Fundamentals of chemical engineering thermodynamics 1st edition themis matsoukas solutions manual by erin.dunbar383

Fundamentals of Chemical Engineering Thermodynamics 1st Edition Themis Matsoukas Solutions Manual Visit to Download in Full: https://testbankdeal.com/download/fundamentals-of-chemic al-engineering-thermodynamics-1st-edition-themis-matsoukas-solutions-manual/

Solutions Manual for Fundamentals of Chemical Engineering Thermodynamics Themis Matsoukas Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Cape town • Sydney • Tokyo • Singapore • Mexico City ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute. Fundamentals of Chemical Engineering Thermodynamics 1st Edition Themis Matsoukas Solutions Manual Visit TestBankDeal.com to get complete for all chapters

Note to the Instructor

An effort was made to update all solutions requiring steam tables to conform with the tables in Appendix E of the book, which are based on IAPWS95). It is possible, however, that some problems may make use of older tables. Be alert as such discrepancies could confuse students even though the final answers are not much different.

Please report any mistakes or typos to Themis Matsoukas: matsoukas@psu.edu.

The author and publisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein.

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This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

ISBN-10: 0-13-269320-8

ISBN-13: 978-0-13-269320-2

Contents 1Introduction 5 2PhaseDiagramsofPureFluids 13 3Energy&theFirstLaw 59 4Entropy&theSecondLaw 103 5CalculationofProperties 125 6BalancesinOpenSystems 143 7VLEofpureFluid 193 8PhaseBehaviorofMixtures 223 9PropertiesofMixtures 245 10VLEofMixture 271 11IdealSolution 291 12Non-IdealSolution 311 13Miscibility,SolubilityandotherPhaseEquilibria 347 14Reactions 371 3 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

1.INTRODUCTION

Problem1.1Solution

Wewillneedthefollowingunitconversions:

(Noteontheconversionmoltolbmol:onemolhasamassequaltothemolecularweighting,whileone lbmolhasamassequalequaltothemolecularweightinlb.)Wealsoneedthemolarmassofammonia

1 ft D 0:3048 m;1 lb D 0:454 kg;1 lbmol D 454 mol

whichis Mm D 17 g/mol D 17 10 3 kg/mol a)SpeciﬁcVolume: V D 1 41:3 lb/ft3 D 0:02421 ft3/lb D 0:00151 m3/kg D 1:51 cm3/g b)MolarVolume Vmolar D VMm D 2:567 10 5 m3/mol D 25:67 cm3/mol D 0:4116 ft3/lb-mol 6 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Problem1.2Solution Firstwewritethegivenequationas TK D B A .ln Pbar/=.ln 10/ C where TK refersto T inkelvin, Pbar referstopressureinbar,andthelogarithmisnatural.Nextweuse TF D 1:8.TK 273:15/ C 32 and Pbar D Ppsi 14:5 andsubstitutethesevaluesintotheaboveequation.Aftersomemanipulationtheresultis TF D B.1:8/.ln 10/ A ln 10 C ln.14:5/ ln Ppsi C.1:8/ .1:8/.273:15/ C 32 Doingthealgebra, TF D 5062:37 13:2153 ln Ppsi 302:217 Therefore, A0 D 13:2153;B0 D 5062:37C 0 D 302:217 7 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Thisresultdependsonlyontemperatureandsinceallthreephasesarethesametemperature,moleculeshave thesamemeanvelocityinallthreephases.

b)Themeankineticenergyis

Thisisthemeankineticenergypermoleculeinallphases.Thenumberofmoleculesin1kgofwateris

Comment: Thisisthe translational kineticenergyofthemolecule,i.e.,thekineticenergyduetothe motionofthecenterofmass.Awatermoleculepossessesadditionalformsofkineticenergythatarise fromtherotationofthemolecule,thebendingofbonds,andthevibrationofoxygenandhydrogenatoms abouttheirequilibriumpositions.ThesearenotincludedinthiscalculationastheMaxwell-Boltzmann distributionrefersspeciﬁcallytothetranslationalkineticenergy.

c)Theabovecalculationshowsthatthemeankineticenergydepends onlyontemperature (itisindependent ofpressureorofthemassofthemolecule).Therefore,oxygenat0.01 ıChasthesamekineticenergyas wateratthesametemperature:

Problem1.3Solution

v D 8kB T m 1=2 where kB D 1:38 10 23 J=K, T D 273:16 K.Themassofthewatermoleculeis m D Mm NAV D 18 10 3 kg=mol 6:022 1023 mol 1 D 2:98904 10 26 kg Themeanvelocityis v D 566:699 m=s D 2040:12 km=h D 1267 mph

1.INTRODUCTION

a)Themeanvelocityis

Ekin D 1 2 mv2 D 1 2 mv2 where v2 isthemeansquaredvelocity, v2 D 3kB T m : Withthisthemeankineticenergyis Ekin D 3 2 kB T D 5:65441 10 21 J

N D 1 kg 18 10 3 kg=mol 6=022 1023 D 3:34556 1025

ıC(regardlessofphase)is Ekin D 5:65441 10 21 J 3:34556 1025 D 189;171 J D 189 kJ

Thetotalkineticenergyin1kgofwaterat0.01

Ekin D 3 2 kB T D 5:65441 10 21 J 8

a)Thepotentialhasaminimumjustabove4

derivativeofthepotentialequaltozerosolveforthevalueof

b)Ifweimagine

3 4 5 6 7 8 9 10 4. 10 22 2. 10 22 0 2. 10 22 4. 10 22 r Joule Potential

Problem1.4Solution

˚ Aorso.Todeterminethisvalueaccuratelywemustsetthe

.Todothiseasily,wedeﬁneanewvariable x D r= andrewritethepotentialas: ˆ D a x 12 x 6 Bychainrulewenowhave: dF dr D dF dx dx dr D 12x 13 C 6x 7 dx dr Settingthistozeroandsolvingfor x wehave: 12x 13 C 6x 7 D 0 ) x D 21=6 Since r D x,thevalueof r thatminimizesthepotentialis r D 21=6 D .4:24964/.3:786 ˚ A/ D 4:25 ˚ A

N moleculestobesituatedatthecenterofcubeswhosesidesareequalto r , L L thevolumeoccupiedis V D N.r /3 These N moleculescorrespondto N=NAv molandtheirtotalmassis M D N NAV Mm where Mm isthemolarmassofmethane(Mm D 16 10 3 kg/mol).Forthedensity,therefore,weobtain thefollowingﬁnalformula: D Mm NAV.r /3 Bynumericalsubstitutionweﬁnallyobtainthedensity: D 16 10 3 kg/mol .6:024 1023 mol 1/.4:25 ˚ A/.10 10 m/ ˚ A/ D 346 kg/m3 D 0:346 g/cm3 9 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

c)SpeciﬁcvolumesofsaturatedliquidmethanearelistedinPerry’sHandbookfromwhichwecancompute thedensities.Wenotice(aswewouldhaveexpected)thatliquidvolumesnearthecriticalpoint(Tc D 190:55 K)varywithpressure,from162.3kg/m3 atthecriticalpointto454kg/m3 around90K.Ourvalue correspondstoPerry’stabulationatabout160K.Ourcalculationisapproximateanddoesnotincorporate theeffectofpressureandtemperature.Noticethatifwepickadistancesomewhatdifferentfrom r the resultwillchangequiteabitbecauseifthethirdpowertowhichthisdistanceisraised.Buttheimportant conclusionisthatthecalculationplacedthedensityrightinthecorrectrangebetweenthelowestandhighest valueslistedinthetables.Thissaysthatourmolecularpictureoftheliquid,howeveridealized,isfairlyclose toreality.

