chapter 3. the second law - seoul national university
TRANSCRIPT
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Chapter 3. The Second LawPhenomena
1. A gas expands and fills the available volume, but does not
spontaneously contract into a smaller one.
2. A hot body cools to the same temperature as its surroundings, but a
body does not spontaneously get hotter than its environment.
3. Heating diamonds yields graphite, but heating graphite does not give
diamonds. [Diamond by chemical vapor deposition]
4. When a ball (the system of interest) bounces on a floor, the ball does
not rise as high after each bounce because there are inelastic losses in
the materials of the ball and floor (that is, the conversion of kinetic
energy of the ball’s overall motion into the energy of thermal motion).
The direction of change leads to the greater dispersal of the total energy.
Conclusions
1
_____The First Law of Thermodynamics = Always Valid
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Figure 3A.3
The Kelvin statement of the Second
Law denies the possibility of the
process illustrated here, in which
heat is changed completely into work,
there being no other change.
The process is not in conflict with
the First Law because energy is
conserved.
2
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Figure 3A.1
The direction of spontaneous change for a ball bouncing on a floor. On each bounce
some of its energy is degraded into the thermal motion of the atoms of the floor, and
that energy disperses. The reverse has never been observed to take place on a
macroscopic scale.
3
Irreversible
Video (time or –time)
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Figure 3A.2
The molecular interpretation of the
irreversibility expressed by the Second Law.
(a) A ball is resting on a warm surface, and the atoms are undergoing thermal motion
(vibration, 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.
4
Macroscopic World
Random Motion
Coordinated Motion
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Measuring Dispersal: the Entropy of a System
Entropy S 1. A thermodynamic function that measures how the
dispersal of energy alters when a system changes from
one state to another.
2. A state function
• Statistical definition
• Thermodynamic definition
Statistical View of the Entropy
The direction of spontaneous change is:
- from a state with low probability of
occurring
- to one of maximum probability.
The direction of spontaneous change for
a gas in a pair of connected vessels
Vf
Vi
5
Irreversible
___-190401(월)
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Consider an Ideal Monatomic Gas:
Number of microstates of one atom being in Vi :
Number of microstates that N atoms are in Vi :
Number of microstates that N atoms are in Vf :
Entropy S (??) Number of Microstates W
S = W not extensive
where k is some constant, Boltzmann constant.
When an ideal gas expands isothermally from Vi to Vf ,
where n: the number of moles, NA: Avogadro’s number
6
ii cVV
N
i
NN
ii VcVV W
N
f
N
f VcV W
),,(ln,, TVNkTVNS W
i
f
if
N
i
N
fif
V
VkNcVcVkN
cVkcVkVSVSS
lnlnln
lnln
RkNV
VnRS
nNN
i
f
A
A
,ln
______
_______Ideal Gas
Isothermal
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
(3A.14)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
[White Board]
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Assumptions of this calculation
1) Ideal gas: neglect the interaction between particles and neglect
rotational and vibrational modes.
2) Isothermal process:
Entropy measures the dispersal of energy, and the natural tendency of
spontaneous change is towards the states of higher entropy.
7
0
TV
U
RNkV
VNk
V
VnRS B
i
fB
i
f A ,lnlnIdeal Monoatomic Gas
Isothermal
),,(ln),,( TVNkTVNS B W________________W Number of Microstates위치, 방향, 진동, 전자분포, 전자춤, etc.
(3A.5)
- - - - - - - - - - - - - - - - - - - - -
Same E
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Thermodynamic View of Entropy
Thermal energy random motion
1. A change in entropy dS is proportional to the amount of heat
added dq. [Extensive Properties]
The ratio of a randomizing influence (dqrev)
to a measure of the amount of randomness
already present (at T)
If the changes are restricted to reversible transfer of heat, then the
quantity dqrev/T is found to be a state function.
2. Impact of dq on the chaos is inversely proportional to T.
For a given dq, the change in entropy is large when the temperature is
low, but small if the temperature is high.
