phys321 fall07 notes 01
TRANSCRIPT
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Thermal PhysicsThermodynamics
Statistical Mechanics
Deals with a collection of a large number of particles It is effectively impossible to follow the motion andtrajectory of each particle In thermal physics, we combine two approaches
to understand and predict behaviorthermodynamics
statistical mechanics
Thermodynamics:
Phenomenological theory of matter Macroscopic theory (i.e. it doesnt contain information about
atoms, molecules, )
Draws its concepts directly from experiment Always correct, cannot be questioned since they are
deduced from experiment
Does not depend upon microscopic details(e.g. 1st & 2nd laws, heat flow from hot to cold,
maximum efficiency of engines)
Statistical Mechanics
Attempt to understand thermodynamical laws frominteractions of atoms, molecules, and other microscopic
degrees of freedom
Statistical theory of a
large collection
of particles
Fundamental laws of interaction(e.g. mechanics, quantum
mechanics, etc.)
Statistical concepts and certainbasic assumptions1
Provides an underlying explanation of thermodynamics1
Some of the basic assumptions cannot be proven easily.
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Two key concepts in Thermodynamics
1.Equilibrium2.Temperature
It is difficult to define these terms rigorously at this level. Thefollowing definitions will have to do for now:
EquilibriumEquilibrium is reached when the thermodynamic state
of the system does not change with time.
e.g. when you bring two objects into contact for long enough
time, they will be in equilibrium
TemperatureTemperature is a thing that is the same for two objects
after they have been in contact for a long enough time; i.e.
they are in thermal equilibrium.
Temperature is a measure of the tendency of an objectto spontaneously give up energy to its surroundings.
An object at higher temperature tends to lose energy to
an object at lower temperature.
Well see shortly that temperature is related to the
internal energy of a system.
Internal energy consists of:atomic kinetic (translation, rotation, vibration) and
potential energy (atomic interactions)
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Temperature scale and units
Choose something that changes with temperaturee.g. thermal expansion
Pick two convenient temperatures and assign them arbitrarynumberse.g. water boiling (100) and freezing (0)
Mark off a number of equally spaced intervals in between thetwo end points and extrapolate above and below
The ideal gas thermometer extrapolates to 0P = at -273.15C
What is the physical meaning of 0P = ?
We define an absolute temperature scale
273.15k cT T=
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Use the ideal gas as an example to demonstrate the relationship
State functions ( )P, ,V T microscopic property (velocity)Thermodynamics Statistical Mechanics
(I) Thermodynamicsof anIdeal Gas:The ideal gas law: PV nRT =
number of moles of gasn =universal gas constantR = -1 -18.31 J mol K=
1 mole = number of C12 atoms in 12 grams of C12
= 6.02 x 1023
Define Avogadros number236.02 10
AN =
Then the total number of molecules in n moles of an ideal
gas is A N nN = molecules.
Define Boltzmanns constant
23 -11.381 10 J KA
Rk
N
= = ( )nR Nk =
Then
( )A APV nRT nkN T nN kT NkT = = = = we can write either
orPV nRT PV NkT = =
1. The ideal gas is the thermodynamic system2. P, V, Tare state functions or thermodynamics parameters:
measurable macroscopic quantities that define the state ofthe system
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3. PV nRT NkT = = This is an example of a thermodynamic equation of state
(which is an equation that states the relationship between
different state functions. In general, it is of the form)( ), , , 0f P V T =K
4. As a typical equation of state in thermodynamics, it isdeduced from experiment (Not derived from theory)
(II) Kinetic theory (from a microscopic model) of and Ideal gas:
Assumptions
x
(i) Smooth walls symmetrical bounces (like a mirror)(ii) Molecules behave like elastic billiard balls bouncing
around xv v = x has constant magnitude, although
periodically changing direction
(iii) Low density each molecule moves as if it is alone,i.e. non-interacting
Goal: To relate the temperature of the gas to kinetic energy
(K.E.) of molecules
Approach: first, find how P is related to K.E., then use the
ideal gas law (PV = NkT) to find the relationship between
. .T K E+
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Consider one molecule
Let velocity of the molecule
component ofx
v
v v x x v
=
= =
Recall( )
dt
vdmdt
vmd
dt
pdF ===
The average force on the piston (averaged o
the period
ver
t between collisions)
vx
-vx
v-
x
xvF mt
=
where2
tim
2
x x
x
v v
t
L
v
=
=
=
(in one collision)
e interval between collisions
Therefore 2 xL v= t
2
22 x
x
x vLm
vLvmF ==
The average pressure (force per unit area)
2 2
x x
F m mP v v
A LA V = = =
If there are many (N) molecules, then the total average pressure is
2
1
2
1 N
i i xi
i x
i
P m vV
vmN
V N
==
=
velocitymoleculestheofcomponentx
moleculethitheofmass
thiix
vi
m
=
=
)(identicalbetomoleculestheassuming mi
m =
N
i=1
2
xPV Nmv= where the averagex-component of the
velocity is 2 21
1
N
x i xN
i
v v=
=
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Now2
xPV Nmv= is a microscopic description (derived from
laws of mechanics + some assumptions)
PV NkT = is a macroscopic thermodynamic
description (derived from experiment)
We can identify2
xkT mv= for an ideal gas; this is a
microscopic interpretation ofT
So21 1
2 2 xkT mv= is the average Kinetic Energy associated
with translational motion in thex-direction
We can derive the same expressions for motion in they- andz-
directions
2 21 1 1 1
2 2 2 2
2
x y zkT mv mv mv= = =
The magnitude of the velocity vector is 2 2 2 2 x y zv v v v+ + =
The average Kinetic Energy of molecules is
( )
( )
2 2 21 1
2 2
1
2
3
2
2
x y zmv mv mv mv
kT kT kT
kT
= + +
= + +
=
Therefore, we interpret temperature as the average kinetic
energy of the molecules divided by k. . .~ K Ek
T
Define Thermal Energy( ) thUThermal energy is the internal energy of a system that consists
of the kinetic and potential energies associated with the motionand interactions of the atoms and molecules in the system.
