Table Of ContentPart 1: Equilibrium
1 The properties of gases
Solutions to exercises
Discussion questions
E1.1(b) Thepartialpressureofagasinamixtureofgasesisthepressurethegaswouldexertifitoccupied
alonethesamecontainerasthemixtureatthesametemperature.Itisalimitinglawbecauseitholds
exactlyonlyunderconditionswherethegaseshavenoeffectuponeachother.Thiscanonlybetrue
inthelimitofzeropressurewherethemoleculesofthegasareveryfarapart.Hence,Dalton’slaw
holdsexactlyonlyforamixtureofperfectgases;forrealgases,thelawisonlyanapproximation.
E1.2(b) The critical constants represent the state of a system at which the distinction between the liquid
and vapour phases disappears. We usually describe this situation by saying that above the critical
temperaturetheliquidphasecannotbeproducedbytheapplicationofpressurealone.Theliquidand
vapour phases can no longer coexist, though fluids in the so-called supercritical region have both
liquidandvapourcharacteristics. (SeeBox6.1foramorethoroughdiscussionofthesupercritical
state.)
E1.3(b) The van der Waals equation is a cubic equation in the volume, V. Any cubic equation has certain
properties, one of which is that there are some values of the coefficients of the variable where the
number of real roots passes from three to one. In fact, any equation of state of odd degree higher
than1caninprincipleaccountforcriticalbehaviorbecauseforequationsofodddegreeinV there
are necessarily some values of temperature and pressure for which the number of real roots of V
passes from n(odd) to 1. That is, the multiple values of V converge from n to 1 asT → T . This
c
mathematicalresultisconsistentwithpassingfromatwophaseregion(morethanonevolumefora
givenT andp)toaonephaseregion(onlyoneV foragivenT andp andthiscorrespondstothe
observedexperimentalresultasthecriticalpointisreached.
Numerical exercises
E1.4(b) Boyle’slawapplies.
pV =constant so p V =pV
f f i i
pV (104kPa)×(2000cm3)
p = i i = = 832kPa
f V (250cm3)
f
E1.5(b) (a) Theperfectgaslawis
pV =nRT
implyingthatthepressurewouldbe
nRT
p =
V
Allquantitiesontherightaregiventousexceptn,whichcanbecomputedfromthegivenmass
ofAr.
25g
n= =0.626mol
39.95gmol−1
(0.626mol)×(8.31×10−2LbarK−1mol−1)×(30+273K)
sop = = 10.5bar
1.5L
not2.0bar.
4 INSTRUCTOR’SMANUAL
(b) ThevanderWaalsequationis
RT a
p = −
V −b V2
m m
(8.31×10−2LbarK−1mol−1)×(30+273)K
sop =
(1.5L/0.626mol)−3.20×10−2Lmol−1
(1.337L2atmmol−2)×(1.013baratm−1)
− = 10.4bar
(1.5L/0.626¯mol)2
E1.6(b) (a) Boyle’slawapplies.
pV =constant so p V =pV
f f i i
p V (1.48×103Torr)×(2.14dm3)
andp = f f = = 8.04×102Torr
i Vi (2.14+1.80)dm3
(b) Theoriginalpressureinbaris
(cid:1) (cid:2) (cid:1) (cid:2)
1atm 1.013bar
p =(8.04×102Torr)× × = 1.07bar
i
760Torr 1atm
E1.7(b) Charles’slawapplies.
V V
V ∝T so i = f
T T
i f
V T (150cm3)×(35+273)K
andT = f i = = 92.4K
f V 500cm3
i
E1.8(b) Therelationbetweenpressureandtemperatureatconstantvolumecanbederivedfromtheperfect
gaslaw
p p
pV =nRT so p ∝T and i = f
T T
i f
Thefinalpressure,then,oughttobe
pT (125kPa)×(11+273)K
p = i f = = 120kPa
f T (23+273)K
i
E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature,
andvolume.Oncethisisdone,themassofthegascanbecomputedfromtheamountandthemolar
massusing
pV =nRT
pV (1.00atm)×(1.013×105Paatm−1)×(4.00×103m3)
son= = =1.66×105mol
RT (8.3145JK−1mol−1)×(20+273)K
andm=(1.66×105mol)×(16.04gmol−1)=2.67×106g= 2.67×103kg
E1.10(b) Allgasesareperfectinthelimitofzeropressure.ThereforetheextrapolatedvalueofpV /T will
m
givethebestvalueofR.