1.I

NTRODUCTION

Problem1.5Solution

Assumingmoleculestobesittingatthecenterofacubiclatticewithspacing

moleculesis NL3 andthedensityis

Theintermoleculardistanceisabout10timeslargerinthevapor.

L,thevolumeoccupiedby N

D Mm NAVL3 where Mm isthemolarmass.Solvingfortheintermoleculardistance, L D Mm NAV a)Forliquidwaterwith D 1000 kg=m3 , L D 18 10 3 kg=mol .6:022 1023 mol 1/.1000 kg=m3/ 1=3 D 3:10 10 10 m D 3:01 ˚ A b)Forsteamwith D 0:4 kg=m3 L D 18 10 3 kg=mol .6:022 1023 mol 1/.0:4 kg=m3/ 1=3 D 4:2 10 9 m D 42 ˚ A

Problem1.6Solution

Theforceneededtoseparatethetwohavesisequaltotheforcethatisexertedbypressureononehemisphere:

where P isthepressuredifferencebetweentheatmosphericpressureandthecontentsofthesphere,and A isthecross-sectionalareaofthesphere( R2).Assumingthespheretobefullyevacuated,thepressure differenceisequaltotheatmosphericpressure( 1 bar D 105 Pa).Theforcenowis

Assuminganaverageweightof80kg/person,itwouldtaketheweightof25people(!)toseparatethe spheres.

1.INTRODUCTION

F P0 P=0 2 R

D PA

F D P0. R2/ D .105 Pa/. 0:52 m2/ D 19;635 N Thisforcecorrespondstoamass M D F g D 39269:9 N 9:81 N=kg D 2;000 kg

2PhaseDiagramsofPureFluids

2.PHASE DIAGRAMSOF PURE FLUIDS

Problem2.1Solution a)25 ıC,1bar:liquid,becausethetemperatureisbelowthesaturationtemperature at1bar(99.63ıC).

10bar,80ıC:liquid,becausethetemperatureisbelowthesaturationtemperatureat10bar(179.88 ıC). 120 ıC,50bar:At120 ıCthevaporpressureis198.54kPa=1.9884bar.Sincetheactualpressureishigher, thestateisliquid.(Or,at50barthesaturationtemperatureis263.91 ıC.Sincetheactualtemperatureis lowerthestateisliquid.)

b)Liquid,becausethepressure(1atm=760mmHg)ishigherthanthevaporpressureofbromobenzene (10mgH)atthesametemperature.

c)Liquid,becausethetemperatureislowerthantheboilingpointatthesamepressure.

Note: Allofthesestatementswillmakebettersenseifyouplotyourinformationonthe PV graph.

Problem2.2Solution a)Fromsteamtablesat40barwecollectthefollowingdata:

600 ıC 650 ıC 40 bar 0:0989 m3=kg 0:1049 m3=kg Byinterpolationat V D 100 g=cm3 D 0:1 m3=kgweﬁnd T D 600 C 650 600 0:1049 0:0989.0:1 0:0989/ D 609:2 ıC

ıC:thephaseisvapor.

VL D 0:0011 m3=kg) andsaturatedvapor(VV D 0:3156):thesystemisavapor/liquidmixture.Thevaporandliquidfractionsare obtainedbyleverrule: xL D 0:3156 0:1 0:3156 0:0011 D 0:314 D 31:4% xV D 1 0:314 D 0:686 D 68:6% 15 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Thesystemisat40bar,609.2

b)At6barweﬁndthatthedesiredvolumeliesbetweenthatofthesaturatedliquid(

2.PHASE DIAGRAMSOF PURE FLUIDS

b)Sincethesystemisamixtureofliquidandvapor,itissaturated.Fromthesaturatedtablesat20barwe ﬁndbyinterpolationthefollowing:

V D 2219:2 cm3/g D 2:2192 m3/kg

3,themassis M D V t V D 3:5 m3 2:2192 m3/kg D 1:577 kg

Problem2.3Solution a)At1bar,210 ıC,thespeciﬁcvolumeofsteamisfoundbyinterpolationtobe

Sincethetotalvolumeofthevesselis3.5m

T D 212:37 ıC;VL D 1:18 cm3/g;VV D 99:54 cm3/g Thespeciﬁcvolumeofthemixtureis V D lVL C vVV D .0:85/.1:18/ C .0:15/.99:54/ D 15:93 cm3/g Thetotalvolumerequiredforstorageis V t D MV ,where V isthespeciﬁcvolumecalculatedabove: V t D MV D .525 kg/.1000 g/kg/.15:93 cm3/g/ D 8363250 cm3 D 8:36 m3 16 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Problem2.4SolutionSolution Initialstate: P D 45 bar;T D 257:41 ıC;V L D 1:269 cm3/g;V V D 44:037 cm3/g Thespecificvolumeisobtainedfromtheleverrulewith X L D 0:25: V D x LV L C .1 x L/V V D .0:25/.1:269/ C .1 0:25/.44:037/ D 33:345 cm3/g a)Weknow P D 80 bar, V D 33:345 cm3/g.Fromthesteamtableswefind: T D 375 ıC;V D 32:222 cm3/g T D 400 ıC;V D 34:31 cm3/g Byinterpolationat V D 33:345 cm3/gweobtain T D 388:5 ıC. b)Fromsaturatedtableswefind: T D 272 ıC;P D 56:802 bar;V D 34:42 cm3/g T D 274 ıC;P D 58:587 bar;V D 33:29 cm3/g Byinterpolationat V D 33:345 cm3/gwefind P D 57:76 bar. c)Thevolumefractionoftheliquid, L,andthemassfractionoftheliquid, xL,arerelatedasfollows: vol.occupiedbyliq. vol.occupiedbyliq.+vol.occupiedbyvap D xLV L xLV L C xV V V fromwhich, x L D V V L V L C .V V V L/ L Thesaturatedvolumesarethoseattheinitialstateandwith L D 0:25 wefind x L D .44:037/.0:25/ .1:269/ C .44:037 1:269/.0:25/ D 0:920429

