at derent temperatures. ese objets ould ten be used to run a eat engne untl tey were agan at te same temperature, ater w te proess ould be repeated ndentely. 1. Yes. In at, te eat delvered to a room s typally to 4 tmes te work done by te eat pump. 14. e law o termodynams most pertnent to ts stuaton s te seond law, w states tat pysal proesses move n te dreton o nreasng dsorder. o derease te dsorder n one regon o spae requres work to be done, and a larger nrease n dsorder n anoter regon o spae. Solutons to Problems and Coneptual Exerses. Pture te Problem: A gas expands, dong 100 J o work wle ts nternal energy nreases by 00 J. Strategy: Use equaton 18- to nd te eat low nto te system. Soluton: Solve equaton 18- or : = W+Δ U = 100 J + 00 J = 00 J Insgt: e work s postve beause te system s dong work on te external world. e eat low nto te system provdes te energy to do ts work plus an addtonal 00 J or te system to store as nternal energy. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 1

2 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 4. Pture te Problem: e temperature o one mole o an deal gas nreases as eat s added. Strategy: Use te rst law o termodynams to determne te work, reognzng tat te ange n nternal energy o a monatom gas s gven by Δ U = nrδ Soluton: Solve equaton 18- or W:. W = Δ U = nrδ ( ) ( ) ( ) = 110 J 1mol 8.1 J/ mol K 76 K 7 K = 1160 J Insgt: O te 110 J o eat low nto te gas, only 50 J was onverted to nternal energy. e rest was energy or te gas to do work on te external world. 6. Pture te Problem: A yle o our proesses s sown on te pressure volume dagram at rgt. Strategy: Set te sum o te anges n nternal energy equal to zero. en solve or Δ U. CD Soluton: 1. Sum te anges n nternal energy: Δ UAB +Δ UBC +Δ UCD +Δ UDA = 0. Solve or Δ U : CD Δ UCD = ΔUAB ΔUBC Δ UDA. Insert te numer values: Δ U CD = 8 J 15 J ( 56 J) = 41 J Insgt: In a omplete yle te system returns to ts orgnal state, w means tat te nternal energy must return to ts ntal value. ereore te net ange n nternal energy must be zero. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18

3 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 8. Pture te Problem: A monatom deal gas undergoes a proess n w eat s absorbed and work s done by te gas. e proess results n a ange n temperature or te gas. Strategy: Use te ange n energy or a monatom gas U nr( ) solve or te nal temperature. Soluton: 1. (a) Replae ΔU n te rst law wt te ange n nternal energy or a monatom gas: ( ) Δ = wt te rst law o termodynams to Δ U = W nr = W. Solve or te nal temperature: ( W) = + nr 80J. Insert te numer values: ( 7J) 1mol ( ) 8.1J/ ( mol K) = + 6 K = 468 K 4. (b) Wen tere are more moleules to sare te energy, te average energy ganed per moleule s smaller, resultng n a derease n te nal temperature ound n part (a). Insgt: Beause te eat added was greater tan te work done, te net ange n nternal energy was postve. A postve ange n nternal energy results n a temperature nrease. I te nternal energy ad dereased, te temperature would ave dropped. 10. Pture te Problem: A monatom gas undergoes a proess n w work s done on te gas, resultng n an nrease n temperature. Durng te proess eat may enter or leave te gas. Strategy: Combne te ange n energy or a monatom gas U nr( ) termodynams (equaton 18-) to solve or te eat low. Δ = wt te rst law o Soluton: Solve equaton 18- or : = W +Δ U = W + nr ( ) ( ) ( ) ( ) = 560 J + 4 mol 8.1 J/ mol K 10 C = 5.9 kj Insgt: It was not neessary to onvert te temperature derene rom Celsus degrees to kelvns n ts problem beause temperature derenes are te same n bot sales. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18

4 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 1. Pture te Problem: e PV plots at rgt sow tree derent mult-step proesses, labeled A, B, and C. Strategy: e work done by a gas s equal to te area under te PV plot as long as te gas s expandng (te volume nreases). In ea o te depted proesses te gas expands so te work s postve n ea ase. Fnd te area under ea plot to determne te rankng o te work done. Soluton: 1. Fnd W A : W = PΔ V + PΔ V + PΔV A 1 1 ( )( ) = 0+ 4 kpa 1 m + 0= 4 kj. (b) Fnd W B : W ( )( ) 1 B P1 V1 P V P V ( )( ). () Fnd W C : WC = P1Δ V1+ PΔ V + PΔ V = 0 + ( 1 kpa)( m ) + 0 = kj = Δ + Δ + Δ = 1 kpa 1 m + 1+ kpa m + 0 = 5 kj 4. (d) By omparng te values o te work done n ea proess we arrve at te rankng W C < W A < W B. Insgt: Even proess B ad begun at 0 kpa o pressure and nreased lnearly to te pont (5 m, kpa) to orm a 1 kpa m = 4.5 kj and te rankng would reman te same. trangle sape, te work done would be ( )( ) 14. Pture te Problem: An deal gas s ompressed at onstant pressure to one-al o ts ntal volume. Strategy: Beause ts s a onstant pressure proess, solve equaton 18-4 or te ntal volume. Soluton: 1. Wrte equaton 18-4 n W PV V = W P V V = W V = P 1 ( ) ( ) terms o ntal and nal volumes:. Insert te numer values: ( ) 790 J V = = 0.01 m Pa Insgt: I te ntal volume were larger tan 0.01 m, more work would be needed to ompress te gas to al ts volume at te same pressure. For nstane, t would requre 100 J to ompress a gas wt an ntal volume o 0.00 m to m at a onstant pressure o 10 kpa. 16. Pture te Problem: A monatom deal gas expands to twe ts orgnal volume, wle te temperature s eld onstant. Strategy: An deal gas undergong an sotermal proess obeys te relaton PV = onstant. Use ts relaton to alulate te rato o nal to ntal pressure. Soluton: 1. Wrte te sotermal relaton n terms o ntal and nal ondtons and solve or P P: PV P = PV P = P V 1. Set V = V and solve: = = P V Insgt: Beause te pressure and volume are nversely proportonal to ea oter, nreasng one by a gven ator wll redue te oter by te same ator. For example, to derease te volume to one-trd ts orgnal volume, te pressure would ave to be nreased by a ator o tree. V V Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 4

