ChapterChapter 2020 -- ThermodynamicsThermodynamics AA PowerPointPowerPoint PresentationPresentation byby PaulPaul E.E. TippensTippens,, ProfessorProfessor ofof PhysicsPhysics SouthernSouthern PolytechnicPolytechnic StateState UniversityUniversity

© 2007 THERMODYNAMICSTHERMODYNAMICS

ThermodynamicsThermodynamics isis thethe studystudy ofof energyenergy relationshipsrelationships thatthat involveinvolve heat,heat, mechanicalmechanical work,work, andand otherother aspectsaspects ofof energyenergy andand heatheat transfer.transfer. Central Heating Objectives:Objectives: AfterAfter finishingfinishing thisthis unit,unit, youyou shouldshould bebe ableable to:to: •• StateState andand applyapply thethe first andand second laws ofof thermodynamics. •• DemonstrateDemonstrate youryour understandingunderstanding ofof adiabatic, isochoric, isothermal, and isobaric processes.processes. •• WriteWrite andand applyapply aa relationshiprelationship forfor determiningdetermining thethe ideal efficiency ofof aa heatheat engine.engine. •• WriteWrite andand applyapply aa relationshiprelationship forfor determiningdetermining coefficient of performance forfor aa refrigeratior.refrigeratior. AA THERMODYNAMICTHERMODYNAMIC SYSTEMSYSTEM

•• AA systemsystem isis aa closedclosed environmentenvironment inin whichwhich heatheat transfertransfer cancan taketake place.place. (For(For example,example, thethe gas,gas, walls,walls, andand cylindercylinder ofof anan automobileautomobile engine.)engine.)

WorkWork donedone onon gasgas oror workwork donedone byby gasgas INTERNALINTERNAL ENERGYENERGY OFOF SYSTEMSYSTEM

•• TheThe internalinternal energyenergy UU ofof aa systemsystem isis thethe totaltotal ofof allall kindskinds ofof energyenergy possessedpossessed byby thethe particlesparticles thatthat makemake upup thethe system.system.

Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules. TWOTWO WAYSWAYS TOTO INCREASEINCREASE THETHE INTERNALINTERNAL ENERGY,ENERGY, U.U.

++UU

WORKWORK DONEDONE HEATHEAT PUTPUT INTOINTO ONON AA GASGAS AA SYSTEMSYSTEM (Positive)(Positive) (Positive)(Positive) TWOTWO WAYSWAYS TOTO DECREASEDECREASE THETHE INTERNALINTERNAL ENERGY,ENERGY, U.U.

WW outout QQout --UU DecreaseDecrease hot hot

WORKWORK DONEDONE BYBY HEATHEAT LEAVESLEAVES AA EXPANDINGEXPANDING GAS:GAS: SYSTEMSYSTEM WW isis positivepositive QQ isis negativenegative THERMODYNAMICTHERMODYNAMIC STATESTATE

TheThe STATESTATE ofof aa thermodynamicthermodynamic systemsystem isis determineddetermined byby fourfour factors:factors: •• AbsoluteAbsolute PressurePressure PP inin PascalsPascals •• TemperatureTemperature TT inin KelvinsKelvins •• VolumeVolume VV inin cubiccubic metersmeters •• NumberNumber ofof moles,moles, nn,, ofof workingworking gasgas THERMODYNAMICTHERMODYNAMIC PROCESSPROCESS Increase in Internal Energy, U.

WW out

QQ in

Initial State: HeatHeat inputinput Final State: P V T n P V T n 1 1 1 1 WorkWork byby gasgas 2 2 2 2 TheThe ReverseReverse ProcessProcess Decrease in Internal Energy, U.

WW in

QQ out

WorkWork onon gasgas Initial State: Final State:

P1 V1 T1 n1 LossLoss ofof heatheat P2 V2 T2 n2 THETHE FIRSTFIRST LAWLAW OFOF THERMODYAMICS:THERMODYAMICS:

•• TheThe netnet heatheat putput intointo aa systemsystem isis equalequal toto thethe changechange inin internalinternal energyenergy ofof thethe systemsystem plusplus thethe workwork donedone BYBY thethe system.system.

