Kinetics and Thermo by Amarendra Vijay

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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August 04, 2006 2

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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August 04, 2006 3

Thermodynamics• Basic concepts and the First Law

• 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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August 04, 2006 4

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

>

< −


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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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August 04, 2006 6

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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August 04, 2006 7

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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August 04, 2006 8

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),,( =−≡ nRT PV T V P f


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August 04, 2006 9

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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August 04, 2006 11

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:

( ) ( )[ ] ( ) ( ) ( ) ∞+−

⎥⎦

⎤

⎢⎣

⎡

∂

∂+−

⎥⎦

⎤

⎢⎣

⎡

∂

∂+−

⎥⎦

⎤

⎢⎣

⎡

∂

∂+=

====

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==

⎥⎦

⎤⎢⎣

⎡

∂

∂=⇒−⎥⎦

⎤⎢⎣

⎡

∂

∂=−

( ) dx x x =−

0

0/

x x

mm x f =∂∂

0 x

0/

x x

mm x f =∂∂ 0/ x x

mm x f =∂∂

dx x f df ⎟ ⎠ ⎞⎜⎝ ⎛ ∂∂=

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

+−⎥⎦

⎤⎢⎣

⎡

∂∂

∂+

−⎥⎦

⎤⎢⎣

⎡

∂

∂+=

== =

= =

=

∑∑

∑

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

−⎥⎦⎤⎢

⎣⎡

∂∂=−⇒

==∑

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 ====

∑∑ ⎥⎦

⎤⎢⎣

⎡

∂

∂=⇒−⎥

⎦

⎤⎢⎣

⎡

∂

∂=∆⇒

11

3

3

2

2

1

1

dx x

f dx

x

f dx

x

f df ⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂=

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 ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ∂∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛ ∂∂=


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August 04, 2006 15

Integration

→ x

↑

y

( ) ( )∫=max

min

y

y

x,ydy f xg• Consider, ymax

ymin

• 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.

∫=∆ final

initial

dE E

E ∆

• 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 ∆


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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

∆

=+

• 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.

• For an isolated system, q = 0 and w = 0. Hence, the change ininternal energy is zero; that is, the energy is conserved.

• 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

⎟ ⎠

⎞

⎜⎝

⎛

−=−=⇒ ∫ VinitialVfinal

lnnRTV

dV

nRT w

Vfinal

Vinitial

∫−=⇒V2

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 ⎟ ⎠

⎞⎜⎝

⎛

∂

∂+⎟ ⎠

⎞⎜⎝

⎛

∂

∂=⇒≡


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August 04, 2006 19

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 ⎟

⎠

⎞⎜⎝

⎛

∂

∂+⎟

⎠

⎞⎜⎝

⎛

∂

∂=

dT CdTT

UdU v

C

V

v

=⎟ ⎠

⎞⎜⎝

⎛

∂

∂=⇒

43421

Heat Capacity at constant volume

∆TC∆U

∆T

∆ULimit

T

UC v

V0∆TV

v =⇒⎟

⎠

⎞⎜

⎝

⎛ =⎟

⎠

⎞⎜

⎝

⎛

∂

∂=

→

TCUR f ∆∆


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August 04, 2006 20

Heat Capacitycontinued

…

TCU :Ref v ∆=∆

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 ⎟

⎠

⎞⎜⎝

⎛

∂

∂+⎟

⎠

⎞⎜⎝

⎛

∂

∂=

dV Π

dVV

U

dU T

Π

T

T

=⎟ ⎠

⎞⎜⎝

⎛

∂

∂=⇒

321

Internal Pressure


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August 04, 2006 22

Internal Energy ( ) ( )TP,UTV,U →

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 dVΠdU dTT

U dV

V

UdU VT +=⇒⎟

⎠

⎞⎜⎝

⎛

∂

∂+⎟

⎠

⎞⎜⎝

⎛

∂

∂=

Internal Pressure Heat Capacity

VTV

P

T

P

C VΠαC

T

V Π

T

U+=+⎟

⎠

⎞⎜

⎝

⎛

∂

∂=⎟

⎠

⎞⎜

⎝

⎛

∂

∂

PT

V

V

1α where, ⎟

⎠

⎞⎜

⎝

⎛

∂

∂=

Expansion CoefficientInternal Pressure

VP ⎠⎝ ⎠⎝ V

T

UC T

U ⎟ ⎞⎜⎛ ∂∂==⎟ ⎞⎜⎛

∂∂


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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.

