Chapter 4 : Slide 1

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Chapter 4 The Second Law of Thermodynamics (I)
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OUTLINE SECTION 4.1 - Dispersal of
energy SECTION 4.2 - Entropy SECTION 4.3 -
Entropy changes for certain processes SECTION 4.4
- The Third Law of Thermodynamics SECTION 4.5 -
Helmholtz and Gibbs energies SECTION 4.6 -
Standard molar Gibbs energies
HOMEWORK EXERCISES 4.4 26 Parts(a)
only. PROBLEMS 4.5, 4.7
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We want to know how systems will change
spontaneously i.e. in which direction they will
move without the addition of outside heat or
work. Example
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Imagine an isolated system The First Law uses
the internal energy to identify permissable
changes (those that conserve energy). The Second
Law uses the concept of "entropy" to identify
which of the permissable changes are also
spontaneous. Systems tend to move towards
equilibrium, so that asking about the spontaneous
direction is equivalent to asking about the
position of equilibrium. We shall see later how
the second law of thermodynamics comes into
determining equilibrium constants.
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THE CARNOT CYCLE Four reversible operations
carried out on an ideal gas - a CYCLE.
A cycle because we end up at exactly the same
state as where we started
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Here are the four steps of the cycle
1) Reversible isothermal expansion from A to B.
DU 0 qrev R Th ln(VB/VA) wrev R Th
ln(VA/VB). 2) Reversible adiabatic expansion
from B to C. DU CV(Tc - Th) qrev 0 wrev
CV(Tc - Th). 3) Reversible isothermal
compression from C to D. DU 0 qrev R Tc
ln(VD/VC) wrev R Tc ln(VC/VD). 4) Reversible
adiabatic compression from D to A. DU CV(Th -
Tc) qrev 0 wrev CV(Th - Tc).
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What are the net changes after going around this
CYCLE once? DU 0 qrev R Th ln(VB/VA) R
Tc ln(VD/VC) wrev R Th ln(VA/VB) R Tc
ln(VC/VD) Remember from ch. 2 Th/Tc
(VD/VA) ?-1 (VC/VB) ? -1 so VD/VA
VC/VB so VB/VA VC/VD. qrev R(Th - Tc)
ln(VB/VA) gt 0 heat is absorbed by the
system. wrev R(Th - Tc) ln(VA/VB) lt 0 work
done on the system lt 0 i.e. positive work done
by the system.
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What we have is some heat qh absorbed at Th, and
work -wrev is extracted from the system, which is
therefore called an ENGINE.
Work is not a function of state so net work can
be done even though the system returns to its
initial state
Contrast to a function of state e.g. Hess's Law
is
because dH is an exact differential.
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What is the THERMODYNAMIC EFFICIENCY of this
engine? Work done by the engine heat absorbed
at Th ? -w/qh R(Th-Tc) ln(VB/VA)
R Th ln(VB/VA) (Th -Tc) / Th. ? can
only be 1 (100) if Tc 0. This is an
alternative definition of absolute zero. By the
conservation of energy (First Law), -w net
heat absorbed qh qc (remember that qc lt
0) so ? (qh qc)/qh (Th - Tc)/Th thus
qc/qh - Tc/Th or equivalently
qc/Tc -qh/Th.
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We have analyzed a particular cycle using a
single working fluid (ideal gas). CARNOT'S
THEOREM is that the efficiency of all reversible
cycles between Th and Tc is (Th-Tc)/Th, for any
working medium. PROOF If this untrue then there
is a contradiction Imagine engine B is a Carnot
engine with an ideal efficiency, say, of 70.
Suppose A is a more efficient engine, say with ?
80. 1. 100 J is absorbed at Th by engine A
qh 100 J in so -w 80 J out. qc is the
difference, -20 J i.e. heat comes out of the
engine. qh qc -w 2. Imagine first that
engine B is run forwards. To get 80 J of work out
-w, qh for B must be 80 J / 0.7 114 J, and qc
is therefore 80 - 114 J -34 J.
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3. Now imagine that engine B runs backwards,
powered by A. The signs of w, qh and qc for B all
are reversed. The net result is that 14 J have
flowed spontaneously from Tc to Th, which is
impossible, and thus the initial assumption that
engine A has a higher efficiency than engine B
must be false. There are various forms of the
second law of thermodynamics. One is It is
impossible to devise an engine which, working in
a cycle, can transfer heat from a cold body to a
hotter body with no other effects (such as
relying on external work).
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For an ideal reversible cycle ?
(Th-Tc)/Th For a practical irreversible cycle
? lt (Th-Tc)/Th Example A car engine operates
between 500 oC and 100 oC ? lt (773-373)/773
52. Thus, other factors permitting, we try to
run engines as hot as possible.
