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Unformatted text preview: The direction of spontaneous ' I I change
3.1 The dispersal of energy
3.2 Entropy ' 13.1 Impact onengineering:
Refrigeration _' ' 3.3 Entropychanges ' " accompanying speciﬁc 13.2 Impact on materials chemistry:'._" ._
Crystal defects. ' = . system . " .' '. " concentrating on the 3.5 The senses and Gibbs: } energies '. '_ a 3.6 Standardmolar Gibbs energi¢s_'__: Combining the First and I  .
"Second Laws 3.7 The fundamental equation . _ 3.8 Propertiesofthe internal _ I. 31.18ng .I . 3.9" Properties of the Gibbs energy . Checklist of key equations ' Further information 3.1: The Born . equation
Further information 3.2: The fugacity: ' Discussion questions
Exercises
Problems The Second Law origin of the spontaneity of p of this chapter is to explain the
sea and show how The purpose
two simple proces ical change. We examine
a property. the entropy. to discuss spontaneous chang introduces a maior subsidiary thermodynamic property express the spontaneity of a process in terms
energy also enables us to predict the maximum non As we began to see in Chapter 2, One application of
between properties that might not be thought to be related. Several relations of t be established by making use of the fact that the Gibbs energy is a state see how to derive expressions for the variation of th d how to formulate expressions that are vaii
later when we discuss the effect of temperatur a system.
expansion work that a pro pressure an d for real gases. These
will prove useful
rium constants. some things don’t. A gas expands to
1? its surroundings,
e aspect of the ' Some things happen naturally;
is to the temperature 0 ' n rather than another. Som mines the spontaneous direction of change, the directio require work to bring it about. A gas can be con
can be cooled by using a refrigerator, and some reactions c
(as in the electrolysis ofwater). However, none of these processes is spon
one must be brought about by doing wo
throughout this text ‘spontaneous’ must be interpreted as a natu
or may not be realized in practice. Thermodynamics is silent on the and some spontaneous proc us change in fact occurs,
aybe so slow that the tendency is f diamond to graphite) m
ch as the expansion of a gas a spontaneo
conversion 0
in practice whereas others (su
instantaneous.
The recOgnition o
summarized by the Second
a variety of equivalent ways. is possible in which the sole r
11 into work. f two classes of process, spontaneous
Law of thermodynamics. This 1 One statement was to No process
voir and its complete conversio For example, it has proved impossible to constr
Fig! 3.1, in which heat is drawn from a hot reserve
work. All real heat engines have both a hot source an always discarded into the cold sink as heat and not conv hysical and chem to define. measure, and use
as quantitatively. The chapter also
, the Gibbs energy. which lets us case can do. thermodynamics is ton‘in function. We also a Gibbs energy with temper e and pressure on equilib ﬁll the available
and a chemiCal n of change that does not ﬁned to a smaller volume,
an be driven in reverse  rk. An important point, th
ral tendency that may esses (such as the into a vacuum) are and nonspontaneous,
aw may be expresse rmulated by Kelvin: uct an engine like th
ir and completely co erted into work. The The Gibbs d relations
his kind can ature and
expressions _ world deter—
an object taneous; each
ough, is that rate at which never realized
almost riverth in d a cold sink; some statement is a generalization of ano her eye d W ba 3 r n
t ' e 0
s r S ry ay obse atlon, that a
ter u facl 6 ha never been Observed to leap spontaneously upwards. An u :vlarfi lesat of
he wo e equiv cut to 6 conversion 0 eat 0m e sur ce 'Pnto workp
bal uld b at th fh fr th fa 1 . Thedirection of spontaneous change What determines the d' ‘
1rect1on of spontan 
the isolated s ste ' eous Change? It 13 not the
in any procesYs anmd. The Flrst Law of thermod},nalnics States that ener tot€11 energy of
towards a stat; of l we camwt dlsregard that law now and Say that EVgYIYISdC19nserved
IS it Perha 5 th ower energy: the total energy of an isolated system i: mg tends
ments show 3m tilinfrgy otf 1:he system that tends towards a minimuniglﬁvast
anno e so. First  a erf t ' argu’
Vacuum et its i . ’ p cc gas exPal‘lds s onta 
of a systein doesnl:emal energy remams constant as it does 50 Segondl Elﬁﬁly into a
surroundings mus:li’::::0 iecriase during a spontaneous change th}: enerfgSlffriy
SB yt esameamoutb ' , 15
ener ofthesu ' . . n ( Ythe First Law .1" ' 
f th rroundmgs is Just as spontaneous a me ) he micrease m
0 t 6 System. P 653 as the decrease 1n energy
When a chan
A ge occurs, the total ener f ‘
It is Parceﬂed outi _ 0 an 1solated system remains c
relatEd to the diggilfterent ways. Can it be, therefore, that the directiOn 3:15:31“ b‘it
Spontaneous Ch a ton of energy? We shall see that this idea is th k C ange1s
anges are always accompanied by a dispersal of e e e eY, and that
n rgy. 3.1 The dispersal of energy Key point Duri ‘
rig a spontaneous chan ' '
I ge In an lsolated s stern th ' 'n '
random thermal motion of the particles in the system Y 6 ﬁnal energy ls dispersed into Wecan begin to understand th
aha1H (the S Stem I l e role of the distribution of ener b ' ‘ after 62rd] boigpplraicmg on a ﬂoor (the surroundings). Thiallrdldgiliildf 1? bent The kineﬁc ener ecafus; there are 1nelastic losses in the materials of the balllSE ad
théfmal moﬁon of ‘ gy o .t e ball 5 overall motion is spread out into th an
I _ its part1cles and those of the ﬂoor that it hits The def Enf'rgy 01f:
. 1rec 10D 0 ‘ rsed into disorderl ' 71.1 111 . y thermal mot ' I
we I u Y1 Bite cor (Fig. 3.2). 10 of olecules 1n the an and of the atoms of spended in a liquid or gas. 3.1 THE DISPERSAL OF ENERGY 95 in Source . Heat
Work Flow 01‘ energy
Engine Fig. 3.1 The Kelvin statement of the Second
law denies the possibility of the process
lllustrated here, in which heat is than ed
completely into work, there being neither '
change. The process is not in conﬂict with
the First Law because energy is. conserved ﬁg. 3.2 The direction of spontaneous
change for a ball bouncing on a ﬂoor On
each bounce some of its energy is de Irad (:1
into the thermal motion of the atomgs ofteh
ﬂoor, and that energy disperses. The 6
reverse has never been observed to take
place on a macroscopic scale. [19.3.3 The molecula irreversibility express
(a) Aball resting on awarm sur ed by the Second Law..
face; the atoms are undergoing thermal motion ' , in this instance), as indicated by
the arrows. (b) For the ball to ﬂy upwards,
some of the random vibrational motion
would have to change into coordinated, directed motion. Such a converSion is
highly improbable. is spread out as thermal motion of th spontaneous because it i
leading to unifor
because to do so t
tion of kinetic energy through
same region of the container, r interpretation of the . e atoms of the ﬂoor. The reverse process is not 5 highly improbable that energy will become localized,
m motion of the ball’s atoms. A gas does not contract spontaneously
which spreads out the distribu— he random motion of its molecules,
out the container, would have to take them all into the thereby localizing the energy. The opposite change,
spontaneous expansion, is a natural consequence of energy becoming more dispersed
An object does not spontaneously as the gas molecules occupy a larger volume.
'5 highly improbable that the jostling become warmer than its surroundings because it i
the surroundings will lead to the localization of ther— of randomly vibrating atoms in
mal motion in the object. The opposite change, the spreading of the object’s energy
into the surroundings as thermal motion, is natural. It may seem very puzzling that the spreading out of energy an
d structures as crystals or proteins. Neve the formation of such ordere
course, we shall see that dispersal of energy and matter accounts for change in all its forms. 3.2 Entropy
of spontaneous change. (a) Entropy change is deﬁned Key points The entropy acts as a signpost
in terms of heat transactions (the Clausius deﬁnition). (b) Absolute entropies are deﬁned in terms of the number of ways of achieving a conﬁguration (the Boltzmann formula). (c) The Carnot cycle
prove that entropy is a state function. (d) The efﬁciency of a heat engine is the basis temperature scale and one realization, the Kelvin scale. reases in a spontaneous change and is used to
of the deﬁnition of the thermodynamic (e) The Clausius inequality is used to show that the entropy inc
with the Second Law. therefore that the Clausius deﬁnition is consistent on of the internal energy, U. The First Law of thermodynamics led to the introducti The internal energy is a state
for which the intern only those changes may occur
remains constant. The law that is used to identify the signp
e expresse the Second Law of thermodynamics, may also b
function, the entropy, S. We shall see that the entropy but is a measure of the energy dispersed in'a proce
is accessible from another by a spontaneous change. The First Law uses identify permissible changes; the Second Law uses the entropy to i
among those permissible changes.
thermodynamics can be expresse eases in the course of a spontane energy to
spontaneous changes The Second Law of d in terms of the entropy The entropy of an isolated system incr
AStot > 0 where Stut is the total e
irreversible processes
free expansion of gases) ar
by an increase in total entropy. ntropy of the system an
(like cooling to the temperature of the surro e spontaneous processes, (a) The thermodynamic definition of entropy The thermodynamic deﬁnition of entropy concentrate
hysical or chemical change (in general, as d5, that occurs as a result of ap
The deﬁnition is motivated by the idea that a change in the exte nds on how much energy is trans
s random motion in the surroun ‘process’l.
energy is dispersed depe
remarked, heat stimulate s assess whether a change is permissible:
al energy of anisolated system
ost of spontaneous change,
d in terms of another state
(which we shall deﬁne shortly, ss) lets us assess whether one state
the internal dentify the ous change“. s on the change in entIOP
a result Of nt to Will ferred as heat. As W h ._
dings. On the other hen  I .55! fﬁfh’diéiﬁfﬁﬁ‘IﬁSZﬁﬁizﬁiﬁwmi a! _ 1 f  V AS=—j dqrev=qrev WOIk SUIIllllates unlfOIm “loll” (ii a i) “S [I he surr‘n'ndjngs and 50 docs not change their entropy.
The thermodynamic deﬁnition of entropy isbased on the expression dam
r d5: where am, is the heat supplied '
. I : reverSIbl . For a me
1 and fthls expressmn integrates to y asurable change between two states Aszjqurev
T (3.2) i
That is, to ' '
calculate the difference in entropy between any two states of a s st
y em, we ﬁnd a reversible path between the ‘
I . in, and integrate the ener s 1'
stage of the path d1v1ded by the temperature at which heatii; d as heat at eaCh Exam 133.1 . . . I _ . _ __ _ ...
p Calculating the entropy change for the isothermal expansion ofa perfect gas
Calculate th all f e entropy change of a sample of perfect gas when it e d '
m y mm a volume Vi to avolume Vf. xpan S nether Method The deﬁnition of entropy instruct "
for a rev r I lslus to ﬁnd the ener su lied
actual m:n:l::eirp:t}llii:)lpngeen the stated initial and ﬁnal stategsyreggfdlessadfhfl:
Sim is iSOthermal so tht e process takes place. A simpliﬁcation is that the expan
imegmi in a... 3.5. Theeiiﬁﬁiriiiiiiii wimiimd may be taken wide the
I _ as eat it ' ‘ '
Eiigpﬁiapf a: perfect gas can be calculated from 136:}: j 11: $3121; :czjthjgijli
The work 0;] win general and therefore that qm = aw: for a reversible eh
_ _ reversible isothermal expansion was calculatch in Section 2 3 angel 5 Arts ' '
I I wer Because the temperature is constant, eqn 3.2 becomes T A brief illustration the volume occu '
pied by 1.00 mol f .
t temperature, Vfl' Vi = 2 and 0 any perfeCt gas melecmes ls doabled at any .00 moi) x (3.3145 I K‘1 moi1) x In 2 =+5.76 I K—1 0
gas i:ct;:lllalte:1he change in entropy when the pressure of a ﬁxed [AS 2 Mg; (isothermally from pi to pf. What is this change due to?
pii'pf); the change in volume when the gas is compressed] 3.2 ENTROPY 97 to eqn 3.2, when the energy
transferred as heat is expressed in joules and the temperature is in
kelvins, the units of entropy are joules
per kelvin (I K'l). Entropy is an '
extensive property. Molar entropy,
the entropy divided by the amount of
substance, is expressed in joules per
kelvin per mole (] K“l moi—1). The
units of entropy are the same as those
of the gas constant, R, and molar heat
capacities. Molar entropy is an
intensive property. I
l

,
i
i 3.2 ENTROPY 99 3.1 to formulate an expression for the change in (b) The statistical] view of entropy Consider an inﬁnitesimal transfer of heat dqSlur to The entry Point into th 
servoir of constant volume; so the dynamics is Boltzmann’ 'e rimlecular Interpretation Of the second Law Of thermm
d With the Change in the internal n OS es HI I s inSight, ﬁrst explored in Section F.5a, that an atom or molecule
ca p s s o y certain values of the energy, called its ‘energy levels’. The continuous We can use the deﬁnition in eqn
entropy of the surroundings, AS”.
the surroundings. The surroundings consist of a re energy supplied to them by heating can be identiﬁe
.2 The intern l ener is a state function, and dU . .
a g 5‘” thermal agitation that molecules experience in a sample at T > 0 ensures th t th a ey are ave seen, these roperties imply that dUsu is independ . . _
P r distributed over the available energy levels. Boltzmann also made the link between the i 'a' r" a c, sisif energy of the surroundings, dUsur
is an exact differential. As we h
ent of how the change is brought about and in particular is independent of whether d. . .
. istrlbutlon of molecules over ener levels and
th
ible. The same remarks therefore apply to dqsur, entropy Ofa System15 given by gY e entropy. He proposed that the the process is reversible or irrevers
to which dUsmr is equal. Therefore, we can a constraint ‘reversible’, and write dapt the deﬁnition in eqn 3.1, delete the S=kan d8 : dalsur rev = ﬁg
sui = 43 h '
T5“: TM tylhere ii 11.381: 10 I K 1 and Wis the number of microsmres, the ways in which
t e mo ecu es 0 a system can be arranged while keeping the total energy constant the surroundings is constant whatever the Bach micro t t 1 
f 1 1 s a e asts onlyfor an instant and corresponds to a certain distribution
0 mo ecu es over the available energy levels. When we measure the properties of
a system, we are measuring an average taken over the many microstates the system Furthermore, because the temperature of
change, for a measurable change qsur ,'
AS : (3.3b) . .
sur Tm If):qu untier the conditions of the experiment. The concept of the number of
, 1 a es nia ‘ ‘  _ . .
