sem5 thermal nota v1 dr wan
TRANSCRIPT
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WEEK 1
Thermal & Statistical Physics
Introduction
Thermodynamics
Definition
-Thermodynamics: is the study of the macroscopic
behaviour of physical systems under the influence of
exchange of work and heat with other systems or their
environment
Thermodynamics can be defined as the science of energy.
Energy can be viewed as the ability to cause changes.
Classical thermodynamics
-thermodynamic states and properties
-energy, work, and heat
-with the laws of thermodynamics
- PV = k, a constant --- R Boyle
The 1st and 2nd laws of thermodynamics
-simultaneously in the 1850s
-W Rankine, R Clausius, and W Thomson (Lord Kelvin).
Statistical thermodynamics
-Late 19th century -- molecular interpretation of thermodynamics
-bridge between macroscopic and microscopic properties of
systems.
-The statistical approach is to derive all macroscopic properties
(T, V, P, E, S, etc.) from the properties of moving constituent
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particles and the interactions between them (including quantum
phenomena)
Note:
-The number of the elements can be very large --- impossible to
keep track of the behaviour of each element.
-A statistical property is a single measurement which gives a
physical picture of what is occurring in all the individual parts.
Chemical thermodynamics
-Is the study of the interrelation of heat with chemical reactions or
with a physical change of state within the confines of the laws of
thermodynamics.
Thermodynamic systems
STATISTICAL
MECHANICS
QUANTUM MECHANICS OF
ATOMS & MOLECULES
MARCOSCOPIC PROPERTIES:
Large number of molecules
TIME
DEPENDENT
BEHAVIOUR:
Chemical
Kinetics
EQUILIBRIUM
PROPERTIES:
Thermodynamics
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-System is the region of the universe under study.
-Surroundings - everything in the universe except the system.
-Boundary -separates the system with the remainder of the
universe.
-fixed, moveable, real, and imaginary
There are five dominant classes of systems:
1. Isolated Systems – matter and energy may not cross the
boundary.
2. Adiabatic Systems – heat must not cross the boundary.
3. Diathermic Systems - heat may cross boundary.
4. Closed Systems – matter may not cross the boundary.
5. Open Systems – heat, work, and matter may cross the
boundary
Thermodynamic parameters
Energy transfers between thermodynamic systems as the result of
a generalized force causing a generalized displacement –conjugate
variables.
SYSTEM
SURROUNDINGS
BOUNDARY
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The most common are
Pressure-volume (mechanical parameters)
Temperature-entropy (thermal parameters)
Chemical potential-particle number (material parameters)
Thermodynamic instruments
Two types: the meter and the reservoir.
A thermodynamic meter is any device which measures any
parameter of a thermodynamic system.
-The zeroth law --it is possible to measure temperature.
-An idealized thermometer--ideal gas at constant pressure.
-a barometer-constructed from a sample of an ideal gas held
at a constant temperature.
-a calorimeter is a device which is used to measure and
define the internal energy of a system.
A thermodynamic reservoir is a very large system does not alter
its state parameters when brought into contact with the test
system.
The earth's atmosphere is an example of heat reservoir.
Thermodynamic states
State - a system is at equilibrium under a given set of conditions
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The state of the system can be described by a number of intensive
and extensive variables.
The properties of the system can be described by an equation of
state which specifies the relationship between these variables.
Thermodynamic processes
-the energetic evolution of a thermodynamic system proceeding
from an initial state to a final state.
The seven most common thermodynamic processes are;
1. An isobaris process occurs at constant pressure.
2. An isochoric process, or isometric/isovolumetric process,
occurs at constant volume.
3. An isothermal process occurs at a constant temperature.
4. An adiabatic process occurs without loss or gain of heat.
5. An isentropic process (reversible adiabatic process) occurs at
constant entropy.
6. An isenthalpic process occurs at a constant enthalpy. Also
known as a throttling process.
7. A steady state process occurs without a change in the internal
energy of a system.
The laws of thermodynamics
Classical thermodynamics is based on the four laws of
thermodynamics:
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• Zeroth law of thermodynamics, stating that thermodynamic
equilibrium is an equivalence relation – about temperature
and temperature scale.
• First law of thermodynamics is about the conservation of
energy -- deals with macroscopic properties, work, energy,
enthalpy, etc
• Second law of thermodynamics is about entropy
• Third law of thermodynamics is about absolute zero
temperature --the determination of entropy.
Thermodynamic potentials
-the quantitative measure of the stored energy in the system.
The five most well known potentials are:
Internal energy U
Helmholtz free energy A + U – TS
Enthalpy H = U + PV
Gibbs free energy G = U + PV – TS
Grand potential
Potentials are used to measure energy changes in systems as they
evolve from an initial state to a final state.
Note: the term thermodynamic free energy is a measure of the
amount of mechanical (or other) work that can be extracted from
a system.
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The Kinetic Theory of Gases
Kinetic theory or kinetic theory of gases -- explain macroscopic
properties of gases, such as P, T, or V, by considering their
molecular composition and motion.
Ideal gases kinetic theory of
-modeling the gases as molecules (or atoms) in constant motion in
space
-A mathematical explanation of the behaviour of gases
-the KE depending on the temperature of the gas
The kinetc theory makes seven assumptions:
The volume occupied by the
gas molecules themselves is
negligible compared with the
volume of space between them
All the particles that
make up the gas are
identical
The distribution of
energy amoung
particles is random
There are sufficent
numbers of molecules for
the statistical average to
be meaningful.
Collisions are
all perfectly
elastic.
The molecules
travel in straight
lines between
collisions
Newtonian
mechanics can be
applied to molecule interactions
Assumptions
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The kinetic theory of gases -- deduced equations that related the
easily observable properties such as P, ρ, V and T to properties
not easily or directly observable -such as the sizes and speeds of
molecules.
Pressure
• The pressure of a gas is caused by collisions of the molecules
of the gas with the walls of the container.
• The magnitude of the pressure is related to how hard and how
often the molecules strike the wall
Absolute Temperature
• The absolute temperature of a gas is a measure of the average
kinetic energy of its' molecules
• If two different gases are at the same temperature, their
molecules have the same average kinetic energy
• If the temperature of a gas is doubled, the average kinetic
energy of its molecules is doubled
Favg
m vx
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Molecular Speed
• All the molecules ⇒average kinetic energy (and therefore an
average speed)
• the individual molecules move at various speeds, ⇒exhibit a
DISTRIBUTION of speeds
• Collisions can change individual molecular speeds but the
distribution of speeds remains the same.
• At the same temperature, lighter gases move on average
faster than heavier gases.
• The average kinetic energy, ν, is related to the root mean
square (rms) speed u
2mu2
1υ =
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Statistical mechanics
Statistical mechanics is the application of probability theory to the
field of mechanics, which is concerned with the motion of
particles or objects when subjected to a force.
-relating the microscopic properties of individual atoms and
molecules to the macroscopic or bulk properties of materials that
can be observed in everyday life
-it can be used to calculate the thermodynamic properties of bulk
materials from the spectroscopic data of individual molecules.
The fundamental postulate in statistical mechanics:
Given an isolated system in equilibrium, it is found with
equal probability in each of its accessiblemicrostates.
This postulate is necessary because it allows one to conclude that
for a system at equilibrium, the thermodynamic state (macrostate)
which could result from the largest number of microstates is also
the most probable macrostate of the system.
• Fundamental concepts: microstate, macrostates, accessible
states, ensemble, the fundamentals postulate
• Microcanonical and canonical ensembles and their
applications
• Open systems, chemical potential. Grand canonical ensemble
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• Distinguishable and indistinguishable particles
• Bose-Einstein and Fermi-Dirac statistics and their
applications
Thermal Physics
Thermal physics is the study of the statistical nature of physical
systems from an energetic perspective.
It include,
i. Thermodynamics – macroscopic theory, essentially
completed by Carnot 1824
ii. Kinetic theory – microscopic theory of molecular
distribution
i. Maxwell
ii. Boltzmann
Classical Mechanics
Quantum Mechanics
Thermodynamics
Statistical
Mechanics
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iii. Ensemble theory of statistical mechanics – all
thermodynamics informations can be derived from the
partition function – formulated by Gibbs in 1902
Macroscopic & Microscopic systems
-macroscopic when it is large enough to be visible in the ordinary
sense
-microscopic if it is roughly of atomic dimensions, or smaller
If N is the number of particles in the system
- A system is macroscopic if
1>>N
- And is microscopic if the other way around.
Microscopic descriptions
Describes the system in highest degree of detail. (e.g. positions
and velocities/momentums of all the atoms in the system).
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Macroscopic descriptions
Overall state of the system: averaging over large number of
entities (atoms or molecules). (e.g. pressure).
Instruments measure averages over a large number of atoms or
molecules.
Understanding of a particular system: to relate macroscopic
measurements to theoretical descriptions of the microscopic
entities.
The Microscopic View
Thermodynamics properties like T & P can be related to
microscopic quantities.
e.g T --in terms of a molecular velocity.
P --collisions of molecules with the walls of a container.
Levels of Description and Observation:
Mechanics: small number of particles (single molecule): single
simulation --one single molecule experiment
Stochastisch: several particles (molecules): several simulations
-- several single molecule Experiments
Statistics: huge number of particles and events: macroscopic
measurement
Wave – particle duality
-light and matter exhibit properties of both waves and particles
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Einstein: the electrons were knocked free of the metal by incident
photons, with each photon carrying an amount of energy E that
was related to the frequency, ν of the light by
E = h ν
h: Plank’s constant (6.626 x 10-34 J seconds).
de Broglie: all matter has a wave-like nature; and related
wavelength, λ and momentum, p:
p
hλ =
note: for photon; p = E / c and λ = c / ν.
where c is the speed of light in vacuum
Uncertainty principle
Heisenberg: certain specific pairs of variables cannot be measured
simultaneously with high accuracy.
e.g: within an atom - possible to measure the position (∆x), or the
momentum (∆p), of a subatomic particle (electron,..) but not
possible to measure both of them at the same time.
-because the measuring process interferes to a substantial degree
with what is being measured.
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Formulation and characteristics
At atomic dimensions
–particle is not like a hard sphere
-the smaller the dimension→the more wave-like it becomes
Using light to identify the location or motion of an electron, the
photon of light will influence the electron's motion and position.
Mathematically, the uncertainty principle looks like:
h∆x∆p ≥
Where ∆x = the uncertainty in position
∆p = the uncertainty in momentum
h= Planck's constant
A sine wave:
λλλλ - wavelength
Momentum is precisely known, p = h / λ
The position, ∆x is unknown
Adding several waves of different λ
-Localize the wave (position is certain)
-The momentum is uncertain (each different wavelength
represents different momentum)
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Application
-to caculate energy which would be required to keep a
particle within a given volume.
Example: energy to keep an electron inside an atom.
Assume: the diameter of the atom = 4 A = ∆x
∆p = h / ∆x
h = 6.626 x 10-34 J-s
∆p = h / ∆x = 6.626 x 10-34 J-s / 4 x 10-10 m
= 1. 6565 x 10-24 kgm/s
E = ½ mV2 = (mV)2/(2m) = p2/(2m)
m= 9.11 x 10-31 kg (electron mass)
1 ev = 1.6 x 10-19 J
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( )( )
ev9.4
J/ev1.6x10kg9.11x102
kgm/s1.6565x10E
1931
24
=
=−−
−
Quantized Energy States
Max Planck suggested that energy is transferred in “packets”
called quanta (plural).
Quantum: the smallest quantity of energy that can be emitted or
absorbed as electromagnetic energy
Bohr model (1913)
-electrons were orbiting the nucleus
-when a charge traveling in a circular path should lose energy
by emitting electromagnetic radiation
-it should end up spiraling into the nucleus (which it does not).
Classical physical laws inadequate to explain the inner
workings of the atom
-idea of quantized energy from Planck
-only orbits of certain radii, corresponding to defined
energies, are "permitted"
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-An electron orbiting in one of these "allowed" orbits:
Has a defined energy state
Will not radiate energy
Will not spiral into the nucleus
Electron energy:
RH = 2.18 x 10-18 J Rydberg constant
n: the principle quantum number, n = 1, 2, 3, …..
n --corresponds to the different allowed orbits for the electron.
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Emission of light corresponds to a transition of the atoms between
two states
Frequency of emitted light
h
EEυ&EE∆Ehυ fi
fi
−=−==
Consequence of Bohr's deduction
1. Each atom can be represented by an energy diagram
2. Distance between two energy levels gives a particular
νννν the energy of atoms is quantized
The implications --only certain photon energies are allowed
when electrons jump down from higher
levels to lower levels
→ producing the spectrum.
Energy
Ei
Ef
Light, ν
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An electron transition between quantized energy levels with
different quantum numbers n yields a photon by emission with
quantum energy Ephoton.
Ephoton = hν = E2 – E1
eVn
1
n
113.6
n
1
n
1
h
me2πhν
2
2
2
1
2
2
2
1
2
42
−=
−=
for transition from n = 2 to n = 1
n = 2
n = 1
E2
E1
n = 5
n = 4
n = 3
n = 2
The hydrogen spectrum
The energy levels
n = 1
n = 2
n = 3
n = 4
n = 5
Ionization
energy
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eV10.24
1113.6h ν =
−=
ν = 10.2 eV / h
1 eV = 1.6 x 10-19 J h = 6.626 x 10-34 J.s
ν = (10.2 x 1.6 x 10-19 J) / (6.626 x 10-34 J.s)
= 121.95 nm (UV)
Planck’s theory:
- Energy is always emitted or absorbed in whole number
multiples of hν (i.e hν, 2hν, 3hν)
-the energy levels that are allowed are ‘quantized.’
(restricted to certain quantities or values)
Quantum Harmonic Oscillator
A diatomic molecule vibrates;
- Potential energy that depends upon the square of the displacement
from equilibrium.
-But the energy levels are quantized at equally spaced values.
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The energy levels of the quantum harmonic oscillator are
π
πω
ω
2
.'
)(2
,...3,2,1,02
1
constsPlanck
frequency
nnEn
=
=
=
+=
h
h
The quantum harmonic oscillator
-the foundation for the understanding of complex modes of
vibration in larger molecules, the motion of atoms in a solid
lattice, the theory of heat capacity, etc.
Statistical Mechanics
-relates the macroscopic thermodynamic properties of a system to
the ensemble behaviour of its components.
-Each thermodynamic state of the system (represented by
particular values of the state functions) is termed a macrostate.
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-For each macrostate, there can be many corresponding
microstates.