Q = U + W final - initial)

•• Conversely,Conversely, thethe workwork donedone ONON aa systemsystem isis equalequal toto thethe changechange inin internalinternal energyenergy plusplus thethe heatheat lostlost inin thethe process.process. SIGNSIGN CONVENTIONSCONVENTIONS +W FORFOR FIRSTFIRST LAWLAW +Wout ++QQ in •• HeatHeat QQ inputinput isis positivepositive U • Work BY a gas is positive --WW in • Work ON a gas is negative U •• HeatHeat OUTOUT isis negativenegative --QQ out

Q = U + W final - initial) APPLICATIONAPPLICATION OFOF FIRSTFIRST LAWLAW OFOF THERMODYNAMICSTHERMODYNAMICS ExampleExample 1:1: InIn thethe figure,figure, thethe gasgas absorbsabsorbs 400400 JJ ofof heatheat andand WW out =120=120 JJ atat thethe samesame timetime doesdoes 120120 JJ ofof workwork onon thethe piston.piston. WhatWhat isis thethe changechange inin internalinternal energyenergy ofof thethe system?system? QQin 400400 JJ Apply First Law:

Q = U + W ExampleExample 11 (Cont.):(Cont.): ApplyApply FirstFirst LawLaw

Q is positive: +400 J (Heat IN) WWout =120=120 JJ W is positive: +120 J (Work OUT) QQ QQ == UU ++ WW in 400400 JJ UU == QQ -- WW

UU == QQ -- WW  == (+400(+400 J)J) -- (+120(+120 J)J) U = +280 J

== +280+280 JJ ExampleExample 11 (Cont.):(Cont.): ApplyApply FirstFirst LawLaw

Energy is conserved: WWout =120=120 JJ TheThe 400400 JJ ofof inputinput thermalthermal energyenergy isis usedused toto performperform QQin 120 J of external work, in 120 J of external work, 400 J increasingincreasing thethe internalinternal 400 J energyenergy ofof thethe systemsystem byby 280280 JJ

The increase in  internal energy is: U = +280 J

FOURFOUR THERMODYNAMICTHERMODYNAMIC PROCESSES:PROCESSES:

••• IsochoricIsochoricIsochoric Process:Process:Process: VVV === 0,0,0, WWW === 000 ••• IsobaricIsobaricIsobaric Process:Process:Process: PPP === 000 ••• IsothermalIsothermalIsothermal Process:Process:Process: TTT === 0,0,0, UUU === 000 ••• AdiabaticAdiabaticAdiabatic Process:Process:Process: QQQ === 000

Q = U + W ISOCHORICISOCHORIC PROCESS:PROCESS: CONSTANTCONSTANT VOLUME,VOLUME, VV == 0,0, WW == 00 0 QQ == UU ++ WW soso thatthat QQ == UU Q QIN QQOUT

NoNo WorkWork -U +U DoneDone

HEAT IN = INCREASE IN INTERNAL ENERGY

HEAT OUT = DECREASE IN INTERNAL ENERGY ISOCHORICISOCHORIC EXAMPLE:EXAMPLE: NoNo ChangeChange inin volume:volume: PP2 BB PP A PP B = TT A TT B PP1 AA

VV1 == VV2 400400 JJ HeatHeat inputinput 400400 JJ heatheat inputinput increasesincreases increasesincreases PP internalinternal energyenergy byby 400400 JJ withwith const.const. VV andand zerozero workwork isis done.done. ISOBARICISOBARIC PROCESS:PROCESS: CONSTANTCONSTANT PRESSURE,PRESSURE, PP == 00 QQ == UU ++ WW ButBut WW == PP VV Q QIN QQOUT

WorkWork OutOut WorkWork -U +U InIn

HEAT IN = Wout + INCREASE IN INTERNAL ENERGY

HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY ISOBARICISOBARIC EXAMPLEEXAMPLE ((Constant Pressure):

A B P VA VB = TA T B

400 J V1 V2

Heat input 400400 JJ heatheat doesdoes 120120 JJ ofof increases V work,work, increasingincreasing thethe with const. P internalinternal energyenergy byby 280280 JJ. ISOBARICISOBARIC WORKWORK