That is, regardless of how the change is brought about in the system, revers1bly or cme dispersal of :aiiarnmgtwe thefjl deﬁned qualitatwe concepts 0f ‘diSOl‘der’ and
of entropy of the surroundings by dividng entropy. a more ‘diso Zn 1 engrgy. that are used WidelY to introduce the concept of
 1‘ ery’ istributlon of energy and matter corresponds to a irreversibly, we can calculate the change
hich the transfer takes lace. .
p greater number of microstates associated with the same total energy the heat transferred by the temperature at w
le to calculate the changes in entropy of the . .
‘ Equation 35 18 known as the Boltzmann formula and the entropy calculated from Equation 3.3 makes it very SimP
surroundmgs that accompany any Process. For mstance, for any adiabatic change, it is sometimes called the statistical entropy We see th t ‘f W 1 h h
. I _. a,i : ,w ic corres onds
to one microstate (only one way of achieving a given energy, all molecules in directly i:fsur : 0’ so
For “’1 adiabatic Change: Assur = 0 . {34) I tie same State), then S = 0 because 1n 1 = 0. However, if the system can exist in '
_ _ _ , . _ t an one microstate the . more
This expressmn is true however the change takes place, reversibly or II‘I'EVEI‘SIblY, pm to a greater number ’Ofenn W: 1 and S > 0. If the molecules m the System have access
med in the surroundings. That is, it is true so long as the total ener th t . th ergy evels’ than there may be more “5’5 0f achieving a given
8% a 15, ere are more microstates for a given total energy, Wis greater vided no local hot spots are for
surroundings remain in intern
energy may subsequently disperse spon al equilibrium. If hot spots do form, then the localized and the entropy is greater th 11
taneously and hence generate more entTOPY tical view of entropy summ:ii;:d:1yffl:el3:tl::are afcceSSﬂlla Therefore, the Stam—
I ' ‘ ann ormu a is consisten ' 15:21:: :p’ijeprigpltethat the entropy is related to the dispersal of energy. In LEE: container expands (sFiin a: :oqﬁiner, the energy levels become closer together as the verify in Chapter 8) Aga 1:ea,"111ti51111502;ecpriiclusion from quantum theory that we shall 1 1 ~ the entropy increases exactly as we infei‘ileﬁtfates'become pOSSIble’ Wincreases’ and Eve S move Clom together and more , mm the thermodynannc deﬁnition of r::ﬁl:1t:1:::::$::t?uthimolecules. AS a DC 1011 increases and 0 A brief illustration
Fig. 3.4 When a box expands, the energy To calculate the entropy change in the surroundings when 1.00 mol H200) is formed
we use AHa = 286 k} from from its elements under standard conditions at 298 K, Table 2.8. The energy released as heat is supplied to the surroundings, now regarded as
being at constant pressure, so qwr =+286 k]. Therefore, _ entrﬁPY
~ _. ' '. I'T e m  . .
2.86x105] ,1 _ theﬁlemlﬂlzcular interpretation of entropy advanced by Boltzmann also suggests so does the entmpy'
Assur: 298K =+9601K  I I'th'a't mol Dulynémlc dEﬁnltlon given by eqn 3.1. To appreciate this point consider
I . avaﬂabl ec es in a system at high temperature can occupy a large number of the
This strongly exothermic reaction results in an increase in the entropy of the surro 16 energy levels= so a Small additional “aﬂSfer Of energy as heat will lead t
ings as energy is released as heat into them. I ' ,geroSfmall Change in the Dunlber of accessible energy levels. Consequently Sh: microstates does not increase appreciably and neither does the entiopy he. system. In co ' .Selftest a change in: thesmrou.n . . . . .y wer energY 13:33:11; Eilgflll’i’iggz:ﬁ:§tl low temperature have access
N204(gl 18 formed from 2.00 mol NOiig) under standard conditions at 298 1] Same quantity Ofenergy bY heating Will increase thl: 33:13:33? and till; mmeeT _ _ _ o acce
[ 192 I 3111:5111: :umber of microstates signiﬁcantly. Hence, the change in eisistlogjiliiirdgﬁr
Sfénéd to agﬁgetttljr Eitherri1 the energy is transferred to a cold body than when it is
$61. .0 y. is argument suggests that the change in entropy should
Y proportional to the temperature at which the transfer takes place, as in in which case we coul 2 Alternatively, the surroundings can be regarded as being at constant pressure, equate dqm to der. 100 3 THE. SECOND LAW Final
state Pressure, p Volume, V Fig. 3.5 In a thermodynamic cycle, the
overall change in a state function (from the initial state to the ﬁnal state and then back
to the initial state again) is zero. isotherm Pressure, ,0 Isothernd‘.. I Volume, V Fig. 3.5 The basic structure of a Carnot
cycle. In Step 1, there is an isothermal
reversible expansion at the temperature Th.
Step 2 is a reversible adiabatic expansion in
which the temperature falls from Tl1 to Tc.
In Step 3 there is an isothermal reversible
compression at Tc, and that isothermal step
is followed by an adiabatic reversible
compression, which restores the system to its initial state. (6) Entropy is a state functio
of dS is independent of path. To do so, eqn 3.1 around an arbitrary cycle is zero,
same at the initial and ﬁnal states of the sys them (Fig. 3.5). where the symbol 39 denotes integration around a do :13: pgiglshoyy that the two volume ratios are related in a very simple way From
C a ion etween temperature and volume for reversible adiabatic roc
(VT = constant, eqn 2.28): P esses ny=tut Wﬁ=nni The entropy as a state function n. To prove this assertion, we need to show that the integral it is sufﬁcient to prove that the integral of
for that guarantees that the entropy is the
tem regardless of the path taken between That is, we need to show that 2examWesaareaetaewmmmmﬂ ' Multiplication of the ﬁrst of these expressions by the second gives
; _ nancznnnc
which,"on cancellation of the temperatures, simpliﬁes to E a VD
VB VC (5.6) sur Slides: 0
r sed path. There are three steps in
With this relation established, we can write the argument:
1. First, to show that eqn 3.6 is true for a special cycle (a ‘Carnot cycle’) involving a V V
perfect gas. ' ‘1: = "1521117: = rtRTc 1117" = #nRTC 111%
2. Then to show that the result is true whatever the working substance. B A
_ I and therefore
3. Finally, to show that the result is true for any cycle.
a = _ T. four reversible stages (Fig. 3.6): where qh is the energy supplie so the change in entropy is zero.
from Th to TC, the temperature of the cold sink. A Carnot cycle, which is named after the French engineer Sadi Carnot, consists of qr _nRTc1n(VB/VA) — c as in eqn 3.7. Xpansion from A to B at Th; the entropy change is qhi’Th,
d to the system as heat from the hot source. c expansion from B to C. No energy leaves the system as heat,
In the course of this expansion, the temperature falls 1. Reversible isothermal e
r In the second step we need to show that eqn 3.6 applies to any material, not just a perfect gas (which is why, in anticipation, we have not labelled it with a °) We begin  2. Reversible adiabati step of the argument by introducing the efﬁciency, 77 (eta), of a heat engine. .~ in work performed _ m m C to D at Tc. Energy is released as heat
_ heat absorbed from hot source _ qh ITC; in this expression qc is 3. Reversible isothermal compression fro
to the cold sink; the change in entropy of the system is ac
negative. 4. Reversible adiabatic compression from D to A.
heat, so the change in entropy is zero. The temperatur The total change in entropy around the cycle is the sum of the changes in each of these ar . . . . .
t P 1 ifeplsmgjmodlpﬁuzsigns to avmd complications with signs: all efﬁciencies are H um ers. e e nition implies that the ‘ ter the work t f '
Isu 1 I , grea ou put or a iven
ppeisisogiezt groin thehot reservorr, the greater is the efﬁciency of the engine. Wge can
. _ I 7) :h e nition in terms of the heat transactions alone, because {as shown in
, 6 energy supplied as work by the engine is the difference between the No energy enters the system as
e rises from T.3 to Th. four steps:
Supplied as he t b th ‘ '
{ids = g; + fi—c a Y e hot reservou and returned to the cold reservoir:
Th Tc = 1 _ :qci
th (3.9) However, we show in the following Justiﬁcation that for a perfect gas __IL
T C h (3.7) ‘1: Substitution of this relation i
is what we wanted to prove. nto the preceding equation gives zero on the right, which Justification 3.1 Heating accompanying reversible adiabatic expansion n ‘ S :frp; 1:13:ng I1th: Slime two reservoirs (Fig. 3.8). The working substances
A is more efﬁd 0 tap two engines are entirely arbitrary. Initially, suppose caﬁses engine B ten an engine B, and that we choose a setting of the con ertain quantity or; acquire energyas heat qc from the cold reservoir and to smore efﬁ I h energy as heat into the hot reservoir. However, because _ (:1th t an engine B, not all the work that A produces is needed for This Justiﬁcation is based on two features of the cycle. One feature is that the two
abat in Fig. 3.6. The second and Tc in eqn 3.7 lie on the same adi temperatures Th
during the two isothermal stages are feature is that‘the energies transferred as heat V
5111: nRTh ln—E qc: nRTc ln—12
Vc 101 3.2 ENTROPY Fig. 3.7 Suppose an energy qh (for example,
20 k1) is supplied to the engine and q is lost
from the engine (for example, qc = —f5 kl)
and discarded into the cold reservoir. The
work done by the engine is equal to qh + q
(for example, 20 k] + (—15 kl) = 5 1(1). Thec
efﬁciency is the work done divided by the
energy supplied as heat from the hot
source. gm J‘f'zzgaasm u; 9:; 'r'ﬁ' I? kit‘21 ’4 tea 3 THE SECOND LAW (a) (b) Flg. 3.8 (a) The demonstration of the
equivalence of the efﬁciencies of all
reversible engines working between the
same thermal reservoirs is based on the
ﬂow of energy represented in this diagram. (b) The net effect of the processes is the
conversion of heat into work without there being a need for a cold sink: this is contrary
_to the Kelvin statement of the Second Law. this process, and the difference can be used to do work. The net result is that the cold
reservoir is unchanged, work has been done, and the hot reservoir has lost a certain
amount of energy. This outcome is contrary to the Kelvin statement of the Second
Law, because some heat has been converted directly into work. In molecular terms,
the random thermal motion of the hot reservoir has been converted into ordered
motion characteristic of work. Because the conclusion is contrary to experience, the
initial assumption that engines A and B can have different efﬁciencies must be false.
It follows that the relation between the heat transfers and the temperatures must also
be independent of the working material, and therefore that eqn 3.10 is always true for
any substance involved in a Carnot cycle.
For the ﬁnal step in the argument, we note that any reversible cycle can be approx
imated as a collection of Carnot cycles and the integral around an arbitrary path is the
sum of the integrals around each of the Carnot cycles (Fig. 3.9). This approximation
becomes exact as the individual cycles are allowed to become inﬁnitesimal. The
entropy change around each individual cycle is zero (as demonstrated above), so the
sum of entropy changes for all the cycles is zero. However, in the sum, the entropy
change along any individual path is cancelled by the entropy change along the path it
shares with the neighbouring cycle. Therefore, all the entropy changes cancel except for those along the perimeter of the overall cycle. That is, qre __ qrev_
E} TV_ 2 TT0 perimeter In the limit of inﬁnitesimal cycles, the noncancelling edges of the Carnot cycles match the overall cycle exactly, and the sum becomes an integral. Equation 3.6 then follows
immediately. This result implies that d8 is an exact differential and therefore that Sis a state function. (d) The thermodynamic temperature
eversibly between a hot source at a tem—
T, then we know from eqn 3.10 that (311) Suppose we have an engine that is working r
perature Th and a cold sink at a temperature T=(1—11)Th This expression enabled Kelvin to deﬁne the thermodynamic temperature scale in terms of the efﬁciency of a heat engine: we construct an engine in which the hot sourca
ect of interest. The temperature is at a known temperature and the cold sink is the obj
' of the latter can then be inferred from the measured efﬁciency of the engine. The
) is Kelvin scale (which is a special case of the thermodynamic temperature scale
deﬁned by using water at its triple point as the notional hot source and deﬁning that temperature as 273.16 K exactly. For instance, if it is found that the efﬁciency of such
an engine is 0.20, then the temperature of the cold sink is 0.80 X 273.16 K: 220K. This result is independent of the working substance of the engine. (e) The Clausius inequality We now show that the deﬁnition of entropy is co begin, we recall that more work is done when a change is reversible than whenillf
irreversible. That is, Idwrevl 2 Idw . Because dw and dwrev are negative when_¢nelr'gl_’
leaves the system as work, this expression is the same as —dwm 2 —dw, andphelnﬁ‘?
dw — dwreV 2 0. Because the internal energy is a state function, its change is for irreversible and reversible paths between the same two states, so we can alsohri W dU = dq + dw = dqrev + dwrev nsistent with the Second Law. . a E; fol/lg??? liq... — do} =de — aw... 2 0, or dqrev 2 dq, and therefore that
gm * ‘1  0W we uset e thermod am d ﬁ . .
d3 = diam/T) to write Yn 1c e much of the entropy (eqn 3.1; assﬂ T (3.12) 1111: expresswn 15 the Clausius inequality. It will prove to be of great importance for
t e iscussmn of the spontaneity of chemical reactions, as we shall see in Section 3.5 o A brief illustration Consider the transfer of energy as heat from one system—the hot source—at a temper
ature Th to another system—the cold sink—at a temperature Tc (Fig. 3.10). When dq
Ilsajes the hot source (so dqh < 0), the Clausius inequality implies that dS > dqh/Th When
.5} enters the cold sink the Clausius inequali im lies th — ' I
td82
Overall, therefore, W p a dqc’T‘ (“nth dqc > 0). However, dqh = —ciqc, so dSE—dq‘+dqc= i_i dq
Th T T Th “ C C which is positive (because dqc > O and Th > Tc). Hence, cooling (the transfer of heat from
hot to cold) 13 spontaneous, as we know from experience. 0 We now suppose that the system is isolated fr ' ' I
' I om 1ts surro =
The Clausrus inequality implies that undmgs, so that dq 0' dSZO (3.13) grand we cinclude that in an isolated system the entropy cannot decrease when a spon
aneous .5 range occurs. This statement captures the content of the Second Law . . . N
13.1 ' Refrigeration Elliesame. argument that we have used to discuss the efﬁciency of a heat engine can be
to diiguss the efﬁciency of a refrigerator, a device for transferring energy as heat
Iroomiglhigﬁixt (thfe'contents of the refrigerator) to a warm sink (typically, the f b e re rigerator stands). The less work we have to do to bring this
whet a out, the more efﬁcrent is the refrigerator. '
en an energy I qcl migrates from a cool source at ~  '
Inkal a temperature Th, the change in entropy is a temperature Tc Into a warmer lc
Luge—I“ (314) 933:: c1s not spontaneous because not enough entropy is generated in the warm
III . ome the entropy loss from the cold source (Fig. 3.11). To generate more
Yzﬁﬁletgy must be added to the stream that enters the warm sink. Our task is to
mmlmum energy that needs to be supplied. The outcome is expressed as the tOfPErformance, c: ‘ energy transferred as heat lqcl' .Ilgrgy transferred as work _ w [3.15] 3.2 ENTROPY 103 Pressure, p Volume, V Fig. 3.9 A general cycle can be divided into
small Carnot cycles. The match is exact in
the limit of inﬁnitesimally small cycles.
