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8/19/2019 Kinetics and Thermo by Amarendra Vijay
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August 04, 2006 1
Thermodynamics and Kinetics
Amrendra VijayDepartment of ChemistryIIT Madras, Chennai 600 036
Lecture 01
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why Thermodynamics ? Broadly speaking, there are two approaches of Science:
Holistic: This approach, also known as Phenomenology, is easy tounderstand and utilize in practice. Here, we attempt to rationalize ourobservations, using a few physically meaningful variables and formulateuniversal Laws. These phenomenological Laws have predictive power,limited to the range of their validity. Classical Thermodynamics hasgrown out of this approach and its utility today is pervasive.
Reductionism: Here, we attempt to determine Laws, governing theultimate constituents of matter (for example, electrons) and we hopethese Laws will eventually help to rationalize everything we observearound us and beyond. This is the Holy Grail of Science. Emergence ofwhat we call Quantum Mechanics is a fine example of Reductionism.Laws of reductionism, as we know today, are, unfortunately, toocomplicated to use and hence its utility is limited for the world at large.
Present Status: Axiomatically, phenomenological Laws (ClassicalThermodynamics) are derivable from the Laws of reductionism(Quantum Mechanics), but synthesis of such a knowledge (QuantumThermodynamics) is not yet complete.
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Thermodynamicsβ’ Basic concepts and the First Law
β’ Suggested Reading: Physical Chemistry--- P.W. Atkins
Chapters 2 and 3
or --- D.A. McQuarrie
Chapter 19
or --- Any other author
and --- Ask Professors at IITM
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System and Surroundings A thermodynamic system is the content of a
geometrical volume of macroscopic size.
System
Surroundings
B o u n d a r y
Examples
A piece of matter, a sample of gas,chemical reactions in a vessel,
a sample of human population etc.
For molecular applications
cmacroscopi : cm1 size
cmicroscopi : cm10 size 6
>
< β
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August 04, 2006 5
Thermodynamic Variables Thermodynamic variables are quantities of a macroscopic
system, which can be measured experimentally.
Examples: Pressure, P; Volume, V; Energy, E; Temperature,T; magnetic field, H and so forth.
Types: (a) Extensive: The value of an extensive variabledepends on the size of the system; that is, total = sum ofparts. For example, Mass, Volume etc. (b) Intensive: Thevalue of an intensive variable does not depend on system size.For example, pressure, temperature etc. Note, an extensivevariable can be converted into an intensive variable, if wescale (say, divide) the extensive variable by the size of thethermodynamic system.
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Thermodynamic States A thermodynamic state is specified by a
set of values of all thermodynamicvariables (temperature, susceptibility,pressure etc) necessary for the descriptionof the macroscopic system.
Example: A sample of gas enclosed in a
box at constant temperature and volume.Here, thermodynamic variables are thetemperature and the volume.
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Thermodynamic Equilibrium A thermodynamic equilibrium prevails
when the thermodynamic state(characterized by thermodynamicvariables, like temperature, pressure etc)
does not change with time. And we say,the system is in thermodynamicequilibrium.
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Equation of State This is a functional relation among thermodynamic
variables for a system in equilibrium.
Example: For an ideal gas,
Equation of state is assumed to be known, as a partof the specification of the thermodynamic system.
( ) 0,, =T V P f (a function of thermodynamic variables)
( ) 0),,( =ββ‘ nRT PV T V P f
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Thermodynamic Transformation This is a change of thermodynamic state of the
system. If the initial state is in equilibrium, atransformation can be brought only by changingexternal conditions of the system (like, heating thesystem applying external pressure and so forth.
Quasi-adiabatic transformation: Here, theexternal conditions change so slowly that at anymoment the system is approximately in equilibrium.
Reversible Transformation: The transformation,here, retraces its history in time, when the external
condition retraces its history in time.
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August 04, 2006 10
Types of Thermodynamic systems Three types:
Energy
Matter system1. Open system:
2. Closed system:
3. Isolated system:
Surroundings Surroundings
Energy
Matter systemSurroundingsSurroundings
Energy
Matter system
SurroundingsSurroundings
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Energy Energy is the capacity to do work. In
thermodynamics, the total energy of a system iscalled the internal energy.
When the work is done on a system, the capacity ofthe system to do work increases and we say, theenergy of the system has increased.
When the work is done by the system, the capacityof the system to do work decreases and we say,the energy of the system has reduced.
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August 04, 2006 12
Heat and Work
Heat is the transfer of energy between the system and thesurroundings that makes use of the thermal motion.
Work is the transfer of energy that between the system andthe surrounding that makes use of the organized motion.
Disordered, random or Thermal motion Organized motion
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August 04, 2006 13
Differentiation Consider the Taylor series expansion of function of 1 variable:
Notice, is independent of x and it is valid for all x,independent of we choose to fix. If is very small,then we may truncate the series as follows:
As we deal with a truncated series, we may as well like tomodify as we change x. That is, we makedependent on x, and write as follows:
( ) ( )[ ] ( ) ( ) ( ) β+β
β₯β¦
β€
β’β£
β‘
β
β+β
β₯β¦
β€
β’β£
β‘
β
β+β
β₯β¦
β€
β’β£
β‘
β
β+=
====
x x x
f
! x x
x
f
! x x
x
f x f x f
x x x x x x x x
L3
03
32
02
2
0000
0 3
1
2
1
( ) ( )[ ] ( ) dx x
f df x x
x
f x f x f
x x x x 00
00==
β₯β¦
β€β’β£
β‘
β
β=βββ₯β¦
β€β’β£
β‘
β
β=β
( ) dx x x =β
0
0/
x x
mm x f =ββ
0 x
0/
x x
mm x f =ββ 0/ x x
mm x f =ββ
dx x f df β β βββ β ββ=
In thermodynamics, these partial derivativesdefine various thermodynamic variables
of our interest.
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August 04, 2006 14
Differentiation
( ) ( )[ ] ( )
( ) LL
LL
2
1
,,
22
1 1
1
,21,21
+ββ₯β¦
β€β’β£
β‘
ββ
β+
ββ₯β¦
β€β’β£
β‘
β
β+=
== =
= =
=
ββ
β
mm
y xk m
N
m
N
k
N
m
mm
y xm y x N N
y x x x
f
y x
x
f x x x f x x x f
k k
k k k k
( ) ( ) ( )mm y x
N
m m
N N y x x
f y y f x x f
mm
ββ₯β¦β€β’
β£β‘
ββ=ββ
==β
1
11 LL
( ) m y x
N
m m
mm
y x
N
m m
dx
x
f df y x
x
f f
mmmm ====
ββ β₯β¦
β€β’β£
β‘
β
β=βββ₯
β¦
β€β’β£
β‘
β
β=ββ
11
3
3
2
2
1
1
dx x
f dx
x
f dx
x
f df ββ
β
ββββ
β
β
β+ββ
β
ββββ
β
β
β+ββ
β
ββββ
β
β
β=
Neglect
for three variables.
Consider the Taylor series expansion of an N-variable function:
for an
infinitesimal
change
constant.isif 23x3
1
x1 22
xdx x
f dx
x
f df ββ
β
ββββ
β ββ+ββ
β
ββββ
β ββ=
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August 04, 2006 15
Integration
β x
β
y
( ) ( )β«=max
min
y
y
x,ydy f xgβ’ Consider, ymax
ymin
β’ If the value of does not depend on the path we choose forintegration and depends only on the initial and the final state,
E is called a state function and dE is called an exact differential.
β«=β final
initial
dE E
E β
β’ If the value of does depend on the path we choose forintegration, then E is called a path function and dE is called an
inexact differential.
E β
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August 04, 2006 16
First Law of Thermodynamics
4 4 34 4 21434214 4 4 34 4 4 21
Uwq
EnergyInternal
inChange
systemaon
donework
heatas
nsferedenergy tra
β
=+
β’ In thermodynamics, the total energy of a system is called itsinternal energy. This is a state function. Only the change ininternal energy involved in a thermodynamic transformation isof any physical significance.
β’ For an isolated system, q = 0 and w = 0. Hence, the change ininternal energy is zero; that is, the energy is conserved.
β’ By convention, if w > 0 and q > 0 we say the energy istransferred to the system as work or heat. If w < 0 and q < 0,we say the energy is lost from the system.