A B P VA VB = TA T B

PPA == PPB 400 J V1 V2

Work = Area under PV curve Work P V ISOTHERMALISOTHERMAL PROCESS:PROCESS: CONST.CONST. TEMPERATURE,TEMPERATURE, TT == 0,0, UU == 00 QQ == UU ++ WW ANDAND QQ == WW

QQ IN QQ OUT WorkWork OutOut WorkWork U = 0 U = 0 InIn

NETNET HEATHEAT INPUTINPUT == WORKWORK OUTPUTOUTPUT WORKWORK INPUTINPUT == NETNET HEATHEAT OUTOUT ISOTHERMALISOTHERMAL EXAMPLEEXAMPLE (Constant(Constant T):T):

A PA B PB

UU == TT == 00 V2 V1

Slow compression at constant temperature: P V =PV A A B B ----- NoNo changechange inin UU. ISOTHERMALISOTHERMAL EXPANSIONEXPANSION ((ConstantConstant T)T):: AA PA BB PA VA = PBVB

PB T = T V V A B U = T = 0 A B

400 J of energy is absorbed Isothermal Work by gas as 400 J of work is done on gas. V WnRT ln B T = U = 0 VA W W

  - - In Work U = U =

  W

 Q = 0 Q = 0   U U or U or    U = - + - -  W = W = = 0  

 Q Work Out W ; W ;

  U

 ADIABATIC PROCESS: ADIABATIC PROCESS: U + U + U   NO HEAT EXCHANGE, NO HEAT EXCHANGE,  W = -

 Q = Q = Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy   ADIABATICADIABATIC EXAMPLE:EXAMPLE:

A PP A B

PP B

VV1 VV2

Insulated ExpandingExpanding gasgas doesdoes Walls: Q = 0 workwork withwith zerozero heatheat loss.loss. WorkWork == --UU ADIABATICADIABATIC EXPANSION:EXPANSION:

A PP A B PP AV VA PP BV VB = PP B TT A TT B Q = 0 VVA VVB

400 J of WORK is done, DECREASING the internal   energy by 400 J: Net heat PVA ABB PV exchange is ZERO. QQ == 00 MOLARMOLAR HEATHEAT CAPACITYCAPACITY OPTIONAL TREATMENT

TheThe molarmolar heatheat capacitycapacity CC isis defineddefined asas thethe heatheat perper unitunit molemole perper CelsiusCelsius degree.degree.

CheckCheck withwith youryour instructorinstructor toto seesee ifif thisthis moremore thoroughthorough treatmenttreatment ofof thermodynamicthermodynamic processesprocesses isis required.required. SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY

RememberRemember thethe definitiondefinition ofof specificspecific heatheat capacitycapacity asas thethe heatheat perper unitunit massmass requiredrequired toto changechange thethe temperature?temperature?

Q c  mt

ForFor example,example, copper:copper: cc == 390390 J/kgJ/kgKK MOLARMOLAR SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY

TheThe ““molemole”” isis aa betterbetter referencereference forfor gasesgases thanthan isis thethe ““kilogram.kilogram.”” ThusThus thethe molarmolar specificspecific heatheat capacitycapacity isis defineddefined by:by: QQ CC == nn TT

ForFor example,example, aa constantconstant volumevolume ofof oxygenoxygen requiresrequires 21.121.1 JJ toto raiseraise thethe temperaturetemperature ofof oneone molemole byby oneone kelvinkelvin degreedegree.. SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY CONSTANTCONSTANT VOLUMEVOLUME

HowHow muchmuch heatheat isis requiredrequired toto raiseraise thethe temperaturetemperature ofof 22 molesmoles o o ofof OO 2 fromfrom 00 CC toto 100100 C?C?

Q = nCv T

QQ == (2(2 mol)(21.1mol)(21.1 J/molJ/mol K)(373K)(373 KK -- 273273 K)K)

Q = +4220 J SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY CONSTANTCONSTANT VOLUMEVOLUME (Cont.)(Cont.) SinceSince thethe volumevolume hashas notnot changed,changed, nono workwork isis done.done. TheThe entireentire 42204220 JJ goesgoes toto increaseincrease thethe internalinternal energy,energy, UU.

