686

-5.13 x 10-a bar'm3'mol-r

-5i.3 J'mol t

Chapter 22

Using Equation 16.41,

2n ot n'

B2v(T):'+ [1 - (r'- t)(e'zk"r - t)ldBru _2ro3No (rr _ t) n,t*"r (.+)dr 3 \RBI-/

Substitute into the equation fbr AZ to find

Lc : -'"oir ^ ,r, - tr#ntkur (Pz- P,) +

For Nr(g) under the conditions specified'

^ I rr2tto No." r _ ly.€_r, ,urlp,LU : ---J-(n Rrr

2r(32'/.7 x 10-12 m1316'022 x - ', (H) aos I roo(lo o - t.oor I-

22-8. Determine Z" - Zu for a gas that obeys the cquation of state P lV - b\ : RT

We can write, from the equation of state'

P(V - b)

/aP\t;l/ay\[*l\0r / p

RTD

-v -bR

v

Now we substitute into Equation 22'23:

e ,_e ,: r (#)"(#)" RT

v-b (;) :^

22-9" The coefficient of thermal expanslon of water at25'C

compressibility is 4.525 x 10 5 bar i Calculate thc value

25'C. The densrty of water aL25'C is 0'99705 g'mL ''

The molar volume of watcr ts

_ 1 \ \/l8.ol5g\/ ldm:,\:o.ol8o68dmlv:(o*6;;;)( '* i \rooo'L)-"

We can now substitute into Equation 22'27 to find C" - Cu' For one mole'

is 2.512 x 10-a

of C ,, - C,, for

atT vC,-C,,:-.K

- (2"572 x 10 * r-')'(z9g'ts r)(0'otg0og amt)

4.525 x 10-s bar r

dm3.bar-r .K-l

and its thet'mimole of water

1.815 x

-r-

LZI

00r>I /J

009 00s 00t

'edo1s e,trltsod e suq eurl prnbrl-pllos eql 'prlos eqt;o flrsuep oqt upql ;eteer8 sr prnbrl eqt yo,{ltsuep oql esnEJeg 'eut1 su3-ptnbrl eqt lcnrrsuoo ol lurod IucrlrrJ puu '1urod Burlroq lerurou

,1urod

eldrr1 eqt puu eull prlos-prnbrl oq] tJnrlsuoJ ot turod 3ur11eu luurrou pue lurod eldul eqt esn e \

6-

pue i3"yg I 'turod Surlroq I€rurou :C.lt t ,1urod Burlleru'.uN Zt'0 pue J.EII 'turod e1du1 :e1ep 8ur,tro11og eql

'prlos Jo ,(lrsuep < plnbrl go flrsuap

Ierurou lutle 91 I puu J.ZtE .1urod lucrlrrc

ue.rr8 z1 ro; urer8erp eseqd eqt qole{S .Z-€i

'ernleredrual lurod e1du1 eql ueql req8rq sr e.tnluredruellurod Suqleur Isturou sll esn€Joq 'ernsserd perlddu uu repun lJetu lou seop ue8,{xo l€ql ees uer el\

>t tJ0rI jzt 00t 08

yoI

ruut8erp aql arnpord a16 .lurod JeorlrJ3 eql le sdols eul

se8-plnbrl eq,I 'ouII su8-prnbrl oqt tcnrlsuoc 01 rurod IeJrtrrc puu 'lurod Burlroq pugou ,ru,oa'"ta,r_oql pue eull pllos-plnbtl eqt lJnJlsuoJ o1 lutod Suqleur JEuJou oql pue lurod a1drr1 eqt esn u'c o.\1

isoop Jele,^A se ernsserd perldde up topun l1aur ue8,(xo seoq .Jo6.Zg l _'1urod Sutlroq pt'rou puu i3"7'g I7- 'tutod Sutrleur lp.urou luol g73 Lt puu x 9'tsl .rurod

lecrru::'uot Vy I pue )I g'?E 'lurod e1drr1 :u1up Surmolyo; eql Sursn ua3,,{xo rog ruer8erp esuqd eql qJle{S . t-