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August 04, 2006 17
Computation of Work Work done by the system to move an object a
distance dZ, against the external force, F:
dW = - F dZ (negative sign means: if the systemdoes the work, its internal energy decreases.)
Force is pressure, P, per unit area, A, and hence,
dW = - P A dZ = - P dV (volume)
If the external pressure is zero, then w = 0. Thatis, no work is done. Example: Free expansion of a
gas in a vacuum.
For an ideal gas, P = n R T/V
β β
β
ββ
β
β=β=β β« VinitialVfinal
lnnRTV
dV
nRT w
Vfinal
Vinitial
β«β=βV2
V1dVP w
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August 04, 2006 18
Internal Energy, U Consider an ideal gas, for which the equation of state is,
P=nRT/V. That is, a thermodynamic state of an ideal gas is
characterized by three thermodynamic variables (pressure, P,(temperature, T, and the volume, V); but, only two variablesare independent, as the third one (for examples, P) can beexpressed in terms of other two (volume, V and temperature,
T). This means, the internal energy, U of an ideal gas may beconsidered as a function of any two thermodynamic variables,appearing in the equation of state. Hence, we have:
Remember, this equation retains only first two term of an
infinite Taylor series, and hence this is only an approximation.
( ) dT TU
dVV
UdU TV,UU β β
βββ
β
β
β+β β
βββ
β
β
β=ββ‘
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Internal Energy (Heat Capacity) Reference:
If V is constant, then dV = 0
For a finite change,
What is the physical significance of specific heat ? Should thisbe temperature dependent or temperature independent ?
dT T
U dV
V
UdU β
β
βββ
β
β
β+β
β
βββ
β
β
β=
dT CdTT
UdU v
C
V
v
=β β
βββ
β
β
β=β
43421
Heat Capacity at constant volume
βTCβU
βT
βULimit
T
UC v
V0βTV
v =ββ
β
ββ
β
β =β
β
ββ
β
β
β
β=
β
TCUR f ββ
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August 04, 2006 20
Heat Capacitycontinued
β¦
TCU :Ref v β=β
At constant volume, the internal energy of a system isproportional to the temperature; that is, internal energy
increases when T is raised.
Heat capacity is an extensive quantity. To make it anintensive one, we divide it by number of moles and call it
molar heat capacity. If we divide it by the mass of thesubstance (in gram), it is called Specific Heat.
A large Heat Capacity implies: if we heat a system, theincrease in temperature will be very small --- that is, thesystem has a large capacity for Heat.
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August 04, 2006 21
Internal Energy (Internal Pressure) Reference:
If T is constant, then dT = 0.
For a perfect gas, the internal pressure is zero.
dT T
U dV
V
UdU β
β
βββ
β
β
β+β
β
βββ
β
β
β=
dV Ξ
dVV
U
dU T
Ξ
T
T
=β β
βββ
β
β
β=β
321
Internal Pressure
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August 04, 2006 22
Internal Energy ( ) ( )TP,UTV,U β
What if we consider internal the internal energy of an ideal gasto be function of temperature, T, and pressure, P ?
Recall:
Divide by dT and impose the condition of constant pressure,
For a perfect gas, the internal pressure is zero and hence,
dTC dVΞ dU dTT
U dV
V
UdU VT +=ββ
β
βββ
β
β
β+β
β
βββ
β
β
β=
Internal Pressure Heat Capacity
VTV
P
T
P
C VΞ Ξ±C
T
V Ξ
T
U+=+β
β
ββ
β
β
β
β=β
β
ββ
β
β
β
β
PT
V
V
1Ξ± where, β
β
ββ
β
β
β
β=
Expansion CoefficientInternal Pressure
VP β β β β V
T
UC T
U β βββ ββ==β βββ
ββ
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August 04, 2006 23
Enthalpy, H It is a state function, defined as H = U + P V, where U is the
internal energy of the system, P is the pressure and V is the
volume. Physical Significance: Let us make a small change in the state
of the system:
and neglect term. We then obtain:
If the heating occurs at constant pressure then,
That is, the change in Enthalpy is the Heat supplied at constant
pressure, so long as the system does no additional work.
( ) ( ) ( ) ( )βVVβPPβUUβHH ++++=+βVβP
PVq
βPVβVPβUβH
β+β=
++=
βVP
βU
βW
βU
βq
+=β=
From the first law, if the systemis in mechanical equilibrium ata pressure, P then
q.βH β=
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August 04, 2006 24
Enthalpy Consider Enthalpy, H as a function of the pressure, P and
the temperature, T of the system.
At constant pressure, dP=0 and hence:
Because, the change in Enthalpy is the Heat supplied atconstant pressure, we have This relation
allows us to design experiment for measuring
( ) dT T
H dP
P
HdH TP,HH β
β ββ
β β
ββ+β
β ββ
β β
ββ=ββ‘
dTCdTT
H
dH P
C
P
P
=β β
β
ββ
β
β
β=
321
dq.dTCdH P ==
pressure.constantatCapacityHeattheisCP
.CP
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August 04, 2006 25
Enthalpy Reference:
At constant temperature, dT=0 and hence,
Thus, we can write:
( ) dT T
H dP
P
HdH TP,HH β
β
βββ
β
β
β+β
β
βββ
β
β
β=ββ‘
dPP
H
dHTβ β
β
ββ
β
β
β
=
dTCdPP
H dH p
T
+β β
βββ
β
β
β=
dTCdPH
dH:Reference +β β
ββ β
=
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August 04, 2006 26
Enthalpy as a function of temperature Divide by dT and impose constant volume:
Use Chain relation, to obtain,
This says as how enthalpy changes as a function oftemperature, if the volume of the system is held constant.
dTCdPP
dH :Reference pT
+β β
ββ β
P
VTV
CT
P
P
H
T
H+β
β
βββ
β
β
ββ
β
βββ
β
β
β=β
β
βββ
β
β
β
( ) 1y
z
z
x
x
y then,yx,zif
xyz
β=ββ β
ββββ
β
β
ββ
β
βββ
β
β
ββ
β
βββ
β
β
β
ΞΊ
Ξ±
P
V VΞ±
P
V
T
V
P
V
V
T
T
P
T
1
T
1
TP
1
T
1
PV
=β₯β¦
β€β’β£
β‘β
β
βββ
β
β
β=β₯
β¦
β€β’β£
β‘β
β
βββ
β
β
ββ₯β¦
β€β’β£
β‘β
β
βββ
β
β
ββ=β₯
β¦
β€β’β£
β‘β
β
βββ
β
β
ββ₯β¦
β€β’β£
β‘β
β
βββ
β
β
ββ=β
β
βββ
β
β
β ββββ
T
TP
V
V
1ΞΊ β
β ββ
β β
βββ=
Isothermal compressibility
P
C
PH
1
P
1
HT
CΒ΅T
H
P
T
H
T
T
P
P
H
P
β=β₯β¦
β€β’β£
β‘β β
βββ
β
β
ββ₯β¦
β€β’β£
β‘β β
βββ
β
β
ββ=β₯
β¦
β€β’β£
β‘β β
βββ
β
β
ββ₯β¦
β€β’β£
β‘β β
βββ
β
β
ββ=β
β
βββ
β
β
β ββ
43421
HP
TΒ΅ β
β
βββ
β
β
β= Joule-Thomson
Coefficient
P
T
P
T
P
VCΞΊ
Ξ±Β΅1CΞΊ
Ξ±
CΒ΅T
H
ββ β
β
βββ
β
β=+β=β β
βββ
β
β
β
β΄
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August 04, 2006 27
Relation between Heat Capacities Definition: and
For a perfect gas, the internal pressure is zero and
V
vT
UC β
β
βββ
β
β
β=
P
PT
HC β
β
βββ
β
β
β=
U: Internal energy
H: Enthalpy = U + P V
[ ] [ ]( ) VΞ PΞ±
VPTCT
U
C PVUTCC
T
P
V
P
V
P
VP
+=
β β
β
ββ
β
β
β
+ββ β
β
ββ
β
β
β
=ββ β
β
ββ
β
+β
β
=β
VPΞ±T
V P
P
=β β
βββ
β
β
β
Expansion
Coefficient
VΞ Ξ± T
Internal Pressure
( )
PT
P TΞ because,
T
P VTΞ±
VΞ PΞ±CC
V
T
V
TVP
ββ β
βββ
β
β
β=β
β
βββ
β
β
β=
+=ββ΄
[ ] R nCC T)R nPV(using nR VP VP =ββ==β
β
β
β β
T P β β β
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August 04, 2006 28
Finally Read a Physical Chemistry book, or any
book which has a chapter on First Law ofThermodynamics.