8
T
dqdS rev______
-
(3A.1)
[White Board]
Never Definition
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For a reversible and isothermal expansion (Vi Vf) of an ideal gas,
On integration:
0: isothermal
0: ideal gas
9
dVV
nRTpdVdq
dwdq
dwdqdU
rev
0 dVV
UdT
T
UdU
TV
i
fif
revf
i
rev
f
i
revf
i
i
frev
V
VnR
T
VVnRT
T
qdq
TT
dqdSS
V
VnRTq
ln/ln
1
ln
(same as pages 6 & 7)
Ideal Gas
Isothermal
(3A.14)
열역학 2법칙적용 (reversible 경우)
Ideal Gas
Isothermal
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Figure 3A.6
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.
10
i
f T
dqdS rev
-
For an isolated system, no heat enters or leaves the system irreversively, i.e. dq=0.
This inequality shows that these spontaneous processes must lead to an
increase in entropy of the universe.
The second law: In an isolated system, spontaneous processes occur in
the direction of increasing entropy.0 S
_ _ _ _ _ _ _ _ _ _
Reversible
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3A.4(b) Entropy Change Arising from a Phase Transition
Under the condition of constant pressure, the latent heat is an
enthalpy of phase transition Ht.
Both melting and boiling are endothermic processes (Ht > 0). So, each
is accompanied by an increase of the system’s entropy.
If the phase transition is exothermic (Ht < 0, as in freezing or
condensing), then the entropy change is negative. This decrease in
entropy is consistent with localization of matter and energy that
accompanies the formation of a solid from a liquid or a liquid from a gas.
If the transition is endothermic (Ht > 0, as in the melting and
vaporization), then the entropy change is positive, which is consistent
with dispersal of energy and matter in the system.
11
t
t
T
HS
Tt : transition temperature___
~0.5 eV/atom
s-Si ↔ l-Si
그림 G, H, & S
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The entropy change of transition:
large entropy change due to H-bonds
A wide range of liquids gives approximately the same entropy of
vaporization. Trouton’s rule
12
113
113
,
867.349
1030)rachloridecarbon tet(
1.838.111
1027.9)methane(
molJKS
molJKT
HS
b
mevap
11 109)water( molJKS
(page 122)
liquid)(solid / 1 ~ / 1 ~
vapor)(liquid / 10 ~ / 10~
moleculekmolRS
moleculekmolRS
B
B_____________________________________________
R = 8.314 J mol-1 K-1
http://bp.snu.ac.krFigure 3A.8
Suppose:
An energy qh (for example, 20 kJ) is supplied to
the engine, and
qc is lost from the engine (for example, qc = 15
kJ), and discarded into the cold reservoir.
The work done by the engine is equal to qh + qc
(for example, 20 kJ + (15 kJ) = 5 kJ).
The efficiency is the work done divided by the
heat supplied from the hot source. = 25%
13
Engine
Cold Reservoir
Engine
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3A.3(a) Reversible Heat Engine: The Efficiency
Since the temperature of a cold reservoir is lower than that of a hot one,
an overall increase of entropy can be produced even if qh is withdrawn
from the hot reservoir, and less than qh is transferred to the cold.
is zero.
Therefore, we are free to use some kind of device, an engine, to draw off
the difference as work.
The work that an engine may produce is:
The Carnot efficiency () is the ratio of the maximum work
generated to the heat absorbed.
14
0h
h
c
ctotal
T
q
T
qS
ch qq
h
chch
T
Tqqqw 1 'max
h
c
h T
T
q
w 1
'max
Reversible Engine
= Maximum Work
(3A.9)
Reversible change by an infinitesimal modification
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Example:
Calculate the Carnot efficiency (reversible heat engine) of an
internal combustion engine where Th = 3200 K and Tc = 1400 K.
Solution:
Compare with the practical efficiency ( 25%) of internal combustion
engines.
15
%5656.03200
14001
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Figure 3A.7
Step 1: Reversible isothermal expansion at temperature Th.
Step 2: Reversible adiabatic expansion in which the temperature falls from Th to Tc .
Step 3: Reversible isothermal compression at Tc.
Step 4: Reversible adiabatic reversible compression,
Restores the system to its initial state.