( ) ( ) ( ) ( )∆VV∆PP∆UU∆HH ++++=+∆V∆P

PVq

∆PV∆VP∆U∆H

∆+∆=

++=

∆VP

∆U

∆W

∆U

∆q

+=−=

From the first law, if the systemis in mechanical equilibrium ata pressure, P then

q.∆H ∆=


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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 ⎟

⎠ ⎞⎜

⎝ ⎛

∂∂+⎟

⎠ ⎞⎜

⎝ ⎛

∂∂=⇒≡

dTCdTT

H

dH P

C

P

P

=⎟ ⎠

⎞

⎜⎝

⎛

∂

∂=

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 ⎟

⎠

⎞⎜⎝

⎛

∂

∂+⎟

⎠

⎞⎜⎝

⎛

∂

∂=⇒≡

dPP

H

dHT⎟ ⎠

⎞

⎜⎝

⎛

∂

∂

=

dTCdPP

H dH p

T

+⎟ ⎠

⎞⎜⎝

⎛

∂

∂=

dTCdPH

dH:Reference +⎟ ⎞

⎜⎛ ∂

=


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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

+⎟ ⎠

⎜⎝ ∂

P

VTV

CT

P

P

H

T

H+⎟

⎠

⎞⎜⎝

⎛

∂

∂⎟

⎠

⎞⎜⎝

⎛

∂

∂=⎟

⎠

⎞⎜⎝

⎛

∂

∂

( ) 1y

z

z

x

x

y then,yx,zif

xyz

−=⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂⎟

⎠

⎞⎜⎝

⎛

∂

∂⎟

⎠

⎞⎜⎝

⎛

∂

∂

κ

α

P

V Vα

P

V

T

V

P

V

V

T

T

P

T

1

T

1

TP

1

T

1

PV

=⎥⎦

⎤⎢⎣

⎡⎟

⎠

⎞⎜⎝

⎛

∂

∂=⎥

⎦

⎤⎢⎣

⎡⎟

⎠

⎞⎜⎝

⎛

∂

∂⎥⎦

⎤⎢⎣

⎡⎟

⎠

⎞⎜⎝

⎛

∂

∂−=⎥

⎦

⎤⎢⎣

⎡⎟

⎠

⎞⎜⎝

⎛

∂

∂⎥⎦

⎤⎢⎣

⎡⎟

⎠

⎞⎜⎝

⎛

∂

∂−=⎟

⎠

⎞⎜⎝

⎛

∂

∂ −−−−

T

TP

V

V

1κ ⎟

⎠ ⎞⎜

⎝ ⎛

∂∂−=

Isothermal compressibility

P

C

PH

1

P

1

HT

CµT

H

P

T

H

T

T

P

P

H

P

−=⎥⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛

∂

∂⎥⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛

∂

∂−=⎥

⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛

∂

∂⎥⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛

∂

∂−=⎟

⎠

⎞⎜⎝

⎛

∂

∂ −−

43421

HP

Tµ ⎟

⎠

⎞⎜⎝

⎛

∂

∂= Joule-Thomson

Coefficient

P

T

P

T

P

VCκ

αµ1Cκ

α

CµT

H

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−=+−=⎟ ⎠

⎞⎜⎝

⎛

∂

∂

∴


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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 ⎟

⎠

⎞⎜⎝

⎛

∂

∂=

P

PT

HC ⎟

⎠

⎞⎜⎝

⎛

∂

∂=

U: Internal energy

H: Enthalpy = U + P V

[ ] [ ]( ) VΠPα

VPTCT

U

C PVUTCC

T

P

V

P

V

P

VP

+=

⎟ ⎠

⎞

⎜⎝

⎛

∂

∂

+−⎟ ⎠

⎞

⎜⎝

⎛

∂

∂

=−⎟ ⎠

⎞

⎜⎝

⎛

+∂

∂

=−

VPαT

V P

P

=⎟ ⎠

⎞⎜⎝

⎛

∂

∂

Expansion

Coefficient

VΠα T

Internal Pressure

( )