-requires a large number of particles (1 mol = 6.022 x 1023
particles)
-water = 55 mol/l, i.e. 3.3 x 1025 water molecules per liter
Thermal Concept
Temperature is the average kinetic energy per molecule of the
molecules in the substance.
Heat is the energy transferred between objects when they
change temperature, and moves from areas of high temperature to
areas of low temperature
Internal energy is the total energy related to the thermal motion
of the molecules in a substance --vibrational & translational
δW – all the levels are raised by δεi. No change in populations
δεI –No change in the energy level εi. Populations are change by
dNi
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Distribution Function (Binomial, Gaussian,..)
Probability (Mean, standard deviations,
permutation,…)
Applied
Statistics
Quantum Statistics
For system of
particles
For Fermions For Bosons
Using:
Density of state
&
Distribution function
Energy
Distribution
Follow:
The Fermi-Dirac Dist.
Appl.ication:
Electrons in metal
Follow:
The Bose-Einstein Dist.
Describe
Bose-Einstein Cond. &
Thermal radiatian
Application:
Superconductivity,
Liquid helium, ..
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Note
ωψχφ
υτσρ
πξν
µλκι
θηζε
δγβα
ΩΨΧΦ
ΥΣ
ΠΞ
Λ
Θ
∆Γ
OmegaPsiChiPhi
UpsilonTTauSigmaPRho
PioOOmicronXiNNu
MMuLambdaKKappaIIota
ThetaHEtaZZetaEEpsilon
DeltaGammaBBetaAAlpha
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Lect 2 SSF 2273 Statistical Mechanics: Introduction Statistical mechanics is a branch of physics concerned with interpreting and evaluating the properties of microscopic systems (electrons, ions, atoms, molecules,…particles) to permit the prediction of the properties of the macroscopic systems embodying these elementry ones
The Microscopic View Thermodynamics properties like T & P can be related to microscopic quantities. e.g T can be expressed in terms of a molecular velocity. P can be understood by picturing the miniature collisions of
molecules with the walls of a container. Wave – particle duality
-light and matter exhibit properties of both waves and of particles
Einstein: the electrons were knocked free of the metal by incident photons, with each photon carrying an amount of energy E that was related to the frequency, of the light by
E = h ν
h: Plank’s constant (6.626 x 10-34 J seconds).
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de Broglie: all matter has a wave-like nature; and related wavelength, λ and momentum, p:
ph
=
note: for photon; p = E / c and = c / .
where c is the speed of light in vacuum
Uncertainty principle Heisenberg: certain specific pairs of variables cannot be measured simultaneously with high accuracy. e.g: within an atom - possible to measure the position (∆x), or the momentum (∆p), of a subatomic particle (electron,..) but not possible to measure both of them at the same time. - because the measuring process interferes to a substantial degree with what is being measured. Formulation and characteristics at atomic dimensions
– particle is not like a hard sphere -because the smaller the dimension -the more wave-like it becomes
using light to identify the location or motion of an electron, the photon of light will influence the electron's motion and position.
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Mathematically, the uncertainty principle looks like:
hxp ≥
Where ∆x = the uncertainty in position ∆p = the uncertainty in momentum h= Planck's constant
A sine wave: λ - wavelength Momentum is precisely known, p = h / λ The position, ∆x is unknown Adding several waves of different λ Localize the wave (position is certain)
The momentum is uncertain (each different wavelength represents different momentum)
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Application -to caculate energy which would be required to
contain a particle within a given volume.
Example: energy to keep an electron inside an atom. Assume: the diameter of the atom = 4 A = ∆x ∆p = h / ∆x h = 6.626 x 10-34 J-s ∆p = h / ∆x = 6.626 x 10-34 J-s / 4 x 10-10 m = 1. 6565 x 10-24 kgm/s E = ½ mV2 = (mV)2/(2m) = p2/(2m) m= 9.11 x 10-31 kg (electron mass) 1 ev = 1.6 x 10-19 J
( )( )ev9.4
J/ev1.6x10kg9.11x102kgm/s1.6565x10
E1931
24
=
= −−
−
Quantized Energy States Max Planck suggested that energy is Quantum
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The electrons in free atoms can be found in only certain discrete energy states. These sharp energy states are associated with the orbits or shells of electrons in an atom, e.g., a hydrogen atom.
The implications of these quantized energy states is that only certain photon energies are allowed when electrons jump down from higher levels to lower levels, producing the spectrum.
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e.g Hydrogen Energy Levels
an electron transition between quantized energy levels with different quantum numbers n yields a photon by emission with quantum energy Ephoton.
Ephoton = hν = E2 – E1
n = 2
n = 1
E2
E1
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eVn1
n1
13.6n1
n1
hme2
h 22
21
22
21
2
42
−=
−=
for transition from n = 2 to n = 1
eV10.241
113.6h =
−=
ν = 10.2 eV / h 1 eV = 1.6 x 10-19 J h = 6.626 x 10-34 J.s ν = (10.2 x 1.6 x 10-19 J) / (6.626 x 10-34 J.s) = 121.95 nm (UV)
Planck’s theory: -υυυ υ
-the energy !"#$
(restricted to certain quantities or values)
Quantum Harmonic Oscillator A diatomic molecule vibrates; - potential energy that depends upon the square of the displacement from equilibrium. -But the energy levels are quantized at equally spaced values.
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8
The energy levels of the quantum harmonic oscillator are
The quantum harmonic oscillator
-the foundation for the understanding of complex modes of vibration in larger molecules, the motion of atoms in a solid lattice, the theory of heat capacity, etc.
quantized State & phase space From uncertainty prin. – can only identify a particle as being within a box of area ∆x∆px = h
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if position --- x, y, z momentum --- px, py, pz the particle coordinate - - in a six dimensional space ---- “phase space” uncertainty of the coordinate ∆x∆px = h ∆y∆py = h ∆z∆pz = h or ∆x∆px ∆y∆py ∆z∆pz = h3 -the particle is located within a six-dim. Quantum box or quantum state of volume h3
∆px
∆x
x
P
∆px
∆x
x
P
∆px
∆x
x
P
or
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Let say position and momentum a particle are, 0 < x < xo 0 < p < po the number of quantum state available to the particle
hpx
stateoneofareaareatotal
stateof(No. oo==
in 6-D no of quantum state = (∆x ∆y ∆z∆px ∆py ∆pz ) / h3 = (Vr Vp ) / h3 = (1/ h3) dxdydz dpxdpydpz
appro =3
33
hpdrd
statequantumno .
∆px
∆x
x
Px h
xo
po
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11
2
1
0 1 2 3 4 5 microstate
E
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FROM MICROSCOPIC
TO MACROSCOPIC BEHAVIOUR
1
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THE GOAL: To understand the properties of MACROSCOPIC
SYSTEMS (systems of many electrons, molecules, photons, or other constituents).
Familiar examples:
1. Air in your room (gas)
2. A glass of water (liquid)
3. A coin (solid)
4. A rubber band (polymer)
2
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Less familiar examples:
1. superconductors
2. cell membranes
3. the brain
4. the stock market
5. neutron stars
3
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Macroscopic system: the typical
questions asked.. 1. How does the pressure of a gas depend on the
temperature and the volume of its container?
2. How does a refrigerator work? How can we make it more efficient?
3. How much energy do we need to add to a kettle of water to change it to steam?
4
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Q1-Q3:
Concerned with macroscopic properties such as Pressure, Volume, Temperature and processes related to heating and work.
Relevant to thermodynamics which provides a framework for relating the macroscopic properties of a system to one another.
Thermodynamics is concerned only with macroscopic quantities and ignores the microscopic variables that characterize individual molecules.
Many of the applications of thermodynamics are to engines (internal combustion engines; steam turbines)
5
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4. Why are the properties of water different from those of steam, even though water and steam consist of the same type of molecules?
5. How and why does a liquid freeze into a particular crystalline structure?
6. Why does helium have a superfluid phase at very low temperatures? Why do some materials exhibit zero resistance to electrical current at sufficiently low temperatures?
7. In general, how do the properties of a system emerge from its constituents?
6
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Q4-Q7:
Relate to understanding the behaviour of macroscopic systems starting from the atomic nature of matter.
e.g: water consists of molecules of hydrogen and oxygen
H2O
The laws of classical mechanics (Newton’s laws) & quantum mechanics determine the behaviour of molecules at the microscopic level.
7
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8. How fast does the current in a river have to be before its flow changes from laminar to turbulent?
9. What will the weather be tomorrow?
8
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Question 8 also relates to a macroscopic system, but temperature is not relevant in this case.
Moreover, turbulent flow continually changes in time.
Question 9 concerns macroscopic phenomena that change with time. Although there has been progress in our understanding of time-dependent phenomena such as turbulent flow and hurricanes, our understanding of such phenomena is much less advanced than our understanding of time-independent systems.
9
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Goal of Statistical Mechanics :
Begin with the microscopic laws of physics that govern the behaviour of the constituents of the system and deduce the properties of the system as a whole.
Statistical Mechanics is a bridge between the microscopic and macroscopic worlds.
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For this reason we will focus our attention on systems whose macroscopic properties are independent of time and consider questions such as those in Questions 1–7.
11
The statistical approach is to derive all
macroscopic properties (T, V, P, E, S, etc.) from
the properties of moving constituent particles and
the interactions between them (including quantum
phenomena)
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Quantum (Quanta- plural)
14
In physics, a quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction. Behind this, one finds the fundamental notion that aphysical property may be "quantized," referred to as "the hypothesis of quantization".[1] This means that the magnitude can take on only certain discrete values.
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A photon is a single quantum of light, and is
referred to as a "light quantum". The energy of
an electron bound to an atom is quantized,
which results in the stability of atoms, and hence
of matter in general.
As incorporated into the theory of quantum
mechanics, this is regarded by physicists as part
of the fundamental framework for understanding
and describing nature at the smallest length-
scales.
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Wave-particle duality
Every quantum object is both a particle, like a tennis ball,
and some sort of wave, like a wave in the ocean. Imagine
a tennis ball that would be in several different places at
the same time. But when you try to locate it with
measuring instruments, the quantum object is suddenly
reduced to one spot. This means that electrons, atoms,
molecules, and even photons (light particles), are small
particles and waves, both at the same time!
This is the basic property of the quantum world.
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PRACTICAL APPLICATIONS
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Quantum mechanics and CCD display
All our digital cameras are manufactured with CCD
(charge-coupled device) image sensors that work
using quantum mechanics. Light is made up of
photons, which behave as particles and waves at
the same time. When light falls upon a CCD sensor,
each photon — when it has enough energy — pulls
out one electron. These electrons are thus detected
and converted into a digital image.
This technology has enabled us to develop ultra-
sensitive cameras.
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IN LABORATORY
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Using electrons to observe matter
In order to overcome the limitations of optical
microscopes, physicists use quantum mechanics.
Since electrons are — like light — waves, why not
use them to light up objects? This is what lies
behind the electron microscope, which has an
amazing magnifying power.
It is one of the most popular microscopes
amongst physicists, who want to observe atoms
and molecules, and amongst biologists, who want
to observe microbes and viruses.
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http://www.youtube.com/watch?feature=player
_embedded&v=wsq7qXr9Hl0
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STATISTICAL MECHANICS Statistical mechanics is a branch of theoretical
physics (and mathematical physics) that studies, using PROBABILITY THEORY, the average behaviour of a mechanical system where the state of the system is uncertain.
Statistical mechanics is a collection of mathematical tools that are used to fill this disconnection between the laws of mechanics and the practical experience of incomplete knowledge.
IT IS A BRIGDE BETWEEN THE MICROSCOPIC AND MACROSCOPIC WORLDS.
23
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24
Note:
-The number of the elements can be very
large --- impossible to keep track of the
behaviour of each element.
-A statistical property is a single
measurement which gives a physical
picture of what is occurring in all the
individual parts.
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2727
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29
Thermodynamic potentials
-the quantitative measure of the stored energy in the system.
The five most well known potentials are:
Internal energy U
Enthalpy H = U + PV
Helmholtz free energy A + U – TS
Gibbs free energy G = U + PV – TS
Grand potential
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30
Potentials are used to measure energy
changes in systems as they evolve from an
initial state to a final state.
Note: the term thermodynamic free energy
is a measure of the amount of mechanical
(or other) work that can be extracted from a
system.
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TSP: WEEK 2
1
WEEK 2
Statistical Mechanics: statistics for small system Probability
Mean value
-single element system
Let PS : probability the system in state S
fS : function when the system in state S
f ≅ mean value of f
Definition:
∑=S
SS fPf
Example: coin 2 states: H: head, T: tail
PH: probability of landing heads = ½
PT: probability of landing tails = ½
Let PS : probability the system in state S
fS : function when the system in state S
f ≅ mean value of f
∑=S
SS fPf
Let say f : the number of heat showing
fS = 1 for H & fS = 0 for T
The mean value for f
= PHfH +PTfT = (1/2) (1) + (1/2) (0) = ½
--the average number of heads per coin showing is ½
Example: dice
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TSP: WEEK 2
2
n: no of dot upward = f
if fn = n, what is the mean value for the system?