Paths cancel in the interior of the
collection, and only the perimeter, an
increasingly good approximation to the
true cycle as the number of cycles increases,
survives. Because the entropy change
around every individual cycle is zero, the
integral of the entropy around the
perimeter is zero too. d8 = —dq,_fTh +dqch Fig. 3.10 When energy leaves a hot reservoir
as heat, the entropy of the reservoir
decreaSes. When the same quantity of
energy enters a cooler reservoir, the
entropy increases by a larger amount.
Hence, overall there is an increase in
entropy and the process is spontaneous.
Relative changes in entropy are indicated
by the sizes of the arrows. 104 3 THE SECOND LAW Fig. 3.11 (a) The flow of energy as heat
from a cold source to a hot sink is not
spontaneous. As shown here, the entropy
increase of the hot sink is smaller than the
entropy decrease of the cold source,
so there is a net decrease in entropy.
(b) The process becomes feasible if work is
provided to add to the energy stream. Then
the increase in entropy of the hot sink can
be made to cancel the" entropy decrease of the cold source. The less the work that is require
of performance and the more e
ment it will prove best to work with 11c. the coefﬁcient d to achieve a given transfer, the greater
f this develop— fficient is the refrigerator. For some 0 and the work Iw'l is added to the en Because lqcl is removed from the cold source,
l= tau] 4 wl. Therefore, ergy stream, the energy deposited as heat in the hot sink is qh 1_lWl_lihl__lic_l=l—Qh—l—l c ac 161.1 lad express this result in terms of the temperatures alone, We can now use eqn 3.7 to
is substitution leads to which is possible if the transfer is performed reversibly. Th and therefore
T (atom C =nﬂn C for the thermodynamically optimum coefﬁcient of performance. 0 A brief illustration For a refrigerator withdrawing heat from ice—cold water (Tc 3 273 K) in a typical
ve 10 k] {enough to freeze 30 g of water), = 293 K), c = 14, so, to remo
ors, of course, have a lower environment (Th
f at least 0.71 k] as work. Practical refrigerat requires transfer 0
coefﬁcient of performance. 0 .3 Entropy changes accompanying specific processes 3
when it expands isothermally. (b) The Key points (a) The entropy of a perfect gas increases
nying a change of state at its transition temperature is change in entropy of a substance accompa
(c) The increase in entropy when a substance is heated calculated from its enthalpy of transition. is expressed in terms of its heat capacity.
determined from measurements of its heat capaci allowing for phase transitions in that range. ny a variety of basic We now see how to calculate the entropy changes that accompa processes. (a) Expansion
We established in Example 3.1 that the chang
isothermally from Vi to Vf is Vf AS = 11R 1n——
V. 1 Because 8 is a state function, the value of A8 of the system is indepen between the initial and ﬁnal states, so this expression app
of state occurs reversibly or irrevers volume is illustrated in Fig. 3.12.
The total change inentropy, however, does depend on how the exp
place. For any process the energy lost as heat from the system is acqui (d) The entropy of a substance at a given temperature is
ty from T: 0 up to the temperature of interest, ' e in entropy of a perfect gas that expands (3.17)D dent of the path
lies whether the Change ibly. The logarithmicdependence of entropy 011 ans'ion talkes
red by the is
{i
if g, 3.3 ENTROPY CHANGES ACCOMPANYING SPECIFIC PROCESSES 105 surroundings, so .qu = ——dq For a reversible I
I “I . change we use the ex re ' ‘
Example 3.1 (gm: nRTln{Vf/Vi)); consequently, from eqn 3.3b p 581011 m _ qsur d. qtev l/
___._—_ R f
nhl any A‘Ssur _ T
. T V' rev 1 This change. is the negative of the change in the system, so we can concl d h Asmf 0, which is what we should expect for a reversible process If on the 0th: lit :t
the isothermal expansion occurs, freely (w = 0), then q = Ebecause All jme ’
Consequently, ASSur = 0, and the total entropy change is given by eqn 3.17 itself— J. Angmmﬁ '
v any 1 In this case, also. > 0, as we expect for an irreversible process. (b) Phase transition The degree of dispersal of matter and energy changes when a substance freezes or b '1 as a result of changes in the order with which the molecules pack to ether and (fits
extent to which the energy is localized or dispersed. Therefore, we Shoguld ex ect the
tranSition to be accompanied by a change in entropy. For example when a sulbsta e
vaporizes, a compact condensed phase changes into a widely disperlsed gas and we me
expect the entropy of the substance to increase considerably. The entropy of a sdlaild _ also increases when it melts to a liquid and when that liquid turns into a gas ConSider a system and its surroundings at the normal transition temperature T
Blip: temperature at which twc phases are in equilibrium at 1 atm. This temperature2t
' .(273 K) for ice in equilibrium with liquid water at 1 atm, and 100°C (373 K) f
liquid water in equilibrium with its vapour at 1 atm. At the transition temperatufer
:ya::;i:1fler of energy as heat between the system and its surroundings is reversible
_ _ e two phases in the system are in equilibrium. Because at constant
q— A H, the change in molar entropy of the system 153' Pressure trs “I: A rsH
lLi:AtTSS=iI— trs .imem " ' , p se tranSition is exothermic (AmiI < 0, as in freezing or condensing), then the 0 ch ' ' '
py ange of the system is negative. This decrease in entropy is consistent with .. th .
Increased order of a sohd compared with a liquid and with the increased order of qti‘: fergliieiygth ai sgpsi The cphaiglge in entropy of the surroundings, however, is
Howl Chan e igyent e ease as eat into them, and at the transition tempera—
ting and vaPgmzatiois ago. If the transition is endothermic (AmH > 0, as
‘HSiStem With dis ersal 0} en e entropy change of the system is positive, which
ﬁes by the sampam matter in the system. The entropy of the surroundings ' cunt, and overall the total change in entropy is zero.’ ﬂ}::;:;clf::u:ntrgpﬁesd0fVfilPOI‘ization of several liquids at their boiling points.
. (lard entro e o E e ata is that a Wide range of liquids give approximately the
s called Trout 01:: 0 glapomatlon (abcfut 35 l K_] 11101—1): this empirical observa
m velum s r e. The explanation of Trouton’s rule is that a comparable e occurs when any liquid evaporates and becomes a gas. Hence, all 0m Secti '
0n 2.6 that A H is an enthalpy change per mole of substance; so A S is also a molar trs
trs ‘le 3. ' ‘ '
1 lists some experimental entropies of transition. Table 3.2 lists in more I 4 1 ___4 1 10 20
W '30 Fl. 3.12 The logarithmic increase in
entropy of a perfect gas as it expands
isothermally. interActivity Evaluate the change in
expansion 0131.00 moi CO2 (g) from
0.001 m3 to 0.010 m3 at 298 K, treated as a
van der Waals gas. 106 3 THE SECOND LAW Standard (and phase transitions, Algalll; EA 11101'1) I f. Ti). : a. .. _ ._.V_épdﬁizaition (at Tb) ' . "'3s.oo.<at2r9.K) . 8.2.1.9, @353 K) 
. g_1_0.9.o(at3_73.15 K)  22.00. (airs15 K). . t , ,p. ._ 54 ’9? ‘ rr m _ U} 00 W
U"!
U!
a
p
r, igénzenéigsﬂs I "
Waiter2:320? .: IE . ... ' 7L3. ‘ﬁmﬁﬁwlﬁxﬂﬂmmﬁﬁﬂrfw 3.3 ENTROPY CHANGES ACCOMPANYING SPECIFIC PROCESSES 107 SEE2:35 3.3 Predict the enthalpy of vaporialatlonoiethane 1ts ' ' [16k1mol‘1] (9) Heating we canduse eqn 3.2 to calculate the entropy of a system at a temperature Tf from a
knowle getof its entropy at another temperature Ti and the heat supplied to change its '  r. ' {4.3 (at1.8Kand.sobar). : .l 1'  19:9: W432 K) '.fue1iu_m,lue 1. y  ; 3 . I
I ' ' temperature from one value to the other: '3 ’* Morelvalues aregiven in the Data section. '. T‘dqm 3‘
T ST =sr.
( f) ( Ni (3.21) T. I We shall be particularly interested in the entropy change when the system is subjected
t: cgngtant presfsure (such as from the atmosphere) during the heating Then from
t e e n1t10n o constant—pressure heat capaci (e n 2 22 ' I j . , wr =
Consequently, at constant pressure: W q men as dqm CPdT)’ we 342%” *zThe star'sa.erases:airsaaaaaawas r. '_‘.;l '_ . .. Tf deT Sle) = Sm) + i (3 22) T. l :I‘h'e same expression applies at constant volume, but with CF replaced by C \Nhen C
15 independent of temperature in the temperature range of interest, it calii be takerli
out51de the mtegral and we obtain Ti
dT
5(T)=S(T,+C _= , Tr ‘ .
f ) 12L T 3(T1)+CplnTi (323) 1 entropies of vaporization. Liquids that show signiﬁcant deviations from Trouton’s rule do so on account of strong
molecular interactions that result in a partial ordering of their molecules. As a result,
there is a greater change in disorder when the liquid turns into a vapour than for a fully disordered liquid. An example is water, where the large entropy of vaporization reﬂects the presence of structure arising from hydrogenbonding in the liquid. Hydrogen bonds tend to organize the molecules in the liquid so that they are less ran—
dom than, for example, the molecules in liquid hydrogen sulﬁde (in which there is no
hydrogen bonding). Methane has an unusually low entropy of vaporization. A part of l K“1 mol'1 at 298 K); the reason is that the entropy of the gas itself is slightly low (186
the entropy of N2 under the same conditions is 192 I K’1 mol‘l. As we shall see in
for light Chapter 12, fewer rotational states are accessible at room temperature
molecules than for heavy molecules. _ wftth a snmlar expression for heating at constant volume. The logarithmic dependence
_ o: entropy on temperature is illustrated in Fig. 3.13. liquids can be expected to have similar standard AS/nH fple 3.2 cease 'g' the entrbpr'criantie’i  alculate the entropy change when argon at 25°C and 1.00 bar in a container of olume 0.500 (11113 ' . .
0 100°C. 15 allowed to expand to 1.000 drn3 and 1s Simultaneously heated Met ' '
Pagplt: Belqause is a state function, we are free to choose the most convenient
. mt e 1n1t1al state. One such path is reversible isothermal expansion to the Volume follo " ' 
. a wed by revers1ble heatlng at constant volume to the ﬁnal tem 1 10 TIT 20 30
i i Flg. 3.13 The logarithmic increase in
entropy of a substance as it is heated at
constant volume. Different curves
correspond to different values of the
constantevolume heat capacity (which is
assumed constant over the temperature
range) expressed as CVam/R. O Abrief illustration There is no hydrogen bonding in liquid bromine and IBrz is a heavy molecule that is unlikely to display unusual behaviour in the gas phase, so it is safe to use Trouton’s Inla To predict the standard molar enthalpy of vaporization of bromine given that it boils at 592°C, we use the rule in the form eor i ' ' ' ' '
thersindas 2R. (The equipartitron theorem is reliable for monatomic gases:
1 31 n , In general, use experimental data like those in Table 2.8, converting
He at constant volume by using the relation C m — CV = R ) r
r rm ' Lg interActivity Plot the change in
entropy of a perfect gas of (a) atoms,
(b) linear rotors, (c) nonlinear rotors as
the sample is heated over the same range
under conditions of (i) constant volume,
(ii) constant pressure. Aware: Tb x (8511(—1 mol‘l) Substitution of the data then gives AWPH9= {332.4 K) x {851161 mol’l) = +2.8 x 103 I mot1 = +23 k] mot—1 The experimental value is +2945 k} mol‘l. O 108 3 THE SECOND LAW A‘ note" on good practice .  sensible to proceed as generally as
possible before inserting numerical data so that, if required, the formula
can be used for other data and to avoid rounding errors.
W approximation cyrand s T. Fig. 3.14 The variation of CPI T with the
temperature for a sample is used to
evaluate the entropy, which is equal to the
area beneath the upper curve up to the
corresponding temperature, plus the
entropy of each phase transition passed. N h
interAetivity lowfort 3 temperature dependence of the heat
capacity by writing C = a + bT+ ciTz, and
plot the change in entropy for different
values of the three coefﬁcients (including negative values of c). From eqn 3.23, the entropy change in the second step, from Ti to Tf at constant volume, is The overall entropy change of the system, the sum of these two changes, is V 312 V T 312
AS:nR1n%+ann[%] =nR1n 1 (We have used In x + 1n y = 1n xy.) Now we substitute 1’1 = ini/RTi and obtain 312
A3=ﬂim _ / T. ‘l T V 1 1 At this point we substitute the data: _ (1.00 x105 Pa) x {0.500 x 10‘3 m3) X1 312
As H 1.000 :7:
298 K 0.500 298 =+0.173 1161 [—0.43 JK‘l] 's'érriiési Calculatethe entropy Ichang'e whenth'eisame pressed to 0.0500 dm3 and cooled to ~25°C. (d) The measurement of entropy
T is related to its entropy at T = O by The entropy of a system at a temperature
mperatures and evaluating the integral in measuring its heat capacity CP at different te
eqn 3.22, taking care to add the entropy of transition (AHst Tm) for each phase trans ition between T: 0 and the temperature of interest. For example, if a substance melts
at T f and boils at Tb, then its molar entropy above its boiling temperature is given by Trc. s,T A H Tbc (1,r
sm(r):sm(0)+ “1 )dT+ “5 + 1”“ )dT
0 T Tr T T
f
A H TC ,T
+ “9 +l wig )dT (3.24)
Tb T r b
can be measured calorimetrically, and the as is now more usual, by ﬁtting a poly
al analytically. The former procedure
Tn1'T against T is the integral IB
is to evaluate the area under a All the properties required, except 8mm),
integrals can be evaluated either graphically or,
nomial to the data and integrating the polynomi
is illustrated in Fig. 3.14: the area under the curve of CP)
quired. Because dTIT= d in T, an alternative procedure plot of Cpam against In T.
