Thermodynamics. Question 1. Question 2. Question 3 3/10/2010. Practice Questions PV TR PV T R

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1 /10/010 Question 1 1 mole of idel gs is rought to finl stte F y one of three proesses tht hve different initil sttes s shown in the figure. Wht is true for the temperture hnge etween initil nd finl sttes? Thermodynmis rtie Questions ) It s the sme for ll proesses. ) It s the smllest for proess 1. ) It s the smllest for proess. ) It s the smllest for proess. E) It s the sme for proesses 1 nd. The temperture for gs t ny point is given y: TR T R Sine ll of the three proesses end t the sme temperture, we need only look t the proess tht strts losest to: 0 0 T R R p 0 1 p 0 p roess 1. R R tully egins t this temperture,. so its hnge is 0 R R. R R F Question 1 mole of idel gs is rought to finl stte F y one of three proesses tht hve different initil sttes s shown in the figure. Wht is true for the work done y the gs? ) It s positive for proess 1 nd, ut negtive for proess. ) It s the smllest for proess 1. p 0 ) It s the smllest for proess. 1 ) It s the smllest for proess. p E) Zero work is done long proess. 0 p F Question n idel gs is rought from S -> F y three different pths: SRF, SF, STF. The temperture t S is the sme s the temperture t F. Whih of the following is true? ) roess SRF ours t onstnt temperrure ) The work done y the gs long SRF is the sme s the work done y the gs long STF. ) Net het into the gs long STF is greter thn the work done y the gs long this pth. ) The hnge in the internl energy is the sme for ll three pths. E) Work done y the gs long STF is greter thn the work done y the gs long SF. The internl energy of n idel gs depends only on temperture S R The internl energy of n idel gs depends only on temperture. If the initil nd finl temp re the sme, there is no hnge in internl energy. U=0 for ll three pths. W= - Q for eh proess, whih elimintes Work done y the gs is equl to the re under the urve T F re onsidertions eliminte nd E Temperture t R is greter thn the temperture t S (gs lw) elimintes 1

2 /10/010 Wht is true for the proess ->? Question 4 Question 5 Wht is true for the two step proess -> ->? ) U=0 Q>0. ) U=0 Q<0 ) W=0 U>0 ) W=0 U<0 E) W=0 Q=0 p 0 ) U=0 Q=0 ) U=0 Q>0 ) W=0 Q>0 ) W=0 Q<0 E) W>0 Q<0 p 0 Sine the re (work) under -> is zero, this elimintes nd p 0 Sine the re (work) under -> -> is not zero, this elimintes nd p 0 E is eliminted euse if Q = 0 nd W=0, then U would hve to e zero, thus no hnge in ould our. The internl energy must hve inresed to doule the pressure, therefore U> is eliminted euse if U=0 nd Q=0 then W must e 0, ut it is not. The idel gs lw tells us tht the temperture does not hnge etween nd, so U=0. nd sine expnding Q> The work done on the gs is negtive (the gs does positive work when expnding), therefore Q>0, this elimintes nd E Question 6 00 enters rnot engine from the hot reservoir, held t 400 K. uring the entire engine yle, 50 of useful work is performed y the engine. Wht is the temperture of the old reservoir? ) 100 K ) 00 K ) 00 K ) 50 K E) 150 K From the definition of effiieny: e W y 1 Q H Wy 50 e 0.5 Q 00 H T e T H Question 7 For n idel gs in ontiner with fixed volume nd onstnt temperture, whih of the following is true? I. ressure results from moleulr ollisions with the wlls. II. The moleules ll hve the sme speed. III. The verge kineti energy is diretly proportionl to the temperture ) I, II, nd III ) I nd II only ) II nd III only ) II only E) I nd III only Sine this is rnot engine, we hve T e 1 T H T T 00K The distriution of speeds mong the moleules, leding to n verge vlue, ut the moleules ertinly do not ll hve the sme speed.

