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Unformatted text preview: Chapter 5 w—r' Balance equations
for a closed system 260 Equilibrium and Stability
in OneComponent
Systems Now that the basic principles of thermodynamics have been developed, and some
computational details considered, we can study, in detail, a fundamental problem of
thermodynamics: the prediction of the equilibrium state of a system. In this chapter
we examine the condiiioiis fortlie existence of a stable equilibrium state in a siﬁgiél"
component system, with particular reference to the problem ot phase equilibrium.
Equilibrium in multicomponent systems will be considered in following chapters. T 51 'l‘HEEBH‘EEIAEQUIUBBIUNL Chapter 3 we established that the entropy function provides a means of math—
ematically identifying the state of equilibrium in a closed, isolated system (i.e., a
system in which M, U, and V are constant). The aim inJhiisegjLonis_to_dey§o_ga means __o_t' identifying the e_quilibriuni state for closed thermodynamic systemssub—
ject to otherconstraints. espeei—aﬁy thosero't‘ constant temperature and volume, and
constant temperature and pressure. This will be done by ﬁrst reconsidering the equi—
librium analysis for the closed isolated system used in Sec. 3.1= and then extending
this analysis to the study of more general systems. The starting point for the analysis is the energy and entropy balances for a closed system: 030’ v
——x_~i. — = )— —  5.11
l f P {it ( )
and
as Q  l
_. .... .. ....__.__.~;._’ _ : _ + l'rw 5_12
i (if. I SM ( ) 5.1 The Criteria for Equilibrium 261 with Second law of
thsrmvdraatuics 2 0 (the squeals.ltslsiasstsssilibrium or for reversiplviamcesseli (5.13) Here we have chosen the system in such a way thatthe only work term is_ that of the
_det__r_)_rlrn_a_tio_n_oi t_h_e___sy_s_tcmboundary. For a constantvolume system exchanging no
heat with its surroundings, Eqs. 5.11 and 2 reduce to so that (or, equivalently, U_f___constant_,__si_nce the total number of moles, or mass, is ﬁxed in a closed, one—component system), and
ds .
a = Sgcu 2 (l (5.14)
Since the entropy function canonly increase in value during the approach to equi
librium (because of the sign of SEC“), the erigopvrnusthcaniaigjniurnat equilibrium.
Thus, the equilibrium criterion for _a__cl_osed_, isolated system is S = maximum
or atequilihrium in a closed systeril'atconstant U and__l/__
S = maximum (5.15) At this point you might ashhow S can achieve a maximum value if Q and__y_ar_e ﬁxed, since the speciﬁcation of any two intensive variahltis comp_le_tely___f1_xc_s__theval _of__'all__othi_e_rs.rThe answer to this question was given inHChapter 3, where it was pointed out that the speciﬁcation of two intensive variables ﬁxes the values of all
other statevariables in the uniform equilibrium state of a singlescomponcnt, single
phase system_.:lThus, the equilibrium criteriori of Eq. 5.15‘b‘3ii'6é' used for identifying
the ﬁnal equilibrium state in a closed isolated system that is initially nonuniform, or
in which several phases or components are present. To illustrate the use of this equilibrium criterion, consider the very simple, initially
nonuniform system shown in Fig. 5.11. There a singlephase, single—component ﬂuid
in an adiabatic, constant—volume container has been divided into two subsystems by
an imaginary boundary. Each of thesg_ﬂb_sy§teﬁs_wil_l_be____assumed.to containthe
same molecular species oliiniform thermodynamic properties. ngvcr, these sub
syst—crns areEma tEFtHé flow—of'heat and mass across the internal lm—uifd'ai'y, and . fvl’ 51‘ Lrl‘ vl N‘II’ lS'Il'.J L111, 1111  I Figure 5.1] An isolated nonequilibrium system. 262 Chapter 5: Equilibrium and Stability in OneComponent Systems their teimJerature andjgressure neeclmriot be the same. For the composite system con—
sisting of {Hausa subsystems, the total mass" (though, in fact, we will use number of
moles), internal energy, volume, and entropy, all of which are extensive variables,
are the sums of these quantities for the two subsystems, that is, _ 3 J: in} iv : Ni. __ v” U = U‘ + U" V s V1 _ v11 ; (5.16)
' 'and' ___ S} __ sll Now considering the entropy to be a function of internal energy, volume, and mole
number, we can compute the change in the entropy of system 1 due to changes in N1,
U1, and D" from Eq. 4.25c. ' I "Jill I I I
  '  as as 3
 as“ : —,—, arﬂ+ ——1 dVI+ 5—, dN‘  . .' . ' i; dU J, I av 3N .. . I  , _ _. ‘ 4 VJ" UENI 01,1/1 In a similar f2:shion_gf§"_£ar_i__hedcomputed. The entropy change of the composite
system is I .. . . 1 PI PI}, 61 G”
~_ 1 11_ I n I II_. 1_._ n
__ ds___—_as _+_ 033 — T—ldU + Wat; + Fay + ﬁat! Fay Tu (W
_ I __ However, the total number ofrnoles, total internal energy, end total volutneﬁtfe eon— " stant by the constraints on the syste—m,sothat_ ' ' ' '
' J (W = N + azv” = n or a'N' = —dN”
’5'" L" f l dU : an” + dU“ = 0 01' dUI = —dU“ (517)
' =' ' ' av = til/1+ av” = 0 or av' : —av“ __ Consequently, .. _ , m 1 1 , .01 P“ , 9'1 Q” 1
:_y.l_'_2:__ I' __ _. ' i  ' _ _ . ..