One problem with the determination of entropy is the difﬁculty of measuring heat capacities near T = 0. There are good theoretical grounds for assuming that the hi?"
capacity is proportional to T3 when T is low (see Section 7.1), and this dependence 15
the basis of the Debye extrapolation. In this method, CP is measured down to as low a
temperature as possible, and a curve of the form aT3 is ﬁtted to the data. That ﬁt deter’ mines the value of a, and the expression CF,m = aT3 is assumed valid down to T= 0 E
{a
at
a
is
N ~. r\ o A briefillustration 1 3.4 THE THIRD LAW OF THERMODYNAMICS The standard molar entropy of nitrogen gas at 25°C has been calculated from the fol lowng data: Debye extrapolation Integration, from 10 K to 35.61 K
Phase transition at 35.61 K
Integration, from 35.61 K to 63.14 K
Fusion at 63.14 K Integration, from 63.14 K to 77.32 K
Vaporization at 77.32 K Integration, from 77.32 K to 298.15 K Correction for gas imperfection
Total Therefore 83,10 Kl marl) 1.92
25.25
6.43
23.38
11.42
11.41
72.113
39.20
0.92 ‘ 192.06 33(29815 K) = 5mm) + 192.1 11cl mol‘1 0 Example 3.3 Calciﬁa‘ting the atlow _ Thegrpolar constantpressure heat capacity of a certain solid at 4.2 K is 0.43 I K‘1
I mol .What 15 1ts molar entropy at that temperature? Method 'Because the temperature is so low, we can assume that the heat ca aci vanes With temperature as aT3, in which case we can use eqn 3.22 to calculalie til:
;; entropy ata temperature T in terms of the entropy at T = O and the constant a
; When the integration is carried out, it turns out that the result can be expressed iri
; terms of the heat capacity at the temperature T, so the data can be used directly to calculate the entropy. " Answer The integration required is T rtTS T '
Ismpr) = 3m(()) + ?dT= sm(0) + a] TZdT
0 D
= 3mm) + §ar3 = 3mm) + gcpmm H om which it follows that sm{4.2 K) = 3mm) + 0.14 I K‘1 mol—1 t3. For metals, there is also a contributiontethe heat ectro ' ' '
I 1:15 that is 11neariy propornonal to T when the temperature 18 low. Find its
1‘ utlon to the entropy at low temperatures. [8(T) =3(0)'+ Gym] 109 110 3 THE SECOND LAW At T — 0, all energy of thermal motion has been quenched, and in a perfect crystal all
ocalization of matter and the the atoms or ions are in a regular, uniform array. The l absence of thermal motion suggest that such materials also have zero entropy. This conclusion is consistent with the molecular interpretation of entropy, because 8 = 0 if there is only one way of arranging the molecules and only one microstateis accessible (all molecules occupy the ground state). (a) The Nernst heat theorem
be consistent with the view that the The experimental observation that turns out to
T = 0 is summarized by the Nernst entropy of a regular array of molecules is zero at
heat theorem: The entropy change accompanying any physical or chemical
transformation approaches zero as the temperature approaches zero: AS —> 0 as T ——> 0 provided all the substances involved are perfectly ordered. 0 Abrief illustration
etween orthorhombic sulfur, 3(a), and mono Consider the entropy of the transition b
d from the transition enthalpy (402 I moltl) clinic sulfur, S(B), which can be calculate
at the transition temperature (369 K): (—402 1 morl) _ Ams = sulﬁ) — 8,..(00 =——————369 K — —1.09 IK—1 mol'1 The two individual entropies can also be determined by measuring the heat capacities
from T: 0 up to T: 369 K. It is found that 3mm) = Sm(ot, 0) + 37 IK‘1 mol*1 and Sm(B)
= Smﬂﬁ, 0) + 38 I K‘1 moi". These two values imply that at the transition temperature AHSS = sum, 0) — sure. 0) = iii<1 marl On comparing this value with the one above, we conclude that Sm(ot, 0) — Smui, 0) = 0, in accord with the theorem. 0 It follows from the Nernst theorem that,
the entropies of elements in their perfect crys
talline compounds also have zero entropy at T: 0 (because accompanies the formation of the compounds, like the entropy 0 at that temperature, is zero). This conclusion is summarized by the T thermodynamics: The entropy of all perfect crystalline substances is zero at T: 0. As far as thermodynamics i matter of convenience. The mo the value S = 0 at T — 0. We saw in Section 3.2b that, accor formula, the entropy is zero if there is only one accessible microstate ( W: 1 at T = 0 because there is only one way of achi lecular interpretation of entropy, however, Cases,
put all the molecules into the same, loWe with the Third Law of thermodynamics. In certain cases, though, W
1 at T = 0. This is the case if there is no energy advant orientation even at absolute zero. For in if we arbitrarily ascribe the value zero to
talline form at T = 0, then all perfect crys—
the change in entropy that
f all transformations
hird Law of s concerned, choosing this common value as zero is a
justiﬁes ding to the Boltzmann
w=1)_1n most .
eving the lowest total energl’: ‘ st state. Therefore, S = 0 at T = 0, in accord _
may differ from ; age in adopting a partiCUlar ‘
stance, for a diatomic molecule AB there mall .I ‘A‘a‘z’atiﬁﬂﬁ’zsﬁvhmﬁsﬁrmmﬁazf ,s’w‘ 3.4 THE THIRD LAW OF THERMODYNAMICS l 11 be almost no energy difference between the arrangements . . . AB AB AB . . . and
BA AB BA. . . , so W> 1 even at T: 0.1fS > 0 at T: 0 we say that the substance has
a reSidual entropy. Ice has a residual entropy of 3.4 I K’l mol‘]. It stems from the arrangement of the hydrogen bonds between neighbouring water molecules: a given . O atom has two short 0—H bonds and two long OH bonds to its neighbours, but
there is a degree of randomness in which two bonds are short and which two are long. (b) ThirdLaw entropies Entropies‘reported on the basis that 8(0) = 0 are called ThirdLaw entropies (and
often Just entropres ). When the substance is in its standard state at the temperature
1‘", the standard (ThirdLaw) entropy is denoted S“(T). A list of values at 298 K is given in Table 3.3.
The standard reaction entropy, 89(T), is deﬁned, like the standard reaction en thalpy, as the difference between the molar entropies of the pure, separated products
and the pure, separated reactants, all substances being in their standard states at the speciﬁed temperature: Arse: 2 v5};— 2 vs; Products Reactants (3.25a) In this expression, each term is weighted by the appropriate stoichiometric
coefﬁc1ent. A more sophisticated approach is to adopt the notation introduced in Section 2.8 and to write Ant: 211,530) (3 25b) 3 I .
Standard reaction entropies are likely to be positive if there is a net formation of gas in
a. reaction, and are likely to be negative if there is a net consumption of gas. 9 A brief illustration To calculate the standard reaction entropy of H2( g) + % 02(g) —e> H200) at 25°C, we use :the data in Table 2.8 of the Data section to write =1  ' A.st = an,0.1) — {snaps +" tsgcopgii
= 69.9 iKl mol’l — {130.7 + 520mm K41 moi1
2 —163.4 I K"1 mol’1  v u . n
. e negative value is conststent With the conversion of two gases to a compact liquid. 0 Table ' StandardThirdLaw ' H :_ entropies at 298'_K._ "  73;! . 30345.; ' :I ..
j , G'faphi'te,_cﬁ{sj 
. D_ialnﬁind,'C'(s) I  SucroSe, 915111220 ;,(s) 360.2 .Ilddln?='12I(5)...II i=3 "'116,1. .‘ '
Liquid5;: i. '_ H _ I I. ‘ til“itaciCalali ms ' _ _
==''Water"HQOmf' 3'3 . . :
Merc'urriiigll). . Anote‘on goodpractme 0 not
make the mistake of setting the
_'standard molar entropies of elements
equal to zero: they have nonzero
values (provided T> 0), as we have already discussed. n' ntropy for the combustion of [—243 l K71 mol‘l] nS1:fthe.discussion of enthalpies in Section 2.8, where we acknowledged that
y 0fcations cannot be prepared in the absence of anions, the standard molar
, ions in solution are reported on a scale in which the standard entropy of i0 ' '
115111 water is taken as zero at all temperatures: 112 3 THE SECOND LAW (ACp/TllmJ K42 moi”1 0 0.4 0.8 1.2
Temperature, TiK Fig. 3.15 Molar heat capacity contributions
of the defects in hydrogen— and deuterium
doped niobium. The area under each curve
is used to calculate the entropy due to the
presence of the defects. (Based on G].
Sellers and A.C. Anderson, Phys. Rev. B. 10,
2771 (1974).) The values based on this choice are listed in Table 2.8 in the Data section.4 Because
the entropies of ions in water are values relative to the hydrogen ion in water, they
may be either positive of negative. A positive entropy means that an ion has a higher
molar entropy than H+ in water and a negative entropy means that the ion has a lower
molar entropy than H+ in water. For instance, the standard molar entropy of Cl‘(aq)
is +57 I K"1 moi;1 and that of Mg2+(aq) is —128 ] K‘1 molil. Ion entropies vary as
expected on the basis that they are related to the degree to which the ions order the
water molecules around them in the solution. Small, highly charged ions induce local
structure in the surrounding water, and the disorder of the solution is decreased more
than in the case of large, singly charged ions. The absolute, ThirdLaw standard molar
entropy of the proton in water can be estimated by proposing a model of the structure
it induces, and there is some agreement on the value —21 I K‘1 molﬁl. The negative
value indicates that the proton induces order in the solvent. ‘ =lMPAC‘F ON MAT TRY [3.2 Crystal defects The Third Law implies that at T = O the entropies of perfect crystalline substances
are characterized by long—range, regularly repeating arrangements of atoms, ions, or
molecules. This regularity, and the accompanying inter and intramolecular inter
actions between the subunits of the crystal, govern the physical, optical, and electronic
properties of the solid. In reality, however, all crystalline solids possess one or more
defects that affect the physical and chemical properties of the substance. In fact,
impurities are often introduced to achieve particular desirable properties, such as the
colour of a gemstone or enhanced strength of a metal. One of the main types of crystalline imperfection is a pomt defect, a location where
an atom is missing or irregularly placed in the lattice structure. Other terms used to
describe point defects include voids, or lattice vacancies, sabstitutional impurity atoms,
dopant sites, and interstitial impurity atoms. Many gemstones feature substitutionai
solids, such as in rubies and blue sapphires where the A13+ ions in the corundum
structure of alumina are replaced with Cr3+ and Fe3+ ions, respectively. Interstitial
solids can result from the random diffusion of dopants in interstices (voids) or from
selfdiffusion, as in ionic crystals, where a lattice ion can migrate into an interstitial
position and leave behind a vacancy known as a Frenkel defect. ' Figure 3.15 illustrates the impact of impurities on the heat capacity and thus entropy
of a pure crystal. Niobium has become the dominant metal in lowdemperature
superconductor alloys because it can be manufactured economically in a ductile form
that is needed for the high critical current of a superconductor. The purity of the
metal, however, is essential to yield superconducting properties. Close to 1 K the heat
capacity of pure niobium follows the Debye T3 law. However, when niobium is
treated by allowing H2 or D2 to diffuse over the sample at 700°C impurities are intro—
duced and the heat capacity diverges from that of the pure metal. To identify the role
of the defects the values of CF for the pure metal are subtracted from those of ihe
doped samples, divided by T, and plotted against temperature. The area under the
resulting curves then represents thecontributions to the entropy from the presaﬂce of the impurities. 4 In terms of the language to be introduced in Section 5.1, the entropies of ions in solution are actuallYPm'
rial molar entropies, for their values include the consequenCes of their presence on the organization 0f the
solvent molecules around them. ...=.: ' Concentrating on the system Entropy is the basic concept for discussing the direction of natural change but to use
it welhave to analyse changes in both the system and its surroundings. we: have seen
that it 15 alWays very simple to calculate the entropy change in the surroundin s and
we shall now see that it is possible to devise a simple method for taking that coitiibu
non into account automatically. This approach focuses our attention on the system and simpliﬁes discussions. Moreover, it is the foundation of all the applications of
chem1cal thermodynamics that follow. 3.5 The Helmholtz and Gibbs energies Key points (a) The Clausius inequality implies a number of criteria for spontaneous change
under a variety of conditions that may be expressed in terms of the properties of the system alone
they are summarized by introducing the Helmholtz and Gibbs energies. (b) A spontaneous pro:
cars at constant temperature and volume is accompanied by a decrease in the Helmholtz energy
(c) The change in the Helmholtz energy is equal to the maximum work accompanying a process
at constant temperature. (d) A spontaneous process at constant temperature and pressure is
accompanied by a decrease in the Gibbs energy. (e) The change in the Gibbs energy is equal to the maximum 11  ' '
on expansron Work accompanying a process at constant temperature and pressure. Consider a system in thermal equilibrium with its surroundings at a temperature T.