V 0 D .0:920429/.1:269/ C .1 0:920429/.44:037/ D 4:6721 cm3/g Ifwehavemass M initiallyandmass M 0 attheend,wecanwriteforthetankvolume: V t D NV D V t M 0 ) M 0 M D V V 0 or M M 0 M D V V 0 1 D 33:345 4:6721 D 6:14 Therefore,wemustadd6.14timestheoriginalmass(that’s614%). 17 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Therefore,thenewspeciﬁcvolumeinthetankis

Problem2.5Solution P.bar/0:515 V.m3=kg/1:1240:031 .kg=m3/0:8896832:2581

Theseresultsarequitedifferent.Whichoneshouldwepick? Forthemolarvolumewehave

Weconcludethatwhile V is inversely proportionalto P , 1=V isproportional,providedthat Z doesnotvary muchinthegivenrange.Therefore,wewouldaccepttheinterpolationin

2.PHASE DIAGRAMSOF PURE FLUIDS

V D

Ifweinterpolatefor

D 25:7681 kg=m3 ) V D 1= D 0:0388077 m3=kg:

Ifweinterpolatefor V weﬁnd

0:257138 m3=kg:

weﬁnd

V D ZRT P : Using D 1=V ,themolardensityis D P ZRT (2.1)

Problem2.6Solution

a)198.3C

b)Weﬁrstobtainthespeciﬁcvolumesofthesaturatedliquidandvaporat15bar:

V L D 0:0015 m3=kg;V V D 0:1317 m3=kg Nextwecalculatethetotalmassofeachphase: m L D .0:5 m3/.0:0015 m3=kg/ D 434:8 kg mV D .11:5 m3/.0:1317 m3=kg/ D 87:3 kg Thetotalmassis m D 434:8 C 87:3 D 522:1 kg c)Thequalityofthesteamis: xV D 87:3 522:1 D 0:167 d)Thespeciﬁcvolumeafter87%ofthemassisremovedis V 0 D 0:13 522:1 12 m3 D 0:1768 m3=kg Thetemperatureisobtainedbyinterpolationat15barbetween250 ıCand300 ıC.Weﬁnd T 0 D 320 ıC 19 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Problem2.7Solution a)Sincethecookercontainsbothvaporandliquid,thestateissaturatedsteam. Therefore, T D T sat D 120:23 ıC.

b)Fromsteamtablesweobtainthespeciﬁcvolumesofthesaturatedphasees:

Thetotalvolumeoftheliquidinthecookeris

Eventhoughthevaporoccupies75%ofthevolume,itonlyrepresents0.4%ofthetotalmass.

d)Thequicksolutionistotakealookatthe PV graph.Theinitialstateisat A andtheﬁnalstate, B,is reachedbyconstant-volumecooling.Thisstateisobviouslyinthetwo-phaseregionbecausetheoriginating statewasalsoinsidethatregionaswell.Weconcludethat

3:166 kPa.(If,however, state A wereinthesuperheatedregion,wewouldbeabletotellif B issuperheatedorvapor/liquidandwe wouldhavetodothesolutioninmoredetailasshownbelow.)

Detailedsolution: Thetotalvolumeofthesystemaswellasthespeciﬁcvolumeremainconstant.The speciﬁcvolumeis

2.PHASE DIAGRAMSOF PURE FLUIDS

V L D 1:061 cm3/g;V V D 885:44 cm3/g

V L;tot D .0:25/.8/ liter=2000cm3

mL D V L;tot V L D 2000 1:061 D 1885:0 g

V L;tot D .0:75/.8/ liter=6000cm3

mV D V L;tot V V D 6000 885:44 D 6:78 g Thetotalmassis m D 1885:0 C 6:78 D 1891:78 g D 1:89 kg

xL),andofthevapor(xV )are xL D 1885:0 1885:0 C 6:78 D 0:9964 D 99:64%;xV D 1 xL D 0:004 D 0:4%

.Therefore,themassof theliquidis

Thevolumeofthevaporinthecookeris

anditsmassisgivenby

c)Themassfractionsoftheliquid(

P D P sat.25 ıC

V D Vcooker Mtot D 8000 cm3 1891:78 g D 4:229 cm3/g Fromthesaturatedsteamtablesat25 ıCweﬁnd, V L D 1:003 cm3/g, V V D 43400 cm3/g.Thespeciﬁc volumeofthesystemisbetweenthesetwovalues,thereforewestillhaveasaturatedsystem.Weconclude that P D P sat.25 ıC/ D 3:166 kPa D 0:03166 bar. 20

Thissolutionismoregeneralandwillworkregardlessofwheretheinitialstateis.

e)Whenthesystemhascooled,theoutsidepressureis1barandtheinsidepressureis0.03166bar.Therefore,thelidremainsclosedundertheactionofthispressuredifference.Theforceis

Toputthisforceintoperspectivewecalculatethemasswhoseweightis12166N:

Ifyoucanlift2700–2800lbthenyoucouldremovethatlid!(Note: Whetheryoutaketheoutsidepressure tobe1baror1atmorsomethingsimilar,theconclusionremainsthattherequiredforceisindeedverylarge.

P V initial state const. V process nal state 2 bar 120.21°C 25 °C 0.003175 bar

F D R2.Pout Pin/ D . /.0:2 m2/.1 0:03166/ bar 105N/bar D 12166 N

M D F g D 12166 N 9:81 m/s2 D 1240 kg D 2732 lb

Problem2.8Solution

Itisveryhelpfultodrawthe PV graph,shownbelow:

a)Theinitialstate(A)iscompressedliquid(80 ıC,1.013bar).Theprocessisconductedunderconstant volume.Assumingtheisothermtobevertical,theﬁnalstate,B,isatthesametemperature and onthe saturationline.Therefore: T D 80 ıC, P D 47:36 kPa.

Noticethatthetemperaturehasnotchanged.Thisisaconsequenceofthefactthatwehaveapproximated theisothermwithaverticalline.Inreality,theisothermisnotverticalandstateBshouldbeatatemperature somewhatbelow80 ıC.However,thesteepnessoftheisothermmeansthatthistemperatureis veryclose to 80 ıC.Ifwehadthevalueof and ˇ wecouldcalculatethistemperatureandwouldverifythatitisindeed extremelycloseto80 ıC.

b)TheinitialstateisaV/Lmixure(stateC).Theprocessisunderconstantvolume,therefore,theﬁnalstate islocatedattheintersectionoftheverticallinethrough C andthesaturationline(stateD).Fromsteam tables:

1 bar; saturated:VL D 1:043;V V D 1693:7

Themassfractionsoftheliquidandthevaporare:

2.PHASE DIAGRAMSOF PURE FLUIDS

1 bar A B 80 C 1 bar C D' C' D E F G H part (a) part (b) part (c) part (d)

l D 0:5=1:043 0:5=1:043 C 0:5=1693:7 D 0:99938 v D 1 l D 0:00062 22

Note1: Eventhoughthemassfractionofthevaporisverynearly0,itwouldnotbecorrecttosetitequalto 0.While v issmall,whenmultipliedbyalarge V V itmakesasignificantcontributiontothespecific volumeofthemixture.Ifwehadset v ,wewouldhaveconcludedthattheinitialstateispractically saturatedliquidwhichmeansthatthefinalpressureisalmost1bar.Clearly,thisapproximationmisses therightpressurebymorethan200bar!!!

