Calormetrc calculatons: dq mcd or dq ncd ( specc heat) Q ml ( latent heat) General Formulas applcable to ALL processes n an Ideal Gas: P nr du dq dw dw Pd du nc d C R ( monoatomc) C C R P Specc Processes: Isothermal: =constant d = 0 du =0 dq=dw nr W Pd d nr ln Adabatc: dq=0 du= -dw dw du W nc d nc P 1 const const Isobarc: P=constant W Pd P Isochroc: =constant d=0 dw=0 du=dq
Entropy: General Reversble Processes: ds=dq/ S nc ln nrln Calormetrc Processes: ( specc heat): dq mcd or dq ncd d d ds mc or nc S mcln or ncln ( latent heat): Q ml ml S remans constant at phase change Non Reversble Processes: (one cannot wrte ds=dq/ or the process) But one can use a surrogate reversble process wth the same ntal and nal states to calculate S. nd Law o hermodynamcs: Stot Ssystem Senvronment 0
Physcs 6/66 - exam#1 1. (5 pts) Answer the ollowng questons. (Use the space provded below and on the next page needed.) a). (9 pts) An deal gas s reversbly expanded rom an ntal state to the nal state through three derent thermodynamc processes (a, b, c). ) Whch process wll result n the largest U? ) Whch process wll result n the largest work done by the gas W? ) Whch process wll absorb the largest amount o heat Q? b). (8 pts) A Carnot engne operates between a hgh temperature reservor at and a low temperature reservor at L. One can ncrease the ecency o the engne by ether ncreasng by C or decreasng L by wll result n a larger ecency or the engne?. Whch change or c). (8 pts) ) An deal gas s beng taken rom an ntal temperature 1 to a nal temperature n two separate processes: (a) sobarc and (b) sochorc, as shown. ) Wll the entropy o the system decrease or ncrease n both processes? ) ) Whch process wll result n a larger change n the system s entropy? SOLUION: P b a 1 a) ) U s a state varable. Snce all three processes (a, b, & c) have the same ntal and nal states, we have Ua Ub Uc. ) Work done by the gas s the area under the curve. From the graph shown, we have the ollowng rankng Wa Wb Wc. /11
Physcs 6/66 - exam#1 ) From the 1 st Law, we have Q U W. Snce all U s are the same, the rankng or Q wll be the same as the rankng or W,.e., Qa Qb Qc. L b) For a Carnot engne, the ecency s gven by e 1 so that both decreasng L and/or ncreasng wll gves a smaller L and makes e closer to 1. As or a xed change n temperature,, the rato L wll generally be L smaller than so that e L L e. o see that, we can consder the ollowng rato: L L L 1 L L L 1 1 L L Both blue terms are postve so that ths rato wll always be less than 1 ndcatng that, L wll always be smaller than L. c) ) In both processes (sobarc and sochorc), heat enters the system as the temperature ncreases. Snce ds dq, the change n entropy or both processes wll also be postve (an ncrease). ) For ths problem, the change n temperature or both processes s the same. Snce Cp Cv, we have dqp dqv and Sp Sv. dqpv, d note : S pv, nc pv, nc pv, ln hus, the sobarc process wll result n a larger entropy ncrease. 4/11
Physcs 6/66 - exam#1. (5 pts) In a calormetry experment, a pece o hot ron ntally at 500 o C was placed n 1.50g o water at 0.0 o C. Ater the system has come to equlbrum, all water was vaporzed and the nal temperature or the steam and hot ron was measured to be at 110 o C? What was the mass o the hot ron? [ L.6 10 J g; c.080 J g C c 4.186 J g C c 0.450 J g C] v steam Let the mass o the ron pece be Q 0 gves, m water, heat released, heat absorbed, heat absorbed, hot ron latent heat absorbed water warmng steam warmng coolng to vaporze at 100 C 0 0 C 100 C 100 C 110 C 500 C 10 1 C hs gves C mc w w(100 C C w v wcs mc 110 C500 0 ) mlm 110 C100 C 0 Rearrangng terms, we then can solve or m, 500 110 mc(100 C0 C) v c 110 100 m w cw(80 C) Lv cs10 C c 90 C 1.50g 4.186 J g C(80.0 C).610 J g.08j g C10.0 C 0.450 J g C90 C 1.50g 4.186 J g C(80.0 C).610 J g.08j g C10.0 C 0.450 J g C90 C mc C C ml m C C m m m w w w w s m.4g.4g o hot ron was ntally used n the carlormetry experment. ron 6/11