QQ == UU == nCnCv TT == 42204220 JJ Thus, U is determined by the U = nCv T change of temperature and the specific heat at constant volume. SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY CONSTANTCONSTANT PRESSUREPRESSURE WeWe havehave justjust seenseen thatthat 42204220 JJ ofof heatheat werewere neededneeded atat constantconstant volumevolume.. SupposeSuppose wewe wantwant toto alsoalso dodo 10001000 JJ ofof workwork atat constantconstant pressurepressure?? Q = U + W Same Q = 4220 J + J

QQ == 52205220 JJ CCp >> CCv HEATHEAT CAPACITYCAPACITY (Cont.)(Cont.)

HeatHeat toto raiseraise temperaturetemperature ofof anan idealideal gas,gas, UU,, isis thethe samesame forfor anyany process.process.

U = nCv T

ForFor constantconstant pressurepressure CCp >> CCv QQ == UU ++ WW C  p nCnCp TT == nCnCv TT ++ PP VV Cv REMEMBER,REMEMBER, FORFOR ANYANY PROCESSPROCESS INVOLVINGINVOLVING ANAN IDEALIDEAL GAS:GAS:

PP AV VA PP BV VB PVPV == nRTnRT == TT A TT B

QQ == UU ++ WW UU == nCnCv TT ExampleExample Problem:Problem:

AA 22--LL samplesample ofof OxygenOxygen gasgas hashas anan initialinitial temptemp-- eratureerature andand pressurepressure ofof 200200 KK andand 11 atmatm.. TheThe gasgas undergoesundergoes fourfour processes:processes: •• AB:AB: HeatedHeated atat constantconstant VV toto 400400 K.K. •• BC:BC: HeatedHeated atat constantconstant PP toto 800800 K.K. •• CD:CD: CooledCooled atat constantconstant VV backback toto 11 atmatm.. •• DA:DA: CooledCooled atat constantconstant PP backback toto 200200 K.K. PVPV--DIAGRAMDIAGRAM FORFOR PROBLEMPROBLEM

BB 400 K 800 K HowHow manymany molesmoles PP B ofof OO2 areare present?present? 200 K 1 atm A ConsiderConsider pointpoint A:A: PVPV == nRTnRT 2 L

PV (101,300Pa)(0.002m3 ) n  0.122 mol RT (8.314J/mol K)(200K) PROCESSPROCESS AB:AB: ISOCHORICISOCHORIC

BB 400 K 800 K WhatWhat isis thethe pressurepressure PP B atat pointpoint B?B? 200 K 1 atm AA PP A PP B == 2 L TA T B

11 atmatm PP B P = 2 atm == B 200200 KK 400400 KK or 203 kPa PROCESSPROCESS AB:AB: QQ == UU ++ WW

Analyze first law BB 400 K 800 K for ISOCHORIC PP B process AB. AA 200 K WW == 00 1 atm

QQ == UU == nCnCv TT 2 L UU == (0.122(0.122 mol)(21.1mol)(21.1 J/molJ/mol K)(400K)(400 KK -- 200200 K)K)

Q = +514 J U = +514 J W = 0 PROCESSPROCESS BC:BC: ISOBARICISOBARIC

What is the volume BB 400 K 800 K PP B CC at point C (& D)? 200 K 1 atm VVB VV C DD == TT B TT C 2 L 4 L

22 LL VV C V = V = 4 L == C D 400400 KK 800800 KK FINDINGFINDING UU FORFOR PROCESSPROCESS BC.BC.

BB 400 K 800 K Process BC is 2 atm ISOBARIC. CC 200 K PP == 00 1 atm

UU == nCnCv TT 2 L 4 L UU == (0.122(0.122 mol)(21.1mol)(21.1 J/molJ/mol K)(800K)(800 KK -- 400400 K)K)

U = +1028 J FINDINGFINDING WW FORFOR PROCESSPROCESS BC.BC.

Work depends BB 400 K 800 K on change in V. 2 atm CC 200 K P = 0 1 atm

Work = P V 2 L 4 L

WW == (2(2 atm)(4atm)(4 LL -- 22 L)L) == 44 atmatm LL == 405405 JJ

W = +405 J FINDINGFINDING QQ FORFOR PROCESSPROCESS BC.BC.