Note2: Inthiscasethespecificvolumeofthevapor-liquidmixturewasveryclosetotheliquidsideandfor thisreasonthefinalstatewas liquid.Inotherwords,underheatingthevaporcondensesandbecomes liquid.If,however,theinitialvolumewasmuchcloserthevaporside(stateC’),thenheatingwould produce vapor.Inthiscase,heatingwouldcausetheliquidtoevaporate.Thatis,afterheatingthe contentsofthevesselthefinalstatemighteithersaturatedvapororsaturatedliquid.Canyouestablish acriterionfortheinitialspecificvolumetodeterminewhetherthefinalstateisvapororliquid?

c)Theﬁnalstatesaturatedvapor(stateF).Theprocessiscoolingunderconstantvolume,therefore,the initialstatemustbesomewhereontheverticallinethroughFandabovepointF(sincecoolingimpliesthat theinitialstateisathigher T ).Weconcludetheinitialstateissuperheatedvapor.

d)Bysimilarargumentsasabove,wedeterminethattheinitialstateis vapor/liquid mixture.Noticethat hereweareheatingavapor/liquidmixtureandasaresultthevaporcondensestoproducesaturatedliquid!

3/g) V D .0:99938/.1:043/ C .0:00062/.1693:7/ D 2:085 Attheﬁnalstate: V D 2:085,andsaturated.Fromsteamtablesweﬁnd, T 367 ıC, P D 203:13 bar (saturatedliquid).

Thespeciﬁcvolumeattheinitialstateis(incm

Problem2.9Solution Thespeciﬁcvolumeofwaterundertheseconditionsinthetankis

V D 12 m3 6:2 kg D 1:935 m3=kg:

Accordingtothesteamtables,thevolumeofsaturatedvaporat1.4barisbetween1.694m3=kg(at1bar) and1.1594m3=kg(at1.5bar).Thecalculatedvalueishigher,therefore,thestateistotherightofthe saturatedvaporanditmustbe superheated.

b)Weneedanentryinthesteamtablessuchthatpressureis1.4barandthespeciﬁcvolumeis 1:935 m3=kg. Tolocatethisstate,weinterpolatedinthesteamtablesbetween1.0barand1.5baratvarioustemperatures andconstructthetablebelow:

c)Ifweaddmoresteamwhilekeepingtemperatureconstantto300.6C,pressurewillincreaseandthe speciﬁcvolumewilldecrease.Thevaporwillbecomesaturatedwhenthespeciﬁcvolumeinthetankisthat ofsaturatedvaporat300.6C.Weobtainthisvaluebyinterpolationinthesaturatedsteamtablesbetween 300 ı

2.PHASE DIAGRAMSOF P

URE FLUIDS

T 150200250300350400450 V.at 1:0 bar/ 1:93672:17252:40622:63892:87103:10273:3342 V.at 1:4 bar/ 1:41581:59011:76231:93352:10402:27422:4442 V.at 1:5 bar/ 1:28561:44451:60131:75711:91232:06712:2217 Wecannowseethatthedesiredvalueisbetween300 ıCand350 ıC.Byinterpolationbetweenthesetwo temperaturesweﬁnd T D 300 ıC C .350 300/ ıC .1:5 1:0/ bar .1:4 1:0/ bar D 300:6 C

Cand302

C: T ıC V m3=kg 3000:02166 300:5930:0211498 3020:01994 Thespeciﬁcvolumewhenthetankissaturatedis V D 0:0211498 m3=kg.Thetotalmassis 12 m3 0:0211498 m3=kg 6:2 kg D 567:4 kg Theamountthatmustbeaddedis .567:4 6:2/ D 561:2 kg 24 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Problem2.10SolutionSolution

Thegraphsbelowshowthe PV invariouscombinationsoflinearandlogarithmiccoordinatesandthe ZP graph.

Comments:

Thevolumesspanaverywiderangeandinordertoseetheshapeofthesaturationline,wemustplot onlyasmallerrange.Intheabovegraph,thevolumeaxisrangesfrom0to100cm3/mol.

Bydoingthe V axisinlogcoordinateswecannowlookataverywiderangeofvalueswithout squeezingthegraphintonothingness.Noticethatinthelogplotthevolumegoesfrom1to100,000 cm3/g.

Thesteamtablesdonotcontaindataforthecompressedliquidregionandsoourisothermsstopatthe saturatedliquid.Wecouldextrapolatethemintotheliquidbydrawingthemasverticallines.

The ZP graphhasthefamiliarlook.Noticethattheisothermsarebetterseparatedonthisgraph.

1.0 0.8 0.6 0.4 0.2 0.0 Compressibility Factor 250 200 150 100 50 0 Pressure (bar) 100 C 200 C 300 C 400 C 0.01 0.1 1 10 100 1000 Pressure (bar) 100 10 1 10 2 103 10 4 105 Specific volume (cm3/g) 100 C 200 C 300 C 400 C 250 200 150 100 50 0 Pressure (bar) 1 2 4 6 10 2 4 6 100 2 4 6 1000 Specific volume (cm3/g) 100 C 200 C 300 C 400 C 250 200 150 100 50 0 Pressure (bar) 100 80 60 40 20 0 Specific volume (cm3/g) 100 C 200 C 300 C 400 C

2.PHASE DIAGRAMSOF PURE FLUIDS

Problem2.11Solution

a)Weﬁndthatforethane,

Thereducedconditionsare

Thedesiredisothermisbetween

thetruncatedvirialequationisvalid.

b)Thereducedconditionsare

Z cannotﬁndat Tr D 0:80 underthegivenpressure.therefore,thetruncatedvirialequationisnotvalid.

c)Sincethetemperatureis 35ıC,lowerthanthatboilingpoint,theethaneisliquid.Thetruncatedvirial equationisnotvalid.