Physcs 6/66 - exam#1. (5 pts) A Carnot cycle operated between two temperature reservors at =500 o C and L =00 o C has one mole o an deal monoatomc gas as ts workng substance. You are gven the ollowng parameters or the cycle: A =10.0L, B =0.0L. Calculate a) the net work done by the gas per cycle, b) the heat absorbed by the gas per cycle, c) the heat expelled by the gas per cycle. d) What s the ecency o ths cycle? [1L 10 m 5, 1 atm= 1.01 10 Pa ] o get the volume at state C and D, we can use the adabatc processes: 1 1 C C B B 1( 1) 500 7.15 C B 0L 41.776L L 00 7.15 Smlarly, 1 1 D D A A 1( 1) 500 7.15 D A 10L 0.888L L 00 7.15 Note that snce we really only need the rato D / C, we can smply dvde the above two equatons (wthout dong the numercal calculatons) to get D A 0.5. C B Pressure A adabatc sothermal D B sothermal olume adabatc C a) Net work done by the cycle: Isothermal branches (AB, CD) ( U 0 ): B WAB nr ln( ) A 1 mol(8.14 J / mol K)(77.15 K)ln(0 /10) 4.456kJ D WCD nrl ln( ) C (work done by) (work done on) 1 mol(8.14 J / mol K)(47.15 K)ln(0.5).77kJ Adabatc branches (BC, DA) (Q = 0): WBC UBC nc ( C B ) (1 mol) (8.14 J / mol K)(47.15 77.15) K.741kJ WDA UDA nc ( A D ) (1 mol) (8.14 J / mol K)(77.15 47.15) K.741kJ hus, 7/11
Physcs 6/66 - exam#1 Wnet 4.456kJ.77kJ.741kJ.741kJ 1.79kJ 1.7kJ (net work s done BY gas). b) Now we calculate the heat transers: AB: sothermal. QAB WAB 4.456kJ BC: adabatc. QBC 0 CD: sothermal. QCD WCD.77kJ DA: adabatc. QDA 0 hus, Qabosorbed QAB 4.456kJ 4.46kJ And the system releases heat durng only the process CD so that Q Q.77kJ.7kJ release CD d) he ecency o ths heat engne s Wnet 1.79kJ e 0.88 8.8% Q 4.456kJ absorbed Shorter Method: hs s a Carnot cycle and ts ecency s gven by: L 47.15K e 1 1 8.8% 77.15K (same as beore) Now, we need to calculate the net work done and t s gven by the two sothermal branches: B D Wnet WAB WCD nr ln nrl ln A C (Work done by the adabatc branches (BC and DA) cancel out each other.) 1 From the adabatc relaton, const, we can wrte: and 1 1 A A D D 1 1 B B C C. Dvdng these two equatons, we have. Snce A B and C D L, ths ratos smply to: 1 1 B B C C 1 1 A A D D 8/11
Physcs 6/66 - exam#1 B A C. Insertng these ratos back nto the work equatons, we have, D W nr ln nr ln 0.5 1mol 8.14 J mol K 77.15K 47.15K ln Wnet net L 1.79kJ hen, we can use the denton o ecency to get the absorbed heat: Wnet Wnet 1.79kJ e Q 4.456kJ Q e 0.88 Q 4.456 kj Fnally, to calculate heat related, we use the relaton: W Q Q Q Q W 4.456kJ 1.79kJ.77kJ net L L net Q.77kJ L All these values match wth the prevous drect but longer calculatons. 9/11
Physcs 6/66 - exam#1 4. (5 pts) a) One mole o a monoatomc gas s beng compressed reversbly rom an ntal state 5 5 1 5.00 L, P1 5.00 10 Pa to a nal state.00 L, P.00 10 Pa. Calculate the entropy change S or ths reversble process. b) What wll the entropy change S be or the same deal gas the process s done rreversbly but wth the same ntal and nal states as n part a)? c) 10.0 g o water s heated slowly rom 0 C to ts bolng pont at 100 C and all o t s converted nto steam at 100 C. What s the total entropy change S or ths heatng and vaporzaton process? [ Lv.610 J g c 4.186, water J g K] a) For any reversble process, the change n entropy s gven by: Sa ncvln nrln We have the volume rato and to get the temperature rato, we can use the deal gas law, P P we have const P hus, we have ( Cv R/ or a monoatomc deal gas) P Sa (1 mole) 8.14 J / molek ln ln P 1 1 1 9 Sa 8.14 J / K ln ln 17.0 J / K 5 5 he change o entropy s negatve snce the process s a compresson. b) Snce entropy s a state varable and the ntal and nal states are the same as n part a), the change n entropy or a nonreversble process between the same state 1 to s the same as n part a): S S 17.0 J / K b a c) In heatng lqud water slowly or a small change n temperature d, the nntesmal heat absorbed by the water s gven by: dq mc d heatng water 10/11
Physcs 6/66 - exam#1 hen, the nntesmal change n entropy s temperature change 0 C 100 C, we have: ds heatng dq heatng. For the ull range o dqheatng mcwaterd Sheatng mcwater ln 7.15K Sheatng 10.0g4.186 J gkln 10.10 J / K 9.15K ncreasng process) (temp has to be n K) (heatng s an entropy Now, the water vaporzes nto steam at the bolng pont, the heat o vaporzaton s absorbed at a xed temperature. he ncrease n entropy s then gven by: ml 10.0g.610 J / g v Svaporze 60.57 J / K 7.15K bolng So, the OAL entropy ncrease s the sum o the two, S 10.10 J / K 60.57 J / K 70.7 J / K tot 11/11