h J " i' Q __ Now sir1ce__.§____—i _rnaggimunror dS 2h _Q for all system variations at constant N, U, and V
‘I. —. '_ .  ?_ (here all variations of the independent variables dUI, dV‘, and dNI at constant total
PPM1”“ ' ‘ number ofmoles, total internal encr ' ,and total volume ,we conclude that
I. _ V_ E)“
I t as .1 TI _
I 3 0, 50 that a : 01: TI _ TH (SJ93)
_ _ _. . ._, . . .. . ' VIPVII . . , 7."? as P1 P“ _  is V1) .= 0, so that —T, = if, or P‘ = P“ (st—9h)
“ _ _ H: UI’NI
and as G‘ G”
——I = 0, so that ,— = ‘,— or _(jl = G” (SJ—9c)
61” Irlvl I ' I ' i 5.1 :l'he Criteria for Equilibrium 263 ._ _Therefore, the equilibrium condition for the system illustrated in _F_i_g.__5_._l_:_l_ is__sa_tis~
\ have'the‘sarne'temperature, same H I base system this implies
that the"composite should be unifﬁﬁﬁi‘i‘ﬁi‘s"teas"a'biéiaus'té'suit,and it'i's re—
_ _ assuringthat it arises so naturally from our development.
""3' ' H i" " d The foregoing discussion illustrates how the condition (13 = 0 may be used to
' identity 3 possible equilibrium state of the system, that isfa'state'for which S =
, _.—:" maximum. From calculus we know that {517,3 = 0 is a necessary but not sufﬁcient—eon;
_, masters to achieve a maximum value. In particular, giS__ﬁ__Q at a minimum value
it L".  ' or an inﬂection point of S, as well as when S is a maximum. The condition (1'28 £ 0,
g _ . when__dS_ = assures us that a maximum value of the entropy, and hence a true " equilibrium state, has been identiﬁed rather than a metastable state (an inﬂection
_ ,_ point) or an unstable state (minimum value of S}. Thus, the sign of (1'28 determines
if. _ .. ——— the stabilitypt: the state found from the condition mate's ﬁé‘i‘iﬁﬁtibu‘uensor
this—stability condition are considered in the next section. __,_ It is also possible to develop the equilibrium and stability conditions for systems subject to other constraints. For a closed system at constant temperature and volume, the energy and. entropy balances are i niolﬁﬁribbs—free meager a singl component, single \ d: *_ 
:.:.. and “u ..
‘ ds _ Q 5.
a; _ T ‘ “35” Eliminating between these two equations, and using the fact that TdS = d(TS),
since It" is constant, gives
d — ‘ ' 
w : 2 _T5 an <_: 0
tiff dt 3 Here we have also used the facts that T 2 0 and Sam 2 0, so that (‘ Y‘Sgen) E 0.
Using the same argument that led from Eq. 5.14 to Eq. 5.1—5 here yields A = minimum
or for equilibrium in a closed system at a constant .T and V
A = minimum {SJ10) It should be noticed that this result is also a consequence of $38,, 2 0.
If we were to use Eq. 5.1—10 to identify the equilibrium state '{iTan'ihitially nonuni— form, single—component system, such as that in Fig. 5.11, but now maintained at a
constant temperature and volume, we would ﬁnd, following the previous analysis,
that at equilibrium the pressures of the two subsystems are equal, as are the mo—
lar Gibbs free energies (cf. Eqs. 5.19 and Problem 5.4); since temperature is being
maintained constant, it would, of course, be the same in both subsystems. ii For a closed system maintained at constant temperature and pressure, we have all] v dV  d
_.c_n ‘ P d: ‘ Q 50”” 264 Chapter 5: Equilibrium and Stability in One—Component Systems and as _Q 
are+%n Again eliminating between these two equations gives (EU (1' d «it? 
_ _ —~ __ Ll : _— z — ‘
d: of: (PV) dr (TS) dr T533“ la'l
c: moan so that the equilibrium criterion here is _ l G = minimum
' '. — or for equilibrium in a closed system at constant T and P
Most Important of the l . . ___—__...  . ——— . . . . . . . Q = minimum
equilibrium criteria (5.11.2) This equation leads to the equality of the molar Gibbs free energies (Eq. 5.1—9c) as
a condition for equilibrium; of course, since both temperature and pressure are held
constant, the system is also at uniform temperature and pressure at equilibrium. Finally, the equilibrium criterion for a system consisting of an element of ﬂuid
moving with the velocity of the ﬂuid around it is also of interest, as such a choice of
system arises in the study of continuous processing equipment used in the chemical
industry. The tubular chemical reactor discussed in Chapter 9 is perhaps the most
common example. Since each ﬂuid element is moving with the velocity of ﬂuid sur—
rounding it, there is no convected ﬂow of mass into or out of this system. Therefore,
each such element of mass in a pure ﬂuid is a system closed to the ﬂow of mass, and
consequently is subject to precisely the same equilibrium criteria as the closed sys
tems discussed above (i.e., Eqs. 5.15, it), or 12, depending on the constraints on the
system). Iligequilibrium and stability criteria for systems and constraints that areot‘ in—
terest in this book are collected in Table 5.11. As we will see in Chapter 6, these
equilibrium and stability criteria are also valid for multicomponent systems. Thus
the entries in this table form the basis for the analysis of phase and chemical equi—
librium problems to be considered during much of the remainder of this book. " Table 5.11 Equilibrium and Stability Criteria System Constraint Equilibrium Criterion Stability Criterion