When a change in the system occurs and there is a transfer of energy as heat between
the system and the surroundings, the Clausius'inequality (d5 2 dry] T, eqn 3.12.) reads T (3.27) W: can develop this inequality in two ways according to the conditions (of constant
vo ume or constant pressure) under which the process occurs. (a) Criteria for spontaneity Elm: COHSIder heating at constant volume. Then, in the absence of nonexpansion
work, we can write (1%,: (1U; consequently
dU dS — —~ 2 0
T (3.28) _ Importance of the mequahty 1n th1s form Is that it expresses the criterion for Spontaneous change solely in terms of th '
'   6 State functions of the s stem h ‘ _
1s easily rearranged into Y . T e Inequa1 TdS 2 dU (constant V, no additional work)5 (3 29) Either constant inter I I 
L nal energy (dU: O) or constant entro ' dS  0 th' — 15 ex 
1011 becomes, respectively, ‘PY ( L Pres U, 20 dU s
v S", 0 ' ' (3.30) 3.5 THE HELMHOLTZ AND GIBBS ENERGIES 113 114 3 THE SECOND LAW
3.5 THE HELMHOLTZ AND GIBBS ENERGIES 115 r menu/:zssamrsasmrseseweasas" nternal energy (such as an isolated system), the entropy increases in a dAT V: 0 (3 )
, .37 and constant i
spontaneous change. That statement is essentially the content of the Second Law. The
ntropy and volume of the second inequality is less obvious, for it says that, if the e The expressions = dU— TdS and CIA < 0 are sometimes interpreted as follows. A
negative value ofdA IS favoured by a negative value of dUand a positive value of TdS. system are constant, then the internal energy must decrease in a spontaneous change. _ . Do not interpret this criterion as a tendency of the system to sink to lower energy. It is Th” Observatlon SuggeSts that the tendenCY Ofa SYStem to move to lower A is due to a disguised statement about entropy and should be interpreted as implying that, if Its tendency .to_move towards States of lower internal energy and higher entrOPY the entropy of the system is unchanged, then there must be an increase in entropy of However’ Fins Interpremﬁon is false (“611 though it is a good rule of thumb for
remembering the expression for dA) because the tendency to lowerA is solely a tend— the energy of the system decreases
ency towards states of greater overall entropy. Systems change spontaneously if in dorng so the total entropy of the system and its surroundings increases, not because
they tend to lower Internal energy. The form of dA may give the impression that the surroundings, which can be achieved only if as energy flows out as heat. When energy is transferred as heat at constant pressure, and there is no work other than expansion work, we can write dqp = dH and obtain  t f 1
5375 ems avour ower energy, but that is misleading: dS is the entro h
Tds 2 dH (constant P, no additional work) (3 .3 1) the system, —dU." T is the entropy change of the surroundings (when the animals? t1:
system is constant), and their total tends to a maximum. At either constant enthalpy or constant entropy this inequality becomes, respectively, to) Maximum work dng 2 0 cng, s o (3.32)
. . . . . . . t t t, '   . . _ _
The mterpretatlons of these inequalities are Similar to those of eqn 3.30. The entropy incirgaiieflrjginilgyalggg:Zilgngrfmﬁmtw:, that Ah calrlnes a greater Slgmﬁo
. . . . aneousc an e:tec '
of the system at constant pressure must increase if its enthalpy remains constant functionis equal to the maximum work accom I g “"33 m the HEIthItz
(for there can then be no change in entropy of the surroundings). Alternatively, the Panymg “Process at 50mm“ temperature:
enthalpy must decrease if the entropy of the system is constant, for then it is essential (1me —— clA (3 38)
to have an increase in entropy of the surroundings. A5 a result, A is sometimes called the ‘maxi fu ' r c
Because eqns 3.29 and 3.31 have the forms dU— TdS S O and dH ~ TdS S 0, respec function, 6 mum work Dawn 1 Of the work
tively, they can be expressed more simply by introducing two more thermodynamic  ' ‘
quantities. one is the Helmholtz energy, A, which is deﬁned as ....................................................................................................................................................... ._
' ' Justification 3.2 Maximum work A: U« TS [333} I  To demonstrate that maximum work can be expressed in terms of the changes
in Helmholtz energy, we combine the Clausius inequality d8 2 dqi T in the form
I _ TdS 2 clq With the First Law, dU: dq + dw and obtain
The other is the Gibbs energy, G: J
 = d U S TdS + dw
G = H # TS [334] '_ . (dUis smaller than the term of the right because we are re 1 ' ‘
.  . . p acm d b TdS, h
in general is larger.) This expression rearranges to g q Y W ICh
dw 2 dU— TclS All the symbols in these two deﬁnitions refer to the system. When the state of the system changes at constant temperature, the two properties It follows that the most negative value of dw and therefore the m '
» ammurn energy that can be obtained from the system as work, is given by change as follows: (a)dA=dUerS (b}dG=dH—Td5 {5.35) I. dwm=dUrds , Erligethatllhis wofk is done OﬂlY When the path is traversed reversibly (because then
 ua 1
I that Elwmaiappdes). Because at constant temperature dA : dU‘_ Tds, we conclude (a) dATivg—O dGT’PSO ' ............................................................................................................ ......................................... .. These inequalities are the most important conclusions from thermodynamics for £11 a macroscopic isothermal change takes place in the system eqn 3 38 becomes
. J U chemistry. They are developed in subsequent sections and chapters.  _ When we introduce eqns 3.29 and 3.31, respectively, we obtain the criteria of spon taneous change as (3.39) (b) Some remarks on the Helmholtz energy
A change in a system at constant temperature and volume is spontaneous if dATyS 0' That is, a change under these conditions is spontaneous if it corresponds to a decreﬂﬁe in the Hehnholtz energy. Such systems move spontaneously towards states of loWﬁI A if a path is available. The criterion of equilibrium, when neither the forWard 110‘
reverse process has a tendency to occur, is U— ms '
. (3.40) 1.3.1116 German word for work; hence the symbol A. 116 3 THE SECOND LAW Fig. 3.16 In a system not isolated from its
surroundings, the work done may be
different from the change in internal
energy. Moreover, the process is
spontaneous if overall the entropy of
the system and its surroundings increases.
In the process depicted here, the entropy
of the system decreases, so that of the
surroundings must increase in order for the
process to be spontaneous, which means
that energy must pass from the system to
the surroundings as heat. Therefore, less
work than AU can be obtained. Fla. 3.11 In this process, the entropy of the
system increases; hence we can afford to
lose some entropy of the surroundings.
That is, some of their energy may be lost as
heat to the system. This energy can be
returned to them as work. Hence the work done can exceed AU. ' on the internal energy as it is converted into work.'T This expression shows that in some cases, depending on the sign of TAS, not all the change in internal energy may be available for doing work. If the change occurs with a decrease in entropy (of the system), in which case TAS < 0, then the righthand side of this equation is not as negative as AU itself, and consequently the maximum work
is less than AU. For the change to be spontaneous, some of the energy must escape
as heat in order to generate enough entropy in the surroundings to overcome the
reduction in entropy in the system (Fig. 3.16). In this case, Nature is demanding a tax
his is the origin of the alternative name ‘IIelmholtz free energy? for A, because AA is that part of the change in internal
energy that we are free to use to do work. Further insight into the relation between the work that a system can do and the
Helmholtz energy is to recall that work is energy transferred to the surroundings as
the uniform motion of atoms. We can interpret the expression A — U— TS as showing
that A is the total internal energy of the system, U, less a contribution that is stored
as energy of thermal motion (the quantity TS). Because energy stored in random
thermal motion cannot be used to achieve uniform motion in the surroundings, only the part of U that is not stored in that way, the quantity U — TS, is available for con version into work. If the change occurs with an increase of entropy of the system (in which case
T A8 > 0), the righthand side of the equation is more negative than AU. In this case,
the maximum work that can be obtained from the system is greater than AU. The explanation of this apparent paradox is that the system is not isolated and energy may ﬂow in as heat as work is done. Because the entropy of the system increases, we can afford a reduction of the entropy of the surroundings yet still have, overall, a spon
some energy (no more than the value of TAS) may leave the taneous process. Therefore,
ribute to the work the change is generating (Fig. 3.17). surroundings as heat and cont
Nature is now providing a tax refund. H ‘ Example 3.4" Caloulating' the; maximumavailable Work I When 1.000 mol C‘slIIZO6 (glucose) is oxidized to carbon dioxide and water at
_ 25°C according to the equation C6H1206(s) + 6 02(g) —> 6 C02(g) + 6 H200),
calorimetric measurements give ArUe = —2808 k] mol‘1 and Ass = +259.1 I K'1
mol‘1 at 25°C. How much of this energy change can be extracted as (a) heat at constant pressure, (b) work? Method We know that the heat released at constant pressure is equal to the value
of AH, so we need to relate ArHﬁ to AU“, which is given. To do so, we suppose that all the gases involved are perfect, and use eqn 2.21 in the form AIH= AIU+ AngT For the maximum work available from the process we use eqn 3.39. Answer (a) Because AVE = 0, we know that ArHe = ATU" = —2808 k] mol—1
Therefore, at constant pressure,_the energy available as heat is 2808 k] Incl1
(b) Because T: 298 K, the value of ATA" is Afar: AIUQ— my: 2sss kjmof1 Therefore, the combustion of 1.000 mol (36le06 can be used to produCE up 10
2885 k] of work. The maximum work available is greater than the change in inter' oﬁhe positive entTOPY Ofreac’inD (which is partly due to the
one). The system Can g their entropy) and nal energy on account
generation of a large number of small molecules from one big therefore draw in energy from the surroundings (so reducin
make it available for doing work. Selftest 3.7 Repeat the calculation for the combustion Iof 1.000 mol (g)under
the same conditions, using data from Tables 2.6 and 2.8. 4
[iqpl = 890 k], wmx = 818 k]] (d) Some remarks on the Gibbs energy The Gibbs energy (the ‘free energy’) is more common in chemistry than the Helmholtz
energy because,. at least in laboratory chemistry, we are usually more interested in
changes cecurring at constant pressure than at constant volume. The criterion
dGIP g 0 carries over into chemistry as the observation that, at constant temperature
and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs ener
Therefore, if we want to know whether a reaction is spontaneous the pressure aiifd
temperature being constant, we assess the change in the Gibbs eneigy If G decrea e
as the reaction proceeds, then the reaction has a spontaneous tendency to convert tshS
reactants into products. If G increases, then the reverse reaction is spontaneous e
The existence of spontaneous endothermic reactions provides an illustration of
the role of G. In such reactions, H increases, the system rises spontaneousl to stat
of higher enthalpy, and dH > 0. Because the reaction is spontaneous we lfnow th:
dG < 0 despite dH > 0; it follows that the entropy of the system increases so much that
TdS outwelghs dH in dG = dH — TdS. Endothermic reactions are therefore driven
by theincrease of entropy of the system, and this entropy change overcomes th
reduction of entropy brought about in the surroundings by the inﬂow of heat i t the
system (c185Dr = —dHr' T at constant pressure). n 0 e (e) Maximum nonexpansion work The analogue of the maximum work interpretation of AA, and the origin of the name
free energy’, can be found for AG. In the following Instiﬁcation, we show that at con
stant temperature and pressure, the maximum additional (non—expansion) w k
waddlmax, is given by the change in Gibbs energy:  or ’ dwadenax = _ The corresponding expression for a measurable change is
AG
(3.4 lb) I _ fghiigfressmn is Particularly useful for assessing the electrical work that may be pro
_, Y fuel cells and electrochemical cells, and we shall see many applications ofit. ' Justification 3.3 Maximum nonexpansion work
ecause H = U+ 131/, for a general change in conditions, the change in enthalpy is
'dH=dq+dw+d(pV)
,hecorresponding change in Gibbs energy (G = H — TS) is
dGLdH— TdS—SdT=dq+dw+d(pV) u rdS—Sdr
When the change is isothermal we can set dT= 0; then
‘G=dq+dw+d(pV)  TdS ...en the change is reversible, dw = dwrev and dq = chiral: yids, so for a reversible 3.5 THE HELMHOLTZ AND GIBBS ENERGIES 117 118 3 THE SECOND LAW 3.6 STANDARD MOLAR GIBBS ENERGIES I 119 reaction Gibbs energy for the formation of a compound from its elements in their refer  ' ' ' ence states. Standard Gibbs energies of formation of the elements in their reference Tab]? 37‘” StaIidaId Gibbleliergiés (if '
states are 261:0, because then formation is a ‘null’ reaction. A selection of values for formath? {at 'ngi'K)‘ I i: i ' I. I. it
compounds 15 gtven In Table 3.4. From the values there, it is a simple matter to obtain I I r I I I the standard Gibbs energy of reaction by taking the appropriate combination The work consists of expansion work, which for a reversible change is given by —pdV, and possibly some other kind of work (for instance, the electrical work of
pushing electrons through a circuit or of raising a column of liquid); this additional
work we denote dwadd. Therefore, with d(pV) = pdV+ Vdp, ' ".mzwzswmwnmawmmr , Hum = .A'fee/(k1ms1%l) . . Defeatgeari ?'?:f_' '1. . 717’: d6 = (—pdV+ dwaddm) + pdV+ Vdp = dwadd'rev + Vdp
If the change occurs at constant pressure {as well as constant temperature), we can
set dp = 0 and obtain d6 = dwadd’tev. Therefore, at'constant temperature and pres— ArGe = 2 vAfGe_ z “A Ge Benzenétcstl): __ I  I I +1243;  sure, dwadd m = dG. HOWever, because the process 15 reversrble, the work done must Products React t f (3.43a) Matham CH (g) ' _  " 7' ' '
.’ . ans .’ :2: ..
now have its mammum value, so eqn 3.41 follows. Cuban; die ";e co ( ) ; ' '  t.  '
._ . .....m. 2g._»394.4__
 ~ In the notat10n introduced in Section 2.8, water, H2001 _; _ _ .y _ 4371;  ' ' '
Ammonia;NH3{g). I‘  I I I I stadium" chloride, NaCifs) .. _—.384;1. 3 ' AIG9= EVJAfG’eU) (3 43b)
I . «asa Ork ore reaetr‘on pie 3.5’ Ca‘ibdiatfhg the friaXi'mu'rn nonexpansion w
_ I 3" MmeYam“ 51f? givenin the Data seetion. ' How much energy is available for sustaining muscular and nervous activity from
f glucose molecules under standard conditions at the combustion of 1.00 mol 0
37°C (blood temperature)? The standard entropy of reaction is +259.1 ] K‘1 mol‘l. Method The non—expansion work available from the reaction is equal to the
change in standard Gibbs energy for the reaction (ArGera quantity deﬁned more fully below). To calculate this quantity, it is legitimate to ignore the temperature
dependence of the reaction enthalpy, to obtain AFH“ from Tables 2.6 and 2.8, and to substitute the data into ArG": AIH" — TArSe. O A brief illustration To calculate the standard Gibbs ener ' 1
gy of the reactio CO + —
25cc, we write n (g) 2 ozlg) —> (302%) M
A,Ge=AfGe(co,,g)  {A,Ge(co,g) +%AFG9(OZ,g)}
= —394.4 k] moi1 — {(—137.2) + %(0)}k] mol1
= —257.2 kl mol’1 0 Selfrestless Calculate the stan ar reaction.