THEPROPERTIESOFGASES 5
m
ThemolarmassisobtainedfrompV =nRT = RT
M
mRT RT
whichuponrearrangementgivesM = =ρ
V p p
The best value of M is obtained from an extrapolation of ρ/p versus p to p = 0; the intercept is
M/RT.
Drawupthefollowingtable
p/atm (pV /T)/(LatmK−1mol−1) (ρ/p)/(gL−1atm−1)
m
0.750000 0.0820014 1.42859
0.500000 0.0820227 1.42822
0.250000 0.0820414 1.42790
(cid:1) (cid:2)
pV
FromFig.1.1(a), m = 0.0820615LatmK−1mol−1
T
p=0
(cid:1) (cid:2)
ρ
FromFig.1.1(b), =1.42755gL−1atm−1
p
p=0
8.20615
8.206
8.204
8.202
m
8.200
0 0.25 0.50 0.75 1.0
Figure1.1(a)
1.4288
1.4286
1.4284
1.4282
1.4280
1.4278
1.4276
1.42755
1.4274
0 0.25 0.50 0.75 1.0
Figure1.1(b)
6 INSTRUCTOR’SMANUAL
(cid:1) (cid:2)
ρ
M =RT = (0.0820615Latmmol−1K−1)×(273.15K)×(1.42755gL−1atm−1)
p
p=0
= 31.9987gmol−1
ThevalueobtainedforRdeviatesfromtheacceptedvalueby0.005percent.Theerrorresultsfrom
thefactthatonlythreedatapointsareavailableandthatalinearextrapolationwasemployed. The
molar mass, however, agrees exactly with the accepted value, probably because of compensating
plottingerrors.
E1.11(b) Themassdensityρ isrelatedtothemolarvolumeV by
m
M
V =
m ρ
whereM isthemolarmass.Puttingthisrelationintotheperfectgaslawyields
pM
pV =RT so =RT
m ρ
RearrangingthisresultgivesanexpressionforM;onceweknowthemolarmass,wecandivideby
themolarmassofphosphorusatomstodeterminethenumberofatomspergasmolecule
M = RTρ = (62.364LTorrK−1mol−1)×[(100+273)K]×(0.6388gL−1) =124gmol−1.
p 120Torr
Thenumberofatomspermoleculeis
124gmol−1
=4.00
31.0gmol−1
suggestingaformulaof P
4
E1.12(b) Usetheperfectgasequationtocomputetheamount;thenconverttomass.
pV
pV =nRT so n=
RT
Weneedthepartialpressureofwater,whichis53percentoftheequilibriumvapourpressureatthe
giventemperatureandstandardpressure.
p =(0.53)×(2.69×103Pa)=1.43¯ ×103Pa
(1.43×103Pa)×(250m3)
son= =1.45×102mol
(8.3145JK−1mol−1)×(23+273)K
orm=(1.45×102mol)×(18.0gmol−1)=2.61×103g= 2.61kg
E1.13(b) (a) Thevolumeoccupiedbyeachgasisthesame, sinceeachcompletelyfillsthecontainer. Thus
solvingforV fromeqn14wehave(assumingaperfectgas)
n RT 0.225g
V = J n =
pJ Ne 20.18gmol−1
=1.115×10−2mol, p =66.5Torr, T =300K
Ne
(1.115×10−2mol)×(62.36LTorrK−1mol−1)×(300K)
V = =3.137L= 3.14L
66.5Torr
THEPROPERTIESOFGASES 7
(b) Thetotalpressureisdeterminedfromthetotalamountofgas,n=n +n +n .
CH4 Ar Ne
n = 0.320g =1.995×10−2mol n = 0.175g =4.38×10−3mol
CH4 16.04gmol−1 Ar 39.95gmol−1
n=(1.995+0.438+1.115)×10−2mol=3.548×10−2mol
nRT (3.548×10−2mol)×(62.36LTorrK−1mol−1)×(300K)
p = [1]=
V 3.137L
= 212Torr
E1.14(b) ThisissimilartoExercise1.14(a)withtheexceptionthatthedensityisfirstcalculated.