5 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 19. Pture te Problem: e pressure volume dagram sows ve tproesses troug w a system passes n gong rom state A to tstate C. e area under te urve s dvded nto our geometr msapes to smply te area measurement. S b S t I s Strategy: e area under te lne on te pressure volume dagram s te work. Break te area nto retangles and trangles and add up ter areas to nd te work. Use te deal gas law to nd te nal temperature and use te rst law o termodynams to nd te eat. Soluton: 1. (a) Break te area nto our WAC = PΔ V = Aret + Atr,1 + Atr, + Asquare regons and sum te areas to get te work:. Fnd ea area: Aret = lw = ( 10 m m )( 00 kpa) = 1600 kj 1 1 A b ( )( ) A 1 b 1 tr, ( )( ) A w ( )( ) tr,1 = = 6 m 4 m 600 kpa 00 kpa = 400 kj = = 8 m 6 m 600 kpa 400 kpa = 00 kj square = = 8 m 6 m 400 kpa 00 kpa = 400 kj. Sum all our areas: W AC = 1600 kj kj + 00 kj kj =.6 MJ 4. (b) Use te deal gas law to nd a rato between volumes and temperatures: PV nr PV nr V nr V P =, = = = 10 m = = 0 K = 1100 K V m 5. Solve te rato or te nal temperature: ( ) 6. () Solve te rst law or te eat: =Δ U+ W = nrδ + W V 7. Use te deal gas law to elmnate nr : PV = Δ + W 8. Solve numerally: ( 00 kpa)( m ) ( ) = + = 0 K K J 5.0 MJ Insgt: e work done n an sobar proess between ponts A and C would be te area o te retangle only. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 5

6 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton Inreasng te pressure as te gas expands nreases te work done by te gas. 0. Pture te Problem: A monatom deal gas expands at onstant temperature. Strategy: Use equaton 18-5 to nd te work done n te sotermal proess. en solve te rst law o termodynams (equaton 18-) or te eat. Soluton: 1. (a) Apply equaton 18-5 dretly: W nrln V = V 4.L = 8.00 mol 8.1 J/ mol K 45 K ln =.0 kj 1.1L. Insert te gven data: W ( ) ( ). (b) Solve equaton 18-or : =Δ U+ W = 0+.0 kj=.0 kj e eat low s postve so ts s eat low nto te gas. 4. () Bot answers nrease by a ator o beause te work done s proportonal to number o moles o gas. Insgt: e nternal energy o a monatom deal gas depends only on te temperature o te gas. In an sotermal proess te temperature remans onstant, and tereore so does te nternal energy. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 6

7 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 5. Pture te Problem: A monatom deal gas expands at onstant temperature. S. I Strategy: Use equaton 18-5 and te rst law o termodynams to alulate te eat between any two volumes o te sotermal expanson. Soluton 1. Solve te rst law or eat: =Δ U + W. Set Δ U = 0 and use equaton 18-5 or W: 0 ln V = + nr V (a) Beause te gas expands ( V > V ) eat must be postve, so eat enters te system. 4. (b) Heat nput equals te work done; work done s equal to te area under te PV plot. e area rom 1.00 m to.00 m s greater tan te area rom.00 m to 4.00 m. 5. () Insert te numer values or te expanson rom 1.00 m to 6. (d) Insert te numer values or te expanson rom.00 m to ( ).00 m : ( ) 4.00 m :.00 m = 400 kpa 1.00 m ln 77 kj = 1.00 m 4.00 m = 100 kpa 4.00 m ln 115 kj =.00 m Insgt: e eat absorbed n te expanson rom 1.00 m to.00 m s equal to te eat absorbed n te expanson rom.00 m to 4.00 m, beause te eat absorbed s a unton o te relatve volume expanson. In bot o tese ases te volume doubles. 4. Pture te Problem: A monatom deal gas s termally solated rom ts surroundngs. As te gas expands ts temperature dereases. Strategy: Use te rst law o termodynams (equaton 18-) to solve or te work done by te gas. Use equaton to nd te ange n nternal energy. Soluton: 1. Solve te rst law or W: W = Δ U. (a) Beause te proess s adabat, = 0.. (b) Substtute Δ U = nrδ (equaton 17-16): W = 0 nrδ 4. Insert te gven values: W = (.9 mol)( 8.1 J mol K)( 05 C 485 C) = 1.7 kj Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 7