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Equipartition Theorem
At temperature T, the average thermal energy of each
quadratic degree of freedom is kT
This is an important theorem in classical statisticalmechanics
It will be proven later in the course (see Chapter 6)Note:
(i) It is not true in quantum statistical mechanics (which isimportant at low temperature)
(ii) It is true for quadratic degrees of freedom only all types of kinetic energy
e.g.21
2mv ,
212
I (translational, rotational)
for potential energy,only good for harmonic oscillator, i.e. ( ) 212V x kx=
Example: ideal gas (Nmolecules)
( )
2 21 1 1Total kinetic energy2 2 2
1 1 1
2 2 2
13
2
2 x y z N mv mv mv
N kT kT kT
N kT
= + +
= + +
=
( ) each degree of freedom (quadratic)13has 1 2 average energy2
3 degrees of freedom for simple (monoatomic)
molecule corresponding to , , translational motion
thermalU N kT kT
x y z
=
In general, we have
1
2thermalU Nf k
= T
where number of degreesof freedom
f =
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Countingfis tricky for complex molecules
(I) ConsiderCO (carbon monoxide)C O A good model is a hard bond connecting C and O
5 3 translational 2 rotationalf = = +
C
O
There should also be two additional degrees of freedom associated
with the stretching of the C bondO
1 kinetic
2 vibrational
1 potential
but it is usually very difficult to stretch molecules. We say that the
stretching mode is note normally excited at room temperature.
We also say that it is frozen out.
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(II) Spring model of a solidf 6= Each atom has 6 degree of f
3 foreedom
r kinetic, 3 for potential
l vibrations, theat
(i.e.
energy
for smal
potential energy is like th
for a harmonic oscillator
quadratic ( ) 212V x kx= )
= kTNUthermal
2
16
This is true only at
his example shows that the equipartition theorem is very
high temperature. At
low temperature, some
of the higher energy
modes will be frozen
out due to quantumeffects. This will be discus
r atomic
separation
U(r)
( )2
102 k r r
0r
sed later.
T
powerful and useful, but it is only strictly correct in the realm
of classical physics and harmonic potentials.
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Heat, Work and the 1st
Law of Thermodynamics
Heat
are two different forms of energy transfer
Work
Heat : the form of energy which is transferred from one object
to another due to a difference of temperature between
the objects
Work : transfer of energy to or from a system by a change of
parameters that describe the system (other thantemperature, e.g. push a piston, stir a cup, run a current
through a resistor, etc.)
The internal energy, U, of a system is defined up to anarbitrary additive constant
The change of internal energy, U, is always unambiguous
The First Law of Thermodynamics
U Q W = +
where
Q is heat added to the systemWis work done on the system
This is really just the Law of Conservation of Energy !
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Heat
Historically, scientists did not always know that heat is a form of
energy, until the famous Joules experiment demonstrated the
equivalence of heat and work. Consequently, we use Joule as
the SI unit for energy.
1 calorie is the amount of heat (Q) needed to raise the temperature
of 1 gram of water by 1C (from 14.5 to 15.5C)1 cal 4.186 J
1 kcal 4186 J
=
=
WorkThere are different kinds of work (e.g. mechanical, electrical,)
We focus on the simplest form of mechanical work.
