Solutions Manual for Thermodynamics and Chemistry

0.53 MB·English

by  Howard DeVoe

Solutions Manual for
Thermodynamics and Chemistry
Second Edition
by
Howard DeVoe
AssociateProfessorofChemistryEmeritus
UniversityofMaryland,CollegePark,Maryland
[email protected]
Copyright2020byHowardDeVoe
ThisworkislicensedunderaCreativeCommonsAttribution4.0InternationalLicense:
https://creativecommons.org/licenses/by/4.0/
Contents
Preface 3
1 Introduction 4
2 SystemsandTheirProperties 5
3 TheFirstLaw 8
4 TheSecondLaw 16
5 ThermodynamicPotentials 19
6 TheThirdLawandCryogenics 24
7 PureSubstancesinSinglePhases 26
8 PhaseTransitionsandEquilibriaofPureSubstances 36
9 Mixtures 41
10 ElectrolyteSolutions 55
11 ReactionsandOtherChemicalProcesses 58
12 EquilibriumConditionsinMulticomponentSystems 77
13 ThePhaseRuleandPhaseDiagrams 94
14 GalvanicCells 104
Preface
Thismanualcontainsdetailedsolutionstotheproblemsappearingattheendofeachchapter
ofthetextThermodynamicsandChemistry.
Eachproblemprintedinthetextisreproducedinthismanual,followedbyaworked-out
solution. If a figure or table accompanies a problem in the text, it is also reproduced here.
Included within a solution may be an additional figure or table that does not appear in the
text. Allfigures,tables,andfootnotesinthismanualarenumberedconsecutively(Figure1,
Figure2,etc.) andsodonotagreewiththenumberinginthetext.
Inmostcasesofanumericalcalculationinvolvingphysicalquantities,thesetupinthis
manual shows the values of given individual physical quantities expressed in SI base units
andSIderivedunits, withoutprefixes. TheresultofthecalculationisthenexpressedinSI
base units and SI derived units appropriate to the physical quantity being evaluated. Since
the factors needed to convert the units of the given quantities to the units of the calculated
quantityallhavenumericalvaluesofunitywhenthisprocedureisfollowed,theconversion
factorsarenotshown.
Ofcourse, thesolutiongiveninthismanualforanyparticularproblemisprobablynot
theonlywaytheproblemcanbesolved;othersolutionsmaybeequallyvalid.
4
Chapter 1 Introduction
1.1 Consider the following equations for the pressure of a real gas. For each equation, find the
dimensionsoftheconstantsaandbandexpressthesedimensionsinSIunits.
(a) TheDietericiequation:
RTe .an=VRT/
p (cid:0)
D .V=n/ b
(cid:0)
Solution:
Sincean=VRT isapower,itisdimensionlessandahasthesamedimensionsas
VRT=n. Thesedimensionsarevolume energy/amount2,expressedinm3Jmol 2. bhas
(cid:0)
(cid:1)
thesamedimensionsasV=n,whicharevolume/amountexpressedinm3mol 1.
(cid:0)
(b) TheRedlich–Kwongequation:
RT an2
p
D .V=n/ b (cid:0) T1=2V.V nb/
(cid:0) C
Solution:
Theterman2=T1=2V.V nb/hasthesamedimensionsasp,soahasthesame
C
dimensionsasT1=2V2pn 2. TheSIunitsareK1=2m6Pamol 2. bhasthesame
(cid:0) (cid:0)
dimensionsasV=n,whicharevolume/amountexpressedinm3mol 1.
(cid:0)
5
Chapter 2 Systems and Their Properties
2.1 LetX representthequantityV2 withdimensions.length/6. GiveareasonthatX isorisnot
anextensiveproperty. GiveareasonthatX isorisnotanintensiveproperty.
Solution:
X isnotanextensivepropertybecauseitisnotadditive: .V(cid:146)/2 .V(cid:147)/2 .V(cid:146) V(cid:147)/2(e.g.,
C ⁄ C
12 12 22).
C ⁄
X isnotanintensivepropertybecauseitisdependentonvolume.
2.2 Calculate the relative uncertainty (the uncertainty divided by the value) for each of the mea-
surementmethodslistedinTable2.2onpage38,usingthetypicalvaluesshown. Foreachof
thefivephysicalquantitieslisted,whichmeasurementmethodhasthesmallestrelativeuncer-
tainty?