8 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 5. () Solve te rst law or te ange n nternal energy: Δ U = W = 1.7 kj Insgt: Wen a monatom deal gas expands adabatally t does work on te external world. e energy to do ts work annot ome rom te absorpton o eat, so te nternal energy, and tereore te temperature, must derease. 6. Pture te Problem: A monatom deal gas undergoes a proess n w ts volume nreases wle te pressure remans onstant. Strategy: Use te area under te PV plot to nd te work done by te gas. Calulate te ntal and nal temperatures rom te deal gas law. Use te ange n temperature to alulate te ange n nternal energy. Fnally, alulate te eat added to te system usng te rst law o termodynams. Soluton: 1. (a) Multply P by V :. (b) Solve te deal gas law or : Δ W P V ( ) = Δ = = PV = nr Pa ( 0.75 m ). Insert ntal ondtons: 5 10 kpa 1.9 m 0.75 m.4 10 J = = 49 mol 8.1 J/ ( mol K) Pa ( 1.9 m ) 4. Insert te nal ondtons: 5. () Use equaton to solve or U :.9 10 K = = 49 mol 8.1 J/ ( mol K) K Δ Δ U = nrδ = ( 49 mol) 8.1 J/ ( mol K) ( 980 K 90 K) = J 5 6. (d) Solve te rst law or eat: =Δ U + W = 60 kj + 40 kj = J Insgt: e work done by te gas s proportonal to te pressure. e temperature, and tus te ange n nternal energy, s also proportonal to te pressure. ereore te eat absorbed must also be proportonal to te pressure. I te ntal pressure was 40 kpa (double tat gven n te problem) ten te work done would be 480 kj, te ange n nternal energy 70 kj, and te eat 100 J. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 8

9 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 8. Pture te Problem: A gas wt ntal volume V eter expands at onstant pressure to twe ts ntal volume, or ontrats to 1/ ts orgnal volume. Strategy: Use te area under te PV plot to nd te work done by te gas. Soluton: 1. (a) Set te work equal to te pressure tmes volume: ( ) ( ) W = P V V = P V V W = PV. Insert te numer values: ( ). (b) Set te work equal to te W = 140 kpa 0.66 m = 9 kj V W P V V P = = V PV = pressure tmes volume: ( ) W = 140 kpa 0.66 m = 6 kj 4. Insert te numer values: ( )( ) Insgt: e work s postve wen te volume nreases and negatve wen te volume dereases. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 9

10 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 0. Pture te Problem: A monatom deal gas s adabatally ompressed. Strategy: Use te deal gas law to relate te ntal and nal ondtons o te gas. en use equaton 18-9 to elmnate te unknown volumes rom te equaton. Soluton: 1 (a) Dong work on te system must nrease te nternal energy and tereore te temperature wen no eat lows nto or out o te system.. (b) Wrte te deal gas law n terms o te ntal and nal states: PV PV PV = nr = nr = P. Solve or te nal temperature: V = P V 4. Now solve equaton 18-9 or te rato o volumes: 5. Insert ts rato nto te equaton or PV γ γ 1 1 γ γ V P P = P V = = V P P : ( ) 5, were γ = γ γ ( 1 5) P P P 140 kpa = = = 80 K = 10 K P P P 110 kpa Insgt: As predted, wen work was done on te gas n an adabat proess te temperature nreased. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

11 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton. Pture te Problem: A gas expands to double ts ntal volume n tree derent proesses: onstant pressure, sotermal, and adabat. Strategy: Use te P V dagram to rate te work beause te work s te area under te urve. Determne te rankng o nal temperatures rom te deal gas law. Soluton: 1. (a) e area under te onstant-pressure urve s te greatest, ene ts proess does te most work.. (b) e area under te adabat urve s te smallest, ene ts proess does te least work.. Solve te deal gas law or temperature and note tat te nal temperature s proportonal to te nal pressure: PV nr V nr 0 = = P 4. () e onstant-pressure expanson as te gest nal temperature beause t as te gest nal pressure. 5. (d) e adabat expanson as te lowest nal temperature, beause at nal volume, t as te lowest pressure. Insgt: In te sotermal proess te temperature remaned onstant. From te grap we see tat te nal temperature n te onstant pressure s ger tan te temperature n te sotermal proess, tereore te temperature nreased durng te onstant pressure proess. In te adabat proess te temperature dereased. 4. Pture te Problem: Heat s added to monatom deal gas to nrease ts temperature. e eat s added eter at onstant volume or onstant pressure. Strategy: Use te molar spe eat at onstant volume 18-7) to solve or eat. Soluton: 1. (a) Wrte n terms o. (b) Wrte n terms o V C (equaton 18-6) and at onstant pressure V C P : 5 nr 5( ) p ( ) ( ) C P (equaton = Δ =.5 mol 8.1 J/ mol K K = 1.7 kj C : nr ( ) ( ) ( ) = Δ =.5 mol 8.1 J/ mol K K = 1.0 kj v Insgt: More eat s requred to nrease te temperature at onstant pressure beause part o te nput eat s onverted to te work o te expandng gas. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

12 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 6. Pture te Problem: Heat s added to a monatom deal gas untl te nternal energy as doubled. s proess s rst done at onstant pressure and ten at onstant volume. Strategy: Use equaton to determne te relatonsp between ntal and nal temperatures wen te nternal energy doubles. en use te molar eat at onstant volume (equaton 18-6) and at onstant pressure (equaton 18-7) to alulate te eat added. Soluton: 1. (a) Dvde equaton by temperature and equate ntal and nal terms: U U = = nr U. Set U = U and solve or : U = = = U U. Wrte te eat n terms o molar 5 eat apaty at onstant pressure: P = nr ( ) 4. Wrte nal temperature n terms o ntal temperature: ( ) = nr = nr 5 5 P 5. Solve numerally: 5 ( ) ( ) ( ) 6. (b) Wrte te eat n terms o molar eat apaty at onstant volume: = P.5 mol 8.1 J/ mol K 5 K = 17 kj = nrδ = nr V 7. Solve numerally: ( ) ( ) ( ) = V.5 mol 8.1 J/ mol K 5 K = 10 kj Insgt: At 5 K te nternal energy o te gas was 10 kj. Addng 10kJ at onstant volume s suent to double te nternal energy, but t s not suent at onstant pressure. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 1