Tc D 305:3K;Pc D 48:72bar

Tr D 298:15 K 305:3 K D 0:976;Pr D 10 bar 48:72 bar D 0:21

Tr D 0:95 and Tr D 1:0.Bygraphicalinterpolationweﬁnd Z 0:92

.the isothermisfairlylinearbetween Pr D 0 andthevalueof Pr correspondingtothegivenpressure,therefore,

Tr D 244:15 K 305:3 K D 0:80;Pr D 10 bar 48:72 bar D 0:21

Problem2.12Solution a)Vapor.

b)Themolarmasscanbeobtainedfromtherelationshipbetweenmolarvolumeanddensity:

Weknowdensityattwopressures,soweneedthemolarvolumeinoneofthem.Wechoosethelowest pressurebecauseat0.01weare justiﬁed (below)toassumeideal-gasstate:

Justiﬁcation: Thecriticalpressureisnotknownbutitmustbehigherthanthesaturationpressureat25C, whichis64.3bar.Thatis,thereducedpressureisatmost

Fromgeneralizedgraphsisitclearthatatsuchlowreducedpressuresthestateisessentiallyideal.

c)Thesecondvirialcoefﬁcientcanbecalculatedfromthetruncatedvirialequation

Wejustifytheuseofthisequationat25 ıC,20barasfollows:

Justiﬁcation: Thereducedtemperatureis

.Thereducedpressureisnotknownbutitmustbe less than

sincethecriticalpressuremustbehigherthan64.3bar.Fromgeneralizedgraphsweseethatfor Tr >1, theisotherminthepressurerange Pr D 0 uptoabout0.31isquitelinear.Thisofcourseisajudgementcall butisasgoodaswecandowiththeinformationwehave.

Solvingthetruncatedvirialfor B:

d)Wewillanswerthisquestionusingthetruncatedvirialequation

Justiﬁcation: Ifthetruncatedvirialisvalidat25 ıC,20bar,asassumedabove,itisvalidforallpressures lessthan20bar.

D Mm V ) Mm D V

0:01 bar 64:3 bar D 0:0002

Themolarvolumeis V D RT P D .8:314 J=mol=K/.298:15 K/ 0:01 105 Pa D 2:47882 m3=mol Themolarmassis Mm D 0:177 kg=m3 .2:47882 m3=mol/ D 43:8 10 3 kg=mol

PV RT D 1 C BP RT

Tr D 0:97

20 bar 64:3 bar D 0:31

B D V RT P where V D Mm= .Usingthedataat20barwith V D 0:00110239 m3=molweobtain: B D 1:370 10 4 m3=mol

2.PHASE DIAGRAMSOF PURE FLUIDS Solvingthetruncatedvirialfor V andusing P D 12 bar: V D RT P B D 0:00192866 m3=mol Thenumberofmolestobestoredis n D 20 kg 43:9 10 3 kg=mol D 456 mol andthevolumeofthetankis V tank D .456 mol/.0:00192866 m3=mol/ D 0:88 m3 28 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Problem2.13Solution

a)Wecollectthedataforthisproblem:

WecalculatethesecondvirialusingthePitzerequation:

Thecompressibilityfactoris

b)Themolarvolumeinthetankis

c)Ifwedoublethenumberofmoles,thenewmolarvolumeinthetankis

Weusethetruncatevirialtosolveforthenewpressure P2 (sincetemperatureisthesameasbefore,the secondvirialdoesnotneedtoberecalculated):

d)Tovalidatetheapplicabilityofthetruncatedvirialwecheckwiththe Z0 graphandnoticethatisotherms around Tr D 1:6 remainlinearuptofairlyhighpressures.Thepressureofthisproblem, P D 38:7 bar correspondsto Pr D 0:84,whichisstillwithinthelinearrangeoftheisotherm,asfaraswecantellby nakedeye.

Tc D 190:56 K Pc D 45:99 bar ! D 0:011

Tr D

Pr D

B0 D

B1 D

B D 0:0000420106 m3=mol

1:5646

0:434877

0:12319

0:112756

Z D BPc RT D

0:966104:

V D ZRT P D 0:0011974 m3=mol Thetotalnumberofmolesis n D V tank V D 835:144 mol

V2 D V tank 2n D V 2 D 0:000598699 m3=mol

PV RT D 1 C BP RT ) P D RT V B D 38:7 bar

2.PHASE DIAGRAMSOF PURE FLUIDS

Summaryofresults

P D 20 bar

T D 298:15vK

Vtank 1:0 m3

Tc D 190:56 K

Pc D 45:99 bar

! D 0:011

Tr D 1:5646

Pr D 0:434877

B0 D 0:12319

B1 D 0:112756

B D 0:0000420106 m3=mol

Z D 0:966104

V D 0:0011974 m3=mol

n D 835:144 mol

n2 D 1670:29 mol

V2 D 0:000598699 m3=mol

P2 D 38:6886 bar

Problem2.14Solution

b)Inthisprocessthespeciﬁcvolumestaysconstant.Forthetwounkowns, P and T ,wehavethefollowing twoequations:

Theseshouldbesolvedbytrialanderror.Forexample,specify T ,solvefor P sat fromthesecondequation, solvefor V fromﬁrstequation,andiftheanswerdoesnotmatchtheknownvolume,tryagain.Thesolution isasfollowing:

Pc D 34:0 bar;Tc D 126:2 K;! D 0:038 Thereducedconditionsare Pr D 7 bar 34:0 bar D 0:206;Tr D 110 K 126:2 K D 0:872 FromEqs(2.28),(2.29)wehave B 0 D 0:083 0:422 0:8721:6 D 0:442395 B 1 D 0:139 0:172 0:8724:2 D 0:166745 B 0 C !B 1 D 0:442395 0:166745 0:038 D 0:448731 ThesecondvirialcoefﬁcientiscalculatedformEq.(2.27) B D RTc Pc .B 0 C !B 1/ D . 0:448731/.8:314 J/molK/.126:2 K/ 34:0 105 Pa D 1:3848 10 4 m3/mol Thecompressibilityfactoris Z D 1 C BP RT D 1 .1:3848 10 4m3/mol/.7 105 Pa/ .8:314 J/molK/.110 K/ D 0:8940 Themolarvolumeisﬁnallycalculatedtobe V D ZRT P D .0:8940/.8:314 J/molK/.110 K/ 7 105 Pa D 1:168 10 3 m3/mol Themassofnitrogeninthetank m D nMw D V tank V Mw D 5 m3 1:168 10 3 m3/mol 28:014 g/mol D 119:92 kg

a)Thecriticalconstantsandacentricfactorofnitrogenare:

P satV RT D 1 C BP sat RT P sat D e 14:9542 588:72 6:6CT

T (K) BP sat (mmHg) ZV (m3/mol) 100-1.6710 10 4 5715.790.8468509.2393 10 4 90-2.0477 10 4 2684.390.9020591.8860 10 3 99-1.7038 10 4 5338.880.8526559.8597 10 4 98-1.7377 10 4 4979.380.8584141.0535 10 3 97-1.7726 10 4 4636.940.8641191.1273 10 3 96-1.8085 10 4 4311.170.8697661.2078 10 3 96.48-1.7911 10 4 4465.490.8670631.1682 10 3 31 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

F

is P D 4465:49 mmHg D 5:95 105 Pa.

c)Thereducedstateinpart(a)is Tr D 0:87 and Pr D

Tr D 0:80 and Tr D 0:90.Fromageneralized Z P graphweseethatisothermsinthisrangearefairlylinear, thereforethetruncatedvirialisacceptable.