Isolated, adiabatic U = constant S : maximum
fixed—boundary system V = constant d5 = 0 r1123 *1 0
Isothermal closed T = constant A = minimum
system with fixed V = constant da 2 0 (FA > t)
boundaries
Isothermal, isobaric T = constant G = minimum
closed system P = constant (16‘ = 0 (EEG > 0
isothermal, isobaric T = constant G — minimum
open system moving P = constant (16' = 0 dQG > 0
with the ﬂuid velocity M = constant __________—————— r . (/52 STA BILITY OF THERMODYNAMIQ 5.2 Stability of Thermodynamic Systems 265 Using the method of analysis indicated here, it is also possible to derive the equi— librium criteria for systems subject to other constraints. However, this task is left to
you (Problem 5.2). ILLUSTRATION 5.11 Proving the Equality of Gibbs Free Energies for Vapor—Liquid Equilibrium Use the information in the Steam Tables of Appendix III to show that Eq. 5.19c is
satisﬁed at 100°C and 0.10135 MPa. SOLUTION From the saturated steam temperature table we have at 100°C and 0.10135 MPa that
it?“ = 419.04 kiikg
3L 2 1.3069 kJikg K H" z 2676.1k17kg
3'" = 7.3549 kLr‘kg K
Since é T§,Iw_e have further that ﬁll—mm}? {1904 — 373.15_;< 1.3069 I] —68.6 kit’ng and C7" = 2676.1 — 373.15 x 7.3549 = —68.4k]il<g which, to the accuracy of the calculations here, confirms that (EL = (EV (or equivalently
that QL = 9") at this vapor—liquid phase equilibrium state. I ’“13 In the previous section we used the result d5 =__0 to identify__t_he__equilibrium state
of an initially nonuniform system constrained to remainat constant _mass,_internal
energy, and'y'ol In this section we explore the information content of the stability
criterion tiZS < 0 _ at constant M, U, and V [The stability analysis for closed systems subject to other constraints (i.c., constant
T and V or constant T and P) is similar to, and, in fact, somewhat simpler than the
analysis here, and so it is left to you (Problem 5.3).] By studying thc sign of the second differential of the entropy we are really cont—i
sidering the following question? ﬂpﬂsilllﬂl.§_§rll§llLﬂL1?.Luatiuu.iﬂ.aﬁeld93299113“: Isaynteinperatureor pressure, occursin some region of a ﬂuid that was initially at 5'
equilibrium; is the character of the equilibriumstate such that dES <_ _OLand theﬂuc— __tuation wrll'Ti‘SsipatcfEi—i— which case the ﬂuctuation grows until _the
§X§t§meyolsesto anew equilibrium state of hlglier engopyjff‘ l a—w_ __ _.__ __...__ / . In fact, since we know that ﬂuids exist in thermodynamically stable states (Exper—
imental Observation 7 of Sec. 1.7), we will take as an empirical fact that (£23 < 0
for all real ﬂuids at equilibrium, and instead establish the restrictions placed on the
equations of state of ﬂuids by this stability condition. We ﬁrst study the problem of
the intrinsic stability of the equilibrium state in a pure, single—phase fluid,.and then
the mutual stability of two interacting systems or phases. 266 Chapter 5: Equilibrium and Stability in One—Component Systems We begin the discussion of intrinsic stability by considering further the example
of the last section, equilibrium in a pure ﬂuid at constant mass (actually we will use
number of moles), internal energy, and volume. Using the (imaginary) subdivision
of the system into two subsystems, and writing the extensive properties N, U, V,
and S as sums of these properties for each subsystem, we were able to show, in
Sec. 5.1, that the condition that as 2 (13' + as“ = 0 (5.21) for all system variations consistent with the constraints (i.e., all variations in dNI,
owl, and do”) led to the requirements that at equilibrium TI : Tn
P1 2 P11
and
GI : 911 Now, continuing, we write an expression for the stability requirement (125 <. O for
this system, and obtain £325 = Straw)? + ZSLvth‘XW‘) + Show)?  ZSbndU‘XdN‘) + 25i«N(dVI)(dN1.)_.+ SisN_(_dN[)2 (5243}
— Si_it:(dUU)2 + Zb'tbtdwlttdv“) + Stuart/“)2 + zsﬁNwo‘ldeN”) + 23{!N(dvu)(dtv”) + .s,[},,.(arv“)2 < 0 In this equation we have used the abbreviated notation that 2 I n t 1
311m : f9_5 3;”, = _‘9_ (ﬂ) __
are W an ,ZN av w, Since the total number of moles, total internal energy, and total volume of the
composite system are ﬁxed, we have, as in Eqs. 5.1—7, that w‘ : —dU” (W: = —dV” .i dN‘ : —dN” and
arts = (Slut: + s'tiuttdv'f + 2tstw + stinth‘ttdt/‘t
+ (Sin! + Sarita“)? + 26in: + Sisttdv‘xcm‘) (5233)
+ 2mm + SEN (avi)(aiv1)+ (5,1,1, + 5;ngde _ Furthermore, since the same ﬂuid in the same state of. aggregation is present in re—
gions I and II, and since we have already established that the temperature, pressure,
and molar Gibbs free energy each have the same value in both regions, the value of
any state property must be the same in the two subsystems. It follows that any ther—
modynamic derivative that can be reduced to combinatiorns of_in_tensive variables
must have the same value in both regions of theﬁuid. The second derivatives
5.2:2'b), as we will 'see' shortly, are combinations of intensive and extensive variables.