13 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 8. Pture te Problem: A monatom deal gas s expanded at onstant pressure by a xed ange n volume. s proess s done twe wt derng ntal volumes. Strategy: Use te area under te PV plot to alulate te work done n ea proess. Soluton: 1. (a) Wrte te work as area under te PV plot: W = PΔ V m = Pa 8600 m 5400 m = 0.51 kj m. Solve numerally: W ( )( ). (b) Work s dretly proportonal to te ange n volume. ereore, te work done by te gas n te seond expanson s equal to tat done n te rst expanson m = Pa 5400 m 00 m = 0.51 kj m 4. () Solve numerally: W ( )( ) Insgt: Beause te work s proportonal to te ange n volume, nreasng te volume by 00 m rom any ntal volume wll produe te same amount o work. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 1

14 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 40. Pture te Problem: A termally solated monatom deal gas s ompressed ausng ts pressure and temperature to nrease. Strategy: Use equaton 18-9 to alulate te nal volume n terms o te ntal and nal pressures. en ombne equaton 18-9 and te deal gas law to derve an equaton or te nal volume n terms o te temperatures. Soluton: 1. (a) Solve equa- γ γ γ ton 18-9 or te nal volume: PV PV V V ( ). (b) Combne te Ideal Gas Law wt equaton 18-9 and solve or nal volume: 1 P 105 kpa = = = m = m P 145 kpa PV = PV γ γ nr nr V = V γ V V γ 1 1/ ( ) γ 95 K ( ) = = m = m 17 K V V /5 Insgt: Wen a monatom deal gas s ompressed adabatally, ts volume dereases wle ts temperature and pressure nrease. Beause te temperature n part (b) s less tan te ntal temperature, te gas ad to expand as t ooled. s s seen n te nal volume beng greater tat te ntal volume. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

15 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 4. Pture te Problem: A monatom deal gas expands at onstant pressure and ten s eated at onstant volume. Strategy: Calulate te tree temperatures (, and ) usng te Ideal Gas Law. Use tese temperatures togeter wt C and C V P to alulate te eat transerred durng ea proess. Calulate te work durng te onstant pressure proess rom te area under te PV plot. Fnally, use te rst law o termodynams to nd te ange n nternal energy. Soluton: 1. (a) Solve te Ideal Gas Law or :. Solve te Ideal Gas Law or. Solve te Ideal Gas Law or ( ) Pa 1.00 m PV = = = nr ( ) 60.0 mol 8.1 J/ mol K PV Pa (.00 m ) : 1.6 K = = = 67.8 K nr 60.0 mol 8.1 J/ ( mol K) PV 1 10 Pa (.00 m ) : = = = K nr 60.0 mol 8.1 J/ ( mol K) 5 4. Fnd te eat or proess : p = nr( ) 5 ( ) ( ) ( ) = 60.0 mol 8.1 J/ mol K 45. K = 50 kj 5. Fnd te eat or proess 4: v = nr( ) ( ) ( ) ( ) 6. (b) Fnd te work rom te area under te PV plot: = 60.0 mol 8.1 J/ mol K 67.8 K = 477 kj ( )( ) W = PΔ V = Pa.00 m 1.00 m = 1 kj 7. () Use te rst law to determne te total ange n nternal energy: Δ U = W = 50 kj kj 1 kj = 795 kj Insgt: e ange n nternal energy s ndependent o te proesses between te ntal and nal states. I te ntal proess ad been at onstant volume and te nal at onstant pressure, te net work and eat absorbed would be greater tan n ts problem, but te ange n nternal energy would stll be te same. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

16 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 44. Pture te Problem: A Carnot engne an be operated wt one o our sets o reservor temperatures Strategy: Use te expresson or te eeny o a Carnot eat engne (equaton 18-1) to determne te rankng o te eenes wen te engne s operated between te stpulated temperature reservors. Soluton: 1. Fnd e A :. Fnd e C : e e 400 K, A A = 1 = 1 = Fnd e B :, A 800 K 800 K, C C = 1 = 1 = Fnd e D :, C 100 K 5. By omparng te eenes we arrve at te rankng e D < e B = e C < e A. e e B D 400 K = = = 600 K, B , B 800 K = = = 1000 K, D Insgt: e larger te temperature derene between te two reservors, te greater te eeny o a Carnot eat engne. A Carnot engne operatng between 00 K and 100 K would ave an eeny o 75%., D 46. Pture te Problem: A eat engne does work as t extrats eat rom te ot reservor and rejets eat to te old reservor. Strategy: Use equaton to alulate W and ten equaton to alulate te eeny. Soluton: 1. (a) Calulate work W = = 690 J 40 J = 60 J rom equaton 18-10:. (b) Calulate e rom equaton 18-11: W 60 J e = = = 690 J 0.8 Insgt: Dereasng wle keepng onstant wll always nrease te work done and nrease te eeny. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