3 /10/010 Question 8 1 mole of idel gs is in ontiner of volume, temperture T, nd pressure. If the volume is hlved nd the temperture douled, the pressure will e: ) ) ) 4 ) ½ E) ¼ Sine n idel gs: nrt TR T R 1 TR 4 4 Question 9 Two identil ontiners ontin 1 mole eh of two different montomi idel gses, gs nd gs, with the mss of gs 4 times the mss of gs. oth gses re t the sme temperture, nd 10 of het is dded to gs, resulting in temperture hnge T. How muh het must e dded to gs to use the sme T? ) 10 ) 100 ) 40 ).5 E) 1600 For the internl Energy of Idel Gs U nrt For the sme numer of moles, T depends only on the energy dded, not the mss of the gs moleules Question 10 The entropy of losed mrosopi system will never derese euse ) Energy wouldn t e onserved if entropy deresed ) For lrge systems, the proility of suh derese is negligile ) Mehnil equilirium ouldn t e sustined with derese in entropy ) Het n never e mde to flow from old ojet to hot ojet E) Moleulr motions reh their minimum only t solute 0 Question 11 uring ertin proess, 600 of het is dded to system. While this ours, the system performs 00 of work on the surroundings. The hnge in internl energy of the system is: ) 00 ) 800 ) 400 ) E) Impossile to determine without knowing the temperture The proility of suh mro stte ourring is essentilly 0 The First Lw of Thermodynmis U U Q W

4 /10/010 Question 1 Whih of the following is not true out therml rdition? ) It s mehnism of energy exhnge etween systems t different tempertures ) It involves tomi exittions ) Ie t will emit therml rdition ) It requires some mteril sustne to trvel through E) It s y produt of the food we et Eletromgneti rdition doesn t need mteril medium to trvel through. In E, we et to mintin ody temperture, nd we emit lot of energy s therml rdition Question 1 To inrese the dimeter of n luminium ring from 50.0 mm to 50.1 mm, the temperture of the ring must inrese y 8. Wht temperture hnge would e neessry to inrese the dimeter of n luminium ring from mm to mm? ) ) 4 ) 8 ) 11 E) 16 We pply the formul for length expnsion: 0.1mm mm 0.1mm mm Therefore 0.1mm T mm 40 L T L o Question 14 gs is enlosed in metl ontiner with movele piston on top. Het is dded to the gs y pling ndle flme in ontt with the ontiner's ottom. Wht of the following is true out the temperture of the gs? ) The temperture must go up if the piston remins sttionry. ) The temperture must go up if the piston is pulled out drmtilly. ) The temperture must go up no mtter wht hppens to the piston. ) The temperture must go down no mtter wht hppens to the piston. E) The temperture must go down if the piston is ompressed drmtilly. Use the First Lw of Thermodynmis, U=Q+W. The temperture dds het to the gs, so Q is positive. Internl energy is diretly relted to temperture, so if U is positive, then the temperture goes up (nd vie vers). Here we n e sure tht U is positive if the work done on the gs is either positive or zero. The only possile nswer is. For, the work done on the gs is negtive euse the gs expnds (Note: euse we dd het to gs does NOT men the temperture utomtilly goes up) Question 15 smll het engine opertes using pn of 10 oiling wter s the high temperture reservoir nd the tmosphere s low temperture reservoir. ssuming idel ehviour, how muh more effiient is the engine on old, dy thn on wrm, dy? ) out 1. times s effiient. ) times s effiient.. ) 0 times s effiient. ) Infinitely more effiient. E) ust s effiient. T Using the Engine effiieny eqution: e 1 TH 0 7 Hot y: e old y: 0 7 e Rtio:

5 /10/010 Question 16 In three seprte experiments, gs is trnsformed from i, i to stte f, f long pths (1,, nd ) illustrted elow. The work done on the gs is? Question 17 1 m ontiner ontins 10 moles of idel gs t room temperture. t wht frtion of tmospheri pressure is the gs inside the ontiner? ) Gretest for pth 1 ) Lest for pth ) The sme for pths 1 nd ) The gretest for pth E) The sme for ll three pths i 1 ) 1/40 tm ) 1/0 tm ) 1/10 tm ) ¼ tm E) ½ tm p 0 f Using the Idel Gs Lw nrt p 0 The work done on the gs during thermodynmi proess is equl to the re of the region in the - digrm ove the xis nd elow the pth the system mkes from its initil to its finl stte. f i nrt 10 mol K mol K 1m N 490 m 0 Rtio: 0 N 490 m 0.5 N m Question 18 n idel gs is ompressed isothermlly from 0 L to 10 L. uring this proess, 5 of work done to ompress the gs. Wht is the hnge of internl energy for this gs? ) -10 ) - 5 ) 0 ) 5 E) 10 Question 19 Through series of thermodynmi proesses, the internl energy of smple of onfined gs is inresed y 560. If the net mount of work done on the smple y its surroundings is 0, how muh het ws trnsferred etween the gs nd its environment? ) 40 sored ) 40 dissipted ) 880 sored ) 880 dissipted E) None of the ove The First Lw of Thermodynmis U Q W The First Lw of Thermodynmis U Q W uring n isotherml proess, U is LWYS ZERO 560 Q 0 Q