Inpart(b)wefound Tr D 0:76 and Pr D 0:175.Thedesiredisothermisbetween Tr D 0:70 and Tr D 0:80. Again,theisothermisfairlylinearbetween Pr D 0 andthevalueof Pr correspondingtothegivenpressure, therefore,thetruncatedvirialequationisvalid.

/mol.Thereforethetempetureis96.48K,andthepressure

2.PHASE DIAGRAMSOF PURE

LUIDS Thevolumesolvedinparta)is1.168 10 3 m3

.Thedesiredisothermisbetween

0:206

Problem2.15Solution a)Thesecondvirialcoefﬁcientisdirectlyrelatedtotheslopeofanisothermon the ZP graph.Speciﬁcally,

Thissuggeststhefollowinggraphicalsolution:calculate Z fromthesteamtablesatconstant T ,plotthem versuspressure,andobtaintheslopeofthelinenear P D 0.Tofacilitatecalculations,supposethat P isin kPa, V

Fromthesteamtablesat200 ıCwehave:

Noticethatwehadtogoupto400kPa(4bar)toseeenoughchangein Z sothatwecanobtaintheslopeof theline.Thegraphisshownbelow:

Thelineshownistangenttothepointsat P D 0 anditsslopeis 5:00554 10 5.Theeasywaytodraw thislineistoﬁtastraightlinethroughthepointsclosesttotheorigin,saybelow200kPa.Asmarterwayis touseallthepointsanda quadratic equation:

Then,theequationofthetangentlineat P D 0 is(why?)

anditsslopeis b.Followingthisprocedureweﬁnd

Z D 1 C BP RT C ) BP RT D @Z @P T ˇ ˇ ˇ ˇP D0

P (kPa) V (cm3

Z 12183500.9991 10218200.9984 20109000.9975 307267.50.9976 405447.80.9971 504356.0.9966 752900.20.9953 1002172.30.9940 2001080.40.9887 250861.980.9861 300716.350.9834 350612.310.9806 400534.360.9780

/g)

1.00 0.99 0.98 0.97 0.96 0.95 Z 500 400 300 200 100 0 Pressure(kPa) 200C

f.P/ D aP 2 C bP C c

bP C c

slope D 5:00554 10 5 kPa 1 D 5:00554 10 8 Pa 1 33

2.PHASE DIAGRAMSOF PURE FLUIDS

b)Usingtheabovevalueof

B D .slope/.RT/ D . 5:00554 10 8 Pa 1/.8:314 J/molK/.473:15 K/ D 1:969 10 4 m3/mol

Thesecondvirialcoefﬁcientis

B wecancalculatethemolarvolumeofwaterat14barasfollows: PV RT D 1 C BP RT ) V D RT P C B D .8:314 J/molK/.473:15 K/ 14 105 Pa 1:969 10 4 m3/mol D 2:613 10 3 m3/mol D 145:15 cm3/g

3/g.Thisagreementisverygoodindicatingthatthetruncated virialequationisvalidattheseconditions. 34 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Thevaluefromthesteamtablesis142.94cm

Problem2.16Solution Weneedthedensityofmethaneundertheseconditions.Wewillcalculateitvia thecompressibilityfactorusingtheLee-Keslertables.Thecriticalparametersofmethaneare:

Wemustperformadoubleinterpolationinthetablessinceneithervalueislisted.Thecalculationissummarizedbelow:

Here,numbersinregularfontareformtheLee-Keslertablesandthoseinboldareinterpolations.Firstwe

Theprocedureissimilarfor Z1 whereweﬁnd

Withthesevalueswecalculatethecompressibilityfactor,themolarvolume,density,andtotalvolumeofthe tank:

Tc D 190:6 K;Pc D 45:99 bar;! D 0:012 Thereducedstateofmethaneis Tr D 298:15 K 190:6 K D 1:564;Pr D 75 bar 45:99 bar D 1:631

Interpolationsfor Z0 Pr 1.5 1.631 2.0 1.5 0.8689 0.8595 0.8328 1.564 0.8811 1.6 0.9 0.8931 0.8738 Interpolationsfor Z1 Pr 1.5 1.631 2.0 0.1345 0.1466 0.1806 0.1433 0.1303 0.1414 0.1729

interpolateatTr D 1:5 tocalculate Z0 at Pr D 1:631 andweobtain0.8595.Next,wedothesameat Tr D 0:6 tocalculate Z0 at Pr D 1:631 fromwhichwefind0.8931.Finally,weinterpolatebetweenthese twovaluestoobtain Z0 at Tr D 1:564 tofindthevalueof Z0 atthedesiredstate.Wefind Z0 D 0:8811

Z1 D 0:1433

Z D 0:8828;V D 2:92 10 4 ; m3/mol D 54:8 kg/m3;V tank D 18:2 m3 35 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

2.PHASE DIAGRAMSOF PURE FLUIDS

Problem2.17Solution a)Atthegivenconditions, Tr D 0:963675, Pr D 0:949281.Thestateisvery closetothecritical,therefore,farremovedfromtheideal-gasstate.

b)UsingtheLeeKeslermethodweﬁnd

0 D 0:381614;

1 D 0:597626;

D 0:247148;

Thenumberofmolesinthetankis

c)At25 ıC,1bar,CO2 isessentiallyintheideal-gasstateanditsmolarvolumeis

Wemustremove 199:294s kg.

D 0:0000860517 m3

mol

n D 200 kg 4410 3 kg=mol D 4545:45 mol

V tank D Vn D 0:391 m3

andthevolumeofthetankis

V2 D RT P2 D 0:0243725 m3 whichmeansthatthemolesinthtankare n2 D V tank V2 D 16:048 mol ) m2 D 0:706 kg

Problem2.18Solution

Outline:

a)Calculate Pr and Tr andﬁndthecompressibilityfactorusingtheLee-Keslertablesorgraphs.

b)Withthecompressibilityfactorknown,calculatethespeciﬁcvolumeandthenthetotalvolumeofthetank (sincethetotalmassisknown).

c)Calculatethenewmolarvolumeafter90%(45kg)isremoved.Calculate Z.Sinceweknow Z and T we shouldbeabletoobtain P .IfweusetheLee-Keslergraphfor Z wemustdoatrial-and-errorprocedure: choose P ,calculate Z,ifitdoesn’tmatchtheknown Z tryanotherpressureandcontinue.