Analyze first BB 400 K 800 K 2 atm law for BC. CC 200 K QQ == UU ++ WW 1 atm

QQ == +1028+1028 JJ ++ 405405 JJ 2 L 4 L QQ == +1433+1433 JJ

Q = 1433 J U = 1028 J W = +405 J PROCESSPROCESS CD:CD: ISOCHORICISOCHORIC

B 400 K 800 K What is temperature PB C at point D? 200 K 1 atm A D PP C PP D == 2 L TT C TT D

22 atmatm 11 atmatm == T D = 400 K 800 K TT D PROCESSPROCESS CD:CD: QQ == UU ++ WW

Analyze first law 400 K 800 K CC for ISOCHORIC PB process CD. 200 K 400 K WW == 00 1 atm DD

QQ == UU == nCnCv TT 2 L U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K)

Q = -1028 J U = -1028 J W = 0 FINDINGFINDING UU FORFOR PROCESSPROCESS DA.DA.

Process DA is 400 K 800 K ISOBARIC. 2 atm PP == 00 AA 200 K 400 K 1 atm DD

UU == nCnCv TT 2 L 4 L

UU == (0.122(0.122 mol)(21.1mol)(21.1 J/molJ/mol K)(400K)(400 KK -- 200200 K)K)

U = -514 J FINDINGFINDING WW FORFOR PROCESSPROCESS DA.DA.

WorkWork dependsdepends 400 K 800 K 2 atm onon changechange inin VV. A 200 K 400 K PP == 00 1 atm D

WorkWork == PPVV 2 L 4 L WW == (1(1 atm)(2atm)(2 LL -- 44 L)L) == --22 atmatm LL == --203203 JJ

W = -203 J FINDINGFINDING QQ FORFOR PROCESSPROCESS DA.DA.

Analyze first 400 K 800 K 2 atm law for DA. AA 200 K 400 K QQ == UU ++ WW 1 atm DD

QQ == --514514 JJ -- 203203 JJ 2 L 4 L QQ == --717717 JJ

Q = -717 J U = -514 J W = -203 J PROBLEMPROBLEM SUMMARYSUMMARY ForFor allall Q = U + W processes:processes: Q = U + W Process Q U W AB 514 J 514 J 0 BC 1433 J 1028 J 405 J CD -1028 J -1028 J 0 DA -717 J -514 J -203 J Totals 202 J 0 202 J

NETNET WORKWORK FORFOR COMPLETECOMPLETE CYCLECYCLE ISIS ENCLOSEDENCLOSED AREAAREA +404+404 JJ BB C B CC 2 atm 2 atm -202 J NegNeg 1 atm 1 atm

2 L 4 L 2 L 4 L

BB CC 2 atm Area = (1 atm)(2 L) 1 atm Net Work = 2 atm L = 202 J 2 L 4 L ADIABATIC EXAMPLE:

Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB ). What is the new pressure and temperature? ( = 1.4)

BB   PP B PP AV VA == PP BV VB AA PP A PA VA PBVB = Q = 0 TT A TT B VVB VVA ADIABATICADIABATIC (Cont.):(Cont.): FINDFIND PPB   B B PP AV VA == PP BV VB PP B 300 K SolveSolve forfor PP B: : 1 atm AA  VA PPBA  Q = 0 V VVB 12VVB B

1.4 12VB PPBA  PB = 32.4 atm VB or 3284 kPa 1.4 PB  (1 atm)(12) ADIABATICADIABATIC (Cont.):(Cont.): FINDFIND TTB

TT =?=? BB B PVA ABB PV 32.4 atm  TTAB 300 K 1 atm AA SolveSolve forfor TT Q = 0 B VVB 1212VVB

(1(1 atm)(12Vatm)(12VB ) ) (32.4(32.4 atm)(1atm)(1 VVB ) ) == (300(300 K)K) TT B

TB = 810 K 3 ADIABATICADIABATIC (Cont.):(Cont.): IfIf VV A == 9696 cmcm 3 andand VVA == 88 cmcm ,, FINDFIND WW