However, the quantities Nwa where x and y denote U, V, or N, are intensive vari— ables. Therefore, it follows that . l ~I _ II ‘II
N s“. _ N 5,, (5.23) 5.2 Stability ol'Thennndynamic Systems 267 Using Eqs. 5.1—7 and 5.23 in Eq. 5.2—2 yields 1 NH .‘ I)
“’25 = NINit"')lN15i,u(dUl)‘ + 2N‘Stvth‘de1)
(5.24:1)
+ iV'1.S‘{,V(dl/I)2 + 2N135N{da)(ml) + 2le{m(dv‘)(daﬂ) + N‘sltmww'lﬂ < 0
The term (NI + N“)i’N1N” must be greater than zero since mole numbers can only be positive. Also, we can eliminate the superscripts from the products NS”, as they
are equal in both phases. Therefore, the inequality (Eq. 5.2—4a) can be rewritten as N51,.“ (dUl)2 —— 2NSUV(dUI)(dV1) — NSW(dVI)2 + 2NSUN(d U1}(aw‘) (5.24b)
—— 2Nsw(dv‘)(dN‘) — Nswwivl)2 < (1 Equations S.2;§__a_nd_é}_rrrust be satisﬁed for all variations in NI, UT, and VI if the
ﬁuidHiTt—opbe stable. In pai'ticulaE‘s‘i‘riEheEﬁTwSQétb mtEt‘EﬁaﬁmEZt'far'arrvsrnrm‘ﬁt
in [271(a'Ul aé 0) at ﬁxed values of Ni and VI (i.e., le = [land (fl/1 = 0), stable ﬂuids
must be such that NSUU < 0 (5.25a) Similarly, by considering variations in volume at ﬁxed internal energy and mole num—
ber, and variations in mole number at ﬁxed volume and internal energy, we obtain NSW < 0 (SJSb) and
NSNN < 0 (5.25c) as additional conditions for ﬂuid stability. “'Mor'e severe restrictions on the equation of'state result from demanding that
Eq. 5.24b be satisﬁed for all possible and simultaneous variations in internal en
ergy, volume, and mole number and not merely for variations in one parameter
while the others are held ﬁxed. Unfortunately, the present form of Eq. 5.24b is
not wellsuited for studying this more general situation since the crossterms (i.e.,
rfUldVI, dUIdNI, and dVIdNI) may be positive or negative, depending on the sign
of the variations (1 U1, ail/l, and dNI, so that little can be said about the coefﬁcients
_ of these terms. By much algebraic manipulation (Problem 5.32) Eq. 5.24b can be written as almxgz + 920ing + 63(dX3}2 < 0 (5.26)
where _  I
a, = NSUU “~
92 = (NSUUNva _ NZSEIVVNSUU
63 # (NSULINSNN _ NESEs) f (NSULINSvN _ NSUVNSUNF NSUU N SU 1.} (NSUUNSVV _ NZSETV) 5 S1.
ax, = dU1+ ﬂail/1 + ﬂaNl
SUU SUU dXE = dvl + (SUUSVN "‘ Sovsun) dNI
(SUUSVV — Silly) 268 Chapter 5: Equilibrium and Stability in OneComponent Systems L and
i, dX3 = dNI The important feature of Eq. 5.2—6 is that it contains only square terms in the sys—
tem variations. Thus, (Xmf, (dX2)2, and (cm)? are greater than or equal to zero
regardless of whether dU‘, dVl, and aINI individually are positive or negative. Con
sequently, if '91 = NSUU E 0 r,.._ 2'2r
;; a, = MW NLMW : 0 (5.2413)
: AMISUU and a, g 0 {5.27e) Eq. 5.2—6, and hence Eq. 5.24b, will be satisﬁed for all possible system variations.
a ,, Equations 5.2? provide more restrictive conditions for ﬂuid stability than Eqs. 5.25.
5 It now remains to evaluate the various entropy derivatives, so that the stability
restrictions of Eqs. 5.2—7 can be put into a more usable form. Starting from (as) _ N(a(m‘)) _ _ N (61")
My 3U N.v 6U ivy T2 3U an: and using that for the open system c?
6U Ii NSUU=N 6P
1"
leads to
8U
(ST W
and
1 Since ’1‘ is positive, one condition for the existence of a stable equilibrium state of a
ﬂuid is that r —" —' I I I . q: l I _ :______.._,:......,.,.___2
I.‘ First or thermal stability " f" _ FCVr > 0 f (5.29) criterion That is, the constant—volume heat capacity must be positive, so that internal energy 1. increases as the fluid temperature increases.
Ii Next, we note that
Luv T T2 CW as . a i as '
(“l N (2) m. av w av m, dU W av and from Eq. 5.2—8 that d
N
(311] NSL’V 5.2 Stability of Thermodynamic Systems 26.9 litital NSUV = (5.2'103)1 to obtain Similarly, but with a great deal more algebra, we can show that '2
1 (JP i
NsW —  ' ,2 r 3P P
T ail T CV1" ar 1, (5.21.0b)1 and _ NSUUJ'VS‘VV "_ ENQSEIV 1 T 9 _ ._.
2 NSUU T 5V Thus, a further stability restriction on the equation of state of a substance, since 1" is
positive, is that Second or mechanical 4 0 stability criterion ' 0'1” r
or
1 6V
2 —‘— — $0 5.2]
“T E (6P) ( 1 l where KT is the isothermal compressibility of the fluid introduced in Sec. 4.2. This
result indicates that if a ﬂuid is to be stable, its yolumgﬁtigﬁquation .of state must be such that the ﬂuid volume decreases as the pressure in_crea_s_es_at__consgiiitmtern— perature: As we will see shortly, “tins—r'ﬁrietioﬁ'liag important implications in the
interpretation o'f'phasebehavior from the equation of state of a ﬂuid. Finally, and with a great deal more algebra (Problem 5.32), one can show that 63
is identically equal to zero, and thus does not provide any further restrictions on the I equation of state.
V. _ _,___ . _' ""' Iheﬁaiifconclusion from this exercise is that if a ﬂuid is to exist iri__a“s_t_able equi— libriurri _s__tat_e_,___that_is, an equilibrium state in which all small _i__1_i_t_c_r__i_i__2_i_l_ fluctuations yyill If dissipate rather than grow, the fluid must be such that :2. t"? l cV 2» 0 " (5.2.12)
“f _ __ I and I I \
. ,. '__,_ Eu]: II. : I .. . .. ______ III
l \ ,.__ l. — < (J or KT > 0 1 (5.213)
L_____I_—— ! ' __ _ ____ . __
7 Alternatively, since all real" ﬂuids exist in thermodynamically stable states, Eqs. 12—12 and 13__must be satisﬁed for___r_e_al___llu_i_ds..1n_ fact, unreal ﬂuidstatc forwhich
eilllﬁl'laplal/L > 0 or CV < 0 has yet been found. 1Note, from these equations that N 5%, NSUV, and N SW are intensive variables as was suggested earlier.
whereas Sun. SW, and SW are proportional to N '1. HI 2’70 Chapter 5: Equilibrium and Stability in One—Component Systems Equations 5.212 and 13 may be thought of as part of the philosophical content
of thermodynamics. In particular, thermodynamics alone does not give information
on the heat capacity or the equation of state of any substance; such information can ' be gotten only from statistical mechanics or experiment. Howeyerltherinoditarn;
19;:Closing.)aidere.stric.tiOiis.or. 90_ns_i§_t_egg}i_ltf§lati_ons that r_n__us_t be satisﬁed..by_.such . data; Eqs. 5.212 and 13 are examples of this. (Several other consistency relations for
mixtures are discussed in later chapters.) Higgins24.