Try to understand some solved problems,
given in the book and solve otherproblems given in the book.
In case of difficulty, come and discusswith me.
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August 07, 2006 1
Thermodynamics and Kinetics
Amrendra VijayDepartment of ChemistryIIT Madras, Chennai 600 036
Lecture 02
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August 07, 2006 2
Preface
Thermodynamic state of a system is specified by the value of aset of thermodynamic variables. For example, the state of anideal gas is specified by pressure, volume and temperature.
Change of one thermodynamic state into another, is calledThermodynamic transformation. Broadly speaking, there arethree types of transformations:
Induced: External work is required here. For example, we might
heat the system or pass an electric current. Spontaneous: No external work is required. It happens all by
itself and is usually irreversible.
Rare events: They are due to fluctuations in thermodynamicvariables.
To understand spontaneous processes in nature, the concept ofenergy is not sufficient and we require an additional conceptcalled Entropy.
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August 07, 2006 3
Notations
is infinitesimal change.
refers to a finite change.
As integration is essentially a summation process, afinite change can be thought of as a sequence of
infinitesimal transformations. we will henceforth use rather loosely and
the meaning should be clear from the context.
( )12F
FFF βF dF
2
1
ββ‘=β« if F is a state function
( ) pathF
FβF dF
2
1
=β«if F is not a state function. Subscript denotes thename of the path, for example, reversible (rev),chosen for thermodynamic transformation.
dF
Fβ
FanddF β
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August 07, 2006 4
Adiabatic Processes Adiabatic Process: Here, the external
conditions change so slowly that no energy istransferred as heat, between the system and thesurroundings.
System
Surroundings
External conditions 0βq =For adiabatic transformations
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August 07, 2006 5
Reversible Processes Reversible Process: Here, a thermodynamic
transformation is able to retrace its history in
time, when external condition retraces its historyin time.
The system is
an Ideal GasVT,,Pgas
extPextP
compression expansion
0βPPP gasext ββ‘βFor a reversible process,
equilibrium
Surroundings
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August 07, 2006 6
Adiabatic vs Reversible Adiabatic processes are reversible. But, not all
reversible processes are adiabatic.
We will get a clear picture of these thermodynamictransformations, from the notion of Entropy and itsvariation with time.
Before we introduce Entropy, we will consider anumber thermodynamic transformations, for theIdeal Gas. We use ideal gas as an example, because
its Equation of state is simple: PV = nRT.
Remember: (a) For adiabatic process,
(b) For reversible process,
0βq =
systemext PP =
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August 07, 2006 7
Work done for Expansion/Compression of an ideal gas Situation-1: Temperature remains constant (Isothermal), and
hence only the volume can change from to
The equation of state for an ideal gas is:
For compression, for example, Theconvention then says: we have done work on the system andits energy will increase.
1V .V2
dW = - F dZ = - P A dZ = - P dV
β«β« β=ββ=2
1
2
1
V
Vgasrev
V
Vext dVPw dVPw β
ββ
β
β
βββ
β
β β
tion.transforma
reversibleafor,PP gasextQ
TR nVPgas =
( ) ]isothermalande[reversibl V/VlnnRT
V
dV nRT w 12
V
Vrev
2
1
β=β=β β«
0.w ,VV rev12 >β<
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August 07, 2006 8
Work done for Expansion/Compression of an ideal gas Situation-2: Both Temperature and Volume change,
from
Consider the process to be adiabatic;
that is,
Then, the first law says,
( ) ( ).T,V toT,V 2211( )11 TV ( )12 TV
( )22 TV.0βq =
( )dTTCdUdw dwdqdU Vrevrevrev ==β+=
( )β«=β΄
2
1
T
T Vrev dT TCβ
w
A relation between volume and
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August 07, 2006 9
A relation between volume and
temperature for an ideal gas Consider an adiabatic transformation; that is,
So, the first Law says,
For an ideal gas, P = n R T/V, and hence
Assume, the Heat capacity to be Temperature independent.
β«β« β=β΄2
1
2
1
V
V
T
T
V
VdV
TdT
nR C
( )V
dVnR TdTTC V β=
VV nR/C
2
112
/nR C
1
2
2
1
VVTT
TTln
VVln ββ
β βββ
β β =βββ
β βββ
β β =ββ
β βββ
β β
We will use this for studyingEntropy change in a Carnot Engine.
.0βq =
( ) PdVdTTC dUdw dwdqdU Vrevrevrev β=β=β+=
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August 07, 2006 10
Adiabatic Process for ideal gases Consider the following thermodynamic transformation.
To go from Blue to Red, we have three different pathways:
We will now compute the change in internal energy, change inwork done and change in heat, using the first law and other
ideas we developed in the first lecture.
( )111 TVP( )321 TVP
( )122 TVP
( )223 TVP
D
EA
CB
β’ Direct path, A
β’ Path B + Path Cβ’ Path D + path E
βV
β
P
( ) dT TCdU V=β’ For ideal gases, we know:
( ) ( )TVPTVP β
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August 07, 2006 11
Adiabatic process for ideal gases Path A Temperature does not change.
Hence, the change in energy,
From the first law then, we have,
We already know the reversible work done for ideal gases,
Hence,
( )111 TVP
( )122 TVPA
βV
β
P
( ) ( )122111 TVPTVP β
( ) 0dT TCdU V ==β΄
revrevrevrevrevrev dwdqdwdq0 dwdqdU β=β+=β+=
ββ β
β
βββ
β β=β
1
2
1rev V
V
lnnRTw
0U =β
ββ β
ββββ
β β=
1
21rev
V
VlnnRTβw
ββ
β
βββ
β
β =
1
21rev
V
VlnnRTβq
β’ Change in internal energy :
β’ Change in heat :
β’ Reversible work done :
( ) ( ) ( )122223111 TVPTVPTVP ββ
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August 07, 2006 12
Adiabatic process for ideal gases Path B+C
( ) ( ) ( )122223111 TVPTVPTVP ββ
As we consider only adiabatic processes.
Temperature changes from
Then, from the first law, Path C has no volume change, so the work,
But the temperature changes from
Then, from the first law,
For total, add contributions from path B and C. Hence,
( )111 TVP
( )223 TVP
CB
βV
β
P
( ) 0βq have,weBrev =
21 TT β
( ) ( )β«=β΄2
1
T
TVB dT TCβU
( ) ( )BBrev β
Uβ
w =
( ) 0βwCrev =
.TT 12 β ( ) ( )dTTCβU1
2
T
TVC β«=β
β’ Change in energy :
β’ Heat Change:
β’ Reversible work: ( )β«=2
1
T
TVrev dT TCβW
( ) ( )CC
rev βUβq =
( )β«=
1
2
T
T Vrev dT TCβ
q
( ) ( ) 0dT TCdT TCβU1
2
2
1
T
TV
T
TV =+= β«β«
( ) ( ) ( )122321111 TVPTVPTVP ββ
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August 07, 2006 13
Adiabatic process for ideal gases Path D+E In path D, temp. changes from
Pressure is constant, so the reversible work,
Then, from the first law, In path E, volume is constant. So, the reversible work,
Temperature changes from
Then, from the first law,
For total, add contributions from path D and E. Hence,
( )111 TVP( )321 TVP
( )122 TVP
D
E
βV
β
P
( ) ( ) ( )122321111 VVV ββ
31 TT β
( ) ( )β«=β΄3
1
T
TVD dTTCβU
( ) ( )121V
V1Drev
VVPdVPβw2
1
ββ=β= β«
( ) ( ) β«+ββ=3
1
T
T V121Drev dTCVVPβq( ) .0βw
Erev =
.TT 13 β ( ) ( )β«=β΄1
3
T
TVE dTTCβU
( ) ( )EE
rev βUβq =
β’ Reversible work :( ) ( )121Totalrev VVPβw ββ= β’ heat Change: ( ) ( )121Totalrev VVPβq β=
β’ Change in energy : ( ) ( ) ( ) 0dTTCdTTCβU1
3
3
1
T
TV
T
TVTotal =+= β«β«
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August 07, 2006 14
Resume For a reversible transformation, we have seen the heat change
and the work done are path functions, whereas the change in
internal energy is a state function. And the First Law is,
Question: Why is the reversible heat change, a path function? Answer: Because the expression for heat change involves both
temperature and volume, and the temperature depends on thevolume, for the ideal gas we have considered. However, if we
divide the heat change by Temperature, the resulting expression
will become a state function, and it is called the ENTROPY, S:That is, the change in Entropy, dS is:
( ) ( )gasidealanfordVV
TR ndTTCdwdUdq Vrevrev +=β=
( ) V
dVR nT
dT TCT
dqdS Vrev +==
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August 07, 2006 15
Entropy physical motivation A bouncing ball eventually comes to rest. This is a spontaneous
and irreversible process. Bouncing occurs because of someordered motion (upward) on the surface and it stops becausethere is no more ordered motion. As this is a natural process,we say: the nature prefers to go towards a disordered state.