B 810 K 32.4 atm SinceSince QQ == 0,0, 300 K WW == -- UU 1 atm AA Q = 0 8 cm3 96 cm3

WW == -- UU == -- nCnCV TT && CCV == 21.121.1 j/molj/mol KK PVPV FindFind nn fromfrom PVPV == nRTnRT nn == pointpoint AA RTRT 3 ADIABATICADIABATIC (Cont.):(Cont.): IfIf VVA == 9696 cmcm 3 andand VVA == 88 cmcm ,, FINDFIND WW PVPV (101,300(101,300 Pa)(8Pa)(8 x10x10-6 mm3)) nn == == RTRT (8.314(8.314 J/molJ/mol K)(300K)(300 K)K)

nn == 0.0003250.000325 molmol && CCV == 21.121.1 j/molj/mol KK

TT == 810810 -- 300300 == 510510 KK BB 810 K 32.4 atm

WW == -- UU == -- nCnCV TT 300 K 1 atm AA W = - 3.50 J 8 cm3 96 cm3 HEATHEAT ENGINESENGINES A heat engine is any Hot Res. TH device which through Qhot a cyclic process: Wout

Engine •• AbsorbsAbsorbs heatheat QQhot

Qcold •• PerformsPerforms workwork WWout

Cold Res. TC •• RejectsRejects heatheat QQcold THETHE SECONDSECOND LAWLAW OFOF THERMODYNAMICSTHERMODYNAMICS

Hot Res. TH It is impossible to construct an engine that, operating in a Qhot Wout cycle, produces no effect Engine other than the extraction of heat from a reservoir and the Qcold performance of an equivalent amount of work. Cold Res. TC Not only can you not win (1st law); you can’t even break even (2nd law)! THETHE SECONDSECOND LAWLAW OFOF THERMODYNAMICSTHERMODYNAMICS

Hot Res. T Hot Res. TH H 400 J 400 J 100 J 400 J Engine Engine 300 J

Cold Res. TC Cold Res. TC

• A possible engine. • An IMPOSSIBLE engine. EFFICIENCYEFFICIENCY OFOF ANAN ENGINEENGINE

TheThe efficiencyefficiency ofof aa heatheat engineengine

Hot Res. TH isis thethe ratioratio ofof thethe netnet workwork done W to the heat input Q . QH W done W to the heat input QH . Engine W Q -Q e = = H C QC QH QH

Cold Res. TC QC e = 1 - QH EFFICIENCYEFFICIENCY EXAMPLEEXAMPLE

AnAn engineengine absorbsabsorbs 800800 JJ andand

Hot Res. TH wasteswastes 600600 JJ everyevery cycle.cycle. WhatWhat 800 J W isis thethe efficiency?efficiency? QC Engine e = 1 - 600 J QH 600 J Cold Res. TC e = 1 - e = 25% 800 J

Question: How many joules of work is done? EFFICIENCYEFFICIENCY OFOF ANAN IDEALIDEAL ENGINEENGINE (Carnot(Carnot Engine)Engine) For a perfect engine, the Hot Res. T H quantities Q of heat gained QH W and lost are proportional to the absolute temperatures T. Engine T -T Q H C C e = TH Cold Res. TC

TC e = 1 - TH ExampleExample 3:3: AA steamsteam engineengine absorbsabsorbs 600600 JJ ofof heatheat atat 500500 KK andand thethe exhaustexhaust temperaturetemperature isis 300300 KK.. IfIf thethe actualactual efficiencyefficiency isis onlyonly halfhalf ofof thethe idealideal efficiency,efficiency, howhow muchmuch workwork isis donedone duringduring eacheach cycle?cycle?