3 CH4(g) at 298 K. Answer Because the standard reaction enthalpy is —2808 kl £1014, it follows that the standard reaction Gibbs energy is
— (310 K) x (259.1 1K"1 mol—1) = —2388 k} moi—1 Therefore, wadd'mg, = —2888 k] for the combustion of 1 mol glucose molecules, and
to 2888 k] of non—expansion work. To place this the reaction can be used to do up
f mass 70 kg needs to do 2.1 k] of work result inperspective, consider that a person 0
to climb vertically through 5.0 m; therefore, at least 0.13 g of glucose is needed to complete the task (and in practice signiﬁcantly more). [#818 k} mol—1] AIG" = —2808 k] mol—1 I East as we did in'Section 2.8, where we acknowledged that solutions of cations can
:13) :1 prﬁpared w1thoutthe1r accompanying anions, we deﬁne one ion, convention
y t e ydrogen 1011, to have zero standard Gibbs energy of formation at all _ temperatures: .expansron Sé . .
combustion of 1.00 mol CH4(g) under standard conditions at 298 K? Use
[81814] ATS”: —243 I K4 moi1. In: I  . .
6f: :gsenge, th1s deﬁninon adjusts the actual values of the Gibbs energies of formation
. ., nS Y a ﬁxed amount that ls chosen so that the standard value for one of them H*(aq), has the value zero. "A brief illustration 0: the reacti0n 3.6 Standard molar Gibbs energies
used to calculate the standard Gibbs ener Key points Standard Gibbs energies of formation are
gies of reactions. The Gibbs energies of formation of ions may be estimated from a thermo dynamic cycle and the Born equation. enthalpies of reaction can be combined to obtain the standﬂl'd
dard reaction Gibbs energy’), ArGe: l
Htghaclzts)thachuaq) ArGe=—131.23kimor1 an write Standard entropies and
Gibbs energy of reaction (or ‘stan Arcs: ArHer rats?! The standard Gibbs energy of reaction
in their standard states at the tempe energies of the pro ducts and reactants
speciﬁed for the reaction as written. As in the case of standard reaction enthalpies;
gies of formation, AEGG, the standard convenient to deﬁne the standard Gibbs ener itis 111,, 120 3 THE SECOND LAW A brief comment
The standard Gibbs energies of formation of the gas~phase ions are unknown. We have therefore used ionization energies and
electron afﬁnities and have assumed that any differences from the Gibbs energies arising
from conversion to enthalpy and the
inclusion of entropies to obtain Gibbs
energies in the formation of H+ are Cancelled
by the corresponding terms in the electron
gain of X. The conclusions from the cycles are therefore only approximate. H+igl + Clig) + e'igl +106
H19) + % Clzig) + e(g) «349 "I.
E
E
a
DJ
i (W, aq) +A,G°(Cl‘, aql}
Hrtaq) + CHaql
(a)
H+igl + 1(9) + 919)
i+10____
I
s:
2
E
E
LU A G°(H+) solv m ‘Htiaqi + I'iaq) Flu. 3.18 The thermodynamic cycles for
the discussion of the Gibbs energies of
solvation (hydration) and formation of
(a) chloride ions, (13) iodide ions in
aqueous solution. The sum of the changes
in Gibbs energies around the cycle sum to
zero because G is a state function. The factors responsible for the magnitude of the Gibbs energy of formation of an ion in solution can be identiﬁed by analysing it in terms of a thermodynamic cycle.
As an illustration, we consider the standard Gibbs energy of formation of Cl‘ in water, which is —131 k] mol‘l. We do so by treating the formation reaction Combining the First and Second Laws ThgtFirst acrlid Second .Laws of thermodynamics are both relevant to the behaviour of
ma . er, an we can bring the whole force of thermodynamics to bear on a robl b
setting up a formulation that combines them. p em Y §H,(g) + % X2(g) —> H+(aq) + x(aq) wn in Fig. 3.18 (with values taken from the as the outcome of the sequence of steps sho
closed cycle is Data section). The sum of the Gibbs energies for all the steps around a 3'7 The fundamental equation zero, so
Afcetcraq) = 1272 k] moi1 + Asolva’ﬂF“) + Amcncn An important point to note is that the value of AfG“ of an ion X is not determined by the properties of X alone but includes contributions from the dissociation, ionization, and hydration of hydrogen.
Gibbs energies of solvation of individual ions may be estimated from an equation derived by Max Born, who identiﬁed ASOWG" with the electrical work of transferring
an ion from a vacuum into the solvent treated as a continuous dielectric of relative
permittivity at. The resulting Born equation, which is derived in Further information Key £01m The fundamental equation, a combination of the First and Second Laws is an expression
0 . 1
fort e change in internal energy that accompanies changes in the volume and entropy of a system W rim ﬁf‘f“al,"r"j{:l my”; We have seen that the First Law of thermo dynamics may be written dU= d + d a reversible change in a closed system of constant composition and in theq b w. For
any additional (nonexpansion) work, we may set dw = —pdl}and (from fhsedn: (if
tion of entropy) dam: TdS, where p is the pressure of trier: system and Tit t e e m
Therefore, for a reversible change in a closed system, 5 temperature. dU= TdS — pdV (3 )
.46 3.1, is
2 2 r“) .
A lG“=—Zie NA lei err) (3.45a) However’ because w 15 an exact differential. its value is inde end t f
sov Maori Er u Therefore, the same value of dU is obtained whether the changepis b an 1:) Pbath
' irrevers'bl ' , mug t 21 out
1 y or revers1bly. Consequently, can 3.46 applies to any change—Hreuersibie or izrﬂeuirsible—of a closed system that does no additional (non—expansion) work We shall
c T; isfcomlllamation of the First and Second Laws the fundamental equation '
Changeesalplta; 1:1: :iiindargngntalelqluation applies to both reversible and irreversible
ing a rst srg t. The reason is that onl in th '
change may TdS be identiﬁed with d Y e case Qfa reversible
._ I q and —pdV with dw Wh h h '
irrever51ble, TdS > City} (the Clausius inc ' I I en t e 9 ange 15
1 quality) and —pdV> dw. The sum of dw and d
_ remams equal to the sum of TdS and ﬂpdV, provided the composition is constant. q where zi is the charge number of the ion and ri its radius (N A is Avogadro’s constant).
ive for small, highly charged Note that AsoNGe < 0, and that ASDIVGG is strongly negat
ions in media of high relative permittivity. For water for which 8, = 78.54 at 25°C, 2
21 x (6.86 x 104k1mor1) (3.451)) — (ti/pm) A G“: solv 0 Abr‘ief illustration To see how closely the Born equation reproduces the experimental data, We Calculate 3.8 Properties of the internal en
 j.;e':;»';.. . ergy the difference in the values of AfGﬁ for C11 and I“ in water at 25°C, given their radii as
181 pm and 220 pm (Table 19.3), respectively, is '    :Kéy'pamrs R 1 
I dynamic ande attiﬁns between thermodynamic properties are generated by combining thermo
_ ma ematical expressions for changes in their values. (a) The Maxwell relations are 1 1 . . I.  .
A Grr (31‘ —A Gr 1 =— _—a—# x 6.86 x104klmol‘1 ’5“ 0f relatmns betwe d   . .
501v ( ) 501v ( ) [181 220] l ) 1 'ther erti b en eﬂvatlves 0f.therm0dynamc pmpemes baSEd on Criteria for
67 k] 1 1 dynamic P t.es epig exact differentials. (b) The MaXWell relations are used to derive the
=— mo equa 1011 o stat  .
I e; e and to determine how the internal energy of a substance varies rence is in good agreement with the experimental difference, which ‘A This estimated diffe is —61klm01—1 ' 3 46 h i
.g  S ows that the internal energy of a closed system changes in a simple n e1 ' 5 su:::t8t§:tl/Uis changed (dU oc d8 and dU oc dV). These simple propor . dome is best regarded as a function of S and V. We could regard U bmﬂie SI r variables, such as S and p or Tand V, because they are all inter
implicny of the fundamental equation suggests that U(S,V) is the = Senl 3 es ._ _ '  E ‘ G C sob,
experimental data and from the Born equation.
(«—26 k] mol‘1 experimental; —29 k] mol’1 c Calorimetry (for AH directly, and for 5 via heat capacities) is only one of ﬁle Walls.
of determining Gibbs energies. They may also be obtained from equilibriu.In Con—
stants and electrochemical measurements (Chapter 6), and for gases they maY be
calculated using data from spectroscopic observations (Chapter 16).  ' emotica ' '
. l consequence of U being a function of S and V is that we can 5 (3.47) \ 3.8 PROPERTIES OF THE INTERNAL ENERGY A brief comment Partial derivatives were introduced in
Mathematical background 2. The type of
result in eqn 3.47 was ﬁrst obtained in
Section 2.11, where we treated Uas a
function of Tand V. 121 122 3 THE SECOND LAW
3.8 PROPERTIES OF THE INTERNAL ENERGY 123 derivatives are the slopes of the plots of U against S and V, respectively. _
Just'f'catm." 3'4 The Thermodynamic equation of state The two partial
d to the thermodynamic relation, eqn 3.46, we see When this expression is compare
that, for systems of constant composition, BU 8U
fa?er ital"? (“8) odynamic deﬁnition of temperature
5 in the internal energy (a First—Law 21):: obtain'an expression for the coefﬁcient 7:1 by dividing both sides of e n 3 47 b
, imposmg the constraint of constant temperature, which gives q I Y ere—sierra ‘ 75 Kt, . . .
Ne we introduce the two relations in eqn 3.48 and the deﬁnition of ET to obtain
! "1" ." I ’eﬁi‘ﬁﬁfﬂfi‘ﬁﬁgﬂﬁﬁﬁﬁﬁﬂﬁ’iﬂﬁw The ﬁrst of these two equations is a purely therm (a ZerothLaw concept) as the ratio of the change
concept) and entropy (a SecondLaw concept) of a constantvolume, closed, constant _ T as
composition system. We are beginning to generate relations between the properties of ‘ 3T “ E _ P
a system and to discover the power offthlermodynarnics for establishing unexpected . h h T a T E t ird Maxwell relation in Table 3 5 
r \J‘ v t '
plates the Proof of eqn 3'51. urns (BS/3V}; into (Bp/BT)V, Wthh com relations. {\ ., i.)
(a) The maxwe“ relations I .................................................................................................................................................. ..
An inﬁnitesimal change in a function ﬂxy) can be written df= gdx + hdy where g and h are functions of x and y. The mathematical criterion for df being an exact differen
tial (in the sense that its integral is independent of path) is that . 
Show thermodynamically that ET: 0 for a perfect gas, and compute its value for a van der Waais gas. Example 36' "Def" M119 if". if I 3y — 3x
53thEIIT10ClYnamIC relations and equations of state, without drawing on molecular
argitimtents (such gs the exrstence of intermolecular forces) We know that for a
per ec gas, p = n T] V so this relation should b ' .
I I, eused1nen3.51S' ill
van der Waals equation is ' ' q . 1m ar Y! the
I . given in Table 1.7, f
‘ tron it should be used in eqn 3.51. and or the se'cond part Of the quesr [3g] (3 49)
x y Method Proving a result ‘thermodynamically’ means basing it entirely on general This criterion is discussed in Mathematical background 2. Because the fundamental
equation, eqn 3.46, is an expression for an exact differential, the functions multiply ing ds and dV (namely T and —p) must pass this test. Therefore, it must be the case
that liilf‘liil. We have generated a relation be be related.
Equation 3.50 is an example of a M
unexpected, it does not look particularly interesti there may be other similar relations that are mor _ Answer For a perfect gas we write (3.50) _ 1 3T V at ,‘7 would not seem to tween quantities that, at ﬁrst sight, axwell relation. However, apart from being ' ‘_ RT
ng. Nevertheless, it does suggest that 1751. = n— _ p = 0
. . V e useful. Indeed, we can use the fact that H, G, and A are all state functions to derive three more Maxwell relations. The _ I 7 equation of State Ofa van der Wa 1 I
argument to obtain them runs in the same way in each case: because H, G, and A are  _ nRT 2 a S gas 15:
state functions, the expressions for dH, dG, and (M satisfy relations like eqn 3.49. All ' V _ a1 — n5 V2 four relations are listed in Table 3.5 and we put them to work later in the chapter. (h) The variation of internal energy with volume ‘
. _ . . . the ’5‘? _ 8(nRT/(V — nb)) mg
The quantity ET — (BUIBV) T, which represents how the internal energy changes as  —— =
volume of a system is changed isothermally, played a central role in the manipulation V 3T V V _ n],
of the First Law, and in Further information 2.2 we used the relation ' ' epifmm eqn 3.51,
WT—é T a—p “P (3'51) _ _ “RT HRT n2 :12
8T — nb P — _ — (1— : __4
V I V — 1111 V _ 111, V2 V2 it for it ' ' ‘
r T implies that the internal energy of a van der Waals gas increases on 0i because it is al I i I . l/ C This relation is called a thermodynamic equati
for pressure in terms of a variety of thermodynamic properties of the system_We,are now ready to. derive it by using a Maxwell relation. 3.9 PROPERTIES OF THE GIBBS ENERGY 125 e n I I
These r latlons Show how the Gibbs energy varies with temperature and pressure
Gibbs separation
(Fig. 3.19). The ﬁrst implies that:
energy, G so the total cles. A larger molar volume, corresponding to a greater average parti
between molecules, implies weaker mean intermolecul'arjattractions,
. ., ,
energy ls greater. 3 \‘  B .
,.  ecause S > 0 f r
g V 0 all substances, G always decreases when the temperatur ‘ '
(at constant pressure and composition) e 15 Tamed  Because 8 ‘
( G181")? becomes more negative as 8 increases, G decreases most for a gas that obeys the virial equation of state
sharply when the entropy Of the system is large. Salterest 3.11 Calculate NT
[nT=Rr2(aB/arjvivfn+   ] [Table 1.7).