Using the Steam Tables to Show Thar the Stability Conditions
Are Satisﬁed for Steam Show that Eqs. 5.2—12 and 13 are satisﬁed by superheated steam. Lawton It is easiest to use Eq. 5.2—13 in the form of (away)? i 0, which requires that the
volume decreases as the pressure increases at constant temperature. This is seen to be
true by using the superheated steam table and observing that V decreases as P increases ! _  at ﬁxed temperature. For example, at 1000°C ' :3 P(MPa) 0.50 0.8 1.0 1.4 1.8 2.5
lite V(m3fkg) 1.1747 0.7340 0.5871 0.4192 0.3260 0.2346 Proving that C V “2 0 or C0 “2 0 is a bit more difﬁcult since i I; “ :: I A C , 2 ___
“ii! .' ¥ (5T ) . v and the internal energy is not given at constant volume. Therefore, interpolation meth
ods must be used. As an example of how calculation is done, we start with the following
data from the superheated vapor portion of the steam tables. .0 = 1.80 MPa P = 2.00 MPa T (°C) if {m3tkg) it 00le f4 (m3t'kg) t} (knkg)
800 0.2742 3657.6 0.2467 3557.0
900 0.3001 3849.9 0.2700 3849.3
1000 0.3260 4048.5 0.2933 4048.0 To proceed, we need values of the internal energy at two different temperatures and
the same speciﬁc volume. We will use P 2A 2. 00 MPa and T = 1000°C as one point; I
1': .' at these conditions V f 02933 m3fl<g and U = 4048.0 th'kg. We now need to ﬁnd the 1.
'i i { temperature at which V : 0.2933 m3r’kg at P = 1.80 MPa. We use linear interpolation : .! _' . for this, i.e., r — 800 f/(r, 1.80 MPa) f {/(SOUC’C, 1.80 MPa) _ 0.2933 — 0.2742 900  800 0(9000c, 1.80 MPa) — {1(8000c, 1.80 MPa) (W301 FEE so that T : 873.75°C. Next we need the ipternal energy I} at T = 8?3.75°C and
P = 1.80 MPa {since at these conditions V : 0.2933 msr‘kg). Again using linear interpolation  .u,..,.._...—.....__,wr.__—.... Important criteria for
phase equilibria under
all constraints 5.2 Stability of Thermodynamic Systems . 271 873.75 —__800 _ 0(873.75c, 1.80 MPa) — 0(800°c,1.80 MP3)
900  800 0(9000C, 1.80 MPa) — 0(8000c, 1.80 MPa) _ 0(878.75°c, 1.80 MPa) — 3657.6
3849.9 — 365?.6 and ﬁnd that
0(873.75°c. 1.80 MPa) 2 0(873.75°c, 0.2933 m3i'kg) = 3799.4 klfkg Finally, replacing the derivative by a ﬁnite difference, and for the average tempera—
ture {i.e., T = (1000 + 87375)}?! = 936.90C), we have 6,.(1‘ = 936.9%,1} : 0.2933 mgﬂtg) g 0(1000°C, 0.2933 m3fkg) w 0(873.75°c, 0.2933 m3rkg)
1000°C — 873.76%: 4048.0 — 3799.4 kl
: —.—— : _ .— >
1000 w 873.75 1 969 kg K 0
Similarly, we would ﬁnd that 6“, > 0 at all other conditions. I Next we consider the problems of identifying the equilibrium state for two inter—
acting phases of the same molecular species but in different states of aggregation,
and of determining the requirements for the stability of this state. An example of
this is a vapor and a liquid in equilibrium. To be general, we again consider a com—
posite system isolated from its environment, except here the boundary between the
two subsystems is now the real interface between the phases. For this system, we
have s = s1 + s” N = NI + N” : constant V = V1 + VII = constant
and U = UI + U" = constant (5.214) Since N, U, and V are ﬁxed, the equilibrium condition is that the entropy should
attain a maximum value. Now, however, We allow for the fact that the states of ag
gregation in regions I and II are different, so that the fluids in these regions may
follow different equations of state. Using the analysis of Eqs. 5.143, 7, and 8, we ﬁnd that at equilibrium (i.e., when
035 : 0), r“! ‘ 5 71 2 Tu {5.215a)
: l PI 2 PH 'I (5.21513) and i (5.215c) 272 Chapter 5: Equilibrium and Stability in One—Component Systems Must be satisﬁed in each
stable phase Here Eqs. 5.215a and b provide the obvious conditions for equilibrium, and, since
two different phases are present, Eq. 5.2—l5c provides a less obvious condition for
equilibrium. Next, from the stability condition (2'25 < O we obtain (following the analysis that
led to Eq. 5.22b) azs = {S{W + ngﬂdvlf + 2{sl,,, + .S’l},.}(aol)(avl)
+ {Sim — Shawl)? + ash; + Sllslth‘XdN‘l {5216}
+ {Sim  Shawl? + asst + stlsitdv'xdw')
Here, however, the two partial derivatives within each of. the bracketed terms need
not be equal, since the two phases are in different states of aggregation, and thus
obey different equations of state, or different roots of the same equation of state. It
is clear from a comparison with Eq. 5.2—2b that a sufﬁcient condition for Eq. 5.216 to be satisﬁed is for each phase to be intrinsically stable; that is, Eq. 5216 is satisfied
if, for each of the coexisting phases, thecquations _ _ : ‘r
. « 61” l "
___,______,__., ‘ CV > 0 and < 0 ‘ \‘r are satisfied. TherefofeTif two phases are"each—intr—in—sically stable, an equilibrium
state involving the thermal and mechanical interaction of these phases is also ther— modynamically stable. Stated differently, a condition [or the mutual stability of two
interacting subsystems is that each subsystem be intrinsically stable. 5.3 PHASE EQUILIBRIA: APPLICATION OF THE
EQUILIBRIUM AND STABILITY CRITERIA
TO THE EQUATION OF STATE Figure 5.31 indicates the shape of various isotherms for a typical equation of state
(for illustration we have used the van cler Waals equation of state). In this ﬁgure the
isotherms are labeled so that T5 } T4 ? T3 “> T2 > T1. The isotherm T3 has a single
point, C, for which (6PI6E)T : 0; at all other points on this isotherm (aPMEhrJE 0.