17 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 5. Pture te Problem: A eat engne does work at a rate o 50 MW as t extrats eat rom wte ot reservor at a rate o 80 MW and rejets eat to te old reservor. S S a 48. I s t Strategy: Solve equaton or and dvde by tme to determne te rate tat eat s dsarded rom te eat engne. en nd te eeny rom equaton Soluton: 1. (a) Solve eq or and dvde by tme: W W t t t = =. Insert te numer values: = 88 MW 5 MW = 585 MW t. (b) Wrte equaton n terms o rates: W W t 5 MW e = = = = 0.0 t 88 MW Insgt: ypally n power plants t s te rate o eat low and power output tat s mportant. As sown n ts problem, te same equatons used or eat and work an be used to determne eat low rates and power output. 50. Pture te Problem: A eat engne produes 700 J o work as t extrats eat rom a ot temperature reservor and rejets eat to a old temperature reservor. Strategy: Use equaton to alulate and ten equaton to alulate : Soluton: 1. (a) Solve equaton or : W = e 700 J. Insert te gven values: = = 15 kj (b): Solve equaton or : 4 = W = J 700 J = 1 kj 4. () Hger eeny means less eat nput s needed to produe te same work. Consequently, less eat s lost to te surroundngs. e answers n parts (a) and (b) wll derease. Insgt: As an example o ger eeny, e = 0.4 and W = 700 J, ten = 10.0 kj and = 7.6 kj. 5. Pture te Problem: A Carnot engne produes 00 J o work as t extrats 500 J o eat rom a t temperature reservor and rejets eat to a old temperature reservor. Strategy: Use equaton to determne te eeny. Solve equaton or te eat exausted at te low temperature. Fnally, solve equaton 18-1 or te rato o reservor temperatures. Soluton: 1. (a) Calulate te eeny rom equaton 18-11: W 00 J e = = = J Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

18 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton. (b) Solve equaton or : = W = 500 J 00 J = 00 J. () Solve equaton 18-1 or temperature rato: e max = 1 1 = 1 e = = 1 8. max Insgt: A eat engne wt 88% eeny would need to ave te ot temperature reservor at least 8. tmes otter tan te old temperature reservor. For nstane, te old reservor s at room temperature (00 K) te ot temperature reservor would ave to be at least 500 K, plenty ot enoug to melt steel! 54. Pture te Problem: A Carnot engne produes work n a yle were t exausts / o te nput eat to te old reservor ( = ). Strategy: Calulate te eeny rom equaton 18-1 and te at tat = en use equaton 18-1 to determne te reservor temperature rato. Soluton: 1. (a) Wrte equaton 18-1:. (b) Solve equaton 18-1 or te temperature rato: e = 1 1 = emax = = = Insgt: In a Carnot engne, te rato o eat exausted to eat nput wll always be equal to te rato o te temperature o old reservor to te temperature o te ot reservor, as n ts ase = =. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

19 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 56. Pture te Problem: A rergerator extrats eat rom te old reservor and ejets te eat to te ot reservor. s proess requres work to be nput nto te system. Strategy: Solve equaton or te eat exausted to te ot reservor. en use equaton to solve or te oeent o perormane o te rergerator. Soluton: 1. (a) Solve equaton or : = + W. Insert te gven values: = 110 J J = 0.59 kj. (b) Insert te values nto equaton 18-15: 110 J 480 J COP = = = 0. Insgt: s rater neent rergerator exausts to te room over ve tmes te amount o eat extrated rom nsde te rergerator. e addtonal eat omes rom te work done by te rergerator. W Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser

20 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 58. Pture te Problem: A eat pump njets 40 J o eat nto a warm room by pullng eat rom te old outsde. s proess requres 45 J o work. Strategy: Use equaton to alulate te eat removed rom te outsde ar. en use te Carnot relaton, =, to determne te outsde temperature. Soluton: 1. (a) Solve equaton or : = W = 40 J 45 J =.90 kj. Solve te Carnot relaton 895 J = = K = 6.8 K 40 J or old temperature: [ ]. Convert to Celsus: = 6.8 K 7.15 K = 10 C Insgt: Dereasng te outsde temperature dereases te oeent o perormane, tereby nreasng te work neessary to produe te same eat output. Wen te external temperature drops to 0 C, te same 40 J o eat output requres 455 J o work. 60. Pture te Problem: A reversble rergerator an operate as a rergerator or as a eat engne. Strategy: Combne equatons 18-10, 18-11, and to solve or eeny n terms o te oeent o perormane. Soluton: 1. Combne equatons and 18-11, ten dvde top and bottom by W:. Replae W usng equaton 18-15: W W 1 e = = = W + 1+ W 1 1 e = = = 1+ COP Insgt: Wrtng te eeny as e = sows tat te larger te oeent o perormane, te smaller te 1 + COP eeny. It also sows tat wle te COP an range rom zero to nnty, te eeny ranges rom zero to one. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 0