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GENERALIZED CYCLES Remember that for the Carnot
cycle qc/qh -Tc/Th. Thus qc/Tc qh/Th 0
or S q/T 0.
We can break up any reversible cycle into
infinitesimal movements along isotherms and
adiabats, with many infinitesimal heat transfers
?q.
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A general cycle can be divided into small Carnot
cycles. Paths cancel in the center. The
perimeter approximates the general cycle. For
the whole cycle
In the limiting case,
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This tells us that dqrev/T is an exact
differential i.e. it is the differential of a
function of state. We call this differential
"dS" and the function "S"
S is called the ENTROPY.
and is path independent. The usual function
state results apply e.g. DSA?B - DSB?A.
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CALCULATION OF DS FOR A HEAT TRANSFER The system
is a pair of reservoirs. Heat transfer, amount q,
from Th to Tc is irreversible, but we only know
how to calculate DS for a reversible process.
We design a reversible process that has the same
initial and final states for the system Imagine
we have some ideal gas in a piston at Th, which
is part of the surroundings. a) Let the gas
expand isothermally and reversibly at Th,
absorbing heat q. DSgas q/Th. DShot reservoir
-q/Th.
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b) Now put the piston inside an insulated box and
let it undergo a reversible and adiabatic
expansion until its temperature drops to Tc. q
0 so DS 0. c) Finally compress the gas
reversibly and isothermally, in contact with the
cooler reservoir, transferring heat q out of the
gas. DSgas -q/Tc and DScool reservoir
q/Tc.
DS for the system is -q/Th q/Tc
Comment this is a positive quantity because
Th gt Tc. But S is a function of state, so this
is the same as DS for the original irreversible
case too.
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General results IRREVERSIBLE (SPONTANEOUS)
CHANGE DSuniverse gt 0 REVERSIBLE
CHANGE DSuniverse 0 Thus, for any change,
DS for the universe is ? 0. DS for the universe
is only zero for reversible changes, and for all
real spontaneous changes, DSuniverse gt 0. THIS
IS ANOTHER FORM OF THE SECOND LAW.
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IRREVERSIBLE PROCESSES The efficiency of an
irreversible Carnot cycle is (qhirr
qcirr)/qhirr lt (Th-Tc)/Th so qhirr/Th qcirr/Tc
lt 0. In general,òdqirr/T lt 0 around any cycle
that is not totally reversible. Consider a
change from A to B in an isolated system, which
occurs spontaneously, followed by a change from B
to A which occurs reversibly.
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But the first integral is zero because dq is zero
for an isolated system
Thus òdqrev/T lt 0 so DSB?A lt 0
so DSA?B gt 0. DS is positive for any
spontaneous change in an isolated system (like
the universe!) "The energy of the universe is a
constant the entropy of the universe tends
always to a maximum".
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CALCULATION OF DS The Clausius inequality is dS ?
dq/T. Generally, DS ? òdq/T. Equality holds
only for reversible changes. How do we handle
irreversible changes? Set up a reversible route
between the same initial and final conditions
remember that S is a function of
state. EXAMPLES CHANGES OF STATE 1) Melting At
constant p, Tm is the "melting point", where
solid and liquid are in equilibrium i.e. changes
are reversible. Addition of heat melts more
material DfusH is the latent heat of fusion. At
constant p, qrev DfusH. DfusS òdqrev/T
qrev/T for constant T DfusH/Tm the
units are J K-1 mol-1.
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2) Evaporation At the boiling point Tb liquid
and vapor are in equilibrium. DSvap
DHvap/Tb. 3) Allotropic changes E.G. Tin has two
forms. Above 286 K "white" metallic tin is the
most stable. Below 286 K "grey" powdery tin is
stable. The enthalpy of transition is 2.09 kJ
mol-1. DS286 2090 J mol-1 / 286 K 7.31 J K-1
mol-1. IDEAL GASES 1) Isothermal expansion from
V1 to V2, reversibly. qrev nRT ln(V2/V1) At
constant T, DS qrev/T nR ln(V2/V1). 2) Expa
nsion where V and T both change. Imagine a
reversible change dqrev dU p dV
nCV,mdT (nRT/V) dV thus dS dqrev/T n CV,m
dT/T n R dV/V
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IF CV is independent of T then DS nCV,m
ln(T2/T1) nR ln(V2/V1). This will be true even
for an irreversible change in V and T. 3) Mixing
of gases Equal T and p on each
side Final volume for each gas is V1
V2. DS1 n1R ln (V1V2)/V1 DS2 n2R ln
(V1V2)/V2 Total DS DS1 DS2.