3.9 Properties of the Gibbs energy
the Gibbs energy of a system suggests that it is best regarded as a Key points (a) The variation of function of pressure and temperature. The Gibbs energy of a substance decreases with tempera d increases with pressure. (b) The variation of Gibbs energy with temperature is related to
lids and liquids are the enthalpy by the Gibbs—Helmholtz equation. (c) The Gibbs energies of so
with the logarithm of the pressure. almost independent of pressure; those of gases vary linearly
for the Gibbs energy giggiqugrzhe Gibbs energy the gaseous phase of a substance, which has a hi h
py, is more sensrtive to temperature than its liquid and solid phasges (Fig. 3.20). Similarly, the second relation implies that; Because if > for all Sub ﬂy I {3812535 I) 'I [e () ' stances G alW S .71
. ' , When the 1:655 15 Increased (at temperature and COIIIpOSlthIl)
' Because a T Will I , is “IUIE Se S1Ve It) pressure hen ( ) V . . W volume of the system is large.
Fig. 3.19 The variation of the Gibbs energy Because the molar volum
its condensed ijhases theerpf 1thigaisjeous phase of a substance is greater than that of
, 0 ar 1 bs energy of a gas is more sensitive to Ufa SyStem Wlth (a) temperature at
pressure constant pressu d (b)
1'6 an pressure at The same arguments that we have used for U can be used _ _
G = H — TS. They lead to expressions showing how G varies with pressure and tem than Its hqmd and Sclid Phases (Fig. 3.21).
I d chemical reactions. constant temperature The 51
. _ _ D  0pe of the
(b) The variation of the Glbbs energy with temperature former 15 equal to the negative ofthe
“HOW 0f the system and that of the latter ing phase transitions an ature that are important for discuss
is equal to the volume. Elligirzirﬁpjgte introduction, because the equilibrium composition ofa system
1 5 energy, to discuss the response of ' '
the co
turlelpveﬁneed to know how G varies with temperature mposmon to tempera—
e I t . o v ' 
Sion AltSOJeIIftion in eqn 3.53, (dGldT)P = —S, 15 our starting point for this discus
itinzc g 1 expresses the varianon of G in terms of the entropy we can ex
erms of the enthalpy by using the deﬁnition of G to write 3 = (PI,— G)/ T Tigris:5 lagl 3T 1: T ‘ (3.54)
t_,.: per (a) General considerations
undergoes a change of state, G may change because H, T, and 8 all ﬁnitesimal changes in each property When the system
1, we write for in change. As in Justiﬁcation 2.
dG=dH—d(TS) = dH— TdS SdT Because H = U+ pV, we know that
dH: dU+ d(pV) : dU+pdV+ Vdp Gibbe energy, G We shall 5 ' ' ' i 1 than. to Giecsgtfe; that the equilibrlum constant of a reaction is related sic; T rather
, an it is easy to deduce from the last equation (see the following lustiﬁcation) that I [EKG/T) _ H 8T "—7 _ _ P T This expression ' ‘
 . 15 called ' _  ' penthalpy Ofthe System £2:in Helmholtz equation. It shows that, if we know the “9 329 The variation of the Gibbs ener , , e ow how GI Tvaries with temperature. with the temperature is determined by ti:
................................................................................................................................ _. entr“PYBecauseﬂmmtropvofthe
"""""""" " gaseous Phase of a substance is greater than ust f' ' ‘
I I Ication 3.5 The Gibbs—Helmholtz equation that of th 1' 'd h . e 1qu1 p ase, and the entropy of
the SOllCl phase is smallest, the Gibbs energy and therefore
dG = dU+pdV+ Vdp # TdS  SdT
For a closed system doing no nonexpansion work, we can replac
mental equation dU = TdS — pdV and obtain
dG .= TdS — pdV+ pdVi— Vdp — TdS a SdT Four terms now cancel on the right, and we conclu absence of non—expansion work and at constant composition e dU by the funda (3‘55)  Temperature, T de that for a closed system in the (3.52) dG : Vdp — SdT
This expression, which shows that a change in G is proportional to a change in p or T, “St! We note that .
suggests that G may be best regarded as a function of p and T. It may be regarded1 35 , 'a(G/T) 1 8G d(1/T) 1 3 Changes m0“ SteePIY for the gas hase
'on of chemical thermodynamics as it is so central to the 3T + G 2—. _G _£_i 3G G followed by theiiquid Phase with 'th
T T TZHT _ __ sold h 1 en 6
P p 3T T 1 p ase of the substance. the fundamental equat
quark application of thermodynamics
tity in chemistrybecause the pressure and temperature are usually the variables uﬂd‘?I our control. In other words, G carries around the combined consequences of the Pi”
and Second Laws in a way th larly suitable for chemical applicatltﬂl15 at makes it particu
The same argument that led to eqn 3.48, when applied to the exact cliffelfeﬂt1
dG = Vdp — SdT, now gives (er—s ew
BTP By T to chemistry: it suggests that G is an important
We use can 3.54 to write ﬁlis e . . . .
Xpl‘eSSIOD 1s substituted in the preceding one, we obtain eqn 3 55 2b We der Ve the [65]] t [h t {h 1 lbnum onstant for a reactlon 15 1‘6 ated to 115 Standard
1 a E C(lll l C I @Wﬁﬁmmﬁumam 126 3 THE SECOND LAW The Gibbs—Helmholtz equation is most useful when it is applied to changes,
including changes of physical state and chemical reactions at constant pressure. Then, because AG = Gf— Gi for the change of Gibbs energy between the ﬁnal and initial states
and because the equation applies to both Gf and G3, we can write
[adorn] AH I (356)
P ar “F? This equation shows that, if we know the change in enthalpy of
d of transformation (such as vaporization or reaction), then we undergoing some kin
know how the corresponding change in Gibbs energy varies with temperature. As we
shall see, this is a crucial piece of information in chemistry. a system that is Gibbe energy, G Pressure, p
n of the Gibbs energy with pressure ms of its value at another pressure, the
qn 3.52, which gives dG : Vdp, and (c) The variatio Fig. 3.21 The variation of the Gibbs energy _ .
To ﬁnd the Gibbs energy at one pressure in ter with the pressure is determined by the _
volume of the sample. Because the volume temperature being constant, we set dT = 0 in e of the gaseous phase of a substance is integrate:
greater than that of the same amount of Pf
liquid phase, and the entropy of the solid _, '
phase is smallest (for most substances), the G(Pf) _ GU) '1) + Vdp (35%)
Gibbs energy changes most steeply for the Pi
For molar quantities, gas phase, followed by the liquid phase,
and then the solid phase of the substance. Pf
(3.57b) Because the volumes of the solid and liquid _
phases of a substance are similar, their Gm(Pf) f Gm(Pi) + L Vm dp molar Gibbs energies vary by similar
amounts as the pressure is changed. This expression is applicable to any phase of matter, but to evaluate it we need to know how the molar volume, Vm, depends on the pressure.
The molar volume of a condensed phase changes only slightly as the pressure volume Actual changes (Fig. 3.22), so we can treat V as a constant and take it outside the integral:
ssumed volume m
' Pr
Gm(Pf) = Gm(Pi) + VmJ dP = Gm(Pi) + (PfP1)Vm
Pi w Sél .. _ __.. alciilate'tlie"'change in _ m . . 0°C,
when the pressure is increased from 1.0 bar to 2.0 bar. Volume, V ditions (pf— pi) Vm is very small and maybe neglected. 7 Hence, we may usually suppose that the Gibbs energies of solids and liquids aIe ‘
independent of pressure. However, if we are interested in geophysical problems, then
5 interior are huge, their effect on the Gibbs energy 6311' because pressures in the Earth’
not be ignored. If the pressures are so great that there are substantial volume change
“9'33: The d‘fference “1 Glbbs energy over the range of integration, then we must use the complete expression, eqn 5.57 of a solid or liquid at two pressures is equal
to the rectangular area shown. We have
assumed that the variation of volume with pressure is negligible.  Under normal laboratory con P' Pressure, p Pr 0 Abrief illustration
AUSV2+L0 cm3 mol‘1 independent transition of a solid
30 x 10“ Pa) from 1.0 bar se in pressure to 3.0 Mbar (
f the transition changes from AUSGU bar) to 1) X (3.0 x 1011Pa— 1.0 ><1.0.5 1’a Suppose that for a certain phase
of pressure. Then for an increa
(1.0 X 105 Pa), the Gibbs energy 0 AmGB Mbar} = AmGU bar) + (1.0 x 10* m3 mol—
= AmGU bar) + 3.0 x 102 k] mot] ) where we have used 1 Pam =1]. 0 3.9 PROPERTIES OF THE GIBBS ENERGY The molar volumesof
gases are large so the Gibbs ener
_ , gy of a as de e
onE ti: pressure. Furthermore, because the volume also varies? nuarlISetlldS St'rthntiy
Pzriect e, :‘re canrgot. treat it as a constant in the integral in eqn'3 57b (Pi 32:; F e
P ga we su stitute Vm = RTI'p into the integral, treat RT as a constant. and ﬁlida Pr
G (129:6 emery 1d — P
m m , p—Gm( , +RT1 —f
Pi p p) 11 pi (3.59)°
This expression shows that when the ‘ ‘
v I , pressure is increased tenf ld t the :Eiolart 1Gibb: energy increases by RT 111 10 z 6 k] moi—1. (it alsdigﬂmotfflfemmﬁé’
:21; ogfa a; 1f we set pl :1)" (the standard pressure of 1 bar), then the molaornGilatls gy p ect gas at a pressure p (set pf: p) is related to its standard value by S GAP):an +Rr1n£
P9 (3.60)° ' tetra? 3. I g I .. _ Eérgagid as taZpgesrfect gas) when the pressure is increased isothefrynally‘lefo: lagbur
denéed 2125‘: (s ltV'ott: thza)t, whereas the change in molar Gibbs energy for a co:r I e — es .1 is a few 'oules er _
a gas is of the order of kilojoules per Jmole p m'Ole, the answer yoljlgullll getlﬂll
. mo ‘ The logarithmic dependence 3
I of the molar Gibb
b I ‘ . ‘ 5 energy on the ress '
quﬁgpciﬁgfis Illlllﬁitrated in Fig. 3.24. This very important airmail: Ptlieedfted
(WhiCh is usuglyi: glediplfolgh in the following chapters, applies to perfect gigs
_ ou a roxirn t' . ' ' '
how to take into account gas imperﬁflzlltions.a 10D) Further Information 3.2 descnbes v= nHT/p Q3E
>:
9
m .
I:
E.)
In
.CI
:9
(5
ii
0
5 Pi Pressure, p pr Viv—Dc Pressu re, p The differ ' '
s at tw:1;::Glbl)S energy for Fig. 3.24 The molar Gibbs energy of a
I balow the Page: is equal to the perfect gas is proportional to in p, and the
—gas isotherm. standard state is'reached at p“. Note that as
p —> 0, the molar Gibbs energy becomes
negatively inﬁnite. interAetivity Show how the ﬁrst
derivative of G, (aG/dp) T, varies with pressure, and plot the resulting expression over a pressure range. What is the physical signiﬁcance of (BG/Bp)T? 127 128 3 THE SECOND LAW Checklist of key equations Property Equation Comment
Thermodynamic entropy d8 = dram/T Deﬁnition
Boltzmann formula 5 = it in W Deﬁnition
Ciausius inequality d5 2 quT
Entropy of isothermal expansion AS = all ln(Vfi' Vi) ' Perfect gas
Entropy of transition A“ S = Aer' Tu5 At the transition temperature
Variation of the entropy with 8(Tf) = S(Ti) + C ln(Tf/Tl} The heat capacity, C, is independent of temperature and
temperature no phase transitions occur
Reaction entropy AS": 2 v83 — 2 v8:1
 Products Resctanu
Helmholtz energy A = U— TS Denition
Gibbs energy G = H  TS Deﬁnition
Maximum work wmax = AA
waddmx 2 AG Constant p and T _ Maximum non—expansion work
(a) 66sz 0 and dUSiV S 0, or Criteria of spontaneity
(b) dAT‘V S O and dGTIP S 0
Reaction Gibbs energy AG“: 2 vAfGﬁ— E 1055“
Producls Renciams
Fundamental equation dU: TdS — pdV
dG = Vdp — SdT damental equation of chemical thermodynamics
(scrap) T: V and (aster)? = —_S (3(G/T)IBT)P=eHIT2
Gm(Pf) : + GlPr) = C(Pi) + ﬂRTlnlpr/Pi) Fun Gibbs—Helmholtz equation
lncornpressible substance Perfect gas see the Road map section of the Resource section. ) For a chart of the relations between principal equations, Further information charge Q1 in the presence of a charge Q2 can be expressed in terms of to equation
the Coulomb potential, (1): fy the Gibbs energy of
ion from a vacuum into the Q2 Further information 3.1 The 80 The strategy of the calculation is to identi solvation with the work of transferring an
solvent. That work is calculated by taking the difference of the work v: QiiD = IT;
of charging an ion when it is in the solution and the work of charging
We model an ion as a sphere of radius ri immersed in a medium 31f
is Q, e the same ion when it is in a vacuum.
The Coulomb interaction between by a distance r is described by the Con permittivity 8. It turns out that, w
electric potential, q), at its surface is the same as the potenti two charges Q, and Q2 separated
point charge at its centre, so we can use the last expression lomhic potential energy: V 3 (2in
41cm _ Q
where e is the medium’s permittivity. The permittivity of vacuum is (p _ 4ﬁ8ri _ _
‘1. The relative ermittivi (formerl .' '
P w Y The work of bringing up a charge dQ to the sphere is ¢dQ Theﬁef‘.’re’  to.  80 = 8.854 x 10'12 r1 c2 or called the :dielectric constant’) of a substance is deﬁned as «Sr = 8/80. .