P1= P2 = P3 = P4 = P5 = P6 = 1/6
∑=S
SS fPf = ∑=S
S nPf =(1/6)(1)+ (1/6)(2)+(1/6)(3)
+ (1/6)(4) + (1/6)(5) + (1/6)(6) = 3 ½
if fn = (n-1)2
∑=S
SS fPf = ( )2
1∑ −=S
S nPf = (1/6)(0) + (1/6)(1)+
(1/6)(4) + (1/6)(9) + (1/6)(16) + (1/6)(25) = 6
19
Note:
f, g : functions of the states of systems
c : constant
Then ( ) gfgf +=+ and fccf =
-few elements system
Let p = the probability the criterion is satisfied
q = the probability the criterion is not satisfied
Example:
i. criterion: a flipped coin lands head-up
p = ½ and q = ½
ii. criterion: a rolled dice lands with 2 dots up
p = 1/6 and q = 5/6
p + q = 1 or q = 1- p
The probability the criterion is satisfied or not is always
one
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TSP: WEEK 2
3
System: 2 identical systems (2 coins)
C1 C2
p1 + q1 = 1 and p2 + q2 = 1
for each element, the criterion is or not satisfied is,
(p1+q1)(p2+q2) = 1x1 = 12 = 1
= p1p2 + p1q2 + q1p2 + q1q2
p1p2 – both elements satisfy the criterion
p1q2 – element 1 –satisfy the criterion
element 2 - not satisfy the criterion
q1p2 - element 1 –not satisfy the criterion
element 2 - satisfy the criterion
q1q2 - both elements not satisfy the criterion
Example
system: a box with 2 air molecules inside
V1 + V2 = V and V1 = V/3
Criterion: all possible configurations
-The probability of either molecule in V1,
p1 = 1/3 & p2 = 1/3
1 2 V1 V2
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TSP: WEEK 2
4
-The probability that each molecule is not in V1,
q1 = 2/3 & q2 = 2/3
-the probability for all possible configuration
(p1+q1)(p2+q2) = 1x1 = 12 = 1
= p1p2 + p1q2 + q1p2 + q1q2
-both molecules in V1: p1p2 = (1/3) (1/3) = 1/9
-mol.1 in V1 & mol.2 in V2: p1q2 = (1/3) (2/3)
= 2/9
-mol.1 in V2 & mol.2 in V1: q1p2 = (2/3) (1/3)
= 2/9
-both molecules in V2: q1q2 = (2/3) (2/3) = 4/9
Example: 2 dice D1 D2
Criterion: one dot up for D1 and D2
prob. For one dot up: p1 = p2 = 1/6
prob. For not landing with one dot up: q1 = q2 = 5/6
(p1+q1)(p2+q2) = 1x1 = 12 = 1
= p1p2 + p1q2 + q1p2 + q1q2 -one dot up for D1&D2: p1.p2 = 1/36
-one dot up for D1& not D2: p1.q2 = 5/36
-one dot up for D2& not D1: q1.p2 = 5/36
- not D1& not D2: q1.q2 = 25/36
Idntical elements: with the same probability
For 2 identical elements
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TSP: WEEK 2
5
p1= p2 = p q1 = q2 = q
The possible configurations;
(p1+q1)(p2+q2) = (p + q)2
= (p2 + 2pq + q2)
p2: all satisfy the criterion
2pq: one satisfy – one does not
q2: none satisfy the criterion
For 3 identical elements
(p1+q1)(p2+q2) (p3+q3) = (p + q)3 = 1
= (p3 + 3p2q + 3pq2 + q2)
p3: all satisfy the criterion
3p2q: two satisfy – one does not
3pq2: one satisfy – two do not
q3: none satisfy the criterion
For N elements
(p1+q1)(p2+q2)…..(pN + qN) = (p + q)N but
( ) 1qp
n)!-(Nn!
N!qp n)(Nn
N
0n
N==+ −
=
∑
…..binomial expansion
PN(n): the probability the system in state of n elements
satisfy the criterion and N-n elements do not
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TSP: WEEK 2
6
nNn
N qpn)!(Nn!
N!(n)P −
−=
)!(!
!
nNn
N
− ≅ Binomial coefficient
-the number of different configurations of the
individual elements, for which n satisfy the
criterion and (N-n) do not
Example: consider 5 molecules in abox
V1 + V2 = V V1 = (1/3) V
(i) find the prob. of 2 mol. in V1 and 3 in V2
p = 1/3 & q = 2/3
N = 5, n = 2
32
53
2
3
1
2)!(52!
5!(2)P
−=
= 80 /243
(ii) number of different possible arrangements for 2
mol in V1 and 3 mol in V2
)!(!
!
nNn
N
−= 5! / (2! 3!) = 10
State of a system relative to two different criteria;
-system of N elements in a state where;
V1 V2
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TSP: WEEK 2
7
n1 elements satisfy 1st criterion
n2 elements satisfy 2nd criterion
the probability PN(n1,n2)
Note: an element’s behaviour with respect to one of the
criteria does not affect its behaviour with respect to the
other ……`statistically independent criteria’
Pij: the system in state i with respect to 1st criterion and also
in state j with respect to the 2nd
Pij = Pi . Pj
Or Pijk….. = Pi . Pj. Pk, …….
Where i, j, k, ….
Example:
System: 5 mol. -find: the probability for 2 mol in V1 and 4 mol in VU
N = 5; ni = 2; nj=4
( ) ( ) nN
2
n
2
nN
1
n
155 qp!n)-(Nn!
N!.qp
!n)-(Nn!
N!.P5(4)2P2.4P −−==
V1 V2
VU
VD
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TSP: WEEK 2
8
051.02
1
2
1
!1!4
!5*
3
2
3
1
!3!2
!5432
=
=
Note:
Binomial expansion – correct for any size
nNn
N qpn)!(Nn!
N!(n)P −
−=
but for large N, N! can be calculated using Stirling’s
formula
ln(m!) ≈ m ln n – m + (1/2) ln (2πm)
Probability Distribution of Discrete Random
Variables
A probability distribution is listing of all the possible
values that a random variable can take along with their
probabilities.
Example: to find out the probability distribution for the
number of heads on three tosses of a coin:
First toss.........T T T T H H H H
Second toss.....T T H H T T H H
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TSP: WEEK 2
9
Third toss........T H T H T H T H
the probability distribution
No. of
heads
X
Probability
P(X)
mean or
expected
value
X. P(X)
0 1/8 0.0
1 3/8 0.375
2 3/8 0.75
3 1/8 0.375
∑X P(X) = 1.5
Mean = E(X) = ∑X.P(X)
where:
E(X) = expected value,
X = an event,
P(X) = probability of the event
Binomial Distribution:
-discrete probability distributions
-Several characteristics underlie the use of the
binomial distribution.
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TSP: WEEK 2
10
Characteristics of the Binomial Distribution:
1. The experiment consists of n identical trials.
2. Each trial has only one of the two possible mutually
exclusive outcomes, success or a failure.
3. The probability of each outcome does not change
from trial to trial, and
4. The trials are independent, thus we must sample with
replacement.
Binomial Equation:
-able to identify three things:
-the number of trials
-the probability of a success on any one trial
-the number of successes desired
nNn
N qpn)!(Nn!
N!(n)P −
−=
Example: binomial distribution:
What is the probability of obtaining exactly 3 heads if a fair
coin is flipped 6 times?
Ans:
N = 6, n = 3, and p = q = .5
363 )5.01(5)!36(!3
!6)3( −−
−=P
3125.0)125.0)(125.()23)(23(
23456== o
xx
xxxx
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TSP: WEEK 2
11
Binomial distributions
or
Properties of Binomial Distribution
Mean np
Variance npq
Standard deviation (npq)1/2
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TSP: WEEK 2
12
Binomial Distribution: Fluctuations
Suppose the following binomial distribution;
• the probability of success: p
• the probability of failure: 1-p = q
The mean n = pN
N: the number of elements per system
n∆ : average fluctuation of n about its mean value, n
0)( =−=−=∆ nnnnn
-positive fluctuations cancel the negative one
The standard deviation
222 )()tan( nndeviatiandards −== σ
[ ] 212)()tan( nndeviatiandards −== σ
(Note: using ∑=S
SS fPf and nNn
N qpn)!(Nn!
N!(n)P −
−=
)
( )222 )( nnPnn
n
n −=−= ∑σ
( )( )∑
=
− −
−=
N
n
nNnnnqp
nNn
N
0
2
!!
!
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TSP: WEEK 2
13
Solving----- Npq=2σ
or Npq=σ
relative fluctuations
NNP
q
Np
Npq
n
1∝==
σ
Example:
-system of 100 flipped coins: what are;
average number of H (head),
the standard deviation,
σ = 8
n =125 n
N=250
PN(n)
σ=2.5
n =12.5 n
N=25
PN(n)
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TSP: WEEK 2
14
the relative fluctuation?
Ans: n= pN = (1/2)(100) = 50
σ = (Npq)1/2 = ((100)(.5)(.5))1/2 = 5
σ/ n = 5/50 = 10 %
- now N = 10,000 flipped coins
n= pN = (1/2)(10 000) = 5 000
σ = (Npq)1/2 = ((10 000)(.5)(.5))1/2 = 50
σ/ n = 50/5 000 = 1 %
note:
A binomial probability distribution must meet each of the
following:
1. There are a fixed number of trials
2. The trials must be independent
3. Each trial must have outcomes classified into two
categories
4. The probabilities remain constant for each trial
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TSP: WEEK 2
15
Gaussian Distribution
The Gaussian distribution is useful where binomial formula
is not
The probability PN(n) ≅ P(n)
P(n) – a continuous function of n
Criteria for increase accuracy;
i) choose as smooth a function as possible
ii) choose reference point, as close to the value of n
To satisfy (i)—expand the logarithm of P(n)
To satisfy (ii) – choose reference as nmax
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TSP: WEEK 2
16
interested in values of P(n) for n near nmax
Using Taylor series expansion
..)()(ln2
1)()(ln)(ln)(ln 2
2
2
+−∂
∂+−
∂
∂+=
==nnnP
nnnnP
nnPnP
nnnn
But 0)(ln =
∂
∂= nn
nPn max. point
2nd derivative
using ...)1(
)()1()()(=
−+
−+=
∆
∆=
∂
∂
nn
nPnP
n
nP
n
nP
22
2 11)(ln
σ−=−=
∂
∂= Npq
nPn nn
therefore 2
2)(
2
1)(ln)(ln nnnPnP −−=
σ
nmax n N
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TSP: WEEK 2
17
or 2
2
2
)((
)()( σ
nn
enPnP
−−
=
to calculate P(n)
-the sum of the probabilities of all possible values of n must
be equal to 1
∑ =n
nP 1)(
∑ ∫∞
−∞=
≈∆=n n
dnnPnnP )()(
( )[ ]
1)2)(()(22
2/ === ∫∞
−∞=
−− πσσnPdnenP
n
nn
Note:
therefore
( )[ ]222/
2
1)( σ
σπ
nn
n
enP−−=∑
Because σπ2
1)( =nP
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TSP: WEEK 2
18
Gaussian Distribution
Properties of Gaussian distribution
( )nnx −=
When nn =
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TSP: WEEK 2
19
-3σ -2σ -1σ 0 1σ 2σ 3σ
X
f(X)
-3σ -2σ -1σ 0 1σ 2σ 3σ
X
f(X)
1σ ≅ half width at half maximum
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TSP: WEEK 2
20
Probability of a measurement falling within -σ to +σ of the
mean is 0.683
Probability of a measurement falling within -2σ to +2σ of
the mean is 0.954
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TSP: WEEK 2
21
Probability of a measurement falling within -3σ to +3σ of
the mean is 0.997
Normal Approximation to the Binomial
The shape of a binomial distribution depends on the values
of n and p.
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TSP: WEEK 2
22
-the shape of a BD will be fairly symmetrical
-for large values of n (say, n > 20 ),
-when p is about 0.05 < p < 0.95 ).
EXAMPLE:
-flip a coin for N =20 and p = 0.5
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TSP: WEEK 2
23
- binomial probability distribution → NDF
A normal density curve
-with mean np = 20(0.5) = 10
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TSP: WEEK 2
24
-standard deviation (np(1-p))1/2 = 2.236
--fit with BDF
Random Walk A random process consisting of a sequence of discrete steps
of fixed length
The random walk is central to statistical physics.
-predicting how fast one gas will diffuse into another,
-how fast heat will spread in a solid,
-how big fluctuations in pressure will be in a small
container,
-and etc………..
Problem: to find the probability of landing at a given spot
after a given number of steps, or to find how far
away the girl is on average from where she
started.
…after many many steps
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TSP: WEEK 2
25
The simplest random walk is a path constructed according
to the following rules:
• There is a starting point.
• The distance from one point in the path to the next is a
constant.
• The direction from one point in the path to the next is
chosen at random, and no direction is more probable
than another.
Consider each step is of length s0,
It can be either forward (right) or backward (left).
Probability of going forward (right) is p.
Probability of going backward (left) is q = 1- p
After N steps, if n are forward (right), the distance traveled
is,
S = [n – (N – n)]s0 = (2n – N) s0
The progression of the walk and it’s different outcomes are
nicely organized in Pascal’s triangle.
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TSP: WEEK 2
26
This is known as a binomial distribution.
Define the probability function
fN (n) -- the probability that in a walk of N steps, end at
point n.
For the nonzero probabilities.
For a walk of no steps, f0(0) = 1.
For a walk of 1 step, f1(−1) = ½, f1(1) = ½.
For a walk of 2 steps, f2(−2) = ¼, f2(0) =2 x ¼ = ½, f2(2) = ¼.
……. f3(−3) = 1/8 , f3(−1) = 3/8 , f3(1) = 3/8, f3(3) = 1/8.
…… f4(4) = 1/16, f4(2) = ¼; f4(0) = 3/8, …….
n −5 −4 −3 −2 −1 0 1 2 3 4 5
f0(n) 1
f1(n) ½ ½
f2(n) ¼ ½ ¼
f3(n) 1/8 3/8 3/8 1/8
f4(n) 1/16 1/4 3/8 1/4 1/16
f5(n) 1/32 5/32 5/16 5/16 5/32 1/32
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TSP: WEEK 2
27
Factor by (1/2) N
n −5 −4 −3 −2 −1 0 1 2 3 4 5
f0(n) 1
2f1(n) 1 1
22f2(n) 1 2 1
23f3(n) 1 3 3 1
24f4(n) 1 4 6 4 1
25f5(n) 1 5 10 10 5 1
This is Pascal’s Triangle—every entry is the sum of the
two diagonally above
These numbers are in fact the coefficients that appear in the
binomial expansion of (p + q)N
Picturing the Probability Distribution
Visualizing this probability distribution →
For 5 steps, it looks like:
-a walk of 100 steps → n steps forward and
100 − n steps backward
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TSP: WEEK 2
28
-the final landing place is n − (100 − n) = 2n − 100 paces in
the forward direction.
Note that this is an even number, and goes from −100 to
+100.
The total number of 100-step walks having just n forward
steps is
(100)!/n!(100 − n)!