(p1, V1)
(p2, V2)
(p3, V3)
(p4, V4)
Isotherm
Isotherm
Adiabat
Adiabat
Pre
ssu
re, p
1
2
3
4
Tcold
Thot
16
Basic Structure of a
Carnot CycleIdeal Gas
Reversible
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The Carnot Cycle
All engines work on a cycle.
Step 1 [Isothermal at qh].
The gas absorbs heat (qh) from
the high-temperature resorvoir
(temperature Th) and expands
isothermally and reversibly
from V1 to V2.
(since isothermal, ideal gas)
(p1, V1)
(p2, V2)
(p3, V3)
(p4, V4)
Isotherm
Isotherm
Adiabat
Adiabat
Volume, V
Pre
ssure
, p
1
2
3
4
Tc
Th
17
0
ln
ln
1
2
1
2
U
V
VRTw
V
VRTq
h
hh (3A.14)
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Step 2 [Adibatic].
The gas expands adiabatically
and reversibly from V2 to V3; in
doing so, its temperature drops
from Th to Tc.
(p1, V1)
(p2, V2)
(p3, V3)
(p4, V4)
Isotherm
Isotherm
Adiabat
Adiabat
Volume, V
Pre
ssure
, p
1
2
3
4
Tc
Th
18
c
h
T
TV dTCUw
q 0
Ideal gas
dVV
UdT
T
UTVdU
TV
),( from
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Step 3 [Isothermal at qc].
The gas is compressed from V3 to
V4 while in thermal contact with
the low temperature resorvoir
(temperature Tc), isothermally
and reversibly.
(p1, V1)
(p2, V2)
(p3, V3)
(p4, V4)
Isotherm
Isotherm
Adiabat
Adiabat
Volume, V
Pre
ssure
, p
1
2
3
4
Tc
Th
19
0
ln
ln
3
4
3
4
U
V
VRTw
V
VRTq
c
cc
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Step 4 [Adibatic].
The gas is compressed
adiabatically and reversibly from
V4 to V1, warming in the process
from Tc to Th.
(p1, V1)
(p2, V2)
(p3, V3)
(p4, V4)
Isotherm
Isotherm
Adiabat
Adiabat
Volume, V
Pre
ssure
, p
1
2
3
4
Tc
Th
20
h
c
T
TV dTCUw
q 0
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Adding these together,
For an ideal gas, and for reversible and adiabatic expansion*
21
0cycle
lnlncycle
lnlncycle
3
4
1
2
3
4
1
2
U
V
VRT
V
VRTw
V
VRT
V
VRTqqq
ch
chch
. ;0*
lnln,
dTCdUpdVdwdUdq
V
VR
T
TC
V
i
f
i
f
mV
(2E.2b)
f
i
c
i
f
mVV
V
V
T
T
R
C
nR
Cc
lnln
,
Step 3
Step 3
Step 1
Step 1
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Apply the relation to the steps 2 and 4:
The efficiency of a Carnot engine is the ratio of the net work (-w)
to the fuel burned to provide the heat qh.
22
1
2
1
2
1
2
1
2
3
4
1
2
1
2
4
3
4
1
3
2
4
1
3
2,
ln
ln)(lnln
lnln
lnlnln
V
VRTq
V
VTTR
V
VRT
V
VRTw
V
V
V
V
V
V
V
V
V
V
V
V
V
VR
V
VR
T
TC
hh
chch
c
hmV
3
4
1
2 lnln)cycle(V
VRT
V
VRTw ch
h
c
h
ch
h T
T
T
TT
q
w1efficiency (3A.10)
(p1,V1)
(p2,V2)
(p3,V3)
(p4,V4)
Isotherm
Isotherm
Adiabat
Adiabat
Volume, V
Pre
ssu
re, p
1
2
3
4
Tc
Th
_
(same as page 14)
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Figure 3A.10
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 energy or entropy change around every individual cycle is zero, the integral
of the energy or entropy around the perimeter is zero too.
Volume, V
Pre
ssure
, p
Cancel
Survive
23
Any Reversible Cycle
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Brief Illus. 3A.6
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.
24
(skip)
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(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 of entropy.