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If the initial p and T are identical then n1/V1
n2/V2. Define the mole fraction of 1 as x1
n1/(n1n2) V1/(V1V2). Etc. for x2. DS for
the gas is then -R n1 ln x1 n2 ln x2
. Per mole of mixture, DS -R(x1 ln x1 x2
ln x2). 4) Heating a material reversibly at
constant p At constant p, dqrev Cp dT, so
If Cp is independent of T, then DS Cp
ln(T2/T1).
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Example The freezing of supercooled water at
-10oC.
We need a reversible path to find DS.
DATA Cp for water is 75.3 J K-1 mol-1. Cp for
ice is 37.7 J K-1 mol-1. DHfreezing at 273 K
-6017 J mol-1.
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Thus DS for the system is 2.81-22.04-1.41
-20.64 J K-1. What is DS for the
surroundings? From the point of view of the
surroundings, the heat transfer is reversible so
here DS for the surroundings is DH/T. DH263
-5641 J mol-1 at 263 K CHECK THIS!, so q
into surroundings 5641 J mol-1. DSsurr q/T
21.45 J K-1 mol-1. Thus DSuniverse
21.45-20.64 0.81 J K-1 mol-1 gt 0 as it must
be. Note for the reversible route DSsurr
-DSsys at each step, and therefore DSuniverse
0.
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So far we have defined changes in entropy, DS,
via DS ò Cp dT/T but not absolute values i.e.
S itself. THE THIRD LAW OF THERMODYNAMICS is
that, at absolute zero, a perfect crystal of any
material has S 0. Then
Any DH arising from phase changes must also be
included.
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Calorimetry provides Cp(T) and enthalpies of
phase changes, from which S can be
evaluated. S298 values are tabulated, so that
DS298 can be calculated for any reaction
from DS298 S ?iSi, 298 i.e.
SSproducts, 298 - SSreactants, 298
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THE ENERGETICS OF REFRIGERATION In order to
transfer heat from cold (Tc) to hot (Th) we must
do work on the system we are running a heat
engine backwards. Heat released to surroundings
-qh qc w Assume reversibility i.e. deal
with a Carnot engine in reverse, so Th/Tc
-qh/qc Suppose qc 1000 J. -qh qc x Th/Tc
1000 J x 298/268 1112 J. Therefore the work
done by the compressor is 112 J. The
"COEFFICIENT OF PERFORMANCE", co, is qc/w
1000/112 ? 9 here. In general, the performance
factor for a refrigerator is co
-qc/(qhqc) Tc/(Th-Tc). This is an upper
limit because real engines cannot run reversibly
e.g. to lose heat rapidly, the condenser is run
hotter than the surroundings.
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HEAT PUMPS Here we are interested in the heat
drawn from the cold end, and rejected at the
higher temperature. qc -qh x Tc/Th 1000 x
268/298 899 J, so w must be 101 J. We paid for
101 J, and obtained 1000 J the performance
factor is -qh/w ?10. An upper limit is Th/(Th-Tc)
for a reversible process. WORK NEEDED TO SUSTAIN
LOW TEMPERATURES No insulation is perfect.
According to Newton's law of cooling, the rate of
heat leakage into a cold body at Tc in
surroundings at Th is proportional to (Th-Tc),
i.e. dqc/dt A(Th-Tc) where A is a
constant. w qc/co therefore dw/dt
A(Th-Tc)2/Tc . IMPLICATIONS
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AN INTERPRETATION OF ENTROPY AT THE MICROSCOPIC
LEVEL S is a measure of the degree of disorder in
a system. In fact Boltzmann derived S kB ln
W where W is the number of different microscopic
arrangements that correspond to the specified
state of the macroscopic system. kB is
Boltzmann's constant R/NA. The more
"restrictions" there are on the particles, the
smaller S. e.g. 2 H has a greater entropy than
H2. e.g. Ssolid lt Sliquid lt Sgas. e.g. gas
expands from V to 2V. DS is often gt0 for
reactions where the number of particles
increases. However, it is negative for reactions
of the type AB (aq) ? A(aq) B-(aq). Why?
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CONDITIONS FOR EQUILIBRIUM We have a pointer for
the spontaneous direction of a process. How do we
know when we actually reach equilibrium? Clue
there are no spontaneous changes away from
equilibrium.
The equilibrium point is one of maximum Suni.
dSuni dSsurr dSsys 0. This definition of
equilibrium considers the whole universe. At
times it is more convenient to have a definition
which focuses only on properties of the system,
and not its environment.