RT
M =ρ [Exercise1.11(a)]
p
ρ = 33.5mg =0.1340gL−1, p =152Torr, T =298K
250mL
M = (0.1340gL−1)×(62.36LTorrK−1mol−1)×(298K) = 16.4gmol−1
152Torr
E1.15(b) This exercise is similar to Exercise 1.15(a) in that it uses the definition of absolute zero as that
temperatureatwhichthevolumeofasampleofgaswouldbecomezeroifthesubstanceremaineda
gasatlowtemperatures.Thesolutionusestheexperimentalfactthatthevolumeisalinearfunction
oftheCelsiustemperature.
ThusV =V +αV θ =V +bθ, b=αV
0 0 0 0
Atabsolutezero,V =0,or0=20.00L+0.0741L◦C−1×θ(abs.zero)
θ(abs.zero)=− 20.00L = −270◦C
0.0741L◦C−1
whichisclosetotheacceptedvalueof−273◦C.
nRT
E1.16(b) (a) p =
V
n=1.0mol
T =(i)273.15K; (ii)500K
V =(i)22.414L; (ii)150cm3
(1.0mol)×(8.206×10−2LatmK−1mol−1)×(273.15K)
(i) p =
22.414L
= 1.0atm
(1.0mol)×(8.206×10−2LatmK−1mol−1)×(500K)
(ii) p =
0.150L
= 270atm (2significantfigures)
(b) FromTable(1.6)forH S
2
a =4.484L2atmmol−1 b=4.34×10−2Lmol−1
nRT an2
p = −
V −nb V2
8 INSTRUCTOR’SMANUAL
(1.0mol)×(8.206×10−2LatmK−1mol−1)×(273.15K)
(i) p =
22.414L−(1.0mol)×(4.34×10−2Lmol−1)
(4.484L2atmmol−1)×(1.0mol)2
−
(22.414L)2
= 0.99atm
(1.0mol)×(8.206×10−2LatmK−1mol−1)×(500K)
(ii) p =
0.150L−(1.0mol)×(4.34×10−2Lmol−1)
(4.484L2atmmol−1)×(1.0mol)2
−
(0.150L)2
=185.6atm≈ 190atm (2significantfigures).
E1.17(b) ThecriticalconstantsofavanderWaalsgasare
V =3b=3(0.0436Lmol−1)= 0.131Lmol−1
c
a 1.32atmL2mol−2
p = = = 25.7atm
c 27b2 27(0.0436Lmol−1)2
8a 8(1.32atmL2mol−2)
andT = = = 109K
c 27Rb 27(0.08206LatmK−1mol−1)×(0.0436Lmol−1)
E1.18(b) Thecompressionfactoris
pV V
Z = m = m
RT V
m,perfect
(a) BecauseVm =Vm,perfect+0.12Vm,perfect =(1.12)Vm,perfect,wehaveZ = 1.12
Repulsive forcesdominate.
(b) Themolarvolumeis
(cid:1) (cid:2)
RT
V =(1.12)Vm,perfect =(1.12)× p
(cid:3) (cid:4)
V =(1.12)× (0.08206LatmK−1mol−1)×(350K) = 2.7Lmol−1
12atm
RT (8.314JK−1mol−1)×(298.15K)
E1.19(b) (a) Vo = =
m p (200bar)×(105Pabar−1)
=1.24×10−4m3mol−1 = 0.124Lmol−1
(b) The van der Waals equation is a cubic equation in V . The most direct way of obtaining the
m
molarvolumewouldbetosolvethecubicanalytically.However,thisapproachiscumbersome,
soweproceedasinExample1.6.ThevanderWaalsequationisrearrangedtothecubicform
(cid:1) (cid:2) (cid:1) (cid:2) (cid:1) (cid:2) (cid:1) (cid:2)
RT a ab RT a ab
V3 − b+ V2 + V − =0 or x3− b+ x2+ x− =0
m p m p m p p p p
withx =V /(Lmol−1).