The probability of landing at 2n − 100 after a random 100-
step walk is proportional to the number of such walks that
terminate there
The probability of this occurring is,
The average distance covered after N steps is,
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TSP: WEEK 2
29
and
note: pNn =
Standard deviation
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WEEK 2
1
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2
Example 1: COIN
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3
HEAD showing
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4
Example 2: DICE
=
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5
Criterion:
-something that is used as a reason for making a judgment or decision
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6
Example 3: 2 Identical Systems (2 COINS)
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Example 4: BOX 1
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Example 5: 2 DICE
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Identical Elements: with the same Probability
A:
B:
( p3 + 3p2 q + 3q2 p + q3 )
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C:
Binomial Expansion
Binomial Coefficient
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Example 6: BOX 2 – Consider 5 molecules in a box
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V U = VD
V 1 = 1/3 V
Let the total volume is V
or V U = 1/2 V
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1 COIN;
3 tosses
8 configurations= 8 microstates
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NO OF
HEADS
FIRST
THROW
SECOND
THROW
THIRD
THROW
MACRO
STATE
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NO OF
HEADS
FIRST
THROW
SECOND
THROW
THIRD
THROW
MACRO
STATE
0
1
2
3
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NO OF
HEADS
FIRST
THROW
SECOND
THROW
THIRD
THROW
MACRO STATE
0 T T T 1
1 T T H 2
T H T
H T T
2 H H T 3
H T H
T H H
3 H H H 4
1+3+3+1 4 MACROSTATES
= 1
= 3
= 3
= 1
= 8
MICROSTATES
TOTAL
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0(1/8)+(1)(3/8)+(2)(3/8)+(3)(1/8) = 0.0 + 0.375 + 0.75 + 0.375 = 1.5
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Binomial distribution
In probability theory and statistics,
the binomial distribution is the discrete
probability distribution of the number of
successes in a sequence of n independent
yes/no experiments, each of which yields
success with probability p.
Therewith the probability of an event is
defined by its binomial distribution.
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The binomial distribution is frequently used to
model the number of successes in a sample
of size n drawn with replacement from a
population of size N.
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If the sampling is carried out without
replacement, the draws are not
independent and so the resulting
distribution is a hypergeometric
distribution, not a binomial one.
However, for N much larger than n, the
binomial distribution is a good
approximation, and widely used.
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Example 8: Binomial Distribution
0.5
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N
N
N
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Example 9: system of 100 COINS
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In statistics and probability theory,
the standard deviation (SD) (represented by
the Greek letter sigma, σ) measures the
amount of variation or dispersion from the
average.
A low standard deviation indicates that the
data points tend to be very close to
the mean (also called expected value); a high
standard deviation indicates that the data
points are spread out over a large range of
values.
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A random walk is a mathematical formalization
of a path that consists of a succession of
ramdom steps.
Example of Modeled Random Walks:
i) the path traced by a molecule as it travels
in a liquid or a gas,
ii) the search path of a foraging animal,
iii) the price of a fluctuating stock
iv) the financial status of a gambler
although they may not be truly random in reality.
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The term random walk was first introduced
by Karl Pearson in 1905. Random walks have
been used in many fields:
ecology, economics, psychology, computer
science, PHYSICS, chemistry, and biology.
Random walks explain the observed
behaviours of many processes in these fields,
and thus serve as a fundamental MODEL for
the recorded STOCHASTIC ACTIVITY.
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A popular random walk model is that of a
random walk on a regular lattice, where at
each step the location jumps to another site
according to some probability distribution.
In a simple random walk, the
location can only jump to
neighbouring sites of the lattice.
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In simple symmetric random walk on
a locally finite lattice, the probabilities of
the location jumping to each one of its
immediate neighbours are the same.
The best studied example is of random
walk on the d-dimensional integer lattice
(sometimes called the hypercubic
lattice)
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One-dimensional random walk
This walk can be illustrated as follows:
A marker is placed at zero on the number line
and a fair coin is flipped. If it lands on heads,
the marker is moved one unit to the right. If it
lands on tails, the marker is moved one unit to
the left.
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After five flips, the marker could now be on
1, −1, 3, −3, 5, or −5.
Example:
With five flips, three heads and two tails,
in any order, will land on 1.
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There are :
10 ways of landing on 1 (by flipping three heads and two
tails),
10 ways of landing on −1 (by flipping three tails and two
heads),
5 ways of landing on 3 (by flipping four heads and one tail),
5 ways of landing on −3 (by flipping four tails and one
head),
1 way of landing on 5 (by flipping five heads), and
1 way of landing on −5 (by flipping five tails).
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See the figure below for an illustration of the
possible outcomes of 5 flips.
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AT
LAST!! THE END….
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TSP: WEEK 3
1
WEEK 3
Internal energy -the energy stored in the motions and interactions of the
elements (e.g, atoms, molecules, electrons, etc) of a system
Degree of freedom
-ways each elements can store energy
Consider the atom of a solid
Energy of each atom = PE + KE
Vibration in 3-D
Potential energy: EP = ½ kx2 + ½ ky2 + ½ kz2
Kinetic energy: EK = ½ mvx2 + ½ mvy
2 +1/2 mvz2
= ½ (mvx)2 /m + ½ (mvy)
2/m +1/2 (mvz)2/m
=222
2
1
2
1
2
1ZYX P
mP
mP
m++
P: momentum
PE & KE in the form
E = b q2 b: constant
Each particle → 6 degrees of freedom
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TSP: WEEK 3
2
N particles → 6N degrees of freedom for energy of the form
E = bq2
Equipartition theorem
the internal energy will be distributed equally among all those
degrees of freedom with energy in the form E=bq2
e.g: ideal gas
PV = mRT
In term of Boltzmann’s constant
PV = NkT
From molecular theory
2
3
1vNmPV =
2v : Average velocity
KE: 2222
2
1
2
1
2
1
2
1zyx vmvmvmvm ++=
Equip. theorem: 222
zyx vvv ==
2222
2
1
2
1
2
1
2
1xxx vmvmvmvm ++=
)2
1(3
2
1 22
xvmvm =
kTvm2
3
2
1 2 =
The average energy per degree of freedom
kTE2
1= k= 1.381 x 10-23 J/K
: Boltzmann’s constant
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TSP: WEEK 3
3
KE → related to temperature
Higher KE, higher temperature
Note: when heat added, but temperature remains constant
→ new degrees of freedom, where the added energy is
stored
Changing the internal energy
- 3 ways / interactions
i. thermal interaction
-adding / removing heat from the system
ii. mechanical interactions
-doing work on the system or letting system doing
work on something else
iii. diffusive interaction
-adding or removing particles that undergo reactions
with the particles of the system
Heat Transfer
-heat transfer between A1 and A2 only
A1 A2
Q
Rigid, thermally conducting membrane
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TSP: WEEK 3
4
heat in: increase particles velocity → increase temperature
heat out: lower the velocity → lower the temperature
→ changing the internal energy
Work (mechanical interactions)
Expansion – molecules strike a receding piston, loosing KE
-system does work – loosing energy
Compression – molecules strike on incoming piston – gain KE
-work done on the system – increases energy
Hot – higher velocity Cold – lower velocity
A1 A2
W W
A1 A2
W
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TSP: WEEK 3
5
Particle transfer
PE: interactions of an element with its neighbours
-measured relative to a reference
-- known as zero-energy reference level
zero energy ref. level – the PE of an isolated particle
Consider: a particle trapped in a harmonic oscillator potential
well
ETH: thermal energy
A particle with energy E0 ossilates in a potential well -
ETH = E0 - µ
For N particles: ETH = E0 - µN
Chemical potential, µµµµ: the energy of the very lowest point in
the particle’s potential well.
EP
x 0
Ep = µ
EP = µ + ½ kr2
Total energy
E = EP + EK
= µ + ½ kr2 + ½ mv
2
= µ + ETH
E0
A1 A2
W
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TSP: WEEK 3
6
Adding ∆N particles without adding thermal energy (constant T
and S), the internal energy of the system
( )
ST
STN
EorNE
,
,
∆∆
=∆=∆ µµ
-introduced in 1876 by the American mathematical physicist
Willard Gibbs
If to any homogeneous mass in a state of hydrostatic stress
we suppose an infinitesimal quantity of any substance to be
added, the mass remaining homogeneous and its entropy
and volume remaining unchanged, the increase of the
energy of the mass divided by the quantity of the substance
added is the potential for that substance in the mass
considered
Note: Chemical potential energy is a form of potential energy
related to the structural arrangement of atoms or molecules --the
breaking and forming of chemical bond.
The chemical potential µ of a thermodynamic system is the
amount by which the energy of the system would change if an
additional particle were introduced, with the entropy and
volume held fixed.
∆N
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TSP: WEEK 3
7
Chemical potential µ is a measure of the free energy available
to do the work of moving a mole of molecules from one
location to another or through a barrier such as a cell
membrane…
Molecules tend to spontaneously move from areas of higher µ
(higher concentration) to areas of lower µ (lower
concentration), thus increasing the entropy of the universe
µA > µB: transformation of substance A into substance B, or
transport from place A to place B
µA = µB: no transformation, no transport, chemical equilibrium
µA < µB: transformation of substance B into substance A, or
transport from place B to place A.
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TSP: WEEK 3
8
First law of thermodynamics
Changing the internal energy
(a) adding heat (∆Q)
(b) doing work (∆W)
(c) adding particles (∆N)
∆∆∆∆U = ∆∆∆∆Q - ∆∆∆∆W + µµµµ∆∆∆∆N --- The First Law of Therm.
Note: -∆W: work is done on the system
Note:
U – property of a system, determined by the two end
points (initial & final)
Q, W – interactions, determined by the routes taken.
Cannot be determined from the end points alone.
dU = δδδδQ - δδδδW + µµµµdN
Or dU = δδδδQ - δδδδW for non diffusive int.
To differentiate between properties and interactions;
(a) (b) (c)
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TSP: WEEK 3
9
-exact differential- property functions
-inexacct diff. – non-property functions (Interactions)
Let F = f(x, y) x, y: variables
Initial state: (xi, yi)
Final state: (xf, yf)
∆F = f(xf, yf) – f (xi, yi)
F is determined by the two end points
Diff. form dyy
Fdx
x
FdF
∂
∂+
∂
∂=
Let a diff. eq. g(x,y)dx + h(x,y)dy -- exact or not?
EXACT – if there is a F(x, y) function which satisfy
hy
Fandg
x
F=
∂∂
=∂∂
Alternatively, use the identity
yx
F
xy
F
∂∂
∂=
∂∂
∂ 22
Note: dependent & independent variables
F = F(x, y)
x
yxF
∂
∂ ),( y is held constant.
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TSP: WEEK 3
10
Or yx
F
∂
∂ y is held constant
Note:
The system
System characteristics
Definite boundaries in space
May be open or closed (respect to transfer of matter)
May or may not be thermally insulated
Thermodynamics state of a system
- at equilibrium, system assumed to occupy one of a set of
abstract, thermodynamics states
- each state is defined by the values of state variables
o e.g: at least two of the variables (T, V, P….)
- system state determines system properties
- two types properties
o Extensive (e.g volume, mass..)
o Intensive (e.g T, P, density…..)
Change of system state
- System traversal from initial to final state.
- Resulting change in some macroscopic system property
- Reversible or irreversible
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TSP: WEEK 3
11
o Reversible – change take a path near to equilibrium
– system changes can be accurately predicted
o Irreversible – changes deviate from equilibrium –
system changes cannot be accurately predicted
Statistical Thermodynamics
- concern: behaviour of a large number of particles
- take a statistical view
- thermodynamics state – the state of system for N particles.
The states of a system
-A microstate is the state defined by specifying in detail the
location and momentum of each particle
-Each atom is an oscillator that contains equally spaced energy
levels. Atoms may be in any of these quantum states
The total number of atoms, N
The total energy the system E
-Consider 3 atoms with a total energy of 3 units.
Microstates 1, 2 & 3
En i
i
i =∑ ε
Nni
i =∑
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TSP: WEEK 3
12
Microstates 4, 5, 6, 7, 8, & 9
Microstate 10
For 3 atoms with E = 3 units → 10 possible microstates
To predicts the number of microstates
(Case: 3 atoms with E = 3 units)
STEP 1
1. The 1st oscillator can be in any one of three states → 3
possible microstates.
2. The 2nd oscillator →in any one of the two states→ 2
additional microstates
3. The 3rd oscillator→only one state→ 1 additional
microstate.
The total number of microstates Ω is
6!3123 ==⋅⋅=Ω STEP 2:
-One oscillator in state 3 and the other two in state 0.
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
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TSP: WEEK 3
13
1. The 1st oscillator → 3 possible microstates.
2. The 2nd oscillator → 2 additional microstates
3. The 3rd oscillator → 1 additional microstate.
→ Total of 6 microstates
Note: but
-reduces the configuration to 3 microstates.
3
!2
!3
2
123==
⋅⋅=Ω
STEP 3
-all three oscillators are in the same state
1. The 1st oscillator → 3 possible microstates.
2. The 2nd oscillator → 2 additional microstates
3. The 3rd oscillator → 1 aditional microstate.
→ Total of 6 microstates
Note: But
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
Same microstate
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TSP: WEEK 3
14
etc
-reduces this configuration to 1 microstates
1
!3
!3
6
123==
⋅⋅=Ω
Then, in general, the number of microstates for any given
configuration is
( )( )( )( ) ( )∏==Ω
!
!
!!!!
!
321 io n
N
nnnn
N
Example:
-the number of microstates for N = 5 & E= 5
-possible configurations
-the total number of microstates is
126 = 1 + 30 + 20 + 20 +30 + 20 + 5
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TSP: WEEK 3
15
Example:
4 molecules in a box
How many ways are there?
-to place 0 in L and 4 in R: 1!4!0
!4=
-to place 1 in L and 3 in R: 4!3!1
!4=
-to place 2 in L and 2 in R: 6!2!2
!4=
-to place 3 in L and 1 in R: 4!1!3
!4=
-to place 4 in L and 0 in R: 1!0!4
!4=
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TSP: WEEK 3
16
Microstate Macrostate Degeneracy
LLLL I 1
RLLL
LRLL
LLRL
LLLR
II
4
LLRR
LRLR
RLLR
LRRL
RLRL
RRLL
III
6
LRRR
RLRR
RRLR
RRRL
IV
4
RRRR V 1
Macrostates: 5 Microstates: 16
Changing the number microstates -by changing the internal energy
-Four ways to increase the number of quantum states of
molecules
1-when a system is heated.
2-when a system expands into a vacuum of fluids mix.
3-when a solute is added to a solvent.
4-when a phase change occurs due to the input of energy
as in the first
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TSP: WEEK 3
17
A system is being heated
Microstates of cold
system
Q Microstates of hot
system
A system is allowed to expand in volume
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TSP: WEEK 3
18
Equilibrium For an isolated system;
When the probability of it being in the various possible
states do not vary with time.
Consider: changing the internal energy by compressing
-First the additional energy is given to particles near the
boundary
–Then shared with the rest of the system – by mutual interaction
- When all possible states are equally probable – the system is
in equilibrium
Relaxation time – the time required for a system to reach
equilibrium after being perturbed.
Quasi-static process – when system interact, the transfer of heat,
work, or particles, the process must be slower compared to the
relaxation time --- the system near equilibrium.