SNOIl-NIOS CNV SIA/]IBOUd

e!rqlllnbl aseLld

PINbII

t(, drrdvHf

724 Chapter 23

23-6. The slope of the melting curve of methane is given by

dP - (0.08446our'a-r es)To8s

dT

from the triple point to arbitrary temperatures' Using the fact that the temperature and pressurc

the triple point are 90.68 K and 0.1 : ',:4bat,calculate the melting pressure of methane at 300 K

Integrating from the triple point to 300 K gives

^ P, 1":t00 K

[ ' d, : l^^. ,.0'08446 bar'K-rssrossdr

Jo.ttTq b", Jq0.o8 K

0.08446 bar'K-r'8s[{:oo r)"t - (90.68 K)"']Pz- 0 l714bar :

1.85

Pr: l556bar

This is the melting pressure of methane at 300 K'

23-7 . Thevapor pressure of methanol along the entire liquid-vapor coexistence curve can be express- -

very accurately by the empirical equation

ln(P/bar) - -10'7?849

* 4.373232x

where x -- T lr,' and { : 512'60 K' Use

methanol is 337.67 K.

+ 16.158201 - 3.603425x

, - 2.38t3jj x3 + 4.5l.2199(1 - x)'io

this formula to show that the normal boiling point;=Hrh

''F

t;r

At the normal boiling point, P - 1 atm : 1.01325 bar. If the normal boiling point of methanol i'

331 .67 K, then the equality below should hold when x :337 '61 1512'60:

, l0 15.8491n(1.01325) : _ '"'' :" + r6Js8201 - 3'603425x

+ 4.313232xt - Z.3gtzll x3 + +.512t99(1 - x)1 70

0.013163 :0.013163 =

-16'323364+|6158201-2.313.|19+1.891112-0.6801220+07351410.0132546

23-8. The standard boiling point of a liquid is the temperature at which the vapor pressure is exactll

one bar. Use the ".pi.I"l formula given in the previous problem to show that the standard boiling

point of methanol is 337'33 K'

we do this in the same way as the previous problem' but substitute x : 331 '33 l512'60 into

" -10'"752849 + rc158201 - 3'603425x

x* 4.373232x2 - z.zgt3ll *' + 4'572t99(1 - x)r 70

ln(1)

?

0

0^,- t6.33g820 + 16.1 58207 - 2.31 1329 + 1 .893892 - 0.678668 + 0'1 31 5'/ 2

-0.000143

r_loru.[{ z'ltOOt

ut091

x9z() 9l'zEt)() 9l'Lzil(,->r'r lotu'f Etlt'8)

'd tJ -'J- Ul ---;- : Ho"tV'rl "t'JU

t l'f,7 uotl€nbg esn u€f, e',rr 'e8uer ernlereduel srql :e'to 'fuea'

lou seop H

ul

o"ny Sut,unsse'ure8Y

.rloru.f)09'Ztslenlu^lElueuuedxaeqJ'uol]EzIrode'r'1o'{dluqtue'IBloulslIelEtullsa

'J"6Llst lurod Surlroq [ELUrou sll Pu€ J.?9 t ]E ]ro1 00t sr cp'{qeplezueq 'Jo ernsse:d ;odun eql 'BZ-g'Z

("'t \ y :3\'l - 'J I Ho"'v 'd

lrol (.)LOl : ull€ 80t'I : zd

ZVt'0:1urre /27) ut

It> Et's8t)() El'tlt)l , x , tn.'t't sltE g : tll ur

L- xor Llotu'foggot 'd

/ .gJ-\ -j :'!u,\-t=) ,'':^,,: z,

:g1'97 uorlenbg

ernleredruel eerSep-ual stql le^o ernlereduel Ol lCedse-r qll'44 luBlsuot suI€uIeJ

asn u€c e,Lr 'e8ue-t

Hoonv Surunssy

eql 'J.0 I I le rela,{\'1urod Surltoq l?tuJou

'rJol ELOl sI enlE^ Ieluetulreoxe

3o e:nsserd rocle.t eql elEInJlEt o1 uottenbe uol'(ede13-snlsn€lJ eql esn

strtE r-lotu't)t99'0?sllelB'^ 3ouotleztrodu't1o'{dltqtuer€lotueq;-'Lz-tz

.Lt-EzurolqoJd ur pe1]o1d eSuer eql Jc^o prlE^ st uotlunbe uor,{ede13-snlsn?lJ eql l€q] sreedde