PT

P TΠ because,

T

P VTα

VΠPαCC

V

T

V

TVP

−⎟ ⎠

⎞⎜⎝

⎛

∂

∂=⎟

⎠

⎞⎜⎝

⎛

∂

∂=

+=−∴

[ ] R nCC T)R nPV(using nR VP VP =−⇒==⎟

⎞

⎜

⎛ ∂

T P ⎠⎝ ∂


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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



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 ∆F dF

2

1

−≡=∫ if F is a state function

( ) pathF

F∆F dF

2

1

=∫if F is not a state function. Subscript denotes thename of the path, for example, reversible (rev),chosen for thermodynamic transformation.

dF

F∆

FanddF ∆


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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 0∆q =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

0∆PPP gasext →≡−For 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,

0∆q =

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

∫∫ −=⇒−=2

1

2

1

V

Vgasrev

V

Vext dVPw dVPw ⎟

⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ≈

tion.transforma

reversibleafor,PP gasextQ

TR nVPgas =

( ) ]isothermalande[reversibl V/VlnnRT

V

dV nRT w 12

V

Vrev

2

1

−=−=⇒ ∫

0.w ,VV rev12 >⇒<


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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.0∆q =

( )dTTCdUdw dwdqdU Vrevrevrev ==⇒+=

( )∫=∴

2

1

T

T Vrev dT TC∆

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.

∫∫ −=∴2

1

2

1

V

V

T

T

V

VdV

TdT

nR C

( )V

dVnR TdTTC V −=

VV nR/C

2

112

/nR C

1

2

2

1

VVTT

TTln

VVln ⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ =⇒⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ =⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

We will use this for studyingEntropy change in a Carnot Engine.

.0∆q =

( ) PdVdTTC dUdw dwdqdU Vrevrevrev −=⇒=⇒+=


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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

• Direct path, A

• Path B + Path C• Path D + path E

→V

↑

P

( ) dT TCdU V=• For ideal gases, we know:

( ) ( )TVPTVP →


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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

→V

↑

P

( ) ( )122111 TVPTVP →

( ) 0dT TCdU V ==∴

revrevrevrevrevrev dwdqdwdq0 dwdqdU −=⇒+=⇒+=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −=∆

1

2

1rev V

V

lnnRTw

0U =∆

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

1

21rev

V

VlnnRT∆w

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ =

1

21rev

V

VlnnRT∆q

• Change in internal energy :

• Change in heat :

• Reversible work done :

( ) ( ) ( )122223111 TVPTVPTVP →→


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August 07, 2006 12

Adiabatic process for ideal gases Path B+C

( ) ( ) ( )122223111 TVPTVPTVP →→

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

→V

↑

P

( ) 0∆q have,weBrev =

21 TT →

( ) ( )∫=∴2

1

T

TVB dT TC∆U

( ) ( )BBrev ∆

U∆

w =

( ) 0∆wCrev =

.TT 12 → ( ) ( )dTTC∆U1

2

T

TVC ∫=⇒

• Change in energy :

• Heat Change:

• Reversible work: ( )∫=2

1

T

TVrev dT TC∆W

( ) ( )CC

rev ∆U∆q =

( )∫=

1

2

T

T Vrev dT TC∆

q

( ) ( ) 0dT TCdT TC∆U1

2

2

1

T

TV

T

TV =+= ∫∫

( ) ( ) ( )122321111 TVPTVPTVP →→


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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

→V

↑

P

( ) ( ) ( )122321111 VVV →→

31 TT →

( ) ( )∫=∴3

1

T

TVD dTTC∆U

( ) ( )121V

V1Drev

VVPdVP∆w2

1

−−=−= ∫

( ) ( ) ∫+−−=3

1

T

T V121Drev dTCVVP∆q( ) .0∆w

Erev =

.TT 13 → ( ) ( )∫=∴1

3

T

TVE dTTC∆U

( ) ( )EE

rev ∆U∆q =

• Reversible work :( ) ( )121Totalrev VVP∆w −−= • heat Change: ( ) ( )121Totalrev VVP∆q −=

• Change in energy : ( ) ( ) ( ) 0dTTCdTTC∆U1

3

3

1

T

TV

T

TVTotal =+= ∫∫


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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 +=−=

( ) 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, it’s easierto heat a colder system, as that increases the Entropy.