Solution:
Mass:
analyticalbalance,0:1 10 3g=100g 1 10 6
(cid:0) (cid:0)
(cid:2) D (cid:2)
microbalance,0:1 10 6g=20 10 3g 5 10 6
(cid:0) (cid:0) (cid:0)
(cid:2) (cid:2) D (cid:2)
Volume:
pipet,0:02ml=10mL 2 10 3
(cid:0)
D (cid:2)
volumetricflask,0:3 10 3L=1L 3 10 4
(cid:0) (cid:0)
(cid:2) D (cid:2)
Density:
pycnometer,2 10 3gmL 1=1gmL 1 2 10 3
(cid:0) (cid:0) (cid:0) (cid:0)
(cid:2) D (cid:2)
magneticfloatdensimeter,0:1 10 3gmL 1=1gmL 1 1 10 4
(cid:0) (cid:0) (cid:0) (cid:0)
(cid:2) D (cid:2)
Pressure:
manometerorbarometer,0:001Torr=760Torr 1 10 6
(cid:0)
D (cid:2)
diaphragmgauge,1Torr=100Torr 1 10 2
(cid:0)
D (cid:2)
Temperature:
gasthermometer,0:001K=10K 1 10 4
(cid:0)
D (cid:2)
mercurythermometer,0:01K=300K 3 10 5
(cid:0)
D (cid:2)
platinumresistancethermometer,0:0001K=300K 3 10 7
(cid:0)
D (cid:2)
opticalpyrometer,0:03K=1300K 2 10 5
(cid:0)
D (cid:2)
Themeasurementoftemperaturewithaplatinumresistancethermometerhastheleastrelative
uncertainty,andthemeasurementofpressurewithadiaphragmgaugehasthegreatest. For
eachphysicalquantity,themeasurementmethodwithsmallestrelativeuncertaintyis
underlinedintheprecedinglist.
2.3 Table1onthenextpagelistsdataobtainedfromaconstant-volumegasthermometercontaining
samplesofvaryingamountsofheliummaintainedatacertainfixedtemperatureT inthegas
2
bulb.1 ThemolarvolumeV ofeachsamplewasevaluatedfromitspressureinthebulbata
m
referencetemperatureofT 7:1992K, correctedforgasnonidealitywiththeknownvalue
1
D
ofthesecondvirialcoefficientatthattemperature.
UsethesedataandEq.2.2.2onpage34toevaluateT andthesecondvirialcoefficientofhe-
2
liumattemperatureT .(Youcanassumethethirdandhighervirialcoefficientsarenegligible.)
2
Solution:
Withthethirdandhighervirialcoefficientssetequaltozero,Eq.2.2.2becomes
(cid:18) B (cid:19)
pV RT 1
m D C V
m
1Ref.[13].
6
Table1 Heliumatafixedtemperature
.1=V /=102molm 3 .p V =R/=K
m (cid:0) 2 m
1.0225 2.7106
1.3202 2.6994
1.5829 2.6898
1.9042 2.6781
2.4572 2.6580
2.8180 2.6447
3.4160 2.6228
3.6016 2.6162
4.1375 2.5965
4.6115 2.5790
5.1717 2.5586
2.8
2:7 bc
K bc
bc
= bc
/
R
= bc
m bc
V
p2 bc bc
. 2:6
bc
bc
bc
2:5
0 1 2 3 4 5 6
.1=Vm/=102molm�3
Figure1
Accordingtothisequation,aplotofp V =Rversus1=V shouldbelinearwithanintercept
2 m m
at1=V 0equaltoT andaslopeequaltoBT . TheplotisshowninFig.1. Aleast-squares
mD 2 2
fitofthedatatoafirst-orderpolynomialyieldsaninterceptof2:7478Kandaslopeof
3:659 10 4Km3mol 1. Thetemperatureandsecondvirialcoefficientthereforehavethe
(cid:0) (cid:0)
(cid:0) (cid:2)
values
T 2:7478K
2
D
3:659 10 4Km3mol 1
B (cid:0) (cid:2) (cid:0) (cid:0) 1:332 10 4m3mol 1
D 2:7478K D(cid:0) (cid:2) (cid:0) (cid:0)
2.4 Discussthepropositionthat,toacertaindegreeofapproximation,alivingorganismisasteady-
statesystem.
Solution:
Theorganismcanbetreatedasbeinginasteadystateifweassumethatitsmassisconstant
7
andifweneglectinternalmotion. Matterenterstheorganismintheformoffood,water,and
oxygen;wastematterandheatleavethesystem.
2.5 The value of (cid:129)U for the formation of one mole of crystalline potassium iodide from its el-
ements at 25 C and 1bar is 327:9kJ. Calculate (cid:129)m for this process. Comment on the
(cid:14)
(cid:0)
feasibilityofmeasuringthismasschange.