)I .prle^ aq ot uortenbe uor,(e<Ie13-snlsn'lJ eql JoJ lutlsuoo eq plnoqs pe11o1d eull eql 1o edols eq1

".tJ

'J lI 's^ 7 u1 lold ot L-EZ urelqold ut uant8 d ul roJ €lnturoJ eql esq

eqt )urqt no,{ oP e8uer

'J/l rsura8e 7 u1 told oi

iplle^ eq 11r,r,r uorle nbe uo'r'(ede1 3-s nIS nE I Je;nlureduel 1€q,{\ Je^o '1Z-EZ Luelqold tuo:J suoll€lncluc rno'( Sursn

L-EZ LuelqoJ4 ut ue,tr8 loueqletu JoJ €]€p e:nsserd rode't eq1 eso '92-tz

,-\ I J/L

l€l r:uqrlrnb3 asrq.l

'J .- J se o <- sounv leql eclloN

'elqlsssJdluocul eJ€ ellqdeJS puE

puouerpqtoqteqtetunssv',{1e.tt1cadser', uc'3 9Z'Zpuv r-tuc'3lS'€eqo1 elrqder8puupuotu€Ip

3o,&rsuep eql e{€I 'J.gZN teqlo qc€e qll,lr tunlJqlltnbe ut ere altqder8 pue puoluulp qslq' l le

ornsserd "qt "rn1nr1i3li'o

pun ";7'y elelncn,)'JE9t't+ 9681 : ,-loru'f/"9'y leql ue'ttg

(puoururp)3 * (etrqdur8)3

e8ueqc esuqd eql rePISuoJ '98-tZ

00ttlJ

09t 009 092

JpldpJpl"dp

st lurod e1drr1

ecurs 'JJol Jo sllun uI eJe seJnsserd ereq,Lr

/,J\rpt-tso:-\) z'9062 / " ,d P

se se,\Jnc se8-prnbrl puu se8-pr1os eql Jo sedols eqt elIJ'^ uec ed\

.1urod e1du1 eql le a^JnJ su8-prnbrl eqt pu€ e,trnc se8-pt1os eql Jo sedols eql Jo oIl€J eql elslnopJ

euqrpnb3 aseqS

xw

tt I

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elgrnrce ,(1e1e1duroc tou st euII se3-prnbrl eql JoJ uotlenbe eqt leql elo51 '1utod Suqleur I?uJou eql

puu 'turod e1dr.q eql'1utod pctluc eql 'su8 pue rode'r oql uee'^Aleq euII eql 'se8 puu pllos eql uee'4d'leq

aurl eql 3ur11o1d rq eprpo, u"dorpiq ;o urerSerp eseqd e ecnpo'd u'r ?1y\ os '>l zz7' sr

'' ;o lurod

3ur11eru l€tuJou eq1 l€ql aou{ osl€ olv.JJol g'gtz: ot, tnql pug e^\ 'o'd

'oJ e^los ol 3ur1nlqsqn5

>r Ev'vlz : d'J

) g'01€ : o'J8vv'r

drJ - 020'61-LL>LI_

>r L'9692 >rz'9062

luroo

e1d'l eqt le os .eur€s eqt eJB sernsserd rodu.t prnbrl pu€ pllos eql eJeq^A pelBsol sr lurod e1du1 eql

'eplpol ueSorp,{q;o urur8erp eseqd eql I{rleIS ol

urelqo-rd snora.erd oql ur utep eqt esn ',(1e,tt1cedser 'ru]€ 0'28 pul 5 VZV'>IZZZ are splpol uaSorp'{q

1o elnsserd letltlrc eql pug 'ernlered1uel IBoIIIJc eql 'lurod Suqleur IeIuJou oql luql ue^IC 'St-tZ

drr.