Fundamental Postulate
An isolated system in equilibrium is equally likely to be in
any of its accessible state
Let Ωo: the number of states accessible to the entire system
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TSP: WEEK 3
19
The probability it will be found in any one of them,
o
PΩ
=1
e.g
i) Calculate all possible microstates for a lattice of 3 non-
interacting particle.
ii) What is the probability that the orientation of the first
particle’s magnetic moment is + and the total magnetization is
+µ?
Ans
(i)
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TSP: WEEK 3
20
Number of microstate: 8
The total magnetization of +1µ is 3 (red box) The first particle’s magnetic moment + is 2 (green box)
(ii)
3
21)1( =
+=++
stateofnumbertotal
beingofwithstatesnumberisstateofP
µµ
The spacing of states
- identify the state of a system by its internal energy, U
- several different states have the same energy –
The state is degenerate or
The energy level is degenerate
E.g three spins ½ particles in magnetic field B
microstate energy
↑ ↑ ↑ (+++) +3µB ↑ ↑ ↓ (++ -) +1µB ↑ ↓ ↑ (+- +) +1µB ↓ ↑ ↑ (-++) +1µB ↑ ↓ ↓ (+- -) -1µB ↓ ↑ ↓ (-+ -) -1µB ↓ ↓ ↑ (- -+) -1µB ↓ ↓ ↓ (- - -) -3µB
The Energy, U = +1µB state is three-times degenerate
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WEEK 3
1
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2
Show how to transform from above to below:
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The Internal energy will be distributed equally among all those degrees of freedom with energy in the form E = bq2
e.g: Ideal Gas PV = mRT
In terms of Boltzmann’s Constant PV =NkT
From molecular theory
3
By assumptions, there are no
preferred directions
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since
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11Commons is a freely licensed media file repository
Wavefunctions of a quantum harmonic oscillator
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Picture of wavefunctions (and energy levels) of quantum
harmonic oscillator, using colour scale for probability density.
Commons is a freely licensed media file repository.
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Change of System State
- System traverses from initial to final state.
-Resulting change in some macroscopic system property.
-Reversible or Irreversible.
• Reversible – change take a path near to equilibrium –
system changes can be accurately predicted.
• Irreversible – changes deviate from equilibrium – system
changes cannot be accurately predicted.
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Statistical Thermodynamics
-concern ; behaviour of a large number of
particles
- take a statistical view
-Thermodynamic state – the state of a
system for N particles
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Denotes the
continued product
over all values of i
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Microstates 1, 2, 3 ,4, 5, 6
Microstates 7, 8, 9
Microstate 10
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3 microstates
2 microstates1 microstate
(3+2+1) microstates
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N = 3 ; 1! ; 2! ; 3!
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Degeneracy
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In physics, two or more different
quantum states are said to be
degenerate if they are all at the same
energy level. Statistically this means that
they are all equally probable of being
filled, and in quantum mechanics it is
represented mathematically by the
Hamiltonian for the system having more
than one linearly independent eigenstate
with the same eigenvalue.
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Conversely, an energy level is said to be degenerate if it contains two or more different states. The number of different states at a particular energy level is called the level's degeneracy, and this phenomenon is generally known as a quantum degeneracy.
In quantum theory this usually pertains to electronic configurations and the electron's energy levels, where different possible occupation states for particles may be related by symmetry.
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To cause (a system) to become altered or imbalanced from a normal state.
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Example: (i) How many possible microstates are there
for a lattice of 3 non-interacting particles .
(ii) What is the probability that the orientation of the
first particle’s magnetic moment is + and the total
magnetization of +1m?
microstates magnetization
+3m
+1m
+1m
+1m
-1m
-1m
-1m
-3m
∑= 8 microstates
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ThaT’s all for now….
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TSP: WEEK 4
1
WEEK 4 Density of state
Let ∆U = energy range of finite width
Expect: number of states ∝ ∆U
If Ω(U, ∆U): number of states between U and U+∆U
Ω(U, ∆U)= g(U) ∆U
g (U): density of state
States available to the
system
∆U
U
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TSP: WEEK 4
2
Definition
The energy density of states (EDOS) function measures the
number of energy states in each unit interval of energy and in
each unit volume of the crystal
volumexEnergy
statesofnumberUg =)(
-figure
∆E = 4 -3 = 1 eV → 4 energy states
The density of states at E= (3 + 4) /2 = 3.5 eV
4
11
4)5.3(
3===
cmxeVvolumexEnergy
statesofnumberg
Example: Suppose a crystal has two discrete states (i.e. single
states) in each unit volume of crystal.
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TSP: WEEK 4
3
The density-of-state function consists of two Dirac delta
functions of the form
( ) ( )21)( UUUUUg −+−= δδ
Integrating over energy gives the number of states in each unit
volume
( ) ( ) 2)( 21
00
=−+−== ∫∫∞∞
UUUUdUdUUgNv δδ
If the crystal has the size 1x4 cm3 then the total number of
states in the entire crystal must given by
∫ ==
4
0
8dVNN v
Microstates of interacting systems
2 interacting systems, A1 and A2 – heat, work & particles
Let U1: internal energy of A1
U2: internal energy of A2
A1 A2 Q
W
N
1 cm
1 cm
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TSP: WEEK 4
4
U0 = U1 + U2 ≅ constant
The number of states
ΩO = Ω1Ω2
Let:
R for A1 = 6 degrees of freedom
R for A2 = 10 degrees of freedom
Therefore Ω1 ∝ (U1)3
Ω2 ∝ (U2)5
Assume U1 + U2 = 5 unit of energy
U1 U2 Ω1 Ω2 ΩO = Ω1Ω2
0 5 0 3125 0
1 4 1 1024 1024
2 3 8 243 1944
3 2 27 32 864
4 1 64 1 64
5 0 125 0 0
Total 3896
Since every state is equaly likely, the most probable energy
distribution
U1= 2 eu
U2 = 3 eu
Note: 1944/3896 = 0.5
-half of the time energy distribution is U1 = 2 eu
and U2= 3 eu
Now consider system with R1 = 12 and R2 = 20
Ω1 ∝ (U1)6
Ω2 ∝ (U2)10
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TSP: WEEK 4
5
Assume U1 + U2 = 5 unit of energy
U1 U2 Ω1 Ω2 ΩO = Ω1Ω2
0 5 0 9.77x106 0
1 4 1 1.05x106 1.05x106
2 3 64 5.90x106 3.78x106
3 2 729 1.02x106 0.744x106
4 1 4.1x103 1 0.004x106
5 0 1.56x104 0 0
Total 5.57 x 106
For U1= 2 eu & U2 = 3 eu
-the accessible states are 68%
-the system will have this energy distribution more than
two-thirds of the time!!!
Let R1 = 120 and R2 = 200
Ω1 ∝ (U1)60
Ω2 ∝ (U2)100
U1 U2 Ω1 Ω2 ΩO = Ω1Ω2
0 5 0 7.9x1069 0
1 4 1 1.6x1060 1.6x1060
2 3 1.2x1018 5.2x1047 6.2x1065
3 2 4.2x1028 1.3x1030 5.5x1058
4 1 1.3x1028 1 1.3x1036
5 0 8.7x1041 0 0
Total 6.2 x 1065
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TSP: WEEK 4
6
For U1= 2 eu & U2 = 3 eu
-the accessible states are 99.997%
-the system will have this energy distribution at any instant in
time!!!
Marcoscopic system
Number of particles ~ 1024
R1 = 6 x 1024 and R2 = 10 x 1024
( )24103
11
xUαΩ and ( )
24105
22
xUαΩ
U1 U2 Ω1 Ω2 ΩO = Ω1Ω2
0 5 0 24
1099.610 x 0
1 4 1 24
1002.610 x 24
1002.610 x 2 3
241081.110 x
241077.410 x
241058.610 x
3 2 2410861.210 x
24100.310 x
241087.510 x
4 1 24
1061.310 x 1 24
1061.310 x 5 0
241019.410 x 0 0
Total 24
1058.610 x
For U1= 2 eu & U2 = 3 eu
-the accessible states are the most probable --- i.e 24
1056.010 x
times more probable than the other distribution.
NOTE:
When 2 interacting macroscopic systems are in equilibrium, the
values of the various system variables will be such that the
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TSP: WEEK 4
7
number of states available to the combined system is a
maximum.
The second law of thermodynamics As 2 interacting macroscopic systems approach
equilibrium, the changes in the system variables will be
such that the number of states available to the combined
system increases. Or, in the approach to equilibrium,
∆Ωo > 0
Note: 1st Law: - reflect inviolable fact
-work for small system
2nd Law: - based on probability
-for large system – there is some infinitesimal
probability the law be violated!!!
Heat flow and energy spreading
The fundamental science behind the second law:
Energy spontaneously disperses from being localized to
becoming spread out if it is not hindered
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TSP: WEEK 4
8
Thermal contact:
-thermal energy flows from the higher occupied levels in
the warmer object into the unoccupied levels of the cooler
one until equal numbers are occupied in both bodies,
bringing them to the same temperature.
The degree of dilution of the thermal energy is
Q /T
Cold body: TCold is low; few thermal energy states are occupied,
so the amount of energy spreading can be very great.
When TCold near THot, more thermal energy states are occupied
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TSP: WEEK 4
9
Entropy Entropy – (Greek for ‘turning into’)
– Is a measure of the degree of disorder of a system.
– In a reversible process entropy remains constant.
– In an irreversible process, entropy MUST increase.
Entropy can NEVER decrease in a closed system.
– In any cyclic process the entropy will either increases or
remains the same
The 2nd law of thermodynamics:
Every time energy is transformed from one state to
another, there is a loss in the amount of that form of
energy, which becomes available to perform work of some
kind. The loss in the amount of ‘available energy’ is known
as ‘entropy’
Entropy –a state variable whose change is defined for a
reversible process at T where Q is the heat absorbed
Entropy – a measure of the amount of energy which is
unavailable to do work
Entropy – a measure of the disorder of a system.
Entropy – a measure of the multiplicity of a system
Definition
Entropy just measures the spontaneous dispersal of
energy: how much energy is spread out in a process, or
how widely spread out it becomes – as a function of
temperature.
Or
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TSP: WEEK 4
10
Entropy = S ≅≅≅≅ k ln ΩΩΩΩ
k: Boltzmann’s constant
Ω very big number – take lnΩ
Combine system (A0 = A1 + A2)
Ω0 = Ω1 Ω2
k lnΩ0 = k ln(Ω1 Ω2) = k lnΩ1 + k lnΩ2
S0 = S1 + S2
2nd law can be stated in terms of the entropy (alternative)
As 2 interacting macroscopic systems have reached
equilibrium, the changes in the system variables will be
such that the entropy of the combined system increases.
∆S0 > 0
When A0 = A1 + A2 have reached equilibrium, then ΩO will be
maximized
∆S0 (equilibrium) = 0
2nd law (alternative)
For any 2 interacting systems (whether in equilibrium or
not) the entropy of the combined system cannot decrease.
∆S0 > or = 0
What is entropy? How is it related to the second law?
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TSP: WEEK 4
11
Entropy measures how much energy is dispersed in a particular process (at a specific temperature).
Energy spontaneously tends to flow only from being
concentrated in one place to becoming diffused or dispersed and
spread out.
e.g A hot frying pan cools down when it is taken off the kitchen
stove.
The energy concentrated inside a chemical like oil or coal
(or food) will spread out.
Cars rust, pendulums run down, people get old.
--Things like this don't happen backwards.
Like the flow of time (time arrow)
--they happen only in one sequence
The 2nd law of thermodynamics (or the law of entropy)
The universe and all of its energy systems will increase in
disorder as time moves forward.
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TSP: WEEK 4
12
Disorder means the breakdown of energy into useless heat,
from which no work can be done.
Entropy is a measure of disorder
Chaos is a state of organized disorder
-entropy gives information about the evolution of an isolated
system with time → gives the direction of "time arrow"
-State which is more disordered →this state came later in time
-The second law of thermodynamics --gives the direction of
heat flow in any thermal process.
Heat naturally flows from higher T to lower T.
No natural process has as its sole result the transfer of
heat from a cooler to a warmer object.
No process can convert heat absorbed from a reservoir at
one temperature directly into work without also rejecting
heat to a cooler reservoir. That is, no heat engine is 100%
efficient.
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TSP: WEEK 4
13
-Reasons -- why nature behaves in that way.
• Any process either increases the entropy of the universe -
or leaves it unchanged.
-Entropy is constant only in reversible processes
-All natural processes are irreversible.
• All natural processes tend toward increasing disorder
-energy is conserved
-but iits availability is decreased.
• Nature proceeds from
-simple to the complex
-orderly to the disorderly
-low entropy to high entropy.
• The entropy of a system is proportional to the logarithm of
the probability of that particular configuration of the
system occuringm(S ≅≅≅≅ k ln ΩΩΩΩ)
-more highly ordered the configuration of a system
-less likely it is to occur naturally
-hence the lower its entropy.
-The laws of Thermodynamics --wide range of applicatios
-Cosmology, History, Economics, Military, and to
almost everything
The 2nd law of Thermodynamics:
-the total entropy in the world is constantly increasing
-decrease in ‘available energy’.
-the unavailable energy-form works as pollution.
-the world is moving towards a dissipated state
-the pollution is constantly increasing
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TSP: WEEK 4
14
-entropy (i.e. the ‘unavailable’ energy or pollution) tends
towards a "maximum"
-In a closed system - ultimately reached a stage where
there is no longer any difference in energy level --‘the
equilibrium state’.
-Maximum entropy -no longer ‘free energy’ is available
for work
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DENSITY OF STATE,
ENTROPY AND
THE SECOND LAW
WEEK 4
1
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2
In solid-state and condensed matter physics,
the density of states (DOS) of a system describes
the number of states per interval of energy at each
energy level that are available to be occupied by
electrons.
Unlike isolated systems, like atoms or molecules in gas
phase, the density distributions are not discrete like
a spectral density but continuous.
A high DOS at a specific energy level means that
there are many states available for occupation.
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A DOS of zero means that no states can be
occupied at that energy level.
In general a DOS is an average over the space
and time domains occupied by the system.
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Density of State
4
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R is the number of degrees of freedom
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eV, J, kJ
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MACROSCOPIC SYSTEM
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its
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PHEWH ! !
30
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TSP-WEEK 5
1
WEEK 5
Thermal Interaction
Axiom 1:
There exist special states of macroscopic physical
system, called equilibrium states, which can be fully
described by the internal energy, U, and set of
extensive parameters, X0, X1, ……XC.