Entropy is a measure of disorder -- Larger the entropy, higherthe disorder -- Nature thus prefers the direction of large
Entropy.
Transfer of energy as Heat makes useof disordered, or Thermal motion Transfer of energy as workmakes use of ordered, motion
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August 07, 2006 16
Entropy Let us check if the definition of entropy change,
is consistent with our daily experiences?
Observations: At low temperature, the system has accessto only few states (at T=0, system occupies only the lowestenergy state); that means, the system has less options todisorder itself. At high temperature, it has access to many
more states and hence more opportunity to disorder itself.
For a fixed amount of Heat supply, a colder system would bemore willing (than a hotter system) to accept Heat, as that willincrease its possibility to disorder --- in other words, itβs easierto heat a colder system, as that increases the Entropy.
/TdqdS rev=
β’ Hence, the Entropy change is proportional to the supply of Heat.
β’ Hence, the Entropy change is inversely proportional to the Temperature.
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August 07, 2006 17
EntropySecond Law of Thermodynamics
The Entropy of an isolated system increases in the course of a
spontaneous thermodynamic transformation. This is also known as the Law of increase of Entropy.
And, there are many more equivalent statements of the secondlaw of thermodynamics.
Entropy, like Energy, is an extensive property --- that is,Entropy of 2 gm of ice would be twice as that of 1 gm of ice. Tomake it an intensive property, we divide it by the number ofmoles and call it molar Entropy.
Entropy, like Energy, is a state function --- that is, the value ofEntropy change is independent of the path of thermodynamictransformations.
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August 07, 2006 18
Entropy an Example Entropy change for an isothermal expansion of an ideal gas.
Answer: Isothermal means a constant temperature. From the
first Law we have,
( ) ( )gasidealanfordVV
TR ndTTCdwdUdq Vrevrev +=β=
ββ β
ββββ
β ===β= β«β« 1
2
V
V
V
Vrev
revVVlnR n
VdVR n
TdqβS /TdqdS
2
1
2
1
Q
β’ Notice, the change in Entropy is positive.
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August 07, 2006 19
Second Law of Thermodynamics A precise statement: Consider the change of Entropy,
Second Law: is never negative. More precisely,
For an isolated system: There is no flow of entropy, and hence
( ) ( )EXTINT
dSdSdS +=
β’ Entropy change due to changes inside the system( ) :dS INTβ’ Entropy flow due to interaction with the exterior( ) :dS EXT
( )INTdS( ) β= 0dS INT
( ) β> 0dS INT
Reversible Process
Irreversible Process
( ) 0dSdS INT β₯=
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August 07, 2006 20
Entropy Suppose, we enclose a system, I, inside a larger system II, so
that the global system containing both I and II is isolated. And,
in both parts, I and II, some irreversible process may takeplace. The second Law would be,
What we postulate, is:
That is, absorption of Entropy in one part, compensated by asufficient production in another part of the system is illegal.Thus, in every macroscopic region of the system the Entropyproduction due to the irreversible processes is positive.
0dSdSdS III β₯+=
( ) ( ) 0dS and 0dS INTII
INT
I
β₯β₯( ) ( ) ( ) excluded.is 0SSd with0dS 0,dS IIIINTIIINTI >+
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August 07, 2006 21
Entropy production due to Heat flow Consider two systems , I and II, maintained respectively at
temperatures For the whole system, we have:
β’ Heat received by system I from system II( ) :dq INTI
.T andT III
III
dSdSdS +=
β’ Heat supplied to system I from the outside( ) :dq EXTI
β’ Heat received by system II from system I =:dq INTII
β’ Heat supplied to system II from the outside( ) :dq EXTII( )EXTIdqβ
=dS totaltheHence,( ) ( )
++ IIEXTII
I
EXTI
T
dq
T
dq ( ) β₯β¦β€
β’β£
β‘β
IIIINT
I
T
1
T
1dq
This change results from the irreversible heat flowinside the system, and is postulated to be positive.
( ) +EXTdS= ( )INTdS
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August 07, 2006 22
Entropy production due to Heat flow Ref: is positive.
In fact,
And, the entropy production can be zero, only whenThat is, when the thermal equilibrium is reached.
Entropy production per unit time,
Thus, the direction of heat flow is determined by the sign of the function,
( ) ( ) β₯β¦β€
β’β£
β‘β=
IIIINT
I
INTT
1
T
1dqds
( ) 0T1
T1 when0dq
IIIINT
I>β₯β¦
β€β’β£β‘ β> ( ) 0
T1
T1 when0dq
IIIINT
Iβ₯β¦
β€β’β£
β‘βββ
β
βββ
β
β =β
β
βββ
β
.
T
1
T
1III β
β
ββ
β
β β
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Aug. 08, 2006 1
Thermodynamics and Kinetics
Amrendra VijayDepartment of ChemistryIIT Madras, Chennai 600 036
Lecture 03
R
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Aug. 08, 2006 2
Recap β¦
First Law:
Entropy:
Example: Isothermal expansion (from volume to ) of anideal gas. Isothermal mean constant temperature, and at
constant temperature the change in internal energy of for anideal gas is zero.
( ) ( )gasidealanfordVV
TR ndTTCdwdUdq Vrevrev +=β=
( ) V
dVR n
T
dT TC
T
dqdS V
rev +==
ββ
β
βββ
β
β ===β= β«β«
1
2V
V
V
V
revrev
V
VlnR n
V
dVR n
T
dqβS /TdqdS
2
1
2
1
Q
1V 2V
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Aug. 08, 2006 3
Second Law of Thermodynamics: A precise statement
Consider the change of Entropy,
Second Law says: is never negative. More precisely,
Corollary: For an isolated system, there is no flow of entropyand hence,
( ) ( )EXTINT dSdSdS +=β’ Entropy change due to changes inside the system( ) :dS INTβ’ Entropy flow due to interaction with the exterior( ) :dS EXT
( )INTdS( ) β= 0dS INT
( ) β> 0dS INT
Reversible Process
Irreversible Process
( ) 0dSdS INT β₯=
E
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Aug. 08, 2006 4
Entropy
Suppose, we enclose a system, I, inside a larger system II, sothat the global system containing both I and II is isolated. And,in both parts, I and II, some irreversible process may takeplace. The second Law would read,
As we know, and what we actually postulate
in the second Law, is:
That is, absorption of Entropy in one part, compensated by asufficient production in another part of the system is illegal.Thus, in every macroscopic region of the system the Entropy
production due to the irreversible processes is positive.
0dSdSdS III β₯+=
( ) ( ) 0dS and 0dS INTIIINTI β₯β₯
) ) ) excluded.is 0SSd with0dS 0,dS IIIINTIIINTI >+
( ) ( )EXTIINTII dSdSdS +=
E d i d H fl
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Aug. 08, 2006 5
Entropy production due to Heat flow
Consider two systems , I and II, maintained respectively attemperatures For the whole system, we have:
Define:
β’ Heat received by system I from system II( ) :dq INTI
.T andT III
III
dSdSdS +=
β’ Heat supplied to system I from the outside:dq EXTI
β’ Heat received by system II from system I =( ) :dq INTII
β’ Heat supplied to system II from the outside( ):dq
EXT
II
EXT
Idqβ
=dS totaltheHence,( ) ( )
++ IIEXTII
I
EXTI
T
dq
T
dq ( ) β₯β¦β€
β’β£
β‘β
IIIINT
I
T
1
T
1dq ( ) +EXTdS
= ( )INTdS
This change results from the irreversible heat flowinside the system, and is postulated to be positive.