On the isotherms T4 and T5, (a Play), < O everywhere, whereas on the isotherms T]
and T2, (away), € 0 in some regions and (splat/)1, > 0 in other regions (i.e., be
tween the points A and B on isotherm T1, and the points A’ and B' on T2). The crite—
riﬂn istllaislstahilittrssuirss that (6 PM Elia. <~ 0, which. is satisﬁes for the iéétlié—riﬁé“ '
rg'iiiia T5, but not in the aforementioned regions of the 3"1L and T2 isotherms. Thus we
conclude that the regions (lEL§.andﬁ3n.B.LQ.f.th§_i89thenns. T1 and T2. respectively,
are not physically realizable; that is, they wil h9_t_be_ohservedin any experiment.
This observation raises some question about the interpretation to be given to the
1'"1 and T2 isotherms. We cannot simply attribute these oddities to a peculiarity of the
van der Waals equation because many other, more accurate equations of state give
essentially the same behavior. Some insight into the physical meaning of isotherms
such as T1 can be gained from Fig. 5.3—2, which shows this isotherm separatelyﬁwfe"
109maar_aabar...(saa§tast pressure line) between Branden. lathe.ﬁsstssush 31—3.
150,, we _sge___tl_1_at it___inter_sects the e'qiiatioii—‘olrstiite three times, corres ond_i1_1_g___to the
ﬂuid'vblumesﬂa,_l_/;_, and 3’3. One of these, 1/3, is on the part9 _ rerm that is unattainable by the stability criterion. However, there is no rE’aiiiiiiEi" think that 5.3 Phase Equilibria: Application of the Equilibrium and Stability Criteria to the Equation of State 273 Figure 5.31 lsotherrns of the van der Waals equation in the
pressure—volume plane. Figure 5.32 A low—temperature isotherm of the van der Waals equation. either of the other two intersections at Ya and E; are physically unattainable. This
suggests that at a given pressure and temperature the system can have We differ ent volumes,_ a conclusion that apparently contradicts the experimental observation of Chapter 1 that two state variables completely determine the state of a single—
component, singlephase system. However, ean___oe_eur if equilibrium can exist between two phases of the same species in different states of aggregation (and hence
density).';l"he equilibrium between liquid water and steam at 100°C and 101.325 kPa
(1 atm) is one such example. One experimental observation in phase equilibrium is that the twocoexisting equi
libr'iﬁrﬁlﬁmagmsve tira'gagejtsgﬁiég—amtgjgua prgggﬁfé. ﬁlleariy, the arguments
given in Sees. 5.1. and 5.2 establish this. Another experimental observation is thatas
thepressure is lowered along an isotherm on which a liquid—vapor phase transition 274 Chapter 5: Equilibrium and Stability in One~Component Systems Maxwell or lever rule ' I''
rt v Figure 5.33 A lowtemperature isotherm ofa real ﬂuid. occurs, the actual volume—pressure behavior is as shown in Fig. 5.3—3 and not as in
Fig. 5.3—2. That is, there is a portion of the isotherm where the speciﬁc volume varies
continuouslvatrllsanganfaﬁire and pressure; this is the twophase or here the
vaporliquid coexistence region. The variation oi the overall (or two~phase) speciﬁc
volume in this region arises from the fact that although the_s_pecific volumes of the VaporandliqmdPhases are ﬁ§e@__(§inss_rhe. tamperature andf.pressur_e__ar_e ﬂxiﬁLin sasssf theses.pbsisfsiisilleiistnssbsyasmsl= the fraction of the mixture that is “3‘ per, xv, can varv'continuous'ly from zero to one. In the two—phase region the speciﬁc
volume of the two—phase mixture is given by
'," : xvyv + nyL : xvi/v + (1 _ xvn/L (5.31a) where xV and IL aﬁé'fr'actio'hs 'or'iﬁapo'r' and liquid, respectively, on a molar basis.