21 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 9. Pture te Problem: As eat lows out o a ouse, te entropy o te ouse dereases and te entropy o te outsde nreases. Strategy: Use equaton to sum te entropy anges or nsde and outsde te ouse. e resultng equaton an be dvded by tme n order to determne te rate o entropy ange. Soluton: Sum te nsde and outsde anges n entropy, and dvde ea term by te tme: Insgt: Altoug te ange n entropy o te nsde was negatve, te net ange n entropy s postve. s wll be true or all pysal stuatons. 6. Pture te Problem: A reversble engne an operate n reverse as a eat pump. Strategy: Combne equatons and to solve or eeny n terms o te oeent o perormane. Soluton: Combne equatons and 18-17: ΔS t t = + t nsde outsde 0.0 kw 0.0 kw = + = K K ( ) 1 1 COP = 4. W = e = 0. = 9.6 W K Insgt: e oeent o perormane o te eat pump s te nverse o te reversble engne s eeny. Beause te eeny ranges rom zero to one, te oeent o perormane an range rom one to nnty. 64. Pture te Problem: An deal gas s expanded slowly wle te temperature s eld onstant. Strategy: Use te denton o entropy ange to answer te oneptual queston. Soluton: 1. (a) e denton o entropy ange Δ S = (equaton 18-18) ndates tat wenever eat s transerred tere s a orrespondng entropy ange. In ts ase eat s added to te gas so tat t mgt mantan a onstant nternal energy wle t expands and does work on te external world. We onlude tat te entropy o te gas wll nrease as t s expanded slowly and sotermally.. (b) e best explanaton s I. Heat must be added to te gas to mantan a onstant temperature, and ts nreases te entropy o te gas. Statement II mstakenly supposes entropy s a unton o temperature only, and statement III s partally true (te temperature wll derease) but gnores te nrease n dsorder and entropy tat results rom te expanson. Insgt: In ts ase te entropy ange o te unverse s zero beause te proess s reversble. Any entropy ganed by te gas s lost by te reservor wen te eat s extrated. 66. Pture te Problem: Heat s added to water at ts bolng pont, onvertng te water entrely to steam. Strategy: Use te latent eat o vaporzaton o water to alulate te eat added to te water. en alulate te ange n entropy rom equaton Soluton: Combne equatons 17-0 and 18-18: Insgt: As eat s added to te water, te entropy o te water nreases. ml 5 ( ) 1.85 kg.6 10 J/kg v 4 Δ S = = = = J/K = 11. kj/k K 68. Pture te Problem: You eat a pan o water on te stove. Strategy: Use = mδ (equaton 16-1) to determne te amount o eat added to te water or ea temperature ange, and ten use Δ S = (equaton 18-18) to determne te entropy ange, were av av s te average temperature o te water durng te eatng proess. Soluton: Combnng equatons 16-1 and we see tat te entropy ange, Δ S = m Δ, nreases lnearly av wt te temperature ange Δ and s nversely proportonal to te average teperature, av. Notng tat proesses C and D nvolve only a 5 C temperature ange, we arrve at te ollowng rankng: D < C < B < A. Insgt: A 10 C temperature nrease near 0 C temperature (proess A) produes a larger entropy nrease tan a Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 1

22 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 10 C nrease near 40 C (proess B) beause te average temperature av s smaller or proess A. Usng av s only an Δ S = mln. A omparson o te two approxmaton; te exat soluton requres alulus and reveals tat ( ) approaes reveals tat tey der by less tan 0.01% over te temperature ntervals gven n ts queston. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18

23 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 70. Pture te Problem: Wle desendng at onstant speed, te parautst s gravtatonal potental energy s onverted to eat, resultng n an nrease n entropy. Strategy: Use equaton to alulate te ange n entropy wt te eat gven by te ange n gravtatonal potental energy. Soluton: 1. Set te eat n equaton equal to te ange n gravtatonal potental energy:. Solve numerally or te ange n entropy: ΔU mg Δ S = = = ( )( ) 88 kg 9.81 m/s 80 m Δ S = = 1100 J/K = 0.11 kj/k K Insgt: In te absene o ar resstane, te parautst s gravtatonal potental energy would ave been onverted to knet energy, not eat, wt no nrease n entropy. 7. Pture te Problem: A eat engne extrats 6400 J rom a ot reservor and perorms 00 J o work. e remander o te eat s exausted to te old reservor. Strategy: Calulate te net ange n entropy by summng te anges n entropy at te ot and old reservors. Use equaton to wrte te entropy ange n terms o te eat transers and temperature. Fnally, use equaton to nd te eat exange at te old reservor. Soluton: 1. Sum te entropy anges: Δ S = +. Wrte n terms rom te ot and old reservors: Δ S = +. Use equaton to elmnate : W Δ S = + 4. Substtute numeral values: 6400 J 6400 J 00 J Δ S = + =.6 J K 610 K 0 K Insgt: s engne operates at less tan maxmum eeny. e Carnot eeny o ts engne s and te atual eeny s I te engne were runnng at maxmum eeny te net ange n entropy would be zero. 74. Pture te Problem: A system s taken troug te tree-proess yle sown n te PV plot sown at rgt. Strategy: Wenever a system expands (volume s nreasng) at nonzero pressure t does postve work on te external world. Lkewse, te system s ompressed, work s done on te system and te work as a negatve value. Use tese prnples to determne te orret sgn o te work or ea o te ndated proesses. Soluton: 1. (a) In te proess A B te volume dereases, work s done on te system, and te work s negatve.. (b) In te proess B C te volume nreases, work s done by te system, and te work s postve.. () In te proess C A te volume dereases, work s done on te system, and te work s negatve. Insgt: e two negatve works our at ger pressure tan te postve work, so te net work done by te entre yle s negatve. Anoter way to remember ts s to note tat or omplete yles depted on a PV plot, ounterlokwse yles orrespond to negatve work and lokwse yles orrespond to postve work. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18