6 /10/010 Question 0 In one of the steps of the rnot yle, the gs undergoes n isotherml expnsion. Whih of the following sttements is true onerning this step? ) No het is exhnged etween the gs nd its surrounding, euse the proess is isotherml. ) The temperture dereses euse the gs expnds. ) This steps violtes the Seond Lw of Thermodynmis euse ll the het sored is trnsferred into work. ) The internl energy of the gs remins onstnt. E) The internl energy of the gs dereses due to the expnsion. Sttement is wrong euse no het exhnged etween the gs nd its surroundings is the definition of diti, not isotherml. Sttement nnot e orret sine the step desried in question is isotherml; y definition, the temperture does not hnge. Sttement is flse, euse lthough the het sored is onverted ompletely to work, it does not inlude eing returned k to it is initil stte nd hving 100% onversion to het ( dereses nd inreses). Isotherml y definition sttes internl energy remins onstnt Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. x10 5 Isotherm t T H dit: Q=0 dit: Q=0 Isotherm t T x10-4x10 - Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00. K nd 00. K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. Using the gs lw nd the given,, nd T.0x10 5 Isotherm t T H R n nt RT dit: Q=0 dit: Q=0 5 N m m Isotherm t T K mol K.0x10-4.0x10-0.4moles Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. Rell: if the temperture is onstnt, then so is the internl energy. long -> isotherm, there is no hnge in internl energy, so the first lw gives us U = 0 = Q + W. ut Q is given s 500, so W = -500 is done ON the gs..0x10 5 dit: Q=0 QIsotherm H 500 t T H Isotherm t T.0x10-4.0x10 - dit: Q=0 6

7 /10/010 Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. Rell: if the proess is diti, there is no het trnsfer (Q=0). Therefore U=W..0x10 5 Now we know the dit onnets the two isotherms where we know the temperture, nd sine the idel gs internl energy depends only upon temperture. U nrt 0.4 moles K 00 K mol K 99 W= - 99 dit: Q=0 Isotherm t T H Isotherm t T.0x10-4.0x10 - dit: Q=0 Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. We re looking for Q.0x10 5 T 00K For the rnot yle: e T H 00K Isotherm t T H dit: Q=0 Q dit: Q=0 lso: e 1 QH Q Isotherm t T 0. 1 Therefore: 500.0x10-4.0x10 - Q Question 1 rnot engine using montomi idel gs s working sustne opertes etween two reservoirs held t 00 K nd 00 K, respetively. Strting t point with pressure nd volume s indited on the grph, the gs expnds isothermlly to point, where the volume is 4.0 x 10 - m. 500 of het energy is sored y the gs in this proess. ) How mny moles of gs re present? ) Find the work done on the gs during -> proess. ) Find the work done on the gs during the proess -> diti expnsion. ) How muh het is expelled in the proess ->? E) etermine the hnge in internl energy in the proess ->. One gin we re onneting two isotherms. Sine the internl energy depends only on temperture, we hve U nrt 0.4 moles K 00 K mol K 99 The opposite of ->.0x10 5 dit: Q=0 Isotherm t T H Isotherm t T.0x10-4.0x10 - dit: Q=0 Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work. ) When the system is returned from stte to stte long the urved pth shown, 60 of het flows out of the system. oes the system perform work on its surroundings or do the surroundings perform work on the system? How muh work is done? ) If the system does 10 of work in trnsforming from stte to stte long pth d, does the system sor or does it emit het? How muh het is trnsferred? ) If U =0 nd U d =0, determine the het sored in the proess d nd d. ) For the proess d, identify eh of the following quntities s positive, negtive, or zero. W = Q = U = d 7

8 /10/010 Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work. ) When the system is returned from stte to stte long the urved pth shown, 60 of het flows out of the system. oes the system perform work on its surroundings or do the surroundings perform work on the system? How muh work is done? We require the vlue of W First let s lulte U U Q W 70 0 U 40 euse U -> does not depend on pth tken from to, then sine U = 40 then U = -U = 40-0 euse system does work U Q W W W 0 Sine the vlue is + 0, the surroundings does the work. d Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work.. If the system does 10 of work in trnsforming from stte to stte long pth d, does the system sor or does it emit het? How muh het is trnsferred? gin, using the ft tht U -> does not depend on the pth tken, we know tht U d = 40 Now we require Q d U Qd Wd d 40 Qd 10 Q 50 d Therefore the system sors 50 of het euse the system does 10 of work d Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work. ) If U =0 nd U d =0, determine the het sored in the proess d nd d. For the proess d, there is no hnge in volume, therefore W d = 0. U Q W U d d d d Q Now, U = 40, nd U = 0, tells us tht U = 40, therefore: U U U d d d Q 10 d For d: W d =W d +W d =W d +0=W d Sine W d =-10, then W d =-10 U Qd Wd d U U Qd 10 d 0 0 Qd 10 Q 40 d d Question When system is tken from stte to stte long the pth shown in the figure elow, 70 of het flows into the system, nd the system does 0 of work.. For the proess d, identify eh of the following quntities s positive, negtive, or zero. W = Q = U = The proess d is yli, so U = zero euse the proess is trversed ounterlokwise in the digrm, we know tht W is positive. U=Q+W therefore Q must e negtive The system is het engine, emitting het on eh yle d 8