(b) The process becomes feasible if work is provided to add to the energy stream. Then
the increase of entropy of the hot sink can be made to cancel the entropy decrease of the
cold source.
25
Reverse of
Carnot Cyclenot spontaneous
Brief Illus.
3A.6
(skip)
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The criterion for natural, spontaneous change solely
in terms of the properties of the system is:
i) Constant volume:
for spontaneous
ii) Constant pressure:
for spontaneous
26
dUdq V
dUTdST
dUdS 0
dHdq p
dHTdST
dHdS 0
3C Concentrating on the System (p. 131)
0T
dqdS
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Now one can define new thermodynamic functions for the criterion of
spontaneous process.
i) Helmholz function:
ii) Gibbs function:
At constant temperature,
A: maximum work function or work function
Consider an isothermal system changing reversibly and delivering
maximum work.
Since TdS = dqrev,
at constant temperature
It follows that if we know A for a process, we also know the
maximum amount of work that the system can do. 27
TSUA
TSHG
TdSdHdGTdSdUdA
TdSdwdqTdSdUdA
dwdqdU
revrev
revrev
revdwdA
Awrev
Chap. 3C.1(c) (skip)
_____________________________________________
____________
_
_ *****
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3C.1(e) Some Remarks on the Gibbs Function
At constant temperature
where dwrev is the maximum work of the system (expansion work
(-pdV) + other kinds of work).
0 (constant pressure)
At constant pressure and temperature, the change of the Gibbs
function in a particular process gives the maximum extra work.
This extra work is called the net work.
28
)()( pVddwdqpVddUdH
GGG
revrev
reactprod
pdVVdpdw
TdSpdVVdpdwdqTdSdHdG
rev
revrev
max,
max,
e
e
dwdG
pdVVdpdwpdVdG
Gwe max,
(skip)
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3B The Variation of Entropy with Temperature
If Cp is independent of temperature in the temperature range of
interest, we obtain
At some temperatures between 0 and T, the materials may change its
phase and absorb heat in the process.
In the vicinity of absolute zero:
29
f
i
p
if
f
iprev
revif
T
dTCTSTS
dTCdqT
dqTSTS ,
i
f
pi
f
ipif
T
TCTS
T
dTCTSTS ln
dTT
C
T
HdT
T
C
T
HdT
T
CSTS
T
T
gas
p
b
evapT
T
liq
p
f
meltT
solid
p
b
b
f
f
0
0
3aTCp
_______After Measuring Cp Experimentally,
We Can Identify S(T), H(T) & G(T).
(constant P)
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Figure 3B.1
The determination of entropy from heat capacity data.
(a) The variation of Cp/T with the temperature for a sample.
(b) The entropy, which is equal to the area [in (a)] beneath the upper curve
up to the corresponding temperature, plus the entropy of each phase
transition passed.
Cp/T
Tf Tb T
Melt
Bo
il
fusS
vapS
(a)
(b)
Tf Tb T
S
S(0)
0
Solid GasLiq
uidD
eb
ye
ap
pro
xim
atio
n
30
3aTCp
temperature axis: not linear
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At absolute zero, all quenchable energy has been quenched. In the
case of a perfect crystal at absolute zero, all the atoms are in a regular,
uniform array, and the absence of disorder and thermal chaos suggests
that the entropy is the same in every case.
The Third Law of Thermodynamics:
All perfect crystals have the same entropy at absolute zero.
If the value zero is arbitrarily ascribed to the entropies of the
elements (in the perfect crystalline form stable at T = 0 K),
then all perfect crystalline compounds also have zero entropies at
absolute zero.
31
Chapter 3B.2
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n: composition
Combining the first law and second law
The first law says
For reversible processes in the absence of any kind of work other than
pV-work,
Master equation or fundamental equation
32
),,( nTpGG
dwdqdU 1
2
pdVTdSdU
TdSdq
pdVdw
rev
rev
3
4
Chapter 3D Combining the First and Second Laws
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Comparing eqs (4) and (5), we obtain relationships.