E d i d H fl
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Aug. 08, 2006 6
Entropy production due to Heat flow
Ref: is positive.
In fact,
And, the entropy production can be zero, only whenThat is, when the thermal equilibrium is reached.
Entropy production per unit time,
Thus, the direction of heat flow is determined by the sign of the function,
( ) ( ) β₯β¦β€
β’β£
β‘β=
IIIINT
I
INTT
1
T
1dqds
( ) 0T1
T1 when0dq
IIIINTI
>β₯β¦β€β’β£
β‘ β> ( ) 0T1
T1 when0dq
IIIINTI
β₯β¦
β€β’β£
β‘βββ
β
ββββ
β =β
β
βββ
β
.
T
1
T
1III β
β
ββ
β
β β
H t R i
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Aug. 08, 2006 7
Heat Reservoir
Definition: A Heat reservoir is a system solarge that the gain or loss of any finite
amount of Heat does not change itsTemperature.
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Second La of Thermodynamics
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Aug. 08, 2006 9
Second Law of Thermodynamics If Kelvin is false Clausius is also false.
Proof: Suppose, Kelvin is false.
Similarly, If Clausius is false Kelvin is also false.
β’ That means we can extract heat from a reservoir a temperature,and convert it entirely into work, with no other effect.
β’ Next, we can convert this work into Heat.
β’ Next, we can deliver this Heat to another heat reservoir at aTemperature, with no other effect.β’ The net result is: we have taken heat from a colder reservoir to
a hotter reservoir, with no effect.
β’ Hence, Clausius is false.
1T
( )122 TTT >
β
β
Carnot Engine
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Aug. 08, 2006 10
Carnot Engine
Definition: A Carnot engine consists of any substance that ismade to go through the Reversible Cyclic thermodynamictransformation, as shown in the diagram
a
b
cd
2T
1T
βV
β’ ab is isothermal, is const and duringwhich the system absorbs Heat
β’ cd is isothermal, is const and during
which the system rejects Heat
2T
1T
.q2
.q1β’ bc and da are adiabaticsβ
2q
β1q
In a cyclic transformation, the temp.
does not change, so the change inInternal energy is zero. Hence, thework done by the system in one cycle
would be (from the first law):
12 qqW
β
P
β=
absorb
reject
12 TT >
Infinitesimal Carnot engine, if transformations are infinitesimal.
Carnot Engine
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Aug. 08, 2006 11
Carnot Engine
Ref: For one Carnot cycle, the work done bythe system is:
Efficiency of an Engine: It is defined as,
12 qqW β=
2
1
2 q
q1q
WΞ· β==
Carnot Engine and Refrigerator
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Aug. 08, 2006 12
Carnot Engine and Refrigerator Show that:
Proof:
Similarly, we can prove: Here, the
Carnot engine operates in reverse and becomes a refrigerator.
.0qand0qthen0,WIf 21 >>>
β’ means the system absorbs heat and rejects nothing; that is, it
converts entirely into work. This violates Kelvin statement, and hence,
β’ Suppose, This means the engine absorbs from the reservoir atand from the reservoir at .
β’ The net amount of Heat converted to work, by the engine, therefore iswhich is equal to because is assumed negative.β’ We can next convert this work into Heat and then deliver it to the
reservoir at with no other effect.β’ The net effect is the transfer of a positive amount of Heat from the
reservoir at to the reservoir at But, by assumption. Hence,this violates the Clausius statement. Thus,
0q1 = 2q
2q.0q1 β :0q1 < 2q 2T
1qβ 1T
( ),qq 12 β+ 12 qq + 1q
,T21q
.T21T 12 TT >0. Whenceand0,q1 >>
.0qand0qthen0,WIf 21
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Aug. 08, 2006 13
Carnot s Theorem
Theorem: No Engine operating between two giventemperatures is more efficient than a Carnot engine.
Proof: Recall, for one carnot cycle, the work done is: 12qqW β=
β’ Consider a Carnot engine (C) and an arbitrary engine (X), operatingbetween the heat reservoirs at ( ).TT TandT 1212 >
X
1
X
2
X
qqW β=C
1
C
2
C
qqW β=β’ Work done is:β’ Let
C
X
X
2
C
2
N
N
q
q= Integers (ratio of integers can approximate a
Floating point number, to any accuracy we desire)
β’ Operate X engine cycles forward and C engine cycles inreverse. At the end of the operation we have:
X
NC
N
( ) 0q Nq Nq C2CX
2
X
total2 =β=
( )C
1
CX
1
X
total1 q Nq Nq β=
(by the definition of integers)
( ) ( ) ( )total1total1total2total
qqqW β=β=β
CCXX
total
W NW NW β=
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Absolute Temperature
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Absolute Temperature
Corollary: All Carnot engines operating between two giventemperatures have the same efficiency.
This gives a definition of a scale of temperature called --- TheAbsolute Scale. We define as follows:
The efficiency of a Carnot engine operating between tworeservoirs of respective absolute temperatures and is
Since the absolute temperature of any reservoir isalways greater than zero.
1ΞΈ 2ΞΈ
2
1
ΞΈ
ΞΈ1Ξ· β=
2
1
2
1
2
11
,
ΞΈ
ΞΈ
q
q
q
qΞ·
Also
=ββ=Q
1,Ξ·0 β€β€
Clausius Theorem
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Aug. 08, 2006 16
Clausius Theorem
Theorem: In any cyclic transformation throughout which thetemperature is defined, the following inequality holds.
where the integral extends over one cycle of the transformation.and the equality holds if the cyclic transformation is reversible.
Proof: Divide the cyclic transformation into N infinitesimal stepsand for each step the temperature is assumed constant.
Imagine, at each step the system is brought in contact withheat reservoirs at temperatures,
Let be the amount of heat absorbed by the system duringthe step, from the heat reservoir of temperature
We need to prove and will recover the theorem.
0T
dq
β€β«
mqthm .Tm
.T,,T,T N21
L
0q N m β€
β
β β
β
ββ
β
ββ N
T1m m β β =We have essentially discretized the integral.
Clausius Theorem
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Aug. 08, 2006 17
Clausius Theorem
Construct a set of N Carnot engines, such that
From the definition of temperature scale,
The net result of one complete cycle:
Clausius says: this is impossible unless we have,
N21 C,C,C L mC1. Operates between
2. Absorbs amount of heat from
3. Rejects amount of heat to
m)allfor,T(T TandT mOOm β₯
( )Omq OT
mq mT( )
( )
m
mO
Om
m
O
m
O
m
TqTq
TT
qq =β=
OT
( )
ββ ====
N
1m m
m
O
N
1m
O
mtotal T
q
Tqq
is the amount of heat absorbed from the
reservoir at and converted entirely intowork, with no other effect.
0q total β€
( ) 0T/q N
1m
mm β€ββ=
QED
Clausius Theorem
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Aug. 08, 2006 18
Clausius Theorem
We have already seen, for a cyclic transformation.
If the cyclic transformation is reversible, we reverse it. The line of arguments remains the same and we arrive at the
same inequality, except the signs of heat, are reversed.That is,
Hence, we finally obtain for reversible cyclic
transformations. Corollary: For a reversible transformation, the integral is
independent of the path of transformations and depends onlyon the initial and the final states of the transformation.
mq
( ) 0T/q N
1m
mm β€β=
( ) 0T/q N
1m
mm β€ββ=( ) 0T/q
N
1m
mm =β=
β« Tdq
Hence, the Entropy is a state function.
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Aug. 11, 2006 1
Thermodynamics and Kinetics
Amrendra VijayDepartment of ChemistryIIT Madras, Chennai 600 036
Lecture 04
Ref: Statistical Mechanics, by Ryogo Kubo
Entropy
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Aug. 11, 2006 2
Entropy Consider a model showing a irreversible process (Heat Flow):
I or II are always internally at equilibrium, but they are notmutually in equilibrium. And, thermodynamically we postulate:
is never negative.