24 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 4

25 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 76. Pture te Problem: Note te tree proesses on te P V plot. Strategy: Equate te work done n te proess wt te area under te PV plot. Soluton: 1. (a) Fnd area o te retangle under proess A:. (b) Sum te areas o te retangle and trangle under proess B:. () Fnd te area o te retangle W = = (4.0 kpa)(1.0 m ) 4.0 kj 1 ( ) (.0 kpa)(.0 m ) W = 1.0 kpa.0 m W = kj W = 1.0 kpa.0 m =.0 kj under proess C: ( ) Insgt: In ea o te proesses sown on te P V plot, te vertal (onstant volume) segments dd not ontrbute to te work done. 78. Pture te Problem: A eat engne does work wle extratng eat rom a ot reservor and exaustng eat to a old reservor. Strategy: Use equaton to determne te eeny. Soluton: Apply equaton dretly: Insgt: Eeny s a rato o te work done to te nput eat. W W 1 e = 0.50 = 4W = 4 = 80. Pture te Problem: A monatom deal gas undergoes our proesses represented by te aompanyng P V plot. Strategy: A gas does work wen t expands. Examne te P V dagram or ea proess to determne weter te gas s expandng (postve work), ontratng (negatve work), or at onstant volume (no work). Use te rst law o termodynams to alulate te work done n proess AB. Use equaton and te Ideal Gas Law to relate te ange n nternal energy to te work. Soluton: 1. (a) e volume s expandng so te work s postve.. (b) e volume s onstant so te work s zero.. () e volume s ontratng so te work s negatve. 4. (d) e volume s ontratng so te work s negatve. 5. Solve te rst law or work: W = Δ U 6. Use equaton to elmnate ΔU and use te Ideal Gas Law to set nrδ = PΔ V: 7. Wrte set pressure tmes ange n volume equal to te work and gater lke terms: W = nrδ = PΔ V W = W W = 5 Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 5

26 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 8. Solve or te work: W ( ) = = 7 J = 11 J 5 5 Insgt: For a monatom deal gas undergong a onstant pressure proess, te work, eat, and ange n nternal energy are all proportonal to ea oter. ereore gven one o te tree quanttes t s possble to solve or te oter two. In ts problem te ange n nternal energy s 16 J. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 6

27 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton Insgt: A small raton o tat avalable work s aptured by te Eart ea day and used to grow plants, warm te Eart, et., and n some way s te soure o almost all o te energy used on Eart. 8. Pture te Problem: wo engnes absorb te same amount o eat rom a ot reservor. Engne B exausts twe as mu eat to te old reservor as does Engne A. Strategy: Use equaton 18-1 to determne a relatonsp between exausted eat and te eeny or ea engne. Set te exausted eat o Engne B equal to twe te exausted eat o Engne A and ombne te equatons to solve or te eeny o Engne B. Soluton: 1. (a) Engne B does less work tan engne A beause t exausts more eat to te low temperature reservor. ereore, Engne A as te greater eeny.. (b) Wrte equaton 18-1 or Engne A and solve or te eat rato:. Wrte equaton 18-1 or Engne B and set B, = A, : 4. Wrte te eat rato n terms o te eeny o Engne A: e = A, A, A 1 1 ea = e B, B = 1 = 1 A, ( ) ( ) e = 1 1 e = e 1= = 0. B A A Insgt: As expeted, Engne B as a lower eeny beause t exausts more eat to te old reservor. 84. Pture te Problem: As eat s added to a monatom deal gas (argon) t expands at onstant pressure. Strategy: Use te molar spe eat at onstant pressure (equaton 18-7) and te nternal energy o a monatom deal gas (equaton 17-15) to wrte te ange n nternal energy as a unton o te eat and to alulate te temperature derene. Use te rst law o termodynams to solve or te work n terms o te ange n nternal energy and eat. en alulate te ange n volume rom te work at onstant pressure. Soluton: 1. (a) Solve te molar eat at onstant pressure or nrδ :. Elmnate nrδ rom equaton = nrδ nrδ = p 5 5 Δ U = nrδ = = 5 5 or te ange n nternal energy: ( p). Solve numerally: U ( ) 4. (b) Solve te equaton or molar eat at onstant pressure or te temperature ange: 5. () Solve te rst law or work, set W = PΔ V and solve or Δ V : Δ = 1800 J = 1080 J = 0.11 kj 5 ( ) ( )( ) p 1800 J Δ = = = 4 K 5nR 5.6 mol 8.1 J mol K W = Δ U = PΔV ΔU 1800 J 1080 J Δ V = = = m P Pa p p Insgt: Wen a monatom deal gas expands at onstant pressure te eat, work, and nternal energy are all proportonal to ea oter. I te amount o eat were to be doubled to 600 J, ea o te oter terms n ts problem would also be doubled: Δ U = 160 J, Δ = 48 K, and Δ V = 1 10 m. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 7

28 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 86. Pture te Problem: e table lsts te eat, work, and ange n nternal energy o a monatom deal gas or ea o te proesses overed n te apter. Strategy: Use te rst law o termodynams to relate te eat, work, and ange n nternal energy. Use equaton 18-5 or te work n an sotermal proess. Soluton: 1. (a) Solve te rst law or te work: 5 W = Δ U = nrδ nrδ = nrδ. (b) Adabat means no eat transer: = 0. () Solve te rst law or U : Δ 0 ( ) 4. (d) Work requres a ange n volume: W = 0 5. (e) Solve te rst law or U : 6. () Solve te rst law or eat usng equaton 18-5 or work: 7. (g) Cange n nternal energy requres a ange n temperature: Δ U = W = nrδ = nrδ Δ U = W = nrδ 0 = nrδ Δ ( ) ( ) =Δ U + W = 0+ nrln V / V = nrln V / V Δ U = 0 Insgt: s table provdes a quk reerene or ea o te proesses overed n ts apter. It would be a good dea to memorze te table, or keep a opy n your notes. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 8