Solution:
(cid:129)m (cid:129)U=c2 327:9 103J=.2:998 108ms 1/2 3:648 10 12kg
(cid:0) (cid:0)
D D(cid:0) (cid:2) (cid:2) D(cid:0) (cid:2)
Thismasschangeismuchlessthantheuncertaintyofamicrobalance(Table2.2),whichdoes
notevenhavethecapacitytoweighonemoleofKI—soitishopelesstotrytomeasurethis
masschange.
8
Chapter 3 The First Law
3.1 AssumeyouhaveametalspringthatobeysHooke’slaw: F c.l l /,whereF istheforce
0
D (cid:0)
exertedonthespringoflengthl,l isthelengthoftheunstressedspring,andc isthespring
0
constant. Findanexpressionfortheworkdoneonthespringwhenyoureversiblycompressit
fromlengthl toashorterlengthl .
0 0
Solution:
l l
w DZl00F dl DcZl00.l (cid:0)l0/dl D 12c.l (cid:0)l0/2(cid:12)(cid:12)ll00 D 12c.l0(cid:0)l0/2
air
water
Figure2
3.2 TheapparatusshowninFig.2consistsoffixedamountsofwaterandairandanincompressible
solidglasssphere(amarble),allenclosedinarigidvesselrestingonalabbench. Assumethe
marblehasanadiabaticouterlayersothatitstemperaturecannotchange,andthatthewallsof
thevesselarealsoadiabatic.
Initiallythemarbleissuspendedabovethewater. Whenreleased,itfallsthroughtheairinto
the water and comes to rest at the bottom of the vessel, causing the water and air (but not
the marble) to become slightly warmer. The process is complete when the system returns to
anequilibriumstate. Thesystemenergychangeduringthisprocessdependsontheframeof
reference and on how the system is defined. (cid:129)E is the energy change in a lab frame, and
sys
(cid:129)U istheenergychangeinaspecifiedlocalframe.
Foreachofthefollowingdefinitionsofthesystem, givethesign(positive, negative, orzero)
ofboth(cid:129)E and(cid:129)U,andstateyourreasoning. Takethelocalframeforeachsystemtobea
sys
center-of-massframe.
Solution:
Becauseqiszeroineachpartofthisproblem,(cid:129)E isequaltow and(cid:129)U isequaltow.
sys lab
WecanuseEq.3.1.4with(cid:129)(cid:0)v2 (cid:1)setequaltozero: (cid:129)U (cid:129)E w w mg(cid:129)z
cm (cid:0) sys D lab(cid:0) D(cid:0) cm
or(cid:129)E (cid:129)U mg(cid:129)z .
sys D C cm
(a) Thesystemisthemarble.
Solution:
(cid:129)U iszero,becausethestateofthesystemisunchanged.
(cid:129)E isnegative,becauseinthelabframethemarbledoesworkonthewater(page83).
sys
Thiscanalsobededucedusing(cid:129)E (cid:129)U mg(cid:129)z andthefactthat(cid:129)U iszero
sys D C cm
and(cid:129)z isnegative.
cm
9
(b) Thesystemisthecombinationofwaterandair.
Solution:
(cid:129)U ispositive,becausethesystem’stemperatureincreasesatconstantvolume.
(cid:129)E ispositive,becauseboth(cid:129)U and(cid:129)z arepositive(thecenterofgravityofthe
sys cm
waterriseswhenthemarbleentersthewater). Thiscanalsobededucedbyconsidering
thatthenetforceexertedbythesinkingmarbleonthewaterandthedisplacementofthe
boundaryatthemarbleareinthesamedirection(downward).
(c) Thesystemisthecombinationofwater,air,andmarble.
Solution:
(cid:129)E iszero,becausew iszero(thereisnodisplacementofthesystemboundaryin
sys lab
thelabframe).
(cid:129)U ispositive,because(cid:129)E iszeroand(cid:129)z isnegative. Thiscanalsobededuced
sys cm
fromthefactthatU isanextensiveproperty,sothat(cid:129)U forthissystemisequaltothe
sumoftheinternalenergychangeofthemarbleandtheinternalenergychangeofthe
waterandair. Inparts(a)and(b)thesechangeswerefoundtobezeroandpositive,
respectively.
p D3:00bar porous
V D0:500m3 plug
piston
T D300:0K
gas
T D300:0K
ext
p D1:00bar
ext
Figure3
3.3 Figure3showstheinitialstateofanapparatusconsistingofanidealgasinabulb,astopcock,
a porous plug, and a cylinder containing a frictionless piston. The walls are diathermal, and
thesurroundingsareataconstanttemperatureof300:0Kandaconstantpressureof1:00bar.