Actual e = 0.5e = 20% TC i e = 1 - T W H e = 300 K QH e = 1 - 500 K W = eQH = 0.20 (600 J)

e = 40% Work = 120 J REFRIGERATORSREFRIGERATORS AA refrigeratorrefrigerator isis anan engineengine operatingoperating inin reverse:reverse: HotHot Res.Res. TT H WorkWork isis donedone onon gasgas Qhot W extractingextracting heatheat fromfrom coldcold in reservoirreservoir andand depositingdepositing Engine heatheat intointo hothot reservoir.reservoir. W + Q = Q Qcold in cold hot

ColdCold Res.Res. TTC WIN = Qhot -Qcold THETHE SECONDSECOND LAWLAW FORFOR REFRIGERATORSREFRIGERATORS

ItIt isis impossibleimpossible toto constructconstruct aa

Hot Res. TH refrigeratorrefrigerator thatthat absorbsabsorbs

Qhot heatheat fromfrom aa coldcold reservoirreservoir andand depositsdeposits equalequal heatheat toto aa Engine hothot reservoirreservoir withwith WW == 0.0. Qcold

Cold Res. TC If this were possible, we could establish perpetual motion! COEFFICIENTCOEFFICIENT OFOF PERFORMANCEPERFORMANCE

TheThe COPCOP (K)(K) ofof aa heatheat Hot Res. T H engineengine isis thethe ratioratio ofof thethe QH W HEATHEAT QQ c extractedextracted toto thethe netnet WORKWORK donedone WW.. Engine Q Q QH C K = C = W Q -Q Cold Res. TC H C

ForFor anan IDEALIDEAL TH K = refrigerator:refrigerator: TH-TC COPCOP EXAMPLEEXAMPLE

500 K A Carnot refrigerator operates between 500 K and 400 K. It extracts 800 J from a cold Hot Res. TH reservoir during each cycle. QH W What is C.O.P., W and QH ? Eng ine TC 400400 KK 800 J K = = TH-TC 500500 KK -- 400400 KK

Cold Res. TC 400 K C.O.P. (K) = 4.0 COPCOP EXAMPLEEXAMPLE (Cont.)(Cont.)

500 K NextNext wewe willwill findfind QQH byby assumingassuming samesame KK forfor actualactual Hot Res. T H refrigeratorrefrigerator (Carnot).(Carnot). QH W Eng QC KK == ine Q -Q 800 J H C 800800 JJ Cold Res. TC 4.0 = 400 K QQ H -- 800800 JJ

QH = 1000 J COPCOP EXAMPLEEXAMPLE (Cont.)(Cont.)

500 K Now,Now, cancan youyou saysay howhow muchmuch Hot Res. TH workwork isis donedone inin eacheach cycle?cycle? 1000 J W Engine WorkWork == 10001000 JJ -- 800800 JJ 800 J Work = 200 J Cold Res. TC 400 K SummarySummary TheThe FirstFirst LawLaw ofof ThermodynamicsThermodynamics:: TheThe netnet heatheat takentaken inin byby aa systemsystem isis equalequal toto thethe sumsum ofof thethe changechange inin internalinternal energyenergy andand thethe workwork donedone byby thethe system.system. Q = U + W final - initial)

•• IsochoricIsochoric Process:Process: VV == 0,0, WW == 00 •• IsobaricIsobaric Process:Process: PP == 00 •• IsothermalIsothermal Process:Process: TT == 0,0, UU == 00 •• AdiabaticAdiabatic Process:Process: QQ == 00 SummarySummary (Cont.)(Cont.)

The Molar Units are:Joules QQ Specific Heat per mole per cc == nn TT capacity, C: Kelvin degree

The following are true for ANY process: PV PV Q = U + W A ABB TTA B

U = nCv T PV = nRT SummarySummary (Cont.)(Cont.)

TheThe SecondSecond LawLaw ofof Thermo:Thermo: ItIt isis Hot Res. TH impossibleimpossible toto constructconstruct anan engineengine Qhot Wout that,that, operatingoperating inin aa cycle,cycle, Engine producesproduces nono effecteffect otherother thanthan thethe extractionextraction ofof heatheat fromfrom aa reservoirreservoir Qcold andand thethe performanceperformance ofof anan equivalentequivalent amountamount ofof work.work. Cold Res. TC

Not only can you not win (1st law); you can’t even break even (2nd law)! SummarySummary (Cont.)(Cont.) The efficiency of a heat engine: T QC C e = 1 - Q e = 1 - H TH

The coefficient of performance of a refrigerator: QQ T K CC K  C WQQin H C TTHC CONCLUSION:CONCLUSION: ChapterChapter 2020 ThermodynamicsThermodynamics