/TdqdS rev=

• Hence, the Entropy change is proportional to the supply of Heat.

• 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,



⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ===⇒= ∫∫ 1

2

V

V

V

Vrev

revVVlnR n

VdVR n

Tdq∆S /TdqdS

2

1

2

1

Q

• 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 +=

• Entropy change due to changes inside the system( ) :dS INT• Entropy flow due to interaction with the exterior( ) :dS EXT

( )INTdS( ) ↔= 0dS INT

( ) ↔> 0dS INT

Reversible Process

Irreversible Process

( ) 0dSdS INT ≥=


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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 ≥+=

( ) ( ) 0dS and 0dS INTII

INT

I

≥≥( ) ( ) ( ) 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:

• Heat received by system I from system II( ) :dq INTI

.T andT III

III

dSdSdS +=

• Heat supplied to system I from the outside( ) :dq EXTI

• Heat received by system II from system I =:dq INTII

• Heat supplied to system II from the outside( ) :dq EXTII( )EXTIdq−

=dS totaltheHence,( ) ( )

++ IIEXTII

I

EXTI

T

dq

T

dq ( ) ⎥⎦⎤

⎢⎣

⎡−

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,

( ) ( ) ⎥⎦⎤

⎢⎣

⎡−=

IIIINT

I

INTT

1

T

1dqds

( ) 0T1

T1 when0dq

IIIINT

I>⎥⎦

⎤⎢⎣⎡ −> ( ) 0

T1

T1 when0dq

IIIINT

I⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜

⎝

⎛ =⎟

⎠

⎞⎜⎝

⎛

.

T

1

T

1III ⎟

⎠

⎞⎜

⎝

⎛ −


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Aug. 08, 2006 1



Lecture 03

R


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Aug. 08, 2006 2

Recap …

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.



( ) V

dVR n

T

dT TC

T

dqdS V

rev +==

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ===⇒= ∫∫

1

2V

V

V

V

revrev

V

VlnR n

V

dVR n

T

dq∆S /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 +=• Entropy change due to changes inside the system( ) :dS INT• Entropy flow due to interaction with the exterior( ) :dS EXT

( )INTdS( ) ↔= 0dS INT

( ) ↔> 0dS INT

Reversible Process


( ) 0dSdS INT ≥=

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 ≥+=

( ) ( ) 0dS and 0dS INTIIINTI ≥≥

) ) ) 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:

• Heat received by system I from system II( ) :dq INTI

.T andT III

III

dSdSdS +=

• Heat supplied to system I from the outside:dq EXTI

• Heat received by system II from system I =( ) :dq INTII

• Heat supplied to system II from the outside( ):dq

EXT

II

EXT

Idq−

=dS totaltheHence,( ) ( )

++ IIEXTII

I

EXTI

T

dq

T

dq ( ) ⎥⎦⎤

⎢⎣

⎡−

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,

( ) ( ) ⎥⎦⎤

⎢⎣

⎡−=

IIIINT

I

INTT

1

T

1dqds

( ) 0T1

T1 when0dq

IIIINTI

>⎥⎦⎤⎢⎣

⎡ −> ( ) 0T1

T1 when0dq

IIIINTI

⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ =⎟

⎠

⎞⎜⎝

⎛

.

T

1

T

1III ⎟

⎠

⎞⎜

⎝

⎛ −

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.


58/257

Second La of Thermodynamics


59/257

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.

• That means we can extract heat from a reservoir a temperature,and convert it entirely into work, with no other effect.

• Next, we can convert this work into Heat.

• Next, we can deliver this Heat to another heat reservoir at aTemperature, with no other effect.• The net result is: we have taken heat from a colder reservoir to

a hotter reservoir, with no effect.

• Hence, Clausius is false.