Iext
I
qd
IT IIT
R
I
O
V
R
E
S
E
R
R
I
O
V
R
E
S
E
R
contact
( ) ( )extint
dSdSdS +=
ext
II
qd
( )intIqd
( ) ( ) ( )
II
ext
II
I
ext
I
extT
qd
T
qddS += ( ) ( ) β₯
β¦
β€β’β£
β‘β=
III
int
I
intT
1
T
1qddS
( )intdS
Entropy flow from the outside Entropy production inside the system
( ) β= 0dS INT( ) β> 0dS INT
Reversible Process
Irreversible Process
II
Second Law of Thermodynamics
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Aug. 11, 2006 3
Second Law of Thermodynamics
Kelvin Statement: There exists no thermodynamic
transformation whose sole effect is to extract aquantity of Heat from a Heat reservoir and toconvert it completely into work.
Clausius Statement: There exists no thermodynamictransformation whose sole effect is to extract a
quantity of Heat from a colder reservoir and todeliver it to a hotter reservoir.
Ref: Statistical Mechanics, by Kerson Huang (chapters 1 and 2)
Carnot Engine
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Aug. 11, 2006 4
g
Definition: A Carnot engine consists of any substance that ismade to go through the Reversible Cyclic thermodynamictransformation, as shown in the diagram
a
b
cd
2T
1T
βV
β’ ab is isothermal, is const and duringwhich the system absorbs Heat
β’ cd is isothermal, is const and during
which the system rejects Heat
2T
1T
.q2
.q1
β’ bc and da are adiabaticsβ2q
β1q
In a cyclic transformation, the temp.does not change, so the change inInternal energy is zero. Hence, thework done by the system in one cycle
would be (from the first law):
12 qqW
β
P
β=
absorb
reject
12 TT >
Infinitesimal Carnot engine, if transformations are infinitesimal.
Carnotβs Theorem
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Aug. 11, 2006 5
Theorem: No Engine operating between two giventemperatures is more efficient than a Carnot engine.
Proof:β’ Consider a Carnot engine (C) and an arbitrary engine (X), operating
between the heat reservoirs at In one forward cycle,C/X engine takes amount of Heat and rejects Heat.
( ).TT TandT 1212 >X
2
C
2 /qqX
1
C
1 /qq
β’ Let us define two integers such that, CX
X
2
C2
N N
qq =
β’ Operate X engine cycles forward and C engine cycles in reverse.
At the end of the operation, the total Heat taken by both machines, is
X N
C N
( )
0
q Nq Nq C2CX
2
X
total2
(this can be done to anyaccuracy we like, as long asintegers are large enough)
=
β=
(by the definition of integers)
β’ This implies, the total work done by both machines,0Wtotal β€ (a la Kelvin)
Carnotβs theorem
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The total Heat rejected by both machines:
So, the total work done by both machines is also,
But, we have already seen Hence,
Corollary: All Carnot engines operating between two given
temperatures have the same efficiency.
( ) C1CX
1
X
total1q Nq Nq β=
( ) ( ) ( )total1total1total2total
qqqW β=β=β12 qqW β=
Recall, for one Carnot cycle,
the work done is:
0q Nq N C1CX
1
X β₯ββ
.0Wtotal β€
C
X
X2
C
2
N
N
q
q=
β
Efficiency of aCarnot engine
β₯ ββ β
ββββ
β β
X
2
X
1
q
q1ββ
β
ββββ
β β
C
2
C
1
q
q1
Efficiency of anarbitrary engine
QED
Use:( ) .0q total1 β₯
Temperature Scale
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p
To fix a Scale, we need (1) a uniform distribution ofpoints, and (2) either lowest or the highest point of
the scale.
Lowest Point
Highest Point
OR
Temperature Scale
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p Consider a series of Carnot engines, all performing the same
amount of work, W ---- such that, the Heat rejected by anyCarnot engine is absorbed by the next one.
Now, we have a set of equidistant points. To set a TemperatureScale, let us define:
W
2nq
+ 1nq + nq 1nq β 2nq β
W W W
CarnotEngine
CarnotEngine
CarnotEngine
CarnotEngine
Wqq n1n =β+
W
W
W
constant)arbitrary:(xx WTT n1n =β+
For example, we may choose: Kelvin1x W =
Temperature Scale
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p To fix the lowest highest point of our Scale, let us define the
efficiency of a Carnot Engine, operating between two reservoirsof temperatures, as,
A relation between q and T: Recall, the efficiency was originallydefined as
x==β++ 1n
n
1n
n
T
T
q
q
1n
n1n
T
T1Ξ·
+
+ β=
1n
n
1n
1n1n
q
q1
q
WΞ·
++
++ β==
ββ€β€ 1Ξ·0 Q
( )n1n1nn TT T and T >++
The absolute temperature of any reservoir is
always greater than Zero. It fixes the lower limit.
Thus, the Temperature scale is now fully defined !
x is independent of n
Clausius Theorem
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Theorem: In any cyclic transformation,throughout which the temperature is
defined, the following inequality holds.
where the integral extends over one cycle
of the transformation. And, the equalityholds if the cyclic transformation isreversible.
0T
dqβ€
β«
0T
dqβ€β«Clausius Theorem
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Proof strategy: Divide the cyclic transformation into Nsmall steps and for each step, the temperature, we assume,is constant. That is, at each step, the system is thought to
have been brought in contact with heat reservoirs attemperatures,
Let be the amount of heat absorbed by the system during
the step, from the heat reservoir of temperature
Then, we need to prove:
And, we will recover the original theorem in the limit
.T,,T,T N21 L
0T
q N
1m m
m β€ββ β
ββββ
β β
=
ββ N
mqthm
NB: We have essentially discretized the integral --- that is, wehave written the cyclic transformation, as a sum of N
infinitesimal cyclic transformations (e.g, Carnot engines).
.Tm
Clausius Theorem
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Proof: Construct a set of N Carnot engines, such that
From the definition of temperature scale, The net result of one complete cycle:
Clausius says: this is impossible unless we have,
N21 C,C,C L mC
1. Operates between
2. Absorbs amount of heat from
3. Rejects amount of heat to
m)allfor,T(T TandT mOOm β₯
( )Omq OT
mq mT ( )
m
O
m
O
m
T
T
q
q=
OT( ) ββ
==
== N
1m m
mO
N
1m
O
mtotalT
qTqq
is the amount of heat absorbed from thereservoir at and converted entirely into
work, with no other effect.0q total β€
( ) 0T/q N
1m mm
β€β
β= QED
( )
m
m
O
O
m T
qTq =β
Clausius Theorem
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We have already seen, for a cyclic transformation.
If the cyclic transformation is reversible, we reverse it. The line of arguments remains the same and we arrive at the
same inequality, except the signs of heat, are reversed.That is,
Hence, we finally obtain for reversible cyclic
transformations. Corollary: For a reversible transformation, the integral is
independent of the path of transformations and depends onlyon the initial and the final states of the transformation.
mq
( ) 0T/q N
1m
mm β€β=
( ) 0T/q
N
1mmm β€β
β=
( ) 0T/q N
1m
mm =β=
β« Tdq rev
Hence, which is called the Entropy , is a state function.Tdq rev
Proof of the Corollary -- Entropy is a state function
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Consider two reversible paths:
For reversible transformations, we have:
That is:
That is, the difference of Entropy, computed along two differentpaths is zero --- so, itβs path independent. Hence, Entropy is astate function.
I
IIPath is reverse of the path II./II
0
T
dq rev =β«0
T
dq
T
dq/II
rev
I
rev =+ β«β« 0T
dq
T
dq
II
rev
I
rev =ββ β«β«
QED
Entropy Reference point.
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Define Entropy:
So, the change in Entropy:
( ) β«=
A
Ref
rev
T
dq
ASRef is some arbitrary
thermodynamic state
( ) ( ) β«β«β«β«β« =+=β=ββ΄B
A
rev
A
Ref
rev
Ref
B
rev
B
Ref
rev
A
Ref
rev
T
dq
T
dq
T
dq
T
dq
T
dqBSAS
β’ Thus, the entropy is defined only up to an arbitrary additive constant,so long as the state A is accessible from the Ref state, through areversible transformation. If not so, we would need the Third Law ofthermodynamics to fix the absolute value of the constant --- and hence,the Absolute value of Entropy.
Ref A
Entropy Properties
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For an arbitrary transformation:
Proof:
Entropy of a thermally isolated system never decreases. Thisis because dq is zero, Equals zero for reversible.