Axiom 2:
For all system in equilibrium there exists a function of
the extensive parameters, called the entropy, S. If
there are no internal constraints on the system, the
extensive parameters can take the values that
maximise S over the possible states with internal
constraints.
The functional relation between S and the extensive
parameters
S = S (U, X0, X1,…..XC)
Note: X0 can be V; X1 can be N,….
2 systems interacting thermally
NVU
S
,
∂
∂ no mechanical interactions
no exchange of particle
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TSP-WEEK 5
2
A0 = A1 + A2
When A1 and A2 are in equilibrium;
The entropy S0 is at it maximum value
0
,1
0 =
∂
∂
NVU
S
If S0 = f (U1)
Using S0 = S1 + S2 and dU1 = -dU2
NVNVNV
U
S
U
S
U
S
,1
2
,1
1
,1
0
∂
∂+
∂
∂=
∂
∂
0
,2
2
,1
1
,1
0 =
∂
∂−
∂
∂=
∂
∂
NVNVNVU
S
U
S
U
S
NVNV
U
S
U
S
,2
2
,1
1
∂
∂=
∂
∂ (*)
For 2 systems in thermal equilibrium,
the porperty NV
U
S
,
∂
∂ is the same— Temperature, T
A1 A2 A0
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TSP-WEEK 5
3
Temperature is defined as TU
S
NV
1
,
≡
∂
∂
From (*) 21
11
TT=
Note: when 2 systems are in thermal equilibrium, their
temperatures are equal.
Therefore, for 2 systems of the same temperature in thermal
contact,
NVNV
U
S
U
S
TT,2
2
,1
1
21
110
∂
∂−
∂
∂=−=
( )
NVNVU
S
U
SS
,1
0
,1
21
∂
∂=
∂
+∂=
Entropy combined system maximum at equilibrium
Let A1 in thermal equilibrium with A3
A2 in thermal equilibrium with A3
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TSP-WEEK 5
4
T1 = T3 and T2 = T3
Therefore T1 = T2
Zeroth Law
If two systems are each in thermal equilibrium with a
third system, then they are in thermal equilibrium with
each other.
Note:
- temperature is the indicator of thermal equilibrium
- all parts of a system must be in thermal equilibrium if
the system is to have a definable single temperature.
Entropy – a measure of the number of accessible states Ω
T
1 - measures the Ω varies with the internal energy, U
( )
NV
NV
UkUU
S
T,
,
)(ln1
Ω∂
∂=
∂
∂=
But 2)()(
R
UconstU =Ω
Therefore
( )
∂
∂=Ω
∂
∂= 2
,)ln()(ln
1R
NVUconstk
UUk
UT
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TSP-WEEK 5
5
( )
=+
∂
∂=
U
kRUconst
U
kR 1
2ln)ln(
2
U
Rk
T
2
1
1= or kRTU
2
1=
U is measured to the zero-energy reference level, Nµ
Uthermal = Utotal - Nµ
Utotal = Uthermal + Nµ
= NkRT µ+2
1
if each particle – v degrees of freedom
R = vN
)2
(
2
µ
µ
+=
+=
kTv
N
NNkTv
Utotal
many common process, µ does not change or very little
)(
2TNk
vU ∆≈∆
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TSP-WEEK 5
6
Heat Flow
A1 and A2 -- not yet in thermal equilibrium
T
UU
U
SS
NV
∆=∆
∂
∂=∆
,
02
2
1
1210 >
∆+
∆=∆+∆=∆
T
U
T
USSS
1st Law ∆U1 = - ∆U2
011
21
10 >
−∆=∆
TTUS
If T2> T1, then ∆U1>0
If T1> T2, then ∆U1<0
Note:
if 2 interacting systems are not yet in thermal
equilibrium, then the 2nd law demands that the energy
must flow from the hotter system to the cooler one,
and not vice versa.
A1 A2 A0
T1 T2
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TSP-WEEK 5
7
Phase Transition (add heat, T unchanged)
NVU
S
T ,
1
∂
∂= at higher T, the
NVU
S
,
∂
∂is smaller
From S = k ln Ω(U)
Where Ω(U) = (const)UR/2
S = ½ Rk lnU
TU
Rk
U
S
NV
1)
1(
2,
==
∂
∂
Undergoing phase transition—
Add heat, temperature remains constant
TU
Rk 1)
1(
2= --- why??
TU
S
NV
1
,
=
∂
∂
S
U (∝ T)
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TSP-WEEK 5
8
2 reasons;
i) as heat is added, R increases, R/U constant
ii) the added heat – releases particles from potential
wells ( the zero-energy reference is higher)
Uthermal = Utotal - Nµ
from Ω(U) = (const)UR/2 , Ω(U) = (const)(U-Nµ)R/2
S = ½ Rk ln (U- Nµ)
TNU
Rk
U
S 1
)(2=
−=
∂
∂
µ
2 phase region U
S
∂
∂ constant
boiling
melting
S
U
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TSP-WEEK 5
9
Heat Reservoirs (add heat, T unchanged)
Heat reservoirs – sufficiently large systems that their
temperatures do not change when a small amount of heat is
added to or removed from them.
∆Q – small amount of heat.
Taylor series expansion
....)(2
1)()()( 2
,
2
2
,
+∆
∂
∂+∆
∂
∂+=∆+ Q
UQ
U
SUSQUS
NVNV
ignoring (∆Q)2 and higher
)()()(
,
QU
SUSQUS
NV
δδ
∂
∂+=+
T
QdS
QT
USQUS
δ
δδ
=
=−+ )(
1)()(
----- for any system, large or small
T
QdS
δ=
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TSP-WEEK 5
10
Thermal interaction with reservoir
Let 1U : average Internal Energy for A1
S0: entropy of the combined system (A1+A2)
Ωo: accessible states for the combined system
From fundamental postulate,
P (U1) = (const) Ωo (U1)
And S0 = k ln Ωo or k
S
e0
0 =Ω
k
US
econstUP
)(
1
10
)()( =
Entropy of the combined system
S0 (U1= 1U +∆U1)
Using Taylor series expansion
...)(2
1)()()( 2
12
1
0
2
1
1
0
1101110
11
+∆
∂
∂+∆
∂
∂+==∆+= U
U
SU
U
SUUSUUUS
UU
at equilibrium, S0 is maximum
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TSP-WEEK 5
11
0
11
0 =
∂
∂
UU
S
And
∂
∂+
∂
∂
∂
∂=+
∂
∂=
∂
∂
1
2
1
1
1
21
1
2
2
1
0
2
)(U
S
U
S
USS
UU
S
−
∂
∂=
∂
∂−
∂
∂
∂
∂=
2112
2
1
1
1
11
TTUU
S
U
S
U
T2, reservoir temperature, -- constant
0
1
21
=
∂
∂
TU
And since TRk
U2
=
2
11111 22
1
U
Rk
U
Rk
UTU−=
∂
∂=
∂
∂
Therefore 2
1
2
1
0
2
21
U
Rk
U
S
U
−=
∂
∂
And,
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TSP-WEEK 5
12
2
12
1
1101110 )(4
)()( UU
RkUUSUUUS ∆−==∆+=
Put this into the probability equation
212
1
10
21
210
)(4
)(
))(4/()(
1
)(
)()(
UU
R
k
US
k
UURkUS
eeconst
econstUPi
∆
∆−
=
=
22
1 2/)()(
σUeconst
∆−=
With
1
2U
R=σ
The probability the system has energy in the range U1 and
U1+dU1 is propotional to the size of the range dU1
1
2/)(
11
221)()( dUeconstdUUP
U σ∆−=
1)(
1
11 =∫U
dUUP
∫∫∆−∆− =
1
1
1
221
11
2/)()()(
U
Ua
U
UdUeconstdUeconst
σ
With 22
1
σ=a
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TSP-WEEK 5
13
aconstdUeconst
U
Ua π)()(
1
1
1 =∫∆−
Therfore 22
1)(
1)(
πσπ
π
==
=
aconst
aconst
For system in equilibrium with reservoir, the probability
that the system has energy in the range U and U+dU is
dUedUUP
U 22 2/)(
22
1)( σ
πσ
∆−=
With UR
2=σ and )( UUU −=∆
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TSP-WEEK 5
14
e.g: 2 systems & let ev1=ε for degree of freedom
R = 8 and R = 800
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TSP-WEEK 5
15
Heat Capacity (The Storage of Thermal Energy)
-a measure of how much heat energy must be added
in order to raise its temperature by one standard unit.
General Definition
• If CT = Total Amount of Heat Energy required to raise
the temperature of some System by 1 oC, therefore Q
= CT ∆T ; where CT is heat capacity (SI: J/ oC)
SPECIFIC HEAT CAPACITY
If c = Amount of Heat Energy per kilogram that is
required to raise the temperature of one kilogram of the
substance 1 oC, therefore q = m c ∆T ; where c is
specific heat capacity (SI: J/kg.oC)
MOLAR HEAT CAPACITY
If c = Amount of Heat Energy per mole that is required
to raise the temperature of 6.022x10 23 molecules of the
substance 1 oC, therefore Q = n c ∆T; where c molar heat
capacity (SI: J/mole/ oC)
VOLUMETRIC HEAT CAPACITY of a SUBSTANCE
If cv = Amount of Heat Energy per unit volume that is
required to raise the temperature of one cubic meter of
the substance 1 oC, therefore Q = V cv ∆T, where cv is
volumetric heat capacity (SI: J/m3.oC )
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TSP-WEEK 5
16
Definition
Xconstantatcapacityheat:
X
XT
QC
∂
∂=
Two types of heat capacity
Heat capacity at constant pressure,
P
PT
QC
∆
∆=
Heat capacity at constant volume
V
VT
QC
∆
∆=
∆Q
∆Q
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TSP-WEEK 5
17
Third law of thermodynamics (Nernst's theorem)
Order & disoder
two different gasses separated by a partition will mix when
the partition is removed, increasing system disorder.
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TSP-WEEK 5
18
Entropy (S) is the thermodynamic state function that
describe the amount of disorder, -- a large value for entropy
means high degree of disorder
Entropy changes
An increase in disorder results in an increase in entropy.
- S increases when solid→liquid, liquid → gas
- S decreases when gas dissolves in a solvent
- S increase as temperature increase
Adding heat to the system,
-increases the number of the accessible states,
-increases disorder,
-increases the entropy.
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TSP-WEEK 5
19
Removing heat from the system,
-decreases the number of accessibles states
-decreases the disorder
-decreases the entropy
remove the heat until,
The number of accessibles state Ω = 1
From S = k ln Ω = k ln (1) = 0
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TSP-WEEK 5
20
third law of thermodynamics the entropy of any pure, perfect crystalline element or
compound at absolute zero (0 K) is equal to zero.
-all molecular motion stops at 0K
-represent perfect order
Absolute zero
Definition: The lowest possible temperature allowed by the
laws of thermodynamics.
At this temperature, molecules would possess the absolute
minimum KE allowed by quantum mechanics.
The temperature is equivalent to -273.15°C or 0K (kelvin).
At absolute zero, the entropy of any system vanished.
Or
The temperature at which all possible heat has been
removed from an object.
Absolute zero cannot be reached experimentally, although
it can be closely approached.
Cornell and Wieman cooled a small sample of atoms down
to only a few billionths (0.000,000,001) of a degree above
Absolute Zero!
Note: entropy T
QdS
δ=
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TSP-WEEK 5
21
δQ = mC dT T
mCdTdS =
∫ ∫∫ ===∆T
dTmC
T
mCdTdSS
if C : constant
heating a sample from absolute zero to finite temperature T,
increases the entropy by,
∫∫∫ ===∆TTT
T
dTmC
T
mCdTdSS
000
= ST – S0 = ST – 0 = ST
Example:
A sample with specific heat c = 5T1/2 J/kg.K is heated to
300 K. Determine the entropy of the sample at 300 K
KkgJT
dTT
T
cdT
T
Q
dSSSSS
TT
T
KK
./1735
300
0
2/1
00
0
0300300
====
=∆=−=
∫∫∫
∫
δ
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TSP-WEEK 5
22
Example:
The reaction A + B → C is carried out at 300 K. How much
heat is released (per kg) of C if cA = 5T1/2 J/kg.K, cB = 8T1/3
J/kg.K and cC = 15 T1/3 J/kg.K.
Ans: ∆Q = T∆S and ∆S = SC – (SA + SB)
SA (300K) = 173 J/kg.K
SB (300K) =
SC (300K) =
∆S =
∆Q =
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WEEK 5
THERMAL INTERACTION
1
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2
AXIOM or Postulate
is a proposition that is not proved or demonstrated but
considered to be either self-evident, or subject to necessary
decision.
Therefore its truth is taken for granted and serves as a
starting point for deducing and inferring other truths theory
(dependent).
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AXIOM 1 /Postulate 1
AXIOM 2/Postulate 2
For all system in equilibrium there exists a function of the extensive parameters, called the entropy, S. If there are no internal constraints on the system, the extensive parameters can take the values that maximize S over the possible states with internal constraints.
3
There exist special states of macroscopic physical
system, called equilibrium states, which can be fully
described by the internal energy, U, and a set of
extensive parameters, X0, X1, X2,….Xc
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4
Macroscopic property
(physical system)Equilibrium State (ES)
fully described by
microscopic properties
(extensive thermodynamic
properties) U , V , N , S , …..
Intensive thermodynamic
properties
Extensive thermodynamic
properties
Independent of the size of the
thermodynamic system : P, T, r
Dependent : V , S, U, H, N, …..
= S maximised
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5
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6
U1 U2
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7
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8
Note: when 2 systems are in equilibrium, their temperatures are equal.
Therefore, for 2 systems of the same temperature in thermal contact,
Entropy of combined system is maximum at equilibrium
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9
T1 = T3T2 = T3
Therefore T1 = T2
T3
T2
A2
A3
T2
T3
T1
A1
A3
T3
Let A1 in thermal equilibrium with A3 A2 in thermal equilibrium with A3
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10
Zeroth Law:
If two systems are each in thermal
equilibrium with a third system, then they
are in thermal equilibrium with each other.
Note:
i. Temperature is the indicator of thermal equilibrium
ii. All parts of a system must be in thermal
equilibrium if the system is to have a definable
single temperature.
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11
Entropy - a measure of the number of accessible states Ω
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12
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13
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14
Note: if 2 interacting systems are not yet in thermal equilibrium, then the
2nd law demands that the energy must flow from the hotter system to the
cooler one, and not vice versa.