1T

( )122 TTT >

⇒

⇒

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

→V

• ab is isothermal, is const and duringwhich the system absorbs Heat

• cd is isothermal, is const and during

which the system rejects Heat

2T

1T

.q2

.q1• bc and da are adiabatics↓

2q

↓1q

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

↑

P

−=

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 −=

2

1

2 q

q1q

Wη −==

Carnot Engine and Refrigerator


62/257

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 >>>

• means the system absorbs heat and rejects nothing; that is, it

converts entirely into work. This violates Kelvin statement, and hence,

• Suppose, This means the engine absorbs from the reservoir atand from the reservoir at .

• The net amount of Heat converted to work, by the engine, therefore iswhich is equal to because is assumed negative.• We can next convert this work into Heat and then deliver it to the

reservoir at with no other effect.• 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 ≠:0q1 < 2q 2T

1q− 1T

( ),qq 12 −+ 12 qq + 1q

,T21q

.T21T 12 TT >0. Whenceand0,q1 >>

.0qand0qthen0,WIf 21


63/257

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 −=

• Consider a Carnot engine (C) and an arbitrary engine (X), operatingbetween the heat reservoirs at ( ).TT TandT 1212 >

X

1

X

2

X

qqW −=C

1

C

2

C

qqW −=• Work done is:• 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)

• 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 =−=

( )C

1

CX

1

X

total1 q Nq Nq −=

(by the definition of integers)

( ) ( ) ( )total1total1total2total

qqqW −=−=⇒

CCXX

total

W NW NW −=


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Absolute Temperature


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Aug. 08, 2006 15

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.

1θ 2θ

2

1

θ

θ1η −=

2

1

2

1

2

11

,

θ

θ

q

q

q

qη

Also

=⇒−=Q

1,η0 ≤≤

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

≤∫

mqthm .Tm

.T,,T,T N21

L

0q N m ≤

⎟

⎟ ⎞

⎜

⎜⎛

∑

∞→ N

T1m m ⎠⎝ =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 ≥

( )Omq OT

mq mT( )

( )

m

mO

Om

m

O

m

O

m

TqTq

TT

qq =⇒=

OT

( )

∑∑ ====

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 ≤

( ) 0T/q N

1m

mm ≤⇒∑=

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 ≤∑=

( ) 0T/q N

1m

mm ≤−∑=( ) 0T/q

N

1m

mm =∑=

∫ Tdq

Hence, the Entropy is a state function.


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Aug. 11, 2006 1



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 += ( ) ( ) ⎥

⎦

⎤⎢⎣

⎡−=

III

int

I

intT

1

T

1qddS

( )intdS

Entropy flow from the outside Entropy production inside the system

( ) ↔= 0dS INT( ) ↔> 0dS INT

Reversible 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

→V

• ab is isothermal, is const and duringwhich the system absorbs Heat

• cd is isothermal, is const and during

which the system rejects Heat

2T

1T

.q2

.q1

• bc and da are adiabatics↓2q

↓1q

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

↑

P

−=

absorb

reject

12 TT >

Infinitesimal Carnot engine, if transformations are infinitesimal.

Carnot’s Theorem


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Aug. 11, 2006 5

Theorem: No Engine operating between two giventemperatures is more efficient than a Carnot engine.

Proof:• 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

• Let us define two integers such that, CX

X

2

C2

N N

qq =

• 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)

=

−=

(by the definition of integers)

• This implies, the total work done by both machines,0Wtotal ≤ (a la Kelvin)

Carnot’s theorem


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Aug. 11, 2006 6

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 −=

( ) ( ) ( )total1total1total2total

qqqW −=−=⇒12 qqW −=

Recall, for one Carnot cycle,

the work done is:

0q Nq N C1CX

1

X ≥−⇒

.0Wtotal ≤

C

X

X2

C

2

N

N

q

q=

⇒

Efficiency of aCarnot engine

≥ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

X

2

X

1

q

q1⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

C

2

C

1

q

q1

Efficiency of anarbitrary engine

QED

Use:( ) .0q total1 ≥

Temperature Scale


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Aug. 11, 2006 7

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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Aug. 11, 2006 8