29 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 88. Pture te Problem: A eat engne operates between reservors at 1010 K and 0 K. e nventor lams tat te engne an do 110 J by extratng 150 J rom te ot reservor. Strategy: Use equaton to alulate te lamed eeny. en use equaton 18-1 to alulate te maxmum possble eeny. Soluton: 1. (a) Solve equaton or te eeny:. (b) Solve equaton 18-1 or te maxmum eeny: W 110 J e = = = 150 J e max K = = = 1010 K Insgt: Do not nvest n ts mane. e lamed eeny exeeds te maxmum possble eeny. 90. Pture te Problem: e water n a ds ools and reezes sold. Strategy: Use equaton to alulate te ange n entropy, were te eat leavng te water s te latent eat o uson. We onentrate only on te entropy ange durng te reezng proess at 0 C. Soluton: 1. (a) e entropy o te water dereases beause eat lows out o te water. You an also vsualze te entropy ange by realzng tat te water goes rom a dsordered lqud state to an ordered rystallne state.. (b) Wrte equaton n terms o : 4 ( ) L ml 0.5 kg.5 10 J/kg Δ S = = = =. () e entropy wll nrease n te objet tat absorbed te eat gven o by te e. ( K) 0.65 kj K Insgt: I te surroundng ar (at 5 C) absorbs te eat, ts entropy wll nrease by Δ S = 0.66 kj K, produng a postve net ange n te entropy o te unverse. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 9

30 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 9. Pture te Problem: A Carnot engne operates between two temperature reservors, wt an eeny determned by te rato o te reservor temperatures. Strategy: Beause e = 1, we need to make smaller to nrease te eeny. We an determne w s smaller, +Δ or Δ, by ekng te sgn o ter derene. Soluton: 1. Subtrat te two ratos, nd te ommon denomnator and smply: Δ ( )( ) ( ) = = ( ) ( ) ( ) Δ +Δ Δ + Δ +Δ +Δ +Δ. Beause Δ > 0, and ( ) > 0, te derene s greater tan zero. ereore, lowerng te temperature o te low temperature reservor would result n te greater ange n eeny. Insgt: o sow ts numerally, we an let = 500 K, = 00 K, and Δ = 0 K. e ntal eeny s Inreasng te ot reservor temperature reates an eeny o 0.4, wle dereasng te old temperature reates an eeny o Clearly, lowerng te old reservor temperature as te greatest eet on te eeny. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 0

31 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 94. Pture te Problem: A monatom deal gas under goes a onstant pressure proess. Durng te proess eat enters te system and te system does work. Strategy: Calulate te rato o work over eat. Use te area under te PV plot or te work and te molar spe eat to alulate te eat. Soluton: 1. Set W equal to te area under te PV plot, and use te Ideal Gas Law to wrte n terms o Δ : W = PΔ V = nrδ 5. Use equaton 18-7 nd te eat: = nrδ W nrδ. Calulate te rato o work to eat: = = 5 p nrδ 5 Insgt: wo-ts o te eat absorbed by a monatom deal gas durng a onstant pressure proess s onverted to work as te gas expands. e remanng /5 o te eat beomes nternal energy. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 1

32 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18

33 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 96. Pture te Problem: A Carnot rergerator and a Carnot engne operate between te same two temperature reservors. Strategy: Solve equaton 18-1 to obtan te rato o reservor temperatures n terms o te eeny. en use equaton and equaton to wrte te oeent o perormane n terms o te rato o eats. Fnally employ te Carnot relaton = to wrte te oeent o perormane n terms o te temperature ratons and ten n terms o te eeny. Soluton: 1. Solve equaton 18-1 or te rato o reservor temperatures: e = 1 = 1. Combne equatons and 18-10: COP = = W. Dvde top and bottom by and substtute n te Carnot relaton: 4. Wrte te temperature rato as (1-e) and smply: COP = = 1 1 e 1 e 1 e COP = = 1 1 e ( e) Insgt: e eeny an range rom zero to one. s range o eenes gves a range to te oeent o perormane o zero to nnty. e eeny s a mnmum wen te COP s a maxmum. 98. Pture te Problem: Durng te operaton o an OEC eat engne, 1500 kg o water at C s ooled to 4.0 C. Strategy: Use = mδ (equaton 16-1) to determne te eat released by te warm water. = mδ = 1500 kg 4186 J/kg/C 4.0 C = J Soluton: Apply equaton 16-1: ( )( )( ) 8 Insgt: I ts mu energy were transerred every seond, and 6.1% o te energy low were onverted nto work, te output o ts eat engne would be 6.9 MW. Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18

34 Capter 18: e Laws o ermodynams James S. Walker, Pyss, 4 t Edton 100. Pture te Problem: A eat engne operates between a ot and old reservor. e ot reservor s te same as n Atve Example 18-, but te old reservor s 10 degrees ooler. Strategy: Solve or te eeny usng equaton Calulate te work rom te eeny and nput eat usng equaton Soluton: 1. e eeny nreases wen te temperature derene between te two reservors nreases. Lowerng te temperature o te old reservor nreases te eeny, so te new eeny s greater tan (b) Use equaton 18-1 to solve or te eeny: C 95K e = = 576K =. () Solve equaton or W: W = e = ( 0.488)( 1050 J) = 51 J Insgt: By nreasng te eeny, t s possble to extrat more work rom te same avalable eat. H Copyrgt 010 Pearson Eduaton, In. All rgts reserved. s materal s proteted under all opyrgt laws as tey urrently exst. No porton o ts materal may be reprodued, n any orm or by any means, wtout permsson n wrtng rom te publser. 18 4