[Equation 5.3—1a could also be written using volumes per unit mass and mass frac~
tions.] These fractions can var)r between zero and one subject to the condition that
xV + xl‘ = '1. Solving for x‘“' yields (5.31b) and XV i—r? _ {5.31c) <l<1 Equations analogous to those here also hold for the _U, Q, S, and gt. Equations of
the form of Eq. 5.31c are called Maxwell’s rules or lever rules. IEJUSTRAQNE31
Comparing the Properties ofa Two—Phase Mixture Compute the total volume, total enthalpy, and total entropy of 1 kg of water at 100°C,
half by weight of which is steam and the remainder is liquid water. SOLUTION From the saturated steam temperature table at 100°C the equilibrium pressure is
0.10135 MPa and 5.3 Phase Equilibria: Application of the Equilibrium and Stability Criteria to the Equation of State 275 {XL = 0.00100411131le i?" = 1.6729 m3ikg
131" = 419.04 mag 13” = 26?6.1 kJikg 3L = 1.3069 kakg K 3*" = 13549 kJi'kg K
Using Eq. 5.3—1a on a mass basis gives V = 0.5 X 0.001004 + 0.5 X 1.6?29 3 0.83645 m31'kg. The analogous equation for enthalpy is 1% = xLﬁrL + xvﬁ" = 0.5 x 419.04 + 0.5 x 2676.] = 1547.6 g
g and for entropy is
s = stL + $5” = 0.5 >< 1.3069 + 0.5 X 7.3549 r 4.3309 é—JK Consequently, the continuous variation of Speciﬁc volume of the vapor—liquid mix
ture at ﬁxed temperature and pressure is a result of the change in the fraction of the
mixture that is vapor. The conclusion their is that an isotherm such as that shown
in Fig. 5.32 is an approﬁmalte"Eepﬁeiitaiion ofthe real phase behavior"(shown in
Fig. 5.3.3) by a relatively simpIe' arsslyrieeqaansn'breath; Iii facﬁ'it'ié impossible to
represent the discontinuities in the derivative (3131639? that occur at EL and 1’" with
any analytic equation of state. By its sigmoidal behavior in the two—phase region,
the van der Waals equation of state is somewhat qualitatively and crudely exhibiting
the essential features of vapor—liquid phase equilibrium; historically, it was the ﬁrst
equation of state to do so. We can improve the representation of the two~phase region when using the van dcr
Waals or other'analytic equations of state by recognizing that all van der Waals loops,
such'as’ those shown in Fig. 5.32, should be replaced by horizontal lines (isobars),
as shown in Fig. 5.3—3. This constrEtiorrensureithatuthe_eqtiiliinum_.phases will
have the__sarne_temperature and pressure (sec Eqs, 5.1921 and b). The question that
remains is at which pressure should__the_i_sob_a_r_he drawnlsinﬂiy pr_ess_t_1_r_e suchthat bagp<a Will yield an isotherm like that in Fig. 5.33. The answer is that the pressure chosen
must satisfy the last condition for equilibrium, that 91 = G“.
To identify the equilibrium pressure we start from Eq. 4.2—8b dg = yap — Sci? and recognize that for the integration between any two points along an isotherm of
the equation of state we have .. pg
l Ag = J 1/ (11" P1
Thus, for a given equation of state we can identify the equilibrium pressure for each
temperature by arbitrarily choosing pressures Pa along the van der Waals loop, until
we ﬁnd one for which P4 Pb Pa
9" —§L = 0 2] yap +J ydP+J yap (5.32) Pu Pb 5
I' 1.4 1.3 ——
4. 1.2 —
!. 1.1 — Critical point 1.0 
l 0.9 — 0.8
coexistence 07 _ region "are 0.6 — as —
0.4 — 0.3 — 0.2 '
U.]. —
0 l
5
F.
2::
Figure 5.34 The van der Waals ﬂuid with the vaporliquid coexistence region identiﬁed.
Here the speciﬁc volume in each of the integrations is to be computed from the equa— tion of state for the appropriate part of the van der Waais loop. Alternatively, we can
ﬁnd the equilibrium pressure graphically by noting that Eq. 5.32 requires that areas Vapor—liquid h I and II in Fig. 5.32 be equal at the pressure at which the vapor and liquid exist in
coexistence pressure equilibrium. This vapor—liquid coexistence pressure, which is a function of tempera—
or vapor pressure ture, is called the vapor pressure of the liquid and will be denoted by P"‘*P(T). We can continue in the manner described here to determine the phase behavior of the ﬂuid for all temperatures and pressures. For the van der Waals ﬂuid this result is shown in Fig. 5.34. An important feature of this ﬁgure is the domeshaped, two— ' phase coexistence region. The inﬂection point C of Fig. 5.3—1 is the peak of this dome, and therefore is the highest temperature at which the condensed phase (the liquid) can exist; this point is called the critical point of the liquid. : It is worthwhile retracing the steps followed in idenifyipg the existence ragga
: i I tion of the two—phase regionin the _B:l{_p_l_ane:_ _ N a " m—mmm‘ "  “1 The stahiiimsssditiaa.(3 FWKJL£_U..w_a.s..used to. identify. tlie._u.ns.tahle...tsgi9n I. _ of an isotherm and thereby establish the existence of a twophase region.