Whenthestopcockisopened,thegasdiffusesslowlythroughtheporousplug,andthepiston
movesslowlytotheright. Theprocessendswhenthepressuresareequalizedandthepiston
stopsmoving. Thesystemisthegas. Assumethatduringtheprocessthetemperaturethrough-
outthesystemdiffersonlyinfinitesimallyfrom300:0Kandthepressureonbothsidesofthe
pistondiffersonlyinfinitesimallyfrom1:00bar.
(a) Which of these terms correctly describes the process: isothermal, isobaric, isochoric,
reversible,irreversible?
Solution:
Theprocessisisothermalandirreversible,butnotisobaric,isochoric,orreversible. Note
thatthepressuregradientacrosstheporousplugpreventsintermediatestatesofthe
processfrombeingequilibriumstates,andkeepstheprocessfrombeingreversible;there
isnoinfinitesimalchangethatcanreversethemotionofthepiston.
(b) Calculateqandw.
Solution:
BecauseT isconstantandthegasisideal,therelationp V p V holds,andthefinal
1 1 2 2
D
volumeisfoundfrom
10
p V .3:00bar/.0:500m3/
V 1 1 1:50m3
2 D p D 1:00bar D
2
Theworkmustbecalculatedfromthepressureatthemovingportionoftheboundary
(theinnersurfaceofthepiston);thisisaconstantpressureof1:00bar:
w Z V2pdV p.V V / .1:00 105Pa/.1:50 0:500/m3
2 1
D(cid:0) D(cid:0) (cid:0) D(cid:0) (cid:2) (cid:0)
V1
1:00 105J
D(cid:0) (cid:2)
q (cid:129)U w 0 w 1:00 105J
D (cid:0) D (cid:0) D (cid:2)
3.4 Consider a horizontal cylinder-and-piston device similar to the one shown in Fig. 3.5 on
page 72. The piston has mass m. The cylinder wall is diathermal and is in thermal contact
withaheatreservoiroftemperatureT . Thesystemisanamountnofanidealgasconfined
ext
inthecylinderbythepiston.
The initial state of the system is an equilibrium state described by p and T T . There
1 D ext
is a constant external pressure p , equal to twice p , that supplies a constant external force
ext 1
on the piston. When the piston is released, it begins to moveto the left to compress the gas.
Maketheidealizedassumptionsthat(1)thepistonmoveswithnegligiblefriction;and(2)the
gasremainspracticallyuniform(becausethepistonismassiveanditsmotionisslow)andhas
apracticallyconstanttemperatureT T (becausetemperatureequilibrationisrapid).
D ext
(a) Describetheresultingprocess.
Solution:
Thepistonwilloscillate;thegasvolumewillchangebackandforthbetweentheinitial
valueV andaminimumvalueV .
1 2
(b) Describehowyoucouldcalculatewandqduringtheperiodneededforthepistonvelocity
tobecomezeroagain.
Solution:
TherelationbetweenV andV isfoundbyequatingtheworkdoneonthegasbythe
1 2
piston, nRT ln.V =V /,totheworkdoneonthepistonbytheexternalpressure,
2 1
(cid:0)
p .V V /,wherep isgivenbyp 2p 2nRT=V . Theresultis
(cid:0) ext 2(cid:0) 1 ext ext D 1 D 1
V 0:2032V ,w 1:5936nRT,q w 1:5936nRT.
2 1
D D D(cid:0) D(cid:0)
(c) Calculatewandqduringthisperiodfor0:500molgasat300K.
Solution:
w .1:5936/.0:500mol/.8:3145JK 1mol 1/.300K/ 1:99 103J,
(cid:0) (cid:0)
D D (cid:2)
q w 1:99 103J.
D(cid:0) D(cid:0) (cid:2)
3.5 Thisproblemisdesignedtotesttheassertiononpage60thatfortypicalthermodynamicpro-
cessesinwhichtheelevationofthecenterofmasschanges,itisusuallyagoodapproximation
tosetwequaltow . ThecylindershowninFig.4onthenextpagehasaverticalorientation,
lab
sotheelevationofthecenterofmassofthegasconfinedbythepistonchangesasthepiston
slidesupordown. Thesystemisthegas. Assumethegasisnitrogen(M 28:0gmol 1)at
(cid:0)
D
300K,andinitiallytheverticallengthl ofthegascolumnisonemeter. Treatthenitrogenas
an ideal gas, use a center-of-mass local frame, and take the center of mass to be at the mid-
point of the gas column. Find the difference between the values of w and w , expressed as
lab
a percentage of w, when the gas is expanded reversibly and isothermally to twice its initial
volume.
Solution:
UseEq.3.1.4: w w 1m(cid:129)(cid:0)v2 (cid:1) mg(cid:129)z .
(cid:0) lab D(cid:0)2 cm (cid:0) cm