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CONCENTRATING ON THE SYSTEM Consider the
situation Tsys Tsurr Now suppose a
spontaneous process occurs in the system and an
infinitesimal amount of heat dq enters the
surroundings. -dqsys dqsurr Because the
surroundings are very (infinitely!) large, and
assuming there are no local hot spots which then
cool irreversibly, we can view the transfer of
heat into the surroundings as occurring
isothermally and reversibly. Thus dSsurr
dqsurr/Tsurr which is also -dqsys/Tsys
. The general equilibrium condition is dSuni 0
i.e. dSsys - dqsys/Tsys 0. We are now
looking just at properties of the system alone
(we drop the sys subscript from now on) dS -
dq/T 0 i.e. dq - T dS 0 is an
equilibrium condition for the system.
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(A) Let us concentrate on conditions of constant
T and p dq dH therefore dH - TdS
0. Gibbs defined a new function of state, now
called the Gibbs energy, G (or sometimes the
Gibbs free energy, F) G H - TS. Why? dG
dH - TdS - SdT. At constant T, dT 0 so
then dG dH - TdS. Thus at constant T and p,
dG 0 at equilibrium. For a spontaneous
process dSuni gt 0 and dG lt 0 (G for the
system). The system tends to move towards a
minimum Gibbs energy.
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(B) Let us concentrate on conditions of constant
T and V dq dU now, so the equilibrium
condition is dU - TdS 0. The Helmholtz free
energy, A, is defined by A U - TS. At
constant T and V, dA 0 at
equilibrium. For a spontaneous process dSuni gt
0 and dA lt 0 (A for the system).
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ASPECTS OF THE GIBBS ENERGY G H - TS. At
constant T and p, DG DH - T DS. The
direction of spontaneity is determined by both H
and S. E.G. 2 H2 (g) O2 (g) ? 2 H2O
(g) DH lt 0 and DS lt 0. At low
temperatures, DH dominates, and the reaction
spontaneously moves from left to right. At very
high temperatures, TDS dominates, and the
reaction moves from right to left i.e. H2O
dissociates. Consider the evaporation of water
DvapH at 100 oC, 1 atm 40.60 kJ mol-1 and
calculate DG and DS At equilibrium, as here, DG
0 so DS DH/T 40600 J mol-1 / 373 K
108.8 J K-1 mol-1.
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For any chemical reaction DGo S ?iDfGio
i.e. S DfGo (products) - S DfGo(reactants). whe
re "DfGio" is the Gibbs energy or Gibbs free
energy of formation. Reactions with DGo lt 0 are
called exergonic and are spontaneous. Reactions
with DGo gt 0 are called endergonic Substances
with negative Gibbs energies of formation are
stable with respect to dissociation to their
elements. NOTE Thermodynamic instability versus
kinetic lability. Stable versus inert.
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WORK Reversible work Consider the reversible
isothermal compression of n moles of ideal gas,
from V1 to V2 DU DH 0. wrev nRT
ln(V1/V2) qrev nRT ln(V2/V1). DS qrev/T
nR ln(V2/V1) lt 0. DG DH - TDS nRT
ln(V1/V2). The reversible work done on the
system is DG.
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CHANGES AT CONSTANT PRESSURE AND TEMPERATURE w
wpV wnon-pV. wnon-pV is e.g. electrical,
osmotic etc. dw dwpV dwnon-pV. G H - TS
U pV - TS. so dG dU pdV Vdp - TdS -
SdT. At constant T and p dG dU pdV -
TdS. Now dU dqp dw. dwpV -pdV, so dw
-pdV dwnon-pV Thus dG dqp dwnon-pV - T
dS. For a reversible process, dqp T dS, so
dG dwnon-pV DG wnon-pV for a
reversible process, const. T and p. Similarly,
you can show that DA w (total work).
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SUMMARY ? ENTROPY (S) defined via the Carnot
cycle dS dqrev/T. ? Thermodynamic efficiency
of engines, refrigerators and heat-pumps. ? For
an IRREVERSIBLE or SPONTANEOUS change DSuniverse
gt 0. For a REVERSIBLE change DSuniverse 0. ?
At the microscopic level, S kB ln W. Disorder
versus order. ? We have seen how to work out DS
for phase changes, temperature changes and gas
expansions and mixing. ? The Gibbs energy G H
- TS DG at const T and p DH - TDS.
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? At constant T and p, for a SPONTANEOUS change
DGsystem lt 0. ? For a REVERSIBLE change DGsystem
0. ? At constant T and V it is useful to
consider the Helmholtz energy A U - TS, where
the criterion for spontaneity is dAsystem lt 0. ?
For a reversible process at constant T and p, DG
wnon-pV. ? For any chemical reaction DGo
S?iDfGio where "DfGio" is the Gibbs energy of
formation.