Alternatively,usethetruncatedvirialequation:estimate B usingthePitzercorrelationanduse

tosolvefor P .Onceyouhaveobtained B,conﬁrmthattheuseofthevirialequationwasjustiﬁed.

Calculations

a)Thecriticalparametersofethyleneare:

Checkingwiththegeneralized Z graphweseethatthestateiswell-removedfromtheideal-gasstate.

b)WecalculatethecompressibilityfactorusingthegeneralizedLee-Keslergraphs(orbyinterpolationin tables):

Asexpected,thecompressibilityfactorisquitebelowitsidealvalue.Themolarvolumeofthegasis

PV RT D 1 C BP RT

! D 0:087 TC D 282:3 K PC D 50:40 bar Thereducedtemperatureandpressureare Tr D 293:15 K 282:3 K D 1:05615;Pr D 80 bar 50:40 bar D 1:5873

Z0 D 0:322979;Z1 D 0:0390341Z D .0:322979/ C .0:087/.0:0390341/ D 0:326

V D Z RT P D 0:00010113 m3=mol Thenumberofmolesinthetankis n D 50 kg 28 10 3 kg=mol D 1785:71 mol andthetotalvolumeis V tot D nV D .1785:71 mol/.0:00010113 m3=mol/ D 0:181 m3 c)Thenumberofmolesleftinthetankis n 0 D .0:1/n D 178:6 mol 37

booklooks“linearenough.”Butwewanttobemoreprecise,sowewillcalculatethecompressibility factorformthevirialequationandfromtheLee-Keslergraphsandwillcompare:

Note:Wecouldhaveusedtheideal-gaslawsubjecttothesamecondition:afterthecalculationwe shouldcheckwhethertheideal-gasassumptioniscorrectornot:

Trialanderror Hereishowtoperformatrial-anderrorsolutionusingtheLee-Keslertables.Weneed astartingguessfor P -wewillusetheresultfotheideal-gascalculation, Pguess D 24:5 bar.Fromthe Lee-Keslertablesweﬁnd

2.PHASE DIAGRAMSOF PURE FLUIDS andthenewmolarvolumeis V 0 D V tot n0 D 0:00101128 m3=mol Wewillusethetruncatedvirialsinceitdoesnotrequireiterations,butwewillhavetojustifyitsuseafterward. P 0V 0 RT D 1 C BP 0 RT ) P 0 D RT V 0 B WecalculatethesecondvirialcoefficientusingthePitzercorrelation: B 0 D 0:083 0:422 D 0:303682 B 1 D 0:139 0:172 T 4:2 r D 0:00226241 B D RTc Pc .B 0 C !B 1/ D 0:0001413 m3=mol Finally,thepressureis P 0 D RT V 0 B D 21:5 bar Checkvalidityoftruncatedvirial Atthefinalstate, Tr D 1:05615, Pr D 0:426709.Iftheisothermat this Tr issufficientlylinearupto Pr D 0:426709,thetruncatedvirialisacceptable.Thegraphinthe

Usingthevirialequation: Z D 1 C BP 0 RT D 0:877 UsingtheLee-Keslertablesweﬁnd Z0 D 0:8677, Z1 D 0:007497 and Z D .0:8677/ C .0:018/. 0:007497/ D 0:867 Prettyclose.

P 0 D RT V D 24:5 bar With Tr D 1:05615, Pr D 24:5=50:40 bar D 0:486

Z D

,theLee-Keslerchartsgive

0:846.Thisvalue ismorethan5%wayfromtheideal-gasstate,therefore,werejectthecalculation.

Z0 D 0:846798;Z1 D 0:00762552;Zguess D 0:846135;Vguess D 0:00085568 m3=mol 38 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

Thecorrectpressuremustmakethisratioequalto1.Sinceitislessthan1(i.e.,

guessa lower pressure,toallowvolumetoincrease.Wechoose

ratioissufﬁcientlycloseto1.Thetablebelowsummarizestheresultsoftheseiterations.

Thesolutionissomewherebetween21and22bar.Wecontinueinthesamemannerifwewantto bracketthesolutionmoreprecisely.

Wecomparethistotheknownvalue V 0 D 0:00101128 m3=molbycalculatingtheratio Vguess V 0 D 0:8461

Vguess <V 0

Pguess D 20 barandrepeatuntilthe

iteration Pguess Z0 Z1 Zguess Vguess Vguess=V 0 124:50:846881 0:007625730:8462170:0008561710:84662 2200:877931 0:007309250:8772950:001087331:0752 3220:864336 0:007541810:863680:0009731390:962283 4210:871173 0:007442490:8705250:001027561:0161

)wemust

Problem2.19Solution Weneedtocalculatedensities,i.e.weneed V or Z.Theideal-gaslawisoutof thequestionbecausethepressureistoohigh.Sameforthetruncatedvirialequation.Wecoulduseeitherthe Lee-Keslertablesoranequationofstate.Bothmethodswouldbeappropriatesincekryptonisanon-polar compound.

a)Weneedthedensityofkrypton,sowewillﬁrstﬁndthecompressibilityfactorattheindicatedconditions.

ıC,110bar,correspondtoreducedconditions

FromtheLee-Keslertablesweﬁnd(noticethatwedon’tneed Z1 sincetheacentricfactoris0)

Thatis,3428kgisthemaximummassthatcanbestoredat25 ıCwithoutexceedingthesafetylimit.Itis, therefore,safetostore2500kg.

2.PHASE

DIAGRAMSOF PURE FLUIDS

Fromtablesweﬁnd Pc D 55:02;Tc D 209:4;! D 0 Thegivenconditions,20

Pr D 110 55:02 D 2:0;Tr D 20 C 273:15 209:4 D 1:4

Z D Z0 D 0:7753 ThemolarvolumeofKris V D ZRT P D .0:7753/.8:314 J/molK/.293:15 K/ 110 105 Pa D 1:72 10 4 m3 Thenumberofmolescorrespondingto2000kgofKr(Mw D 83:8)is n D 2000 kg 83:8 10 3 kg/mol D 23866:3 mol andtherequiredvolumeofthetankis V tank D nV D .23866:3 mol/.1:72 10 4 m3/ D 4:1 m3

ıC, wehave Pr D 3:272;Tr D 1:42 1:4 FromLee-Keslerbyinterpolation: Z D Z1 D 0:7202 C 0:7761 0:7202 5:0 3:0 .3:272 3:0/ D 0:7278 Thespeciﬁcvolumeis ZRT P D .0:7278/.8:314 J/molK/.298:15 K/ 180 105 Pa D 1:0 10 4 m3 andthenumberofmolesofKrinthetankis n D V tank V D 4:1 m3/mol 1:0 10 4 m3 D 40907:2 mol Thecorrespondingmassis M D nMW D .40907:2 mol/.83:8 10 3 kg/mol/ D 3428 kg

b)Wewillcalculatethemassinthetankwhenthepressureisthemaximumallowable.At180bar,25

Problem2.20 Wecollectthefollowinginformationforn-butane:

a)Weneedthemolarvolumeoftheliquid.Ouroptionsare:Lee-Kesler,andRackett.WechoosetheRackett equationbecauseitisknowntobefairlyaccuratewhiletheaccuracyoftheLee-Keslerisnotverygoodin theliquidside.Still,ifyoudidtheproblemusingL-KIwillconsiderthesolutioncorrect.