Ions do not interact as strongly in a solvent of high relative here from 0 to Zi‘a 15
permittivity (such as water, with Er = 80 at 293 K) as they do in a solvent of lower relative permittivity (such as ethanol, with .EI = 25 at
ential energy of a 293 K). See Chapter 17 for more details. The pot the total work of charging the sp 2e 2 W 1 ' 222
0 41CEri 0 litteri hen the charge of the sphere _. .
al duets a write This electrical work of charging, when multiplied 1) Ave d ’
constant, is the molar Gibbs energy for charging th: ions ga m S The work of charging an ion in a vacuum is obtainedb sett'
g = £0, the vacuum permittivity. The corresponding valueifo mg
charging the‘ion in a medium is obtained by setting 8: 8 r h
pr 15 the relative permittivity of the medium. It follows thief}?! Elic
in molar Gibbs energy that accompanies the transfer of ion f8 c ange
vacuum to a solvent is the difference of these two quantitiess mm a AGa solv E I' 2 2 2 2 .e . 2 21 NA 218 N. :ciezNA zilezNA ziZeZNA 1
fitteri Susan Snefeori Susan 8ﬂ£ofi 1 which is eqn 3.45. Further information 3.2 The fugacr'ry At various stages in the development of physical chemist it '
necessary to switch from a consideration of idealized s sfdhi If 1
systems. In many cases it is desirable to preserve the fofm of Stho tea
expressions that have been derived for an idealized s stem Th e
dev1ations from the idealized behaviour can be exprzssed en
Simply. For instance, the pressure dependence of the molar Gs'hb
energy of a real gas might resemble that shown in Fig 3 25 T l clS
eqn_.’jl.60 to this case, we replace the true pressure p4 an . ffo a’ apt
pressure, called the fugacity? f, and write 1 a Y 6 ethe 1
P6 The fugaCity, a function of the pressure and temperature, is deﬁned so Gm=G$+RTln [3 61] th . . .
at this relatlon is exactly true. Although thermodynamic expressions m Molar Gibbs energy, G VLOO Pressure, p e} gy r1232: glbbs energy of a real gas. As p +9 0, the molar
e). When attfas with the value for a perfect gas (shown by the
I h m 1 ctive forces are dominant (at intermediate
0 ar Gibbs energy 15 less than that of a perfect gas ' FURTHER INFORMATION 129 in re . . . .
userif fpgaclictliles derived from this expression are exact, they are
y 1 we ow how to inter ' ' '
pret fugactties in terms of a
i _ ctu
pressures. To develop this relation we write the fugacity as a1 f: .
W [3.62] 131$; 3: 1: Itlhtrle1 dimensmnless fugacity coefﬁcient, which in general
Equation 3 gimperature, the pressure, and the identity of the gas.
Ex r ‘ I . 15 true for all gases whether real or perfect.
p essmg it in terms of the fugac1ty by using eqn 3.61 turns it into P
Jp’depz Gmfp) — Gm(p’) ={G;+RTmi} —{G° +RTln~—f’}
P9 “1 e
P In this expression, f is the fugacity when the pressure is p and f’ is the ' fugacit when th  i
write Y e pressure is p . If the gas were perfect, we would P
P
l
J Vperfectmdp = RTJ —dP = RT in L
p’ PI P p’
The difference between the two equatiOns is 'P (Vin _ Vperfect,m)dp = RT — [ J P’ 1" (p/p’) which can be rearranged into f p’ I 1 P [— X — = — (V131 m Vperfect,1n)d.lJ p f’ RT .
1323:; pa 3 (if 1tlhe gfas bichaves perfectly and f’ becomes equal to the
i , . ere ore, ’r'p' —> 1 as p’ —> 0 If we take this 1' '
‘ I I . t,
which means setting f [p = 1 on the left and p’ = 0 on the the last equation becomes f 1 P '
in — =
P R T; 0( Vm _ VPEFfectm) dp Then: with ¢= ftp. 1 P
ll] = — __
9) RT frf vm' VPBTfecl.m)dP For a perfect gas V _ . , ' perfect,m — RT/p. For a real gas, V : 13sz
Z is the compressmn factor of the gas (Section 1 Ball1 With tiff, Where
substittitions, we obtain '  ese two P
in ¢=J P (3.63} 0 Pro . . .
int :1de live know how Z varies With pressure up to the pressure of
8 es , t is expressmn enables us to determine the fugacity 1 
CDC ﬁclent and hence till: ()ugll eqn 3 62, ti) relate llle lugactty t0 the Prgfsigii: fgopicﬁiigélltl that for most gases Z < I up to moderate
, u a > 1 at higher pressures If Z < 1 h
range of integration then the inte ' I t mughout the
I ' , II grand in eqn 3.63 is negative and
(i) < 1. This value implies that f < p (the molecules tend to stick 130 3 THE SECOND LAW Flg. 3.26 The fugacity coefﬁcient of a van
der Waals gas plotted using the reduced
variables of the gas. The curves are labelled with the reduced temperature TI 2 TITS. interActivity Evaluate the fugacity
coefﬁcient as a function of the
reduced volume of a van der Waals gas and
plot the outcome for a selection of reduced
temperatures over the range 0.8 S Vr _<. 3. 179
M
01 Fugacity coefficient, 45 Reduced pressure, pIr = pip: and that the molar Gibbs energy of the gas is less than that
higher pressures, the range over which Z > 1 may over which Z < 1. The integral is then positive, qb > 1 and f > p (the repulsive interactions are dominant and tendto
drive the particles apart). Now the molar Gibbs energy of the gas is
the same pressure.
eater than that of the perfect gas at I
gr Figure 3.26, which has been calculated usmg the full van der Wials
state, shows how the fugacity coefficient depends on t e
e reduced variables (Section 1.4)._Because ' ' bl 1.5, the raphs can be used
ailable in Ta e g able 3.6 together)
of a perfect gas. At
dominate the range equation of
pressure in terms of th
critical constants are av  . I
for quick estimates of the fugacrties of a wide range of gases. T gives some explicit values for nitrogen. Discussion questions ires the organization of a very large number of 3.1 The evolution of life requ nisms violate molecules into biological cells. Does the formation of living orga the Second Law of thermodynamics? State your conclusion clearly and present detailed arguments to support it. 3 2 You received an unsolicited proposal from a selifdeclgredciengitpses heat
I ' ‘ lopment of his latest i ea: a ev1 .
seeking investors for the deve ‘ I t is
extracted from the ground by a heat pump to borl Eat: into :;:a}1;:::1;ump
steam engine t at rives  .
used to heat a home and to power a I ' I .
This procedure is potentially very lucrative because, after an initial extraction of energy from the ground, no I .
I running indeﬁnitely. Would you invest in '
clearly and present detailed arguments to support it. ‘ ' d to establish cr‘teria for
. Th f llowm expressions have been use
33 e 0 g >0and dUSyS 0, dAnVSO, and ontaneous change: AS‘ t> 0, dSwa  . . I . I
iipG < 0 Discuss the origin, signiﬁcance, and applicability of each criterion
T,p — ' r I this idea? State your conclusion 1.5 Exercises w” 47 Aﬁﬁast’ﬁ‘féiﬁmmnm" EXERCISES 131 Assume that all gases are perfect and that data refer to 298.15 K unless
otherwise stated. 3_1(a) Calculate the change in entropy when 25 k] of energy is transferred
reversibly and isothermally as heat to a large block of iron at (a) 0°C,
(b) 100°C. 3.10:) Calculate the change in entropy when 50 k] of energy is transferred
reversibly and isothermaily as heat to a large block of copper at (a) 0°C,
(b) 70°C. P
01 Fugacity coefficient, p = f/p 3,2(a) Calculate the molar entropy of a constantvolume sample of neon at
500 K given that it is 146.22] K4 mol‘l at 298 K. 3,233) Calculate the molar entropy of a constantvolume sample of argon at is 20
12 250 K given that it is 154.34 1 1c1 moi1 at 293 K. so 100 0 4 3 Reduced pressure, pr = p/pc 3.3(a) Calculate A8 (for the system) when the state of 3.00 mol of perfect gas
atoms, for which Cle = %R, is changed from 25°C and 1.00 arm to 125°C and
5.00 atm. How do you rationalize the sign of A3? 3.30)) Calculate AS (for the system) when the state of 2.00 mol diatomic
perfect gas molecules, for which Cpim = %R, is changed from 25°C and
1.50 atm to 135°C and 7.00 atm. How do you rationalize the sign of AS? 'aAia) A sample consisting of 3.00 mol of diatomic perfect gas molecules at
200 K is compressed reversibly and adiabatically until its temperature reaches r who is fossil fuels would be required to keep the device 250 K. Given that CMm = 27.5] K’1 mol‘l, calculate :1, w, AU, AH, and AS. 3.4(b) A sample consisting of 2.00 mol of diatomic perfect gas molecules at
250 K is compressed reversibly and adiabatically until its temperature reaches
300 K. Given that CWT] = 27.5 I K'I mol‘l, calculate q, w, AU, AH, and AS. 3.5(al Calculate AH and ASK“ when two copper blocks, each of mass 10.0 kg,
one at 100°C and the other at 0°C, are placed in contact in an isolated . container. The speciﬁc heat capacity of copper is 0.385 I K’1 g“1 and may be
assumed constant over the temperature range involved. 3.50;) Calculate AH and ASIm when two iron blocks, each of mass 1.00 kg, one
at 200°C and the other at 25°C, are placed in contact in an isolated container.
The speciﬁc heat capacity of iron is 0.449 I K‘1 g’1 and may be assumed
constant over the temperature range involved. 35(8) Consider a system consisting of 2.0 mol C02(g), initially at 25°C and 10 at}?! and conﬁned to a cylinder of crosssection 10.0 cmz. It is allowed to exPanti adiabatically against an external pressure of 1.0 atrn until the piston has filmed Outwards through 20 cm. Assume that carbon dioxide may be iIOI'isrdered a perfect gas with CV,“ = 28.8] K"1 mol‘1 and calculate (a) g, (b) w,
. (ciliU. (d) AT, (e) as. ' 6 have been used to establish criteria for I I
' : 9.0 tm' d ‘ reSSions I _ , .
3‘4 The fouowmg exp < 0. Discuss the origin, Significance. spontaneous change: dATHV < 0' and dGTSP
and applicability of each criterion. _'cler a system consisting of 1.5 mol C02(g), initially at 15°C and
conﬁned to a cylinder of crosssection 100.0 cmz. The sample is
39 i9 exPand adiabatically against an external pressure of 1.5 attn until
3t0_§1'has moved outwardsthrough 15 cm. Assume that carbon dioxide
Epnsi'dered a perfect gas with C1,,2m = 28.8 I K’1 rnol‘l, and calculate
‘(cl AU. (:1) AT, (e) as. ' i the
3.5 Discuss the physical interpretation of any one Maxwell relation. I . 3.6 Account for the dependence of 71:] of a van der Waals gas in terms 0 I , _ signiﬁcance of the parameters a and b. I
tation of the dependence of the Gibbs_ enersi' millillPY of vaporization of chloroform (CHCla) is 29.4 k] mol’1 at
1mg point of 334.88 K. Calculate (a) the entropy of vaporization
I'm at this temperature and (b) the entropy change of the 3.7 Suggest a physical interpre on the pressure. ‘ encr
3.8 Suggest a physical interpretation of the dependence of the Gibbs, I I thiﬂlfh’ Ofvaporization of methanol is 35.27 k] mol"I at its
813011“ of 64.1“C. Calculate (a) the entropy of vaporization
at this. temperature and (b) the entropy change of the on the temperature. 3.8(a) Calculate the standard reaction entropy at 298 K of (a) 2 CH3CHO(g) + 02(g) —> 2 CH3COOHU)
(b) 2 AgCl(s) + Br2(l) —> 2 AgBr(s) + C12(g)
(c) Hg(i) + C12(g) ——> HgC12(s) 3.80:) Calculate the standard reaction entropy at 298 K of (a) Zn(s) + Cu1+(aq) —> Zn2+(aq) + Cu(s)
(b) cunnoun) + 12 02(3) —> 12 co,(g}+11 H200) 3.9(a) Combine the reaction entropies calculated in Exercise 3.8a with the reaction enthalpies, and calculate the standard reaction Gibbs energies at'
298 K. 3.90)] Combine the reaction entropies calculated in Exercise 3.81) with the
reaction enthalpies, and calculate the standard reaction Gibbs energies at 298 K. 3.10(a) Use standard Gibbs energies of formation to calculate the standard
reaction Gibbs energies at 298 K of the reactions in Exercise 3.8a. 3.1003) Use standard Gibbs energies of formation tov‘c'alculate the standard
reaction Gibbs energies at 298 K of the reactions in Exercise 3.8b. 3.11(a) Calculate the standard Gibbs energy of the reaction 4 HCl( g) + 02(g) —9 2 C12(g) + 2 HZOU) at 298 K, from the standard entropies and enthalpies of
formation given in theDnta section. 3.1103) Calculate the standard Gibbs energy of the reaction CO( g) +
CH3OHU) —> CHsCOOHG) at 298 K, from the standard entropies and
enthalpies of formation given in the Data section. 3.12(a) The standard enthalpy of combustion of solid phenol (CéHSOH) is
—3054 k] rnol‘1 at 298 K and its standard molar entropy is 144.0 I K‘I mol‘l.
Calculate the standard Gibbs energy of formation of phenol at 298 K: 3.12M The standard enthalpy of combustion of solid urea {CO(NH2)2) is
—632 k] incl"1 at 298 K and its standard molar entropy is 104.60 I K’] mol‘l.
Calculate the standard Gibbs energy offormau'on of urea at 298 K. 3.13(a) Calculate the change in the entropies of the system and the
surroundings, and the total change in entropy, when a sample of nitrogen gas
of mass 14 g at 298 K and 1.00 bar doubles its volume in (a) an isothermal reversible expansion, (b) an isothermal irreversible expansion against prm : 0,
and (c) an adiabatic reversible expansion. 3.13(b) Calculate the change in the entropies of the system and the
surroundings, and the total change in entropy, when the volume of a sample
of argon gas of mass 21 g at 298 K and 1.50 bar increases from 1.20 dirt3 to
4.60 dni3 in (a) an isothermal reversible expansion, (b) gin isothermal
irreversible expansion against pEX = 0, and (c) an adiabatic reversible expansion. 3.14{a) Calculate the maximum nonexpansion work per mole that may be 7 obtained from a fuel cell in which the chemical reaction is the combustion of methane at 298 K. 3.14(b} Calculate the maximum nonexpansion work per mole that may be
obtained from a fuel cell in which the chemical reaction is the combustion of
propane at 298 K. 3.15(a) (a) Calculate the Carnot efﬁciency of a primitive steam engine
operating on steam at 100°C and discharging at 60°C. (b) Repeat the calculation for a modern steam turbine that operates with steam at 300°C and
discharges at 80°C. 3.1509) A certain heat engine operates between 1000 K and 500 K. (a) What is
the maximum efﬁciency of the engine? (b) Calculate the maximum work that ...
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This note was uploaded on 04/07/2012 for the course CHEM 340 taught by Professor Staff during the Spring '08 term at Ill. Chicago.
 Spring '08
 STAFF

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