>r L'9692rrlr i :

-:

>17'9062

eql lE sedols eql Jo OIIEJ eql '1urod eldrrt eql lE ,f, -- sd

/,t\"lrI '* l,,l : -

Pue\xlsoszl - dP

LVt

742 Chapter 23

To find the standard molar Gibbs entropy, we use the Maxwell relation (Equation 22.46)/^

A. s" : _ eA : _3.:03 J.mor I' \ ar /,Substitutinginto A,Go: A,E _ TL,s'(Equation 22.13),wefindthatA.H': 1g95J.mol-r.Because both graphite and diamond are incompressible, we can write (as in Problem 23-18\

and

Gi.upr' : ci.*n + V *urn(P - Po)

4,u' : do,"' * {,",{e - ro)

Combining these two equations gives

A,G - l,d + (%,", - T r_r)( p * po)

When graphite and diamond are in equilibrium, A,G : 0. Substituting into the equation given inthe problem, we see that at 25'C, L,G' : 2898 J.mol-t. Then

0: A,G'. (#-;= - t-h;=) rt,ot g mol-r)1P - l bar)

:2898 J.mol-r - (1.916 x 10-3 dm3.mol-r)1p - 1 bar)

P- | /?RqRr.-^r-r1/o'o82o6dml- t.916 x l0 r 6r:--^1-r

(zals l'mot- ) ( ,ffi !1) +ru-: 15 000 bar

23-37. Use Equation23.36 to calculate k" - En for Kr(g) at298.15 K. The literature value is

-42.12 kJ.mol-r.

lt" - Eo: -RZ ln

We do this in the same way we found p' - Eofor A

qo(V.Tl (2rmk"T\3'2-r-lv\r'/

l({\wlL\y/ P" )r(g) in Section 23-5. First,

(23.36)

_l (2n)(1.391 x 10-25 kg.mol-')(t.381 x l0-23 J.K-')(298.15 K)lt/'_|l

L (6.626 x 10-34 J.s;2 |

:7 .422 x 1032 m-3

kuT _ (1.381 x 10-r3 J.K-')(298.15 K)Po 1"0 x 105 Pa

Substituting into Equation 23.34 giv es

: 4.716 " 16*26 --3

ti" - En: -Rr," If4) g-l' u L\Y/ P"): -R(298.15 K) lnt(7 422 x 1032 m-,) {+.t t6 x 10-26 m-3)l

: -4.212 x 104 J.mol-l

^a, rn - ft- +).ur- = ^?.b

^arcn r ^a,$ =ft-tl

"$e- --11,!

,'P tJ-*={.P;15-(4.

Ap F=nP

S '!4,"

/A I q'.r-n ltu-n\,Fb * lfl -\lju Y = !*'

H'n(*)^1w\ r/^e)

,..d- \;alr

tlrte t a/ l€ )Ap'r- = /tp \Te) +,f \Te /=nP

(fo*rrttotl) o J_ (",

'^$Fy{^M? ffi J=SF

ao',unr wpwb,, --'ft)"-ffi=d 0il)t,&U,o), 1:tE {k}"- n 3t08 Yoq0

nemt

t\)rsv= -f It p dYJU,

o.=nZ-T -o/t-\'rffi3/r,rr"v = -.*\t +il

dV + "'t: -L&V

r,ortv = -nrll r^(#*J tanP(# - h)

$*u= -"nl(t-;.)+nRrr,r.(ffi) 4^nz(# -b)

$*u= nfrrhtr#

@ cp-cv =ry , n=L(#),) K-- + (ry),

Jt"u: V=nBr 50 eI= t ar,d (#\= *F

Ce-Cv=fJ .$'N--r-- .ff ", \v Vt ft.-F) Ylt =oL

flq-oitrl = wbb'tlql =r-l

(r)(# W w6) $+ilb)=d--! ?sset*se

\BJnd g'TggQLofro)=dy

6>t St '$tC ) ft.ur. f, ooo,rrblrgr$.9o*tloo,1, )