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 − 2nq −

W W W

CarnotEngine

CarnotEngine

CarnotEngine

CarnotEngine

Wqq n1n =−+

W

W

W

constant)arbitrary:(xx WTT n1n =−+

For example, we may choose: Kelvin1x W =

Temperature Scale


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Aug. 11, 2006 9

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==⇒++ 1n

n

1n

n

T

T

q

q

1n

n1n

T

T1η

+

+ −=

1n

n

1n

1n1n

q

q1

q

Wη

++

++ −==

⇒≤≤ 1η0 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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Aug. 11, 2006 10

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

dq≤

∫

0T

dq≤∫Clausius Theorem


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Aug. 11, 2006 11

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 ≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ∑

=

∞→ 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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Aug. 11, 2006 12

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 ≥

( )Omq OT

mq mT ( )

m

O

m

O

m

T

T

q

q=

OT( ) ∑∑

==

== 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 ≤

( ) 0T/q N

1m mm

≤⇒

∑= QED

( )

m

m

O

O

m T

qTq =⇒

Clausius Theorem


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Aug. 11, 2006 13

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 ≤∑=

( ) 0T/q

N

1mmm ≤−

∑=

( ) 0T/q N

1m

mm =∑=

∫ 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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Aug. 11, 2006 14

Consider two reversible paths:

For reversible transformations, we have:

That is:

That is, the difference of Entropy, computed along two differentpaths is zero --- so, it’s path independent. Hence, Entropy is astate function.

I

IIPath is reverse of the path II./II

0

T

dq rev =∫0

T

dq

T

dq/II

rev

I

rev =+ ∫∫ 0T

dq

T

dq

II

rev

I

rev =−⇒ ∫∫

QED

Entropy Reference point.


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Aug. 11, 2006 15

Define Entropy:

So, the change in Entropy:

( ) ∫=

A

Ref

rev

T

dq

ASRef is some arbitrary

thermodynamic state

( ) ( ) ∫∫∫∫∫ =+=−=−∴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

• 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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Aug. 11, 2006 16

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

−≤∫

A B

reversible

irreversible

• Consider the cyclic transformation, madeup of (1) Irreversible, plus (2) reverse ofReversible. From Clausius’ theorem:

∫∫∫∫ ≤⇒≤− REVIRR REVIRR Tdq

T

dq 0

T

dq

T

dq( ) ( )ASBS

T

dq

IRR −≤⇒ ∫

QED

( ) ( ) .0ASBS ≥−

• For reversible transformations,

it’s obvious by definition.

Thermodynamic Relations


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Aug. 11, 2006 17

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 ⎟ ⎠

⎞⎜⎝

⎛

∂

∂+=

Heat capacity Internal pressure

dVPV

U

T

1

T

dTC

T

dqdS

T

V ⎥⎦

⎤⎢⎣

⎡+⎟

⎠

⎞⎜⎝

⎛

∂

∂+==

Rules:


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Aug. 11, 2006 18

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, Maxwell’s relations).

yxxh

yg ⎟

⎠ ⎞⎜

⎝ ⎛ ∂∂=⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ ∂∂

wwwz

x

z

y

y

x

⎟ ⎠

⎞⎜⎝

⎛

∂

∂=⎟ ⎠

⎞⎜⎝

⎛

∂

∂⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂ 1

zzxy

yx

−

⎭⎬⎫

⎩⎨⎧ ⎟

⎠ ⎞⎜

⎝ ⎛ ∂∂=⎟⎟ ⎠

⎞⎜⎜⎝ ⎛ ∂∂

1x

z

z

y

y

x

yxz

−=⎟ ⎠

⎞⎜⎝

⎛

∂

∂⎟ ⎠

⎞⎜⎝

⎛

∂

∂

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

∂

∂

Derivation of TdS equations:


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Aug. 11, 2006 19

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

⎥⎦

⎤⎢⎣

⎡+⎟

⎠

⎞⎜⎝

⎛

∂

∂+

⎭⎬⎫

⎩⎨⎧

⎟ ⎠

⎞⎜⎝

⎛

∂

∂

T

C

V

V

T

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟

⎠

⎞⎜⎝

⎛

∂

∂⎟ ⎠

⎞⎜⎝

⎛

∂

∂P

V

U

T

1

T TV

,T

U CV

V ⎟ ⎠ ⎞⎜

⎝ ⎛ ∂∂=

P

T

P T

V

U

VT

−⎟

⎠

⎞⎜

⎝

⎛

∂

∂=⎟

⎠

⎞⎜

⎝

⎛

∂

∂

0.PV

nR T

V

U

T

=−=⎟ ⎠

⎞⎜⎝

⎛

∂

∂

Hence, the internal pressure of of an ideal gas is zero,and we already knew it !!!