m
THEPROPERTIESOFGASES 9
Thecoefficientsintheequationareevaluatedas
b+ RT =(3.183×10−2Lmol−1)+ (8.206×10−2LatmK−1mol−1)×(298.15K)
p (200bar)×(1.013atmbar−1)
=(3.183×10−2+0.1208)Lmol−1 =0.1526Lmol−1
a = 1.360L2atmmol−2 =6.71×10−3(Lmol−1)2
p (200bar)×(1.013atmbar−1)
ab = (1.360L2atmmol−2)×(3.183×10−2Lmol−1) =2.137×10−4(Lmol−1)3
p (200bar)×(1.013atmbar−1)
Thus,theequationtobesolvedisx3−0.1526x2+(6.71×10−3)x−(2.137×10−4)=0.
Calculatorsandcomputersoftwareforthesolutionofpolynomialsarereadilyavailable.Inthiscase
wefind
x =0.112 or V = 0.112Lmol−1
m
Thedifferenceisabout15percent.
E1.20(b) (a) V = M = 18.015gmol−1 = 31.728Lmol−1
m ρ 0.5678gL−1
pV (1.00bar)×(31.728Lmol−1)
Z = m = = 0.9963
RT (0.083145LbarK−1mol−1)×(383K)
RT a
(b) Usingp = − andsubstitutingintotheexpressionforZaboveweget
V −b V2
m m
V a
Z = m −
V −b V RT
m m
31.728Lmol−1
=
31.728Lmol−1−0.03049Lmol−1
5.464L2atmmol−2
−
(31.728Lmol−1)×(0.08206LatmK−1mol−1)×(383K)
= 0.9954
Comment. BothvaluesofZ areveryclosetotheperfectgasvalueof1.000, indicatingthatwater
vapourisessentiallyperfectat1.00barpressure.
pV
E1.21(b) ThemolarvolumeisobtainedbysolvingZ = m [1.20b],forV ,whichyields
RT m
V = ZRT = (0.86)×(0.08206LatmK−1mol−1)×(300K) =1.059Lmol−1
m p 20atm
(a) Then,V =nV =(8.2×10−3mol)×(1.059Lmol−1)=8.7×10−3L= 8.7mL
m
10 INSTRUCTOR’SMANUAL
(b) AnapproximatevalueofBcanbeobtainedfromeqn1.22bytruncationoftheseriesexpansion
afterthesecondterm,B/V ,intheseries.Then,
m
(cid:1) (cid:2)
pV
B = V m −1 =V ×(Z−1)
m RT m
= (1.059Lmol−1)×(0.86−1)= −0.15Lmol−1
E1.22(b) (a) Molefractionsare
n 2.5mol
x = N = = 0.63
N n (2.5+1.5)mol
total
Similarly,x = 0.37
H
(c) Accordingtotheperfectgaslaw
p V =n RT
total total
n RT
sop = total
total V
(4.0mol)×(0.08206Latmmol−1K−1)×(273.15K)
= = 4.0atm
22.4L
(b) Thepartialpressuresare
p =x p =(0.63)×(4.0atm)= 2.5atm
N N tot
andp =(0.37)×(4.0atm)= 1.5atm
H
E1.23(b) ThecriticalvolumeofavanderWaalsgasis
V =3b
c
sob= 1V = 1(148cm3mol−1)=49.3cm3mol−1 = 0.0493Lmol−1
3 c 3
Byinterpretingbastheexcludedvolumeofamoleofsphericalmolecules,wecanobtainanestimate
of molecular size. The centres of spherical particles are excluded from a sphere whose radius is
the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro
constantisthemolarexcludedvolumeb
(cid:3) (cid:4) (cid:1) (cid:2)
4π(2r)3 1 3b 1/3
b=N so r =
A 3 2 4πN
A
(cid:3) (cid:4)
r = 1 3(49.3cm3mol−1) 1/3 =1.94×10−8cm= 1.94×10−10m
2 4π(6.022×1023mol−1)
Thecriticalpressureis
a
p =
c 27b2
soa =27p b2 =27(48.20atm)×(0.0493Lmol−1)2 = 3.16L2atmmol−2
c