Equation (6) enables a temperature to be expressed solely in terms of
extensive thermodynamic quantities. If the volume is constant, the
relation states that the ratio of the change in energy (a First Law concept)
to the corresponding change in entropy (a Second Law concept) is equal
to the temperature of the system, whatever its nature or composition.
33
(5)
),(
dVV
UdS
S
UdU
VSUU
SV
(7)
(6)
pV
U
TS
U
S
V
pdVTdSdU
pdVTdSdU
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3.8(a) Various Maxwell Relationships
U : exact differential
By using the relation No. 4,
Maxwell relation
G : exact differential
(4)
Maxwell relation
34
pdVTdSdU
(8) VS S
p
V
T
(9)
SdTVdpdG
pdVTdSdUSdTTdSVdppdVdUdG
TSHG
(10) Tp p
S
T
V
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Note:
35
http://bp.snu.ac.kr 36
relation Maxwell
aldifferentiexact :
(11) and (4) equations From
VT T
p
V
S
A
SdTpdVdA
SdTTdSdUdA
TSUA
(11)
(12) Maxwell relation
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relation Maxwell
aldifferentiexact :
(13), and (4) equations From
pSS
V
p
T
H
VdpTdVdH
TdS-pdVdU
VdppdVdUdH
PVUH
(14)
(13)
Maxwell relation
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3D.1(b) The Variation of Internal Energy with Volume
Derivation
Dividing equation (5) by dV and imposing constant temperature,
Combining equations (6), (7) and (15), one can obtain
Substitution of equation (12) into equation (16)
38
(3D.6)
pT
pT
V
U
VT
5 ),( dVV
UdS
S
UVSdU
SV
15 STVT V
U
V
S
S
U
V
U
16 pV
ST
V
U
TT
VT T
p
V
S
pT
pT
V
U
VT
____________________
(Sec. 2D.2)
Maxwell relation
____________________
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Example 3D.2
Show thermodynamically that is zero for an ideal gas,
and compute its value for a van der Waals gas.
Solution:
For an ideal gas, pV=nRT.
Combining two equations,
The equation of state of a van der Waals gas
39
TV
U
pT
pT
V
U
V
nR
T
p
VTT
0
p
V
nRT
V
U
T
2
2
V
an
nbV
nRTp
2
2
2
2
V
an
V
an
nbV
nRT
nbV
nRTp
nbV
nRT
V
U
nbV
nR
T
p
T
V
pT
pT
V
U
VT
ppt 2-61
____________________
Definition of internal pressure
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3D. 2 Properties of the Gibbs Free Energy
Master equation
40
20 ,
obtains one (19), and (18) eqs Combining
19
,
18
(17), and (4)equation Combining
17
ST
GV
p
G
dTT
Gdp
p
GdG
TpGG
SdTVdp
SdTTdSVdppdVpdVTdSdG
SdTTdSVdppdVdU
SdTTdSdHdG
pVUH
TSHG
pT
pT
(3d.8)
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Figure 3D.1
The variation of the Gibbs energy of a
system:
- with temperature at constant P
- with pressure at constant T
The slope of the former is equal to the
negative of the entropy of the system
and that of the latter is equal to the
volume
41
Curvature for T
Curvature for P
G(T,P)
PT
i
iidnVdpSdTdG
(Chap. 5)
X
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Figure 3D.2
The variation of Gibbs energy with the
temperature is determined by the entropy.
Because the entropy of the gaseous phase of
a substance is greater than that of the liquid
phase, and the entropy of the solid phase is
smallest, the Gibbs energy changes most
steeply for the gas phase, followed by the
liquid phase, and then the solid phase of the
substance.
42
X
ST
G
p
curvature
curvature
X
i
iidnVdpSdTdG
S
L
G
X
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3D.2(b) The Change of Gibbs Energy with Temperature
Equation (20) implies that, as S is always a positive quantity, G must decrease
when the temperature is raised at constant pressure (see Figure 3D.1).