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Add Heat, temperature remains constant
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Ū1
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probability
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where cv is volumetric heat capacity
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cv
cp
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Order & Disorder
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Entropy changes
An Increase in disorder results in an increase in
entropy.
- S increases when solid – liquid, liquid – gas
- S decreases when gas dissolves in a solvent
- S increases as temperature increases
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J/K
Example:
A sample with specific heat c = 5T ½ J/kg/K is heated to 300 K from
0 K. Determine the entropy of the sample at 300 K.
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161J/kg.K
301 J/kg.K
301 – (173 + 161)= - 32.74 J/kg.K
-32.74 x 300 = - 9.8 kJ/kg
(heat is given out by the system)
J/K
kJ
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THAT’S ALL FOR NOW!!
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TSP: WEEK 6
1
WEEK 6
Mechanical Interaction (Work)
The system exerts force F against the piston of area A, and the piston moves distance ds, The work done by the system,
δW = F. ds = (P/A) ds = P (A ds) = PdV
Note: δW is inexact, can be changed into exact diff. dV
δW. (1/P) ….. exact
And δQ .(1/T) ….. exact , dS ,entropy 1st Law (exact differential form)
dU = TdS – PdV + µdN (*) Where U = U (T,S,P)
µ = µ(P,V, N)
T = T (P,µ, N) ……..etc..
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TSP: WEEK 6
2
From (*) NVS
UT
,
∂
∂=
NSV
UP
,
∂
∂=
VSN
U
.
∂
∂=µ
change in entropy, dN
TdV
T
PdU
TdS
µ−+=
1 form (*)
if S = S(U, V, N)
dNN
SdV
V
SdU
U
SdS
VSNSNV ,,,
∂
∂+
∂
∂+
∂
∂=
and NVU
S
T ,
1
∂
∂= ,………..thermal interaction
NUV
S
T
P
,
∂
∂= , ……..mechanical (work) int.
VUN
S
T ,
∂
∂=−
µ ……….diffusive (particles) int.
mechanical interaction,
NUV
S
T
P
,
∂
∂=
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TSP: WEEK 6
3
NUNU V
kTV
kTP,,
lnln
∆
Ω∆=
∂
Ω∂=
or
1
212 lnlnlnln
Ω
Ω=Ω−Ω=Ω∆=
∆
kT
VP
or kTVP
e/
1
2 ∆=Ω
Ω
1st Law: ∆U =∆Q - p∆V + µ∆N
but ∆U = ∆N = 0, then ∆Q = P∆V
kTQ
e/
1
2 ∆=Ω
Ω
Expansion When heated – materials tend to expand
Gases: increase the force & frequency of collisions with the container wall, -- pushing outward
Solid & liquid: osscillate with greater amplitudes, increase the intermolecular spacing --- expand !!!
Coefficient of volume expansion, β ---- a measure of the relative increase in volume per unit increase in temperature.
PT
V
V
∂
∂≡
1β
Note: generally β = β(T, P) , but most expansions are at atmospheric pressure, --- const. pressure.
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TSP: WEEK 6
4
For small change in temperature, V = V (T)
TVTT
VV
P
∆=∆
∂
∂=∆ β
α, coefficient of linear expansion a measure of relative increase in length (X)
α = α (T, P)
TXTT
XX
P
∆=∆
∂
∂=∆ α
relation : α and β
V′= V + ∆V = V ( 1 + β ∆T ) (*)
But V′= X′Y′Z′=X(1+α∆T) Y(1+α∆T) Z(1+α∆T)
= XYZ (1+α∆T)3
= V(1+α∆T)3 = V(1 + 3α∆T +… ) (**)
for small ∆T, … ∆T2 ≅ 0
therefore , from (*) & (**), β = 3α
Isothermal compressibility (symbol: κ ) -a measure of relative change in volume per unit increase in pressure (temperature: constant)
TP
V
V
∂
∂−≡
1κ κ = κ (T, P) and V = V( P, T )
PVPP
VV
T
∆−=∆
∂
∂=∆ κ
bulk modulus :TV
PV
∂
∂−=
κ
1
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TSP: WEEK 6
5
The Diffusive interaction
Chemical Potential (symbol: µ) (in many cases, Free Energy is used instead of chemical potential)
The chemical potential : the change in the energy of the sytem when an additional constituent particle is introduced, with the entropy and volume held fixed.
VSN
U
,
∂
∂=µ
• It express how eager system is for particles.
• In equilibrium, it is equal in two systems placed in diffusive contact.
• Particles move form a region of high chemical potential to a region of low chemical potential.
• It can be found by differentiating themodynamic potentials with respect to N.
∆N
∆U is realesed when ∆N energyless particles are added
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TSP: WEEK 6
6
∆N : a known number of particles are added into the system
The gain in thermal energy is ∆U
µ ∆N = - ∆U
note: - ∆U : inside the potential well
N
U
∆
∆−=µ
µ and temperature
+
−
+
−
+ −
+ −
+ −
− +
− +
− +
− +
− +
+
−
−
+
+
−
+
−
+
−
+ −
+
−
cold
hot
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TSP: WEEK 6
7
µ and pressure at higher P, particles are closer , increase the strength of
interactions :attractive (reduce µ ) or repulsive (increase µ)
µ and particles concentration will increase or decrease --- depending on the nature of the interaction. the chemical potential increases as
• the internal energy, U, of the phase increases,
• the entropy, S, of the phase decreases at a given temperature, T.
• the volume, V increases for a given pressure, P. components that possess HIGHER U are destabilized relative to those with LOWER U components with LOWER S are destabilized relative to those with HIGHER S. Equilibrium Conditions
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TSP: WEEK 6
8
Let A1 and A2 : interacting – thermally,mechanically and diffusively
1st Law NVPSTU ∆+∆−∆=∆ µ
or NT
VT
PU
TS ∆−∆+∆=∆
µ1
law of conservation
12
12
12
NN
VV
UU
∆−=∆
∆−=∆
∆−=∆
the change in entropy due to redistribution of energy, volume, or number of particles,
∆S0 = ∆S1 + ∆S2
1
2
2
1
11
2
2
1
11
21
0
11N
TTV
T
P
T
PU
TTS ∆
−−∆
−+∆
−=∆
µµ..(*)
at equilibrium, entropy is maximum
1
2
2
1
11
2
2
1
11
21
0
110 N
TTV
T
P
T
PU
TTS ∆
−−∆
−+∆
−==∆
µµ
therefore
011
21
=−TT
02
2
1
1 =−T
P
T
P
02
2
1
1 =−TT
µµ
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TSP: WEEK 6
9
and
T1 = T2 P1 = P2 µ1 = µ2 For 2 systems interacting thermally, mechanically and diffusively are in equilibrium. Approach to equilibrium (not yet in equilibrium)
∆S0 > 0 Let A1 and A2 : interacting – thermally,mechanically and diffusively
1st Law: 111111 NVPQU ∆+∆−∆=∆ µ
using (*)
( )
1
2
2
1
11
2
2
1
1
11111
21
0
11
NTT
VT
P
T
P
NVPQTT
S
∆
−−∆
−
+∆+∆−∆
−=∆
µµ
µ
( ) ( ) 01111
121
2
121
2
1
21
0 ⟩∆−−∆−+∆
−=∆ N
TVPP
TQ
TTS µµ
therefore
011
1
21
⟩∆
− Q
TT , ( ) 01
121
2
⟩∆− VPPT
( ) 01
121
2
⟩∆− NT
µµ
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TSP: WEEK 6
10
from the three equations;
1. if T1> T2, ∆Q1 < 0, and if T2 >T1, ∆Q1>0 interacting thermally: heat flow from hotter towards the cooler, NOT vice versa
2. if P1> P2, ∆V1 > 0, and if P2 >P1, ∆V1 < 0 int. Mechanically: volume is gained by the system having higher pressure at the expense of the other, and NOT vice versa.
3. if µ1> µ2, ∆N1 < 0, and if µ2 >µ1, ∆N1>0 int. diffusively: particles flow from the system with
higher µ toward the one with lower µ, and NOT vice versa.
µ and Ω Let A1 and A2 : interacting – thermally,mechanically and diffusively
dNPdVTdSdU µ+−=
and
VSN
U
,
∂
∂=µ
1st Law
dNT
dVT
PdU
TdS
µ−+=
1
if U and V constants
VUN
S
T ,
∂
∂=−
µ or
VUN
ST
,
∂
∂=− µ
but S = k ln Ω
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TSP: WEEK 6
11
VUNkT
,
ln
∂
Ω∂=− µ ………….($)
Ω, accessible states – increasing function of number of particles.
µ neg. or dS/dN pos. – attract particles
µ pos. or dS/dN neg. – release particles
let ∆N – energyless particles added to a system at constant volume. From ($)
N
kT∆−=Ω∆
µln
but 1
212 lnlnlnln
Ω
Ω=Ω−Ω=Ω∆
)/(
1
2 kTNe
∆−=Ω
Ω µ
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TSP: WEEK 6
12
Ideal Gas
a hypothetical gas with molecules of negligible size that exert no intermolecular forces their energies are entirely kinetic for 1 molecule, mass m and momentum p
222
2
2
1
2
1
2
1
2ZYX p
mp
mp
mm
p++==ε
each molecule : 3 degrees of freedom for N molecules : 3N degrees of freedom number of quantum states available (6 dim.)
31h
pppzyx zyx
particles
∆∆∆∆∆∆=Ω
= (const) ∆x∆y∆z∆px∆py∆pz for N particles
iziyixiii
N
i
N
i
iparticlesN dpdpdpdzdydxh
∏ ∏= =
=Ω=Ω
1 13
1
limit: volume V and let U = total energy of the gas
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TSP: WEEK 6
13
∫∏=
=Ω
N
i
iziyix
N
particlesN dpdpdpVh 1
3
1……(*)
since mUppppP Nzxzyx 2.... 22
2
2
1
2
1
2
1 =+++++
-integral in (*) is equivalent to over the surface of a 3N dimension sphere of radius (2mU)1/2
-surface area of 3N-dim. Sphere ∝ (radius)(3N-1) or
∝ (radius)3N
NN
particlesN mUV3
)2(∝Ω
2/32/3)2)(( NNN
particlesN UVmconst=Ω
2/3)( NN
particlesN UVconst=Ω
UNkVNkconstkS gasidealgasideal ln2
3ln)(ln ++=Ω=
using
TU
S
V
1=
∂
∂ and T
P
V
S
U
=
∂
∂
V:constant
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TSP: WEEK 6
14
TU
Nk
U
S
V
12
3
00 =++=
∂
∂
NkTU
2
3=
U:constant
T
P
V
Nk
V
S
U
=++=
∂
∂00
NkTPV = ………..ideal gas law …….. gas model Real gas
- degrees of freedom may be larger than 3
NkTU2
υ=
where υ is degrees of freedom (we will use R for gas constant)
- mutual interactions and sizes cannot be ignored le v : molar volume R = NAk : gas constant Pv = RT
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TSP: WEEK 6
15
Compress gas, will become liquid, and cannot compress anymore b = molar volume of liquid phase – limit on the molar volume
v → v-b mutual interaction – reduce the velocity of molecules hitting the wall or pressure sensor real pressure is higher than the measured one
P →( P + mutual attraction)
Mutual attraction ∝ 1/(v2 ) Incorporated the two effect into the gas law
( ) RTbv
v
aP =−
+
2
van der Waals equation of state Liquid No good model for liquid yet Can use van der Waals equation with modification
-liquid phase, the volume ≅ b -pressure term
)(22
bfv
aP
v
aP +
+→
+
where f(b) : due to other interactions (liquid)
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TSP: WEEK 6
16
since v ≅ b : constant for liquid
.)()(22
constbfb
abf
v
a≅+≈+
the modified van der Waals eq, (P + const.) b = RT Solids Models for solids – have more than one components One property of one component at a time Ex. The heat capacity of the lattice alone Solid – a lattice of atomic masses coupled by spring
Potential energy, εP = (1/2) kx2 When one atom vibrating – send wave down the lattice Quantum energy of the wave: called phonon Phonon travel throughout the solid’s volume – phonon gas Metal – conduction band: electrons in the conduction band are mutually shared by all atoms – known as electron gas
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TSP: WEEK 6
17
Example of application of the models 1. heat capacity
let N: constant dQ = dU + PdV
υ: degree of freedom
U = (1/2) υNkT + Nµ
dQ = (1/2) υNk dT + N dµ+ P dV
change in N dµ is small
dQ = (1/2) υ R dT + P dV
per mol: dq = (1/2) υ R dT + P dv …. (*)
R = NA k -- gas constant Molar heat capacity at constant volume
RT
qc
V
V2
υ=
∂
∂=
since v= v(T,P) --- molar volume
dPP
vdT
T
vdv
TP
∂
∂+
∂
∂=
from (*), at constant P ------ dP = 0
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TSP: WEEK 6
18
dTT
vPRdTdq
P
∂
∂+=
2
υ
PP
PT
vPR
T
qc
∂
∂+=
∂
∂=
2
υ
or
P
VPT
vPcc
∂
∂+=
molar heat capacity at constant pressure or
P
VPT
vPcc
∂
∂=−
for gases, change in volume at constant pressure is large For solid & liquid, change in volume at constant pressure is smaller -----cp & cv nearly the same. Example
1. Calculate cp-cv for an ideal gas Ans: Pv = RT or v = RT/P
P
VPT
vPcc
∂
∂=−
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TSP: WEEK 6
19
P
R
T
v
P
=
∂
∂
RP
RP
T
vPcc
P
vp ==
∂
∂=−
2. Calculate cp-cv for va der Waals gas
( ) RTbvv
aP =−
+
2
a and b are constant Ans:
( ) RdTdvv
aPbvdv
v
adP =
++−
−
23
2
−+
−−=
12
1
)(
2v
b
Pv
a
dPbvRTPdv
−+
∂
∂=−
12
12 v
b
Pv
a
R
T
vPcc
p
vp
note: for P >> a/v2 and v >> b
Rcc vp =− ……ideal gas
for P << a / v2 and v ≅ b
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TSP: WEEK 6
20
0
2
≅=−
pva
Rcc vp
………liquid
3. Find an expression for the isothermal compressibility of an ideal gas
Ans:
TP
v
v
∂
∂−=
1κ compressibility
Pv = RT Pdv = RdT – vdP
vP
vP
T
−=
∂
∂ or PP
v
v T
11=
∂
∂−=κ
4. 4. Find an expression for the isothermal compressibility of a van der Walls gas
Ans: ( ) RTbv
v
aP =−
+
2
−+
−−=
12
1
)(1
2 v
b
Pv
aPv
dPbvRTdv
v
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TSP: WEEK 6
21
−+
−
=
∂
∂−=
12
1
111
2v
b
Pv
a
vb
PP
v
v T
κ
Assigment 1 Suppose the equation of state for some system was
beTPaV =
−3
12
where a and b are constants
i. write this in differential form, expressing dV in terms of dT and dP
ii. Express the isothermal compressibility of this system in terms of (T, V, P)
iii. Express the coefficient of volume expansion for this system in terms of (T, V, P)
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MECHANICAL & DIFFUSIVEINTERACTIONS
WEEK 6
1
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from (*)
U,N U,V
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Comparing the above 2 equations:
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g
But most expansions are at atmospheric pressure , - constant pressure.