That means, for a thermally isolated system, the equilibriumstate is the state of maximum entropy.
( ) ( )ASBST
dqB
A
ββ€β«
A B
reversible
irreversible
β’ Consider the cyclic transformation, madeup of (1) Irreversible, plus (2) reverse ofReversible. From Clausiusβ theorem:
β«β«β«β« β€ββ€β REVIRR REVIRR Tdq
T
dq 0
T
dq
T
dq( ) ( )ASBS
T
dq
IRR ββ€β β«
QED
( ) ( ) .0ASBS β₯β
β’ For reversible transformations,
itβs obvious by definition.
Thermodynamic Relations
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Recall the first law:
And, assuming the internal energy being a function of volumeand temperature,
So,
Using this, we can derive several important thermodynamicrelations --- and much more.
PdVdUdq +=
dVV
UdTC dU
T
V β β
βββ
β
β
β+=
Heat capacity Internal pressure
dVPV
U
T
1
T
dTC
T
dqdS
T
V β₯β¦
β€β’β£
β‘+β
β
βββ
β
β
β+==
Rules:
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df = g(x,y)dx + h(x,y)dy is an exact differential, if
Let x, y and z are quantities satisfying a functional relation,f(x,y,z)=0. Let w be a function of any two of x, y, z then
Given an exact differential, identify allthe variables. This relation then allowsus to obtain a host of useful relations,(for example, Maxwellβs relations).
yxxh
yg β
β ββ
β β ββ=ββ
β βββ
β β ββ
wwwz
x
z
y
y
x
β β
βββ
β
β
β=β β
βββ
β
β
βββ β
ββββ
β
β
β 1
zzxy
yx
β
ββ¬β«
β©β¨β§ β
β ββ
β β ββ=ββ β
ββββ β ββ
1x
z
z
y
y
x
yxz
β=β β
βββ
β
β
ββ β
βββ
β
β
β
β
β
β
β
β
β
β
β
β
β
Derivation of TdS equations:
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Ref:
dS is an exact differential and RHS is a function of V and T, so
Use the definition, differentiate and rearrange.
For ideal gas, P=nRT/V and hence,
T
dqdS ==
T
dTCV dVPV
U
T
1
T
β₯β¦
β€β’β£
β‘+β
β
βββ
β
β
β+
ββ¬β«
β©β¨β§
β β
βββ
β
β
β
T
C
V
V
T
=ββ¬β«
β©β¨β§
β₯β¦
β€β’β£
β‘+β
β
βββ
β
β
ββ β
βββ
β
β
βP
V
U
T
1
T TV
,T
U CV
V β β ββ
β β ββ=
P
T
P T
V
U
VT
ββ
β
ββ
β
β
β
β=β
β
ββ
β
β
β
β
0.PV
nR T
V
U
T
=β=β β
βββ
β
β
β
Hence, the internal pressure of of an ideal gas is zero,and we already knew it !!!
VT T
P TP
V
U β
β
ββ
β
β
β
β=+β
β
ββ
β
β
β
ββ
TdS Equations
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Recall:
Similarly, we can show:
We can also write TdS equations, purely in terms ofexperimental observables:
dVPV
U
T
1
T
dTCdS
T
V β₯β¦
β€β’β£
β‘+β
β
βββ
β
β
β+=
VT T
P TP
V
U β
β
βββ
β
β
β=+β
β
βββ
β
β
β
dTCV dVT
PT
V
β β
βββ
β
β
β+TdS =β
dTCP dPT
VT
P
β β
βββ
β
β
ββTdS =
TdS Equations with exptal observables
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Define:
We can then show:
PT
V
V
1Ξ± β
β
βββ
β
β
β=
T
TP
V
V
1
ΞΊ β β
β
ββ
β
β
β
β=
S
SP
V
V
1ΞΊ β
β
βββ
β
β
ββ=
Coefficient of thermal expansion
Isothermal compressibility
adiabatic compressibility
T
1
TP
1
TPV ΞΊ
Ξ±
P
V
T
V
P
V
V
T
T
P=β₯
β¦
β€β’β£
β‘β β
βββ
β
β
ββ β
βββ
β
β
ββ=β₯
β¦
β€β’β£
β‘β β
βββ
β
β
ββ β
βββ
β
β
ββ=β
β
βββ
β
β
β ββ
TdS =Q dTCV dVTPT
V
β β ββ
β β ββ+
TdS =β dTCV dVΞΊ Ξ±T
T
+
TdS = dTCP
Ξ±TVdPβSimilarly,
Entropy of Phase Transition
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Normal transition temperature: This is the temperature at whichtwo phases are in equilibrium at 1 atmosphere pressure. Forexample: Liquid water and vapur at 100 degree C.
Solution: (1) At the transition point, there is an equilibrium andhence any transfer of heat between two phases is reversible. (2)The pressure is constant by definition, and at constant pressure,the change is Heat at constant pressure is the definition ofEnthalpy. Therefore,
Troutonβs rule: Standard entropy of vaporization is amost thesame for a wide range of liquids, and the value is 85 J/(K mol).
( ) ( )
Transition
TransitionTransition
T
βHβS =
( ) ( )
( ) ( )meltingc,endothermi 0βH
freezing,exothermic 0βH
Transition
Transition
e.g.,
e.g.,
>
<
Entropy as a function Temperatured
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Recall:
If only temperature changes, then we can directly integrate:
If we fix the pressure, then at constant P,
If the heat capacity is independent of Temperature:
T
dqds =
( ) ( ) β«+=2
1
T
T
rev12
T
dqTSTS
( )dT.TCdq Prev =
( ) ( ) ( )T
dT TCTSTS
2
1
T
T
P12 β«+=
( ) ( ) ( )ββ β
β
βββ
β +=+=
β« 12
P1
T
TP12 T
TlnCTS
T
dTCTSTS
2
1
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Thermodynamics and Kinetics
Amrendra VijayDepartment of ChemistryIIT Madras, Chennai 600 036
Lecture 05
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Happy Independence Day
Syllabus --- Equilibrium Thermodynamics
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Basic Concepts
The Second Law of Thermodynamics; Entropy; ClausiusInequality.
Entropy Changes for reversible and irreversible processes.
Third Law of Thermodynamics and Absolute Entropy
Definition of Free Energy and Spontaneity; Maxwellβs Relations;
Free Energy and Chemical Equilibria; Gibbs-Helmholtz andvanβt Hoff Equations.
Thermodynamic properties from EMF measurements; NernstEquation
Phase Equilibria: Clausius-Clapeyron equation; phase rule
Phase diagrams: One component systems
Eutectic systems
References
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β’ Physical Chemistry ThroughProblems
--- S.K. Dogra and S. Dogra
Reference: Statistical Mechanics by Kerson Huang
Recap β¦
W d Cl i I li 0dqβ«
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We proved Clausius Inequality:
We then defined Entropy as:
We then observed Entropy is defined only up to an arbitraryadditive constant:
This implies: Even though, we defined Entropy only for areversible transformation, but we can correctly compute thechange in Entropy of a system by choosing any arbitrary path !
0T
dq β€β«
( )
β«=
A
Ref
rev
T
dqAS
( ) ( ) β«β«β«β«β« =+=β=βB
A
rev
Ref
A
rev
B
Ref
rev
A
Ref
rev
B
Ref
rev
Tdq
Tdq
Tdq
Tdq
TdqASBS
That is, S is an exact differential.
Assuming there is acontinuous path oftransformation fromRef to A.
What is the implication ?
Change of Entropy
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Consider: One mole of an ideal gas expandsisothermally from volume by two different
routes:
We will compare the change of Entropy of the gasand of the surroundings, computed by both routes.
21 VtoV
1.Reversible Isothermal Expansion
2.Irreversible Free Expansion
Entropy Change: Reversible Isothermal Expansion
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Model:
gas
Reservoir, T
Piston
Spring
β
P
βV1V 2V
T constant line
Work done
Entropy Change: Reversible Isothermal Expansion
Computation: Temperature is constant so the change in
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Computation: Temperature is constant, so the change inInternal Energy for ideal gas will be zero. The First law thensays: the amount of heat absorbed, is equal to the work
done, W.