Combining eqs (20) and (21),
Combining eqs (22) and (23), one obtains
(23)
Gibbs-Helmholtz equation 43
(3D.10)
21 T
GHSTSHG
ST
G
p
T
G
T
G
T
TTG
T
G
TT
TG
T
HG
T
G
p
ppp
p
1
11/
22
24 /
2T
H
T
TG
p
____________________G(T) → H(T)
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For a chemical reaction,
initial state (reactants)final state (products)
44
(3D.11)
25
/
/
2
22
T
H
T
TG
T
H
T
H
T
TG
p
if
p
if GGG
2
/
T
H
T
TG
p
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3D.2(c) The Change of Gibbs Free Energy with Pressure
For a reaction in which Gi changes to Gf ,
where V= Vf Vi.
Integration of the above equation results in
For liquid or solid, volume depends only very weakly on the pressure:
Except at very high pressure, is very small and virtually no
error is introduced.
For solids and liquids,
G is virtually independent of pressure for solid or liquid.
(*)
45
Vp
G
T
,Vp
G
T
f
i
p
pif dppVpGpG
mifimfm VpppGpG
.imfm pGpG
mif Vpp
(3D.12b)
~ ( ≠ 지구과학)
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Figure 3D.3The variation of the Gibbs energy with the
pressure is determined by the volume of
the sample.
Because the volume of the gaseous phase
of a substance is greater than that of the
same amount of liquid phase, and the
volume of the solid phase is smallest (for
most substances), the Gibbs energy
changes most steeply for the gas phase,
followed by the liquid phase, and then the
solid phase of the substance.
Because the volumes of the solid and
liquid phases of a substance are similar,
they vary by similar amounts as the
pressure is changed.
46
Vp
G
T
T < Tm
Curvature
~5×1022 atoms / cm3
for typical solid or liquid
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Figure 3D.4
The difference in Gibbs energy 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.
47
pVG
VpppGpG
dppVpGpG
mm
mifimfm
p
pif
f
i
Solid or Liquid
i
iidnVdpSdTdG (Chap. 5)
http://bp.snu.ac.kr
For an ideal gas,
Inserting the above equation into equation (*) yields
(*)
48
p
nRTV
f
i
p
pif dppVpGpG
f
i
p
pif dp
p
nRTpGpG
i
f
ifp
pnRTpGpG ln Ideal Gas
constant T(3D.15)_________
(3D.12b)
Ideal Gas + Constant T
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Figure 3D.5
The difference in Gibbs energy for
a perfect gas at two pressures is
equal to the area shown below the
perfect-gas isotherm.
/V nRT p
Vo
lum
e, V
Pressure, ppi pf
Vdp
49
Isothermal Compression
i
f
ifp
pnRTpGpG ln)()(
Ideal Gas
constant T
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3D.2(c) Chemical Potential of an Ideal Gas
Standard state of an ideal gas = 1 atm
Gibbs function at 1 atm = G 0
At any other pressure p,
For one mole of material,
Molar Gibbs Free Energy
We write : Chemical Potential
50
(3D.15)
i
f
ifp
pnRTpGpG ln)()(
)/ln()( 0 atmpnRTGpG
)/ln()( 0 atmpRTGpG mm
pRTTTp ln)(),( 0
pGm
For a single component, chemical potential = Gm
_____________ Ideal Gas
iii pRTTppppTp ln),(,...),,,,( 0321
(later)____________Ideal Gas
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
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Figure 3D.6
The molar Gibbs free energy of a
perfect gas is proportional to ln p.
51
pRTTTpGTp m ln)(),(),( 0
)(0 T
p = 1 (unitless)
Single Component + Ideal Gas
___
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Real Gases: the Fugacity
: applicable to ideal gases
For real gases, we are to determine the pressure dependence of the
volume of a sample of real gas and to calculate G(p) from
by numerical integration.
The quantity f plays the role of the pressure, but it has a value which
ensures that the chemical potential is given by the last equation
whatever the pressure. Here f is called as the fugacity of gas.
The following equation is true for all substances.