\M)w =dV

)S/'Ele ')uQ ='tt W 4'hV?l5l'8?e =)"9- = lL W (trlrt urtr.{=utsV -- dV

.-!- 'rf SnttV ,dfT,rJorE =dprJ

?o rlq 1 n*o.v'' iF nW =dP

"N 7u/ ffi=4H= Tl =-dl €oe \ndilov + ,ruqwr vt t =:'ttul 'u/ [ 'L\ *n"uttu't]

=d

@ VP*r,a.= a,b4 To-utVPario. = 0,3,63T0-wVPrr^. [1,q3 Tq,rYP{= 553bTma"

AHffr= a\\8.

J= -llLo (T= -n6.f,"C Fnot L\lsubl

T= -luD'L Al]r*, Al-lpos #1= -tA\ T^,1r4" pt,

A\\eus - fu Alwpt b\s,a 1"'AS*p : llaq

'n(k)=

CAr,r^ /CW*%4\:*Aygp_H"/!_f\

L-- t T, ar)

0, lr3b4

7b lorrt\__l

a,swle.

h ti lu3 #) = +#?S * ( #"-* - #,,,)

A f\rrn s Tl)eUa^au |C}ryo1l.,o'^€,6- ?

//n( As.rn\.. t-Y - t \(E,sl4sr/r,n"t) \ lbl,lSK l4ln;rs K )

^[g')= +-\"tu\\"(+

+)

- A$*\\o

Asro\'= +

e^sl\.., % kxlrrhrf3 tJS

AKrlp"L

-A!-) P t,, T. !/( Pt!tr'e44-

T,WPt"" Fn *e*daa," fW W". { k*,p' A fuyt' :px

i^/A= A$rbll--StPr/- &lT T/

-t^ (p) -,e/^(p'') = +t( + })A(p.)=i/ncq)-ry(+

+)

,o\L U^ {*',rL #,^ %rdry -l)iu""

-l,n(p,)= l^cg) ^y(+ {)A+ i-rfi" p'ar, p,

N-,,u Sifs - pr .#^ tlG'

,q.,"tpJ-+p(+-k)-,t^Q.) - brye D

lt. ro5 +31_CU t; A13,0\l "l1

tst,ll'lr+ryu_4il'ww

T1

jrn [0,affi)- 3J;wYr - J; =,bn [ss'a6) -U4 ! !9' l+ L\

r3l4-L rr l4t l'i)- ' t'3i4s (" lry,tsz

-.33,3?3*;q$03 r tt 0AY4

Tr

r, N \?osR

@ [,,Jo uJa^^l AG t-S'c). Sr*, fu rnsr,,r,,v,al waL,ata Oo(tAbLO"C)=C\/y'^ {

- at{T,

AG Cs'c) -qbt,tf, K

'tlA*- Ce.0b\ \.\4[,h^hl4 WzI a trtu/'\I-\ )T)p

fM=-au\Ep }t\ ['h{, bstr,,vwo l

A6(r?) _at

AG ta6?,trK) = + Nt\ (o,o rt:os)U!-

^ail, A\{ , 1la. $" Ucrp.r- ?"* ?

VPun @ -s'C = 99 * tO-3 bouV Ps"r @ -S"C = e[- lo-t boo-

,L'''falJp3)= AU{,p/ \ - \-\ e/noLtlffir/- T-tem-Fens7

bnxnru- w c\F- L'oLilcd fum ,Ato -- \G{*r"All = A[\+,e.,r.

[C,r rr.qsc\ L[\t"*y = -ARluu = - Af\trn u AS"p

UJr* Fr,* hV,,fuwr,x^tr?

"lrnteg^tD-) 4/vr(P*r,r) = Ak+(-0, oo@68ant)

/in (A.bxtu=) --tu" (Pil,uD= All"uu (-0, w)obrz&)

Alw aul'+1ar+ f,t^c !if^a,

$n (09rtD{)-.0h (4l,* lD-) : p,Wq.TAB(AHuop - AHsr)

- Afl$,ro =

G(a6gJsK)