Hence, the change in Entropy of the gas is:
But, the reservoir has supplied the heat and so the change inEntropy of Reservoir will be:
βqV
VlnRTW
1
2 β‘=
( )1
2gas
V
VlnR
T
βqβS ==
βq
( )1
2
Reservoir V
VlnR
T
βqβS β=β=
Entropy Change: Reversible Isothermal Expansion
h l h
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So, the total change in Entropy is Zero:
Question: Where did the work done, by the gas go ?
Answer: Well, the work gets stored in the spring (spring gets
compressed) connected to the piston, which can be used tocompress the gas back, by reversing the transformation.
1
2
VVlnRTW =
( ) ( ) ( ) 0VVlnR
VVlnR βSβSβS
1
2
1
2Reservoir GasTotal =β=+=
Entropy Change: Irreversible Free Expansion
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Model:
Before After
Water Bath, T constant Water Bath, T constant
Gas occupies volume,1V Gas occupies volume, 2V
Jouleβs free-expansion experiment
Entropy Change: Irreversible Free Expansion
C t ti Fi t f ll h t li d b th
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Computation: First of all, no heat was supplied by theReservoir, and hence the change in Entropy for the Reservoirwould be zero here.
But. the initial and the final states are the same, and the
Entropy has turned out to be a state function, and thereforethe change in Entropy for the gas here will be the same as thatwe found for Reversible Thermal Expansion. That is:
So the total change in Entropy here is:( ) 1
2
gas V
V
lnR βS =
( ) 0βS Reservoir =
( ) ( ) ( )1V
2Reservoir GasTotal
VlnR βSβSβS =+=
Entropy Change: Comparison
Reversible Isothermal Expansion Irreversible Free Expansion
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Reversible Isothermal Expansion Irreversible Free Expansion
We henceforth say: Irreversibility is wasteful, and is marked byan increase of Entropy of the total system under consideration.For this reason, the Entropy may be viewed as a measure of theunavailability of useful energy.
( )
1
2gas
V
VlnR
T
βqβS ==
( )1
2Reservoir
V
VlnR
T
βqβS β=β=
( ) 0βSTotal
=β΄
1
2
V
VlnRTW =
Gets storedin the Spring
( )
1
2gas
V
VlnR
T
βqβS ==
( ) 0βS Reservoir =
( )TotalβS TW = Gets wasted
( )1
2Total
V
VlnR βS =
Entropy as a function Temperature
Recall:dq
d
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Recall:
If only temperature changes, then we can directly integrate:
If we fix the pressure, then at constant P,
If the heat capacity is independent of Temperature:
T
qds =
( ) ( ) β«+=2
1
T
T
rev12
T
dqTSTS
( )dT.TCdq Prev =
( ) ( ) ( )T
dT TCTSTS
2
1
T
T
P12 β«+=
( ) ( ) ( )β
β
β
β
β
β
β
β +=+=
β« 12
P1
T
T
P12
T
TlnCTS
T
dTCTSTS
2
1
(From Atkins)
Standard State
Definition: The standard state of a substance at a specified
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Definition: The standard state of a substance at a specifiedTemperature is its pure form at one bar pressure.
Conventionally, the Temperature at which thermodynamic dataare reported is 298.15 K (Kelvin) = 25 degree C (Centigrade).
Reason: Frequently, we are interested in computing the changein thermodynamic quantities, like Enthalpy, in which the initialand the final substances are in their standard state.
Example: (1) The standard state of liquid ethanol at 200 K ispure liquid ethanol at 200 K temperature and 1 bar pressure.
Units: atm)(1 bar1Pa10 Pa;1m N1 5-2 ==
Entropy of Phase Transition
Normal transition temperature: This is the
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Normal transition temperature: This is thetemperature at which two phases are in equilibrium
at 1 atmosphere pressure. For example: Liquid waterand vapor at 100 degree C.
Solution: (1) At the transition point, there is anequilibrium and hence any transfer of heat betweentwo phases is reversible. (2) The pressure is constantby definition, and at constant pressure, the change inHeat at constant pressure is the definition of
Enthalpy. Therefore,
( ) ( )
Transition
TransitionTransition
T
βHβS =
( ) ( )
( ) ( )melting c,endothermi 0βH
freezing ,exothermic 0βH
Transition
Transition
e.g.,
e.g.,
>
<
Troutonβs Rule
Standard entropy of vaporization is almost the same for a wide
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Standard entropy of vaporization is almost the same for a widerange of liquids, and the value is
This rule allows us to compute either the Heat of Vaporization
or the Boiling Point of a liquid approximately.
The rule is less valid for liquids, in which molecules areexpected to be arranged in an orderly manner (probably due to
strong intra-molecular interactions) and a great change ofdisorder is expected to occur when the liquid evaporates.
.molK J87.822 11 ββ
( ) ( ) ( ) ( ) 11-Point-Boiling
onVaporizati
onVaporizati
Transition
TransitionTransition molK J85
TβHβS
TβHβS ββ=β΄=Q
Troutonβs Rule
Example: The normal boiling point of Chloroform is C61.5Β°
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Example: The normal boiling point of Chloroform is .Calculate the molar heat of vaporization, from Troutonβs rule.
Answer:
C61.5
( ) ( )K 61.5273molK 85J 11 +Γββ
( ) ( )
( ) Point-Boiling11-
onVaporizati
11-
Point-Boiling
onVaporizati
onVaporizati
TmolK J85βH
molK J85TβHβS
Γ=β
β=
β
βQ
Example
Calculate the increase in the Entropy if one mole of Krypton is
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Calculate the increase in the Entropy if one mole of Krypton isheated from to at constant volume and at pressure(the molar heat capacity at constant volume is 1.5 R and
R=8.314 ).
At constant volume:
At constant pressure:
C Β°27 C Β°227
11
molK J ββ
( )( ) ( )( )K 227273K 27273lnmolK J 8.3141.51mol
TTlnCnβS 11
1
2V
++Γ== ββ
( ) [ ]( ) ( )( )K 227273
K 27273lnmolK J 8.3141.01.51mol
T
TlnCnβS 11
1
2P
+
+Γ+== ββ
Example
One mole of the ideal gas at 3 atm and 300 K is expanded (1)
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One mole of the ideal gas at 3 atm and 300 K is expanded (1)isothermally to double its initial volume against an externalpressure of 1.5 atm, (2) isothermally and reversibly to twice its
original volume. Calculate the work done. (1)
(2)
( )P
TR nVand VVPβVPW 112extext =β==
( )( )( )β
β
β
β
β
β
β
β =
β
β
β
β
β
β
β
β = ββ
1
111
1
2
V
2Vln300K mol8.314JK 1mol
V
VlnRTnW
Example
A reversible Carnot cycle does work equivalent to 150 kJ per
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e e s b e Ca ot cyc e does o equ a e t to 50 J pecycle. If the heat supplied by cycle is 225 kJ at 227 degree Cper cycle, calculate (1) the temperature at which the heat is
rejected, and (2) the thermal efficiency of the engine. (1) Thermal Efficiency:
(2)
( )
( ) 3
2
kJ225
kJ150
qabsorbedheat
WworkΞ·
2
===
( ) ( )K 273227321TΞ·1T TT1Ξ· 2121
+β β
βββ
β β=β=β΄β=Q
Thermodynamic Potentials
Helmholtz Free Energy, A: STUA β= Entropy
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gy,
Gibbs Free Energy, G:
Note: We introduce these potentials to rationalize the direction
of natural processes, by focusing our attention only to thesystem. In contrast, if we use the concept of Entropy, we alsoneed to take into account the changes taking place in thesurroundings
STUA β=
VPAG +=TSHVPTSUG β=+β=β
Enthalpy
Internal Energy Temperature
Entropy
Potentials --- Physical Meaning
Helmholtz Free Energy, A: Consider an arbitrary Isothermal
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gy, ytransformation from state A to state B. Clausius theorem says:
Because Temperature, T is constant, we have:
From the First Law we know, and hence,
( ) ( )ASBS TdqB
Aββ€β«
S ββ€ββ€ Tβq βST
βq
ationtransformtheduring
absorbedHeat:βq
( ) ( )ASBSβS β=
WdUβq +=
( )STUW STWU βββββ€βββ€+β
AβW ββ€ A: Helmholtz Free Energy
Potentials --- Physical Meaning
Ref: For an arbitrary isothermal transformation.
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Physica