9 /10/010 Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 From the Idel Gs Lw: T nr T R T R R T R R T R Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 roess -> ours t onstnt pressure. So we n use the First Lw U Q W Q U W nrt W nrt T W 4 R W R R 0 Sine Work done on the gs is negtive of the re under grph R R 5 0 Q R The re under the proess -> W 0 The derese in volume orresponds to positive work done on the gs The sign of the net work is negtive for yles tht run lokwise

10 /10/010 Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess -> E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 roess -> ours t onstnt volume. So we n use the First Lw U Q W QU nrt nrt T R R R Sine t onstnt volume, no work is done 0 Q Question 1 mole of montomi idel gs is rought through yle ->->->-> s depited elow. ll proesses re performed slowly. Respond to the following in terms of 0, 0, nd R. ) Find the temperture t eh vertex. ) Find the het dded to the gs for the proess ->. ) Find the work done on the gs for the proess ->. ) Find the het dded to the gs fro the proess ->. E) Find the hnge in internl energy for i. roess -> ii. The entire yle 0 i) Using: U nrt U RT T 4 R R R ii) For the entire yle, you end up k in the sme stte, so U = 0 0 Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) Wht is the temperture of: mol K i. Stte ii. Stte iii. Stte ) etermine the hnge in internl energy of the gs for: 0.6 i. Step. ii. Step. iii. Step ) How muh work, W is done y the gs during step? (x10 - m ) ) Wht is the totl work done over the yle? E) Is het sored or disrded step? F) If so how muh? G) Wht is the mximum possile effiieny (without violting the Seond Lw of Thermodynmis) for ylil het engine tht opertes etween the tempertures of stte nd? Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln.4 ) Wht is the temperture of: i. Stte ii. Stte iii. Stte 0.6 From the Idel Gs Lw: T nr m T 870K 0.4mol 8.1 mol K T 870K Stte is on the Isotherm with stte m T 0K 0.4mol 8.1 mol K 1 48 (x10 - m ) 10

11 /10/010 Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) etermine the hnge in internl energy of the gs for: mol K i. Step. ii. Step. iii. Step. 0.6 Sine step tkes ple long n isotherm, U = 0 U Q W npt 1 48 (x10 - m ) mol K U U nr n vt T U U U U U or mol mol K K 0 0 K K mol mol K K U 5400 U U mol 9.1 0K 870K m m Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) How muh work, W is done y the gs during step? mol K W nrt ln 4810 m 0.4mol K ln 1 10 mol K m (x10 - m ) Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K ) Wht is the totl work done over the yle? mol K Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: (x10 5 ) W nrt ln 9.1 mol K E) Is het sored or disrded step? mol K F) If so how muh? The totl work done over yle is equl to the sum of the vlues of the work done over eh step. W W W W yle (x10 - m ) y the First Lw of Thermodynmis U Q W Q 0 Q W W 4000 U Q W 0.6 Sine Q is positive, this represents het sored y the gs (x10 - m ) 11

12 /10/010 Question mol smple of n idel ditomi gs undergoes slow hnges from stte to stte to stte nd k to long the yle shown. th is n isotherm nd the work done y the gs s it hnges isothermlly from stte to stte is given y the eqution: W nrt ln G) Wht is the mximum possile effiieny (without violting the Seond Lw of Thermodynmis) for ylil het engine tht opertes etween the tempertures of stte nd? (x10 5 ) The mximum possile effiieny is the effiieny of the rnot engine..4 T e 1 T H 0K K 75% (x10 - m ) Question 5 You hve just purhsed up of offee (45 g), ut you notie tht the offee is old, so you ll over your host. It s 45 o, you sy, ut I like it 65 o.. The offee in the pot is 95 o, how muh offee should the host pour to rise the temperture of the 45g of offee in your up? Look t the het lost y the pot s offee nd the het gined y your offee offee in up: offee in pot: Q m T T f i Q m T T f Q mt i m Now, Q1 Q m g0 m 0 m 0g 1