Coefficient of volume expansion, - a measure of the relative increase in
volume per unit increase in temperature.
When heated – materials tend to expand
Gases: - increase the force & frequency of collisions with the container wall,
-- pushing outward.
Solid & liquid: - oscillate with greater amplitudes, increase the intermolecular
spacing – expand!!
Expansion
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• Particles move from a region of high chemical potential to a region of
low chemical potential.
• It express how eager a system is for particles.
•In equilibrium, it is equal in two systems placed in diffusive contact.
•It can be found by differentiating thermodynamics potentials with
respect to N.
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DU is released when DN energyless
particles are added
DN : a known number of particles
are added into the system.
The gain in thermal energy is DU.
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> < 0
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Six-dimensional space
Six-dimensional space is any space that has six
dimensions, that is, six degrees of freedom, and that
needs six pieces of data, or coordinates, to specify a
location in this space. There are an infinite number of
these, but those of most interest are simpler ones that
model some aspect of the environment. Of particular
interest is six-dimensional Euclidean in which 6-
polytopes and the 5-sphere are constructed.
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Thermal
interaction
Mechanical
interaction
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Let
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but
hence
so
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van
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Assignment (individual) Exercise
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Week 8
1
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ENSEMBLE : Synonyms
1. totality, entirety, aggregate
ahn-sahm-buh l,
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THERE ARE 3 TYPES OF ENSEMBLES:
MICROCANONICAL
CANONICAL
GRAND CANONICAL
3
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U constant;
N, V, E constant;
Isolated
U constant;
T not constant;
In contact with
Heat reservoir
T , m constant;
U, N not constant;
In contact with Heat and
Particle reservoir
SUMMARY
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Consider TWO cases:
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Ui : ground stateUj: excited state& Pi: probability in the ground state i
Pj: probability in the excited state j
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In solid-state physics, the electronic
band structure (or simply band
structure) of a solid describes those
ranges of energy that an electron within
the solid may have (called energy
bands, allowed bands, or simply bands),
and ranges of energy that it may not
have (called band gaps or forbidden
bands).
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Band theory derives these bands and band gaps
by examining the allowed quantum
mechanical wave functions for an electron in a
large, periodic lattice of atoms or molecules.
Band theory has been successfully used to explain
many physical properties of solids, such as
electrical resistivity and optical absorption, and
forms the foundation of the understanding of
all solid-state devices (transistors, solar cells, etc.).
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- Normally a system can have many excited
states and many ground states.
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-how many particles can be in any quantum state? Total N is fixed
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Energy and Quadratic Forms
A simple model of a crystal--masses m on springs with coefficients k.
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separation
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A LITTLE CAPTION OF
TANGKUBAN PARAHU, INDONESIA (2009)
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THE END
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Week 8
1
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2
Statistical mechanics
a branch of theoretical
physics and chemistry (and mathematical physics)
that studies, using probability theory, the average
behaviour of a mechanical system where the state
of the system is uncertain.
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A common use of statistical mechanics is in
explaining the thermodynamic behaviour of large
systems.
Microscopic mechanical laws do not contain concepts
such as temperature, heat, or entropy,
however, statistical mechanics shows how these
concepts arise from the natural uncertainty that arises
about the state of a system when that system is
prepared in practice.
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The benefit of using statistical mechanics
is that it provides exact methods to
connect thermodynamic quantities (such
as heat capacity) to microscopic
behaviour,
whereas in classical thermodynamics the
only available option would be to just
measure and tabulate such quantities for
various materials.
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Statistical mechanics also makes it possible
to extend the laws of thermodynamics to cases
which are not considered in classical
thermodynamics,
for example microscopic systems and other
mechanical systems with few degrees of freedom.
This branch of statistical mechanics which treats
and extends classical thermodynamics is known
as statistical thermodynamics or equilibrium
statistical mechanics.
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ENSEMBLE : Synonyms
1. totality, entirety, aggregate
ahn-sahm-buh l,
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THERE ARE 3 TYPES OF ENSEMBLES:
MICROCANONICAL
CANONICAL
GRAND CANONICAL
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NVE ensemble)
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In statistical mechanics, a microcanonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system which has an exactly specified total energy.
The system is assumed to be isolated in the sense that the system cannot exchange energy or particles with its environment, so that (by conservation of energy) the energy of the system remains exactly known as time goes on.
The system's energy, composition, volume, and shape are kept the same in all possible states of the system.
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The macroscopic variables of the microcanonical ensemble are quantities such as
(i) the total number of particles in the system (symbol: N),
(ii) the system's volume (symbol: V) each which influence the nature of the system's internal states,
(iii) as well as the total energy in the system (symbol: E).
This ensemble is therefore sometimes called the NVE ensemble, as each of these three quantities is a constant of the ensemble.
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In simple terms, the microcanonical ensemble is defined by assigning an equal probability to every microstate whose energy falls within a range centered at E. All other microstates are given a probability of zero. Since the probabilities must add up to 1, the probability P is the inverse of the number of microstates W within the range of energy,
P= 1/W The range of energy is then reduced in width until it
is infinitesimally narrow, still centered at E. In the limit of this process, the microcanonical ensemble is obtained.
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2. CANONICAL ENSEMBLE:
The system is in thermal equilibrium with a heat reservoir at temperature T.
Hence the temperature of the system is constant; but the energy of the system is not constant.
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In statistical mechanics, a canonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system which is in thermal equilibrium with a heat bath.
The system is said to be closed in the sense that the system can exchange energy with a heat bath, so that various possible states of the system can differ in total energy. The system's composition, volume, and shape are kept the same in all possible states of the system.
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The thermodynamic variable of the canonical ensemble is the absolute temperature (symbol: T).
The ensemble is also dependent on mechanical variables such as
(i) the number of particles in the system (symbol: N)
(ii) the system's volume (symbol: V), each which influence the nature of the system's internal states.
This ensemble is therefore sometimes called the NVT ensemble, as each of these three quantities is a constant of the ensemble.
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In simple terms, the canonical ensemble assigns a probability P to each distinct microstate given by the following exponential:
where E is the total energy of the microstate, and k is Boltzmann's constant.
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The number A is the free energy (specifically, the Helmholtz free energy) and is a constant for the ensemble.
However, the probabilities and A will vary if different N, V, T are selected.
The free energy A serves two roles: to provide a normalization factor for the probability distribution (the probabilities, over the complete set of microstates, must add up to one); and, many important ensemble averages can be directly calculated from the function A(N, V, T).
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3. GRAND CANONICAL ENSEMBLE:
The system is in contact with both a heat reservoir
and a particle reservoir.
(i) the U and N of the system are not
constant.
(ii) the T and the m are constant.
(the m is the energy required to add a particle to the
system)
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In simple terms, the grand canonical ensemble assigns a
probability P to each distinct microstate given by the
following exponential:
where N is the number of particles in the microstate
and E is the total energy of the microstate.
k is Boltzmann's constant.
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In statistical mechanics, a grand canonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system of particles that is being maintained in thermodynamic equilibrium (thermal and chemical) with a reservoir.
The system is said to be open in the sense that the system can exchange energy and particles with a reservoir, so that various possible states of the system can differ in both their total energy and total number of particles. The system's volume, shape, and other external coordinates are kept the same in all possible states of the system.
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The thermodynamic variables of the grand canonical ensemble are chemical potential (symbol: µ) and absolute temperature (symbol: T).
The ensemble is also dependent on mechanical variables such as volume (symbol: V) which influence the nature of the system's internal states.
This ensemble is therefore sometimes called the µVT ensemble, as each of these three quantities are constants of the ensemble.
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U constant;
N, V, E constant;
Isolated system
U not constant;
T constant;
In contact with
Heat reservoir
Closed system
T , m constant;
U, N not constant;
In contact with Heat and
Particle reservoir
Open system
SUMMARY
µVT ensembleNVT ensembleNVE ensemble
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Consider TWO cases:
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Ui : ground stateUj: excited state& Pi: probability in the ground state i
Pj: probability in the excited state j
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In solid-state physics, the electronic
band structure (or simply band
structure) of a solid describes those
ranges of energy that an electron within
the solid may have (called energy
bands, allowed bands, or simply bands),
and ranges of energy that it may not
have (called band gaps or forbidden
bands).
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Band theory derives these bands and band gaps
by examining the allowed quantum
mechanical wave functions for an electron in a
large, periodic lattice of atoms or molecules.
Band theory has been successfully used to explain
many physical properties of solids, such as
electrical resistivity and optical absorption, and
forms the foundation of the understanding of
all solid-state devices (transistors, solar cells, etc.).
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- Normally a system can have many excited
states and many ground states.
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-how many particles can be in any quantum state? Total N is fixed
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Energy and Quadratic Forms
A simple model of a crystal--masses m on springs with coefficients k.
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separation
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A LITTLE CAPTION OF
TANGKUBAN PARAHU, INDONESIA (2009)
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THE END
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Week 10
1
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S
S
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= =
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Let’s say :
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Total no of accessible states?
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Total no of accessible states?
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TO WRITE MAXWELL-BOLTZMANN DISTRIBUTION FUNCTION:
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THE END….
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WEEK 11
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Partition functions describe the statisticalproperties of a system in thermodynamic equilibrium. It is a function of temperatureand other parameters, such as the volumeenclosing a gas.
Most of the aggregate thermodynamicvariables of the system, such as the total energy, free energy, entropy, and pressure, can be expressed in terms of the partition function or its derivatives.
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There are actually several different types of partition functions, each corresponding to different types of statistical ensemble (or, equivalently, different types of free energy.)
The canonical partition function applies to a canonical ensemble, in which the system is allowed to exchange heat with the environment at fixed temperature, volume, and number of particles.
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The grand canonical partition functionapplies to a grand canonical ensemble, in which the system can exchange both heat and particles with the environment, at fixed temperature, volume, and chemical potential.
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This is the reason for calling Z the "partition function": it encodes how the probabilities are partitioned among the different microstates, based on their individual energies.
The letter Z stands for the German word Zustandssumme, "sum over states". This notation also implies another important meaning of the partition function of a system: it counts the (weighted) number of states a system can occupy. Hence if all states are equally probable (equal energies) the partition function is the total number of possible states. Often this is the practical importance of Z.
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From
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The End
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WEEK 12
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occupied by
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1.FERMI-DIRAC DISTRIBUTION
2.BOSE-EINSTEIN DISTRIBUTION
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THE FERMI-DIRAC DISTRIBUTION
Assumptions:
1. The particles are identical and indistinguishable
2. They obey the Pauli exclusion principle.
This means that no quantum state can accept more than
one particle, taking spin into account. Such particles have
half-integer spin and are given the generic name
FERMION.
Examples of fermions: electrons, positrons, protons,
neutrons, and muons.
An important application of Fermi-Dirac statistics is to the
behaviour of free electrons in metals and semi-conductors.
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The Pauli exclusion principle is
the quantum mechanical principle that no
two identical fermions (particles with half-
integer spin) may occupy the same quantum
state simultaneously.
A more rigorous statement is that the
total wave function for two identical fermions
is anti-symmetric with respect to exchange of
the particles. The principle was formulated by
Austrian physicist Wolfgang Pauli in 1925.
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Integer spin particles, bosons, are not subject
to the Pauli exclusion principle: any number of
identical bosons can occupy the same
quantum state, as with, for instance, photons
produced by a laser and Bose–Einstein
condensate.
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BOSE-EINSTEIN DISTRIBUTION
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For the continuous energy spectrum,
Bose-Einstein
distribution
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PHEW…!!
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BOSE EINSTEIN CONDENSATION
WEEK 13
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Here we shall be concerned with a gas of non-interacting
particles (atoms or molecules) of comparatively large mass
such that quantum effects only become important at very low
temperatures. The particles are assumed to comprise an
ideal Bose-Einstein gas. The discussion is relevant to ,
which undergoes a remarkable phase transition known as
Bose-Einstein condensation.
This phenomenon is intimately related to the
superfluidity of liquid helium at low temperatures.
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As noted earlier, bosons are particles of integral spin that obey Bose-Einstein statistics. There is no limit to the number of bosons that can occupy any single particle state.
Consider an ideal boson gas consisting of N
bosons in a container of volume V held at
absolute temperature T. The Bose-Einstein
continuum distribution is
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THE END
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WEEK 14
BLACKBODY RADIATION
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Statistical thermodynamics is applicable to
(i)radiant energy (ii) material particles.
Familiar observation:
a hot body loses heat by radiation. The energy loss is
attributable to the emission of electromagnetic waves
from the body.
The distribution of energy flux over the wavelength spectrum does
not depend on the nature of the body but does depend on its
temperature.
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Now: concern with the thermodynamic properties of electromagnetic radiation
in thermal equilibrium.
The radiation can be regarded as a photon gas. Consider an enclosure or
cavity of volume V at a constant temperature T. The walls of the cavity are
thermally insulated and perfectly reflecting. Since the system is isolated, it has
a fixed energy U.
However, the photons emitted by one energy level may be absorbed at
another, so the number of photons is not constant. This means that the
restriction does not apply.
Photons are bosons of spin 1 and hence obey Bose-Einstein statistics.
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What happens to this radiation?
• The radiation is absorbed in the walls of the cavity.
• This causes a heating of the cavity walls.
• Atoms in the walls of the cavity will vibrate at frequencies
characteristic of the temperature of the walls.
• These atoms then re-radiate the energy at this new
characteristic frequency.
• The emitted “thermal” radiation characterizes the
equilibrium temperature of the black-body.
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EXAMPLE:
Assume that the radiation from the Sun can be regarded as blackbody
radiation. The radiant energy per wavelength interval has a maximum at
480 nm. Estimate the temperature of the sun.
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AT LAST!
THANK YOU
FOR YOUR
PATIENCE
GUYS…