VT T

P TP

V

U ⎟

⎠

⎞⎜

⎝

⎛

∂

∂=+⎟

⎠

⎞⎜

⎝

⎛

∂

∂⇒

TdS Equations


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Aug. 11, 2006 20

Recall:

Similarly, we can show:

We can also write TdS equations, purely in terms ofexperimental observables:

dVPV

U

T

1

T

dTCdS

T

V ⎥⎦

⎤⎢⎣

⎡+⎟

⎠

⎞⎜⎝

⎛

∂

∂+=

VT T

P TP

V

U ⎟

⎠

⎞⎜⎝

⎛

∂

∂=+⎟

⎠

⎞⎜⎝

⎛

∂

∂

dTCV dVT

PT

V

⎟ ⎠

⎞⎜⎝

⎛

∂

∂+TdS =⇒

dTCP dPT

VT

P

⎟ ⎠

⎞⎜⎝

⎛

∂

∂−TdS =

TdS Equations with exptal observables


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Aug. 11, 2006 21

Define:

We can then show:

PT

V

V

1α ⎟

⎠

⎞⎜⎝

⎛

∂

∂=

T

TP

V

V

1

κ ⎟ ⎠

⎞

⎜⎝

⎛

∂

∂

−=

S

SP

V

V

1κ ⎟

⎠

⎞⎜⎝

⎛

∂

∂−=

Coefficient of thermal expansion

Isothermal compressibility

adiabatic compressibility

T

1

TP

1

TPV κ

α

P

V

T

V

P

V

V

T

T

P=⎥

⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛

∂

∂⎟ ⎠

⎞⎜⎝

⎛

∂

∂−=⎥

⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛

∂

∂⎟ ⎠

⎞⎜⎝

⎛

∂

∂−=⎟

⎠

⎞⎜⎝

⎛

∂

∂ −−

TdS =Q dTCV dVTPT

V

⎟ ⎠ ⎞⎜

⎝ ⎛ ∂∂+

TdS =⇒ dTCV dVκ αT

T

+

TdS = dTCP

αTVdP−Similarly,

Entropy of Phase Transition


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Aug. 11, 2006 22

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,

Trouton’s 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

∆H∆S =

( ) ( )

( ) ( )meltingc,endothermi 0∆H

freezing,exothermic 0∆H

Transition

Transition

e.g.,

e.g.,

>

<

Entropy as a function Temperatured


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Aug. 11, 2006 23

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 =

( ) ( ) ∫+=2

1

T

T

rev12

T

dqTSTS

( )dT.TCdq Prev =

( ) ( ) ( )T

dT TCTSTS

2

1

T

T

P12 ∫+=

( ) ( ) ( )⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ +=+=

∫ 12

P1

T

TP12 T

TlnCTS

T

dTCTSTS

2

1


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August 14, 2006 1



Lecture 05


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August 14, 2006 2

Happy Independence Day

Syllabus --- Equilibrium Thermodynamics


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August 14, 2006 3

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; Maxwell’s Relations;

Free Energy and Chemical Equilibria; Gibbs-Helmholtz andvan’t 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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August 14, 2006 4

• Physical Chemistry ThroughProblems

--- S.K. Dogra and S. Dogra

Reference: Statistical Mechanics by Kerson Huang

Recap …

W d Cl i I li 0dq∫


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August 14, 2006 5

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 ≤∫

( )

∫=

A

Ref

rev

T

dqAS

( ) ( ) ∫∫∫∫∫ =+=−=−B

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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August 14, 2006 6

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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August 14, 2006 7

Model:

gas

Reservoir, T

Piston

Spring

↑

P

→V1V 2V

T constant line

Work done


Computation: Temperature is constant so the change in


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August 14, 2006 8

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:

∆qV

VlnRTW

1

2 ≡=

( )1

2gas

V

VlnR

T

∆q∆S ==

∆q

( )1

2

Reservoir V

VlnR

T

∆q∆S −=−=


h l h


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August 14, 2006 9

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 ∆S∆S∆S

1

2

1

2Reservoir GasTotal =−=+=

Entropy Change: Irreversible Free Expansion


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August 14, 2006 10

Model:

Before After

Water Bath, T constant Water Bath, T constant

Gas occupies volume,1V Gas occupies volume, 2V

Joule’s 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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August 14, 2006 11

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 ∆S =

( ) 0∆S Reservoir =

( ) ( ) ( )1V

2Reservoir GasTotal

VlnR ∆S∆S∆S =+=

Entropy Change: Comparison

Reversible Isothermal Expansion Irreversible Free Expansion


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August 14, 2006 12

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

∆q∆S ==

( )1

2Reservoir

V

VlnR

T

∆q∆S −=−=

( ) 0∆STotal

=∴

1

2

V

VlnRTW =

Gets storedin the Spring

( )

1

2gas

V

VlnR

T

∆q∆S ==

( ) 0∆S Reservoir =

( )Total∆S TW = Gets wasted

( )1

2Total

V

VlnR ∆S =

Entropy as a function Temperature

Recall:dq

d


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August 14, 2006 13

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 =

( ) ( ) ∫+=2

1

T

T

rev12

T

dqTSTS

( )dT.TCdq Prev =

( ) ( ) ( )T

dT TCTSTS

2

1

T

T

P12 ∫+=

( ) ( ) ( )⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ +=+=

∫ 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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August 14, 2006 14

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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August 14, 2006 15

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

∆H∆S =

( ) ( )

( ) ( )melting c,endothermi 0∆H

freezing ,exothermic 0∆H

Transition

Transition

e.g.,

e.g.,

>

<

Trouton’s Rule

Standard entropy of vaporization is almost the same for a wide


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August 14, 2006 16

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 −−

( ) ( ) ( ) ( ) 11-Point-Boiling

onVaporizati

onVaporizati

Transition

TransitionTransition molK J85

T∆H∆S

T∆H∆S −≈=∴=Q

Trouton’s Rule

Example: The normal boiling point of Chloroform is C61.5°


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August 14, 2006 17

Example: The normal boiling point of Chloroform is .Calculate the molar heat of vaporization, from Trouton’s rule.

Answer:

C61.5

( ) ( )K 61.5273molK 85J 11 +×−−

( ) ( )

( ) Point-Boiling11-

onVaporizati

11-

Point-Boiling

onVaporizati

onVaporizati

TmolK J85∆H

molK J85T∆H∆S

×=⇒

≈=

−

−Q

Example

Calculate the increase in the Entropy if one mole of Krypton is


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August 14, 2006 18

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 °27 C °227

11

molK J −−

( )( ) ( )( )K 227273K 27273lnmolK J 8.3141.51mol

TTlnCn∆S 11

1

2V

++×== −−

( ) [ ]( ) ( )( )K 227273

K 27273lnmolK J 8.3141.01.51mol

T

TlnCn∆S 11

1

2P

+

+×+== −−

Example

One mole of the ideal gas at 3 atm and 300 K is expanded (1)


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August 14, 2006 19

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 VVP∆VPW 112extext =−==

( )( )( )⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ = −−

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

Wworkη

2

===

( ) ( )K 273227321Tη1T TT1η 2121

+⎟ ⎠

⎞⎜⎝

⎛ −=−=∴−=Q

Thermodynamic Potentials

Helmholtz Free Energy, A: STUA −= 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 −=

VPAG +=TSHVPTSUG −=+−=⇒

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

A−≤∫

S ∆≤⇒≤ T∆q ∆ST

∆q

ationtransformtheduring

absorbedHeat:∆q

( ) ( )ASBS∆S −=

WdU∆q +=

( )STUW STWU ∆−∆−≤⇒∆≤+∆

A∆W −≤ A: Helmholtz Free Energy

Potentials --- Physical Meaning

Ref: For an arbitrary isothermal transformation.


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Physica

Kinetics and Thermo by Amarendra Vijay

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