Note: ThisproblemcouldalsobedoneusingtheLee-Kesler.Thesolutionrequiresmorecalculationsand theﬁnalresultisveryclosetotheabove.Thiscalculationisgivenattheendofthissolution.b)Forthe volumeofthevaporweuseLee-Kesler.Therequiredinterpolationisshownbelow.

Pc D 37:96 bar;Tc D 425:1 K;! D 0:2;Vc D 255 cm3/mol;Zc D 0:274

V D .255 cm3/mol/.0:274/.1 293:15=425:1/0:2857 D 100:9 cm3/mol Themolesis ML D 107 cm3 100:9 cm3/mol D 99:1 103 mol

Tr D 0:7 Pr D 0:05Pr D 0:0545Pr D 0:1 Z0 0:9504 0.9455 0:8958 Z1 0:0507 0:0566 0:1161

Z D 0:9455 C .0:2/. 0:0566/ D 0:9342 Themolarvolumeis V D ZRT P D .0:9342/.8:314/.293:15/ 2:07 105 D 1:1 10 2 m3/mol Themolesofthevaporare MV D 10 m3 1:1 10 2 m3/mol D 909 mol c)Toanswerthisquestionwetakealookatthe PV graph. A A' B B' T T + d 41 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.

fromwhichweobtainthecompressibilityfactor:

Bothsidesofthetankundergoconstant-volumeprocessesasindicatedbythedashedlines.Thegraph showsthetwostatesattheinitialtemperature T ,aswellasthestatesatsomehighertemperature, T C ı Itisobviousthatthepressureintheliquidsidewillalwaysbe higher thanthepressureofthevaporside. Therefore,thepressureof40barwillbereachedﬁrstintheliquidside,causingthatalarmtogooff.

d)Tocalculatethetemperatureatthestatewerecallthatforliquidswithconstant ˇ and ,wehave

Thealarmwillsoundat T D 20

whilethatofthevaporwillbenotmuchhigherthan2bar!

Calculationofliquid V usingLee-Kesler:

IfyouoptedtodothecalculationusingtheLee-Keslertables,thecorrectsolutionisshownbelow.Firstwe calculatethereducedtemperatureandpressure.

Note:becausethephaseisliquid,onemust

TheanswerisveryclosetothatobtainedusingtheRackettequationbuttheLee-Keslermethodrequiresmore calculations.

2.PHASE DIAGRAMSOF PURE FLUIDS

ln V2 V1 D ˇ T P

D V2 andsolvingfor T weﬁnd T D P ˇ D 3:4 10 4 bar 1.40 2:07/ bar 2:54 10 3 K 1 D 5 ıC

SIncevolumeisconstant, V1

C 5 D 25 ı

C.Atthatpointthepressureoftheliquidsidewillbe40bar

Tr D 0:69 0:7;Pr D 0:545

extrapolate to Pr D 0:0545 fromthelistedvaluesfortheliquid (showninthetablesinitalics): Tr D 0:7 Pr D 0:0545Pr D 0:2Pr D 0:4 Z0 9:45 1 3 0:03440:0687 Z1 4:1785 10 3 0:0148 0:0294 Withthesevaluesweobtainthefollowing: Z D .9:45 1 3/ C .0:2/. 4:1785 10 3/ D 8:614 10 3 V D ZRT P D .8:614 10 3/.8:314 J/molK/.293:15 K/ 2:07 105 Pa D 1:01 10 4 m3/mol ML D 10 m3 1:01 10 4 m3/mol D 98:5 103 mol

Problem2.21SolutionSolution a) Filledwithxenon

Weneedthevolumeofthetankwhichwillobtainbyﬁrstcalculatingthemolarvolumeofxenon.Wewill dothiscalculationusingthePitzermethodandtheLeeKeslertables.Forxenon:

At200 ıC,thesaturatedvolumesofwaterare1.156and127.2cm3/g.Sincethespeciﬁcvolumeliesbetweenthetwovalues,thesteamisasaturatedvapor/liquidmixtureandthepressureisequaltothesaturation pressureat200 ı

P D 15:45 bar.

b)Ifthemassinthetankisreducedtohalf,thespeciﬁcvolumedoubles:

Thisvalueisstillbetweenthatofthesaturatedvaporandliquid,thereforethepressureremainsconstant.

Tc D 289:7 KI Pc D 58:4 barI ! D 0I Mw D 131:30 g/mol Thereducedtemperatureandpressureare Tr D 132 C 273:15 289:7 D 1:39852 1:4;Pr D 82 58:4 D 1:4: Interpolatingat Tr D 1:4 betweenPr=1.2andPr=1.5weﬁnd Z0 D 0:836436 Since ! D 0, Z D Z0 D 0:836436.UsingSIunits,themolarvolumeofxenonis V D ZRT P D .0:836436/.8:314/.405:15/82 105 D 3:436 104 m3/mol Sincethetankcontains10,000kg,or n D 10;000 kg 131:30 10 3 kgmol D 76;161 mol thevolumeofthetankis V t D nV D 26:16 m3 Filledwithsteam Thespeciﬁcvolumeofsteaminthetankis V D 26:16 m3 10000 kg D 2:616 cm3/g

V D .2/.2:616/ D 5:232 cm3/g

DIAGRAMSOF PURE FLUIDS

Problem2.22Solution a)Vapor.

b)At0.1bar,200C,o-xyleneisessentiallyintheideal-gasstate(why?).

c)At44.9bar,200C,thereducedtemperatureanpressureis

2.PHASE

V D RT P D 0:393377 m3=mol Thevolumeofthetankis V tank D .200 mol/.0:393377 m3=mol/ D 39:3 m3 : (2.2)

Tr D 0:75;Pr D 1:2 FromtheLee-Keslertablesweﬁnd Z0 D 0:84435;Z1 D 0:0453;Z D 0:8585 Themolarvolumeis V2 D ZRT P2 D 0:000752133 m3=mol andthenumberofmoles n D V tank V2 D 52301:5 mol 44 ThistextisassociatedwithThemisMatsoukas,FundamentalsofChemicalEngineeringThermodynamics0-13-269306-2(978-0-13-269306-6).Copyright©2013PearsonEducation,Inc.Donotredistribute.