Hence, for a real gas,
52
(skip)
)/ln()( 0 atmpRTGpG mm
dppV )(
)/ln( )/ln( 00 atmfRTatmfRTGG mm
dppVpGpGdppVpGpGf
i
f
i
p
pmim
p
pfmif )()()( ;)()()(
i
f
i
real
f
realp
p
real
mf
fRTTpTpdpV
f
i
ln),(),(
http://bp.snu.ac.kr
For an ideal gas,
The difference of the two equations
53
(skip)
i
f
i
real
f
realp
p
real
mf
fRTTpTpdpV
f
i
ln),(),(
p
RTV
dppVpVRTp
f
p
fp
dppVpVRTpf
pf
p
pRT
f
fRTdppVpV
ideal
m
pideal
m
real
m
i
ii
p
p
ideal
m
real
m
ii
ff
i
f
i
fp
p
ideal
m
real
m
f
i
f
i
0)()(
1ln
1 0
)()(1
/
/ln
lnlnln)()(
http://bp.snu.ac.kr
Figure 3.25
The molar Gibbs free energy of a
real gas. As p 0, the molar Gibbs
free energy coincides with the value
for a perfect gas. When attractive
forces are dominant (at intermediate
pressures), the molar Gibbs free
energy is less than that of a perfect
gas and the molecules have a lower
‘escaping tendency’. At high
pressures, when repulsive forces are
dominant, the molar Gibbs free
energy of a real gas is greater than
that of a perfect gas. Then the
‘escaping tendency’ is increased.
54
(skip)
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The Real Volume-Pressure Dependence
Compression factor
For a real gas, z =z (p, T )
55
(skip)
p
zRTV
RT
pVz real
m
real
m
dppVpVRTp
f pideal
m
real
m
0
1ln
p
p p
dpp
Tpz
dpp
Tpz
RT
RTdp
p
RT
p
zRT
RTp
f
0
0 0
1,
1,1ln
p
dpp
Tpzpf
0
1,exp
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Standard States of Real Gases
For an ideal gas, p = 1atm: standard state
For a real gas, f = 1atm: standard state
g : fugacity coefficient
The chemical potential of
an ideal gas
Deviation from ideality
56
(skip)
pf
atmfRTp /ln0
ln/ln0 RTatmpRTp
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Open System and Changes of Composition
At constant pressure and temperature, Gibbs function has no changes.
However, when composition change, Gibbs function changes.
where n1, n2, … are the amount of substances 1, 2, …present.
Let the composition be fixed and permit only the temperature and
pressure to change.
(1)
(2); (3)
57
Derivation in
Chap. 5
VdpSdTdG
,,,,,, 321 nnnTpGTpG
2
,,,2
1
,,,1
,,,,
3132
2121
dnn
Gdn
n
G
dTT
Gdp
p
GdG
nnTpnnTp
nnpnnT
Vp
G
nnT
21 ,,
ST
G
nnp
21 ,,
i
iidnVdpSdTdG ________
i
iinG _____
(선행학습 )
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Suppose now that substance 1 is the only substance present.
G the amount of material present
At constant temperature and pressure,
Therefore, the chemical potential of species 1 is the measure of how the
Gibbs function of species 1 changes when the amount present is varied.
More than One Component:
Partial Molar Gibbs Free Energy for atom i
= Chemical Potential for atom i
(4)
58
1111 nGnG m
11dndG
1
,1
Tpn
G
jnpTi
jin
GnpT
,,
,,
___________ (as a definition)
in Chap. 5
(선행학습 )
_
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Substituting eqs (2), (3) and (4) into eq (1), one obtains
More general form
the master equation of chemical thermodynamics
59
(later)
(reversible)
2211 dndnVdpSdTdG
2
,,,2
1
,,,1
,,,,
3132
2121
dnn
Gdn
n
G
dTT
Gdp
p
GdG
nnTpnnTp
nnpnnT
___________i
iij dnVdpSdTnpTdG ),,(
(선행학습 )
진짜물질
Derivation in
Chap. 5
http://bp.snu.ac.kr 60
Problems from Chap. 3
3A.2
3A.1(b) 3A.2(b) 3A.3(b) 3A.5(b) 3A.7(b)
3A.2
3B.8
3D.2
3D.1(b) 3D.2(b) 3D.3(b)
3.2
Entropy (Boltzmann Formula) + 열역학제2법칙 짱중요
Binary alloy에서 A and B 측정방법: 선행학습 (Chap. 5)_____________