3 ' 2. The conditions_f._'_"I q; T“ and P1 = P“ weret'h'en'tts'ea"to establish th_e shape I! I I ...__._._...._._.._. ...v. 1 _ .  h (but not the location) of the horizontal coexrstence hne in the P—V plane. 3. Finally, the equilibrium condition Q1 = Q” was position of g _ the coexistence line. l A more detailed representation of phase equilibrium in a pure ﬂuid, including
' the presence of a single solid phase,2 is given in the isometric flﬂ'__p_h_a_se__tha_gram region is pentitioncd into several regions. 211' several solid phases occur corresponding to different crystal structures as is frequently the case, the solid 1
 5.3 Phase Equilibria: Application of the Equilibrium and Stability Criteria to the Equation of State 27‘? Pressure, P ———+ Figure 5.3—5 The PVT phase diagram for a substance with a single solid
phase. (Adapted from J. Kestin, A Course in Thermodynamics, vol. 1. G?) I 1966 by Blaisdell Publishing Co. (John Wiley & Sons, Inc.) Used with
I permission of John Wiley & Sons, Inc.) of Fig. 5.3—5. Actually, such complete phase diagrams are rarely available, although
data may be available in the form of Fig. 5.3—4, which is a projection of the more
complete diagram into the P—V plane, and Fig. 5.3—6, which is the projection into the
P—T plane.  The concept of phase equilibrium and the critical point can also be considered
from a somewhat different point of View. Presume it were possible to compute the l Melting (fusion) curve
Critical point (To Pc) P 501m Vapor Vaporpressure
CUTVE Triple point (T7, PT) Sublimation pressure curve . Figure 5.36 Phase diagram in the P—T
1" plane. __,_.  ._....—u 278 Chapter 5: Equilibrium and Stability in OneComponent Systems Liquid I
 i t a

I Vapor
I
 P : constant ll : Figure 5.37r The molar Gibbs free energy as a
_' .— 1" function of temperature for the vapor and liquid
1P phases of the same substance. Gibbs free energy as a function of temperature and pressure for any phase, either
from an equation of state, experimental data, or statistical mechanics. Then, at ﬁxed
pressure, one could plot Q as a function of T for each phase, as shown in Fig. 5.3—?' for
the vapor and liquid phases. From the equilibrium condition that (3 be a minimum,
one can conclude that the liquid is the equilibrium phase at temperatures below TP,
that the vapor is the equilibrium phase above TP, and that both phases are present
at the phase transition temperature TP. If such calculations are repeated for a wide range of temperatures and pressures, it
is observed that the angle ofinterseetion 6‘ between the liquid and vapor free energy
curves decreases as the pressure (and temperature) at which the intersection occurs
increases (provided P :1 PC). At the critical pressure, the two Gibbs free energy
curves intersect with 6 x D; that is, the two curves are collinear for some range of T
around the critical temperature TC. Thus, at the critical point 8T 5T i
P P  Further, since (39*an = —._S‘, we have that at the critical point
SL(TC’ PC) = §V(7‘ci PC) Also, for the coexisting phases at equilibrium we have
QL(TCr PC) = GV(TC* PC) by Eq. 5.215c. Since the molar Gibbs free energy, molar entropy, temperature, and
pressure each have the same value in the vapor and the liquid, the values of all other
state variables must be identical in the two equilibrium phases at the critical point.
Consequently, the vapor and liquid phases should be indistinguishable at the critical
point. This is exactly what is experimentally observed. At all temperatures higher
than the critical temperature, regardless of the pressure, only the vapor phase exists.
This is the reason for the abrupt terminus of the vapor—liquid coexistence line in the
pressure—temperature plane (Fig. 5.36). [Th us, we have two ways of recognizing the
ﬂuid critical point: ﬁrst, as the peak in the vapor—liquid coexistence curve in the P—V
plane (Fig. 5 .34), and second as the abrupt terminus of the vapor—liquid coexistence
curve in the PFT plane] Also interesting is the ﬂuid triple point, which is the intersection of the solid—
liquid, liquid—vapor, and solid—vapor coexistence curves. It is the only point on the
phase diagram where the solid, liquid, and vapor coexist at equilibrium. Since the 5.4 The Molar Gibbs Free Energy and Fugacity of a Pure Component 279 solid—liquid coexistence curve generally has a steep slope (see Fig. 5.36), the triple
point temperature for most ﬂuids is close to their normal melting temperature, that
is, their melting temperature at atmospheric pressure (Problem 5.10). Although, in general, we are not interested in equilibrium states that are unsta
ble to large perturbations (usually called metastable states), superheated liqg__i_ds__and subcooled vapors do occur and are sufﬁciently familiar that we will brieﬂy relate
these states to the equilibrium and stability discussions of this chapter. For conve
nience, the van der Waals equation of state and Fig. 5.32 are the basis for this dis~
cussion, though the concepts involved are by no means restricted to this equation.
We start by noticing that although the liquid phase is thermodynamically stable along
the isotherm shown in Fig. 5.3—2 down to a pressure P“, the phase equilibrium analy—
sis indicates that the vapor, and not the liquid, is the equilibrium phase at pressures
below the vapor pressure PC, = P"3P(.Tl). If care is taken to avoid vapor—phase nucle
ation sites, such as by having only clean, smooth surfaces in contact with the liquid,
it is possible to maintain a liquid at fixed temperature below its vapor pressure (but
above its limit of stability, Pa), or at ﬁxed pressure at a temperature higher than its
boiling temperature, without having the liquid boil. Such a liquid is said to be su
perheated. The metastability of this state is illustrated by the fact that a superheated
liquid, if slightly perturbed, may vaporize with explosive violence. (To avoid this oc—
currence “boiling stones” are used in chemistry laboratory experiments.) It is also
possible, if no condensation nucleation sites, such as dust particles, are present, to
prepare a vapor at a pressure higher than the liquid—vapor coexistence pressure or
vapor pressure at the given temperature, but below its limit of stability (i.c., between
PC, = P““P(T1) and Pb in Fig. 5.3—2) 01', at lower than the liquid boiling temperature.
Such a vapor is termed subcooled and is also metastable. (See Problem 5.8.) At sufﬁciently low temperatures, the van der Waals equation predicts that the limit
of stability of the liquid phase occurs at negative values of the pressure, that is, that
a liquid could support a tensile force. In fact, such metastable behavior has been
observed with water in capillary tubes and is thought to be important in the vascular
system of plants.3 5.4 THE MOLAR GIBBS FREE ENERGY AND
FUGACITY OF A PURE COMPONENT 1n this‘snectippnwgconsider how one pscsan pquation of__state to identify the statesof Vapor—liquid squihrium in—a elite Illsid} The starting paint..isilteequaliiy of molar
Gibbs free energies in the coexisting phases I '1' ' _ _ _  i QL(T,P) = 9W1?) } ' (5.19c)
To proceed, we note that from Eq. 4.2—8b
_ _. dQ = —SdT +_VdP sothat I p _ __ __ I (E) 2—,; (5.4—1)
__ —  a? P 3For a review of water under tension, especially in biological systems, see P. If. Scholander, Am. Sci. 60,
534 (1972). ...
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