Consider a simple, idealized situation: ideal gas compression
Assumptions:
(i) Quasi-static, which means that the motion of the piston isso slow that the system always has time to reach
equilibrium
(ii) No friction, which means that the piston does work on thegas only; no work is wasted in overcoming friction
(iii) The change in volume is small so that the pressure remainsnearly constant
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Calculate the work done on the gas when the piston is displaced a
distance x to the left
( )
( )
W F x
PA x
P A x
P V
=
=
=
=
0 final initialV V V = < which means that 0W >
Usually, our assumption (iii) ( )constantP = is not valid. IfPchanges with volume, then a more general expression for Wis
dVPdW =
f
i
V
VW P= dV where ( )P P V= is a function ofV
Wis the area under the
curve
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Ideal Gas Compression: Two Kinds of Process
1) Isothermal: Tremains constantfor this to happen, the gas cylinder must be in contact
with a heat bath (reservoir) of fixed temperature T, andthe process must be slow enough that heat can be
transferred to or from the reservoir much faster than work
is being done on the gas.
2) Adiabatic: No heat can enter or leave the gas (Q = 0) ;(i) it is perfectly insulated, or(ii) the process happens so quickly that heat does not have a
chance to leave of enter
1 Isothermal process
PV NkT
NkTP
V
=
=
( )ln ln
ln
f
i
f
i
V
V
V
V
f i
i
f
W PdV
dVNkT
V
NkT V V
V
NkT V
=
=
=
=
For compression, 0i f
V V W> >
(the work done on the gas is positive)
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Isotherm pression (continued)al com
Lets find the heat added or removed during an isothermal
compression of an ideal gas
Use the 1st Law
U Q W = + recall is heat added to the system
is work done on the system
Q
W
( )
( )
12
12
equipartition theorem
0 since 0 (isothermal)
U N f kT
U N f k T
T
=
=
= =
Therefore
ln i
f
VQ W NkT
V
= =
For compression is negativei f
V V Q>
Heat must leave the system during an isothermalcompression.
The amount of heat that leaves is equal to the work doneon the system.
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2 Adiabatic process
1 Lawst
U Q W = +0
For an ideal gas ( )12U Nf kT =
( )12dU Nf k dT = eq. (1)
For adiabatic process U W = dU dW PdV = = eq. (2)
Substitute eq. (1) into eq. (2)
12
Nf kdT PdV =
Use the ideal gas law ( )PV NkT =
12
NkT Nf kdT dV
V
=
2
f dT dV
T V
=
Solve this by integrating
=
=
f
i
f
i
f
i
f
i
f
V
V
T
T
V
V
T
Tf
2
lnln2
2 2f f
f f i iV T V T =
2or constantfVT =
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Substitute for Tfrom the ideal gas lawPV
TNk
=
Adiabatic compression (continued)
2
2
2 2
2
constant
or constant
constant
f
f
f
f
f
PVV Nk
P V
PV
+
+
=
=
=
2
constant is the adiabaticexponent
f
fPV
+
= =
(PV= constant)
Isotherms:
For an adiabatic compression,W
T
0
0
U W = >
>
PV = nRTi
PV = nRTf
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1 Heat capacity at constant volume, VC
Since volume is constantP
VTi
Tf
== 0dVPW
VV
VT
U
T
UC
=
= we can also call this
the Energy capacity
this means at
constant volume
For an ideal gas
( )12
2V
V
U Nf kT
U NCT
=
= = fk
e.g. for a monoatomic gas 3f = 32
32
VC N
nR
k=
=
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2 Heat capacity at constant pressure, PC
At constant pressure,
volume is not constant and
work is not zero
=== VPdVPdVPW
P
VTi
Tf
P
P P
U W U P V C
T T
U VPT T
+ = =
= + 14243 14243
This term describes the part of
the heat added that causes the
internal energy to increase
This term describes the part of
the heat added that is used to
do work of expansion
P VC C>
For ideal gas (PV NkT = ), the second term is
P
VP Nk
T
=
and the CP is
2 2
2 2P V
f fC C Nk Nk nR
+ + = + = =
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For solids and liquids,V
T
is small, so that we can take CVand CP
to be approximately equal
V PC C
For a solid, 6f = (K.E. and P.E. inx, y, andz)
63 3
2VC N k Nk n= = = R (or 3 24.9 J mol Kc R = =
)
C n
This is called the Dulong and Petit rule. It is true for all solids,
provided the temperature is high (compared to something called
the Debye temperature every material has a different unique
Debye temperature). At low temperature, Quantum effects
become important!
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Latent Heat
As a phase transition (e.g. ice water, water stream), thesystem takes in or releases heat, but the temperature remains
constant.
During the transition, latent heat is used to break chemical bonds
rather than increasing the kinetic energy of atoms. So, at the phase
transition temperature, Tc, the heat capacity diverges
0
Q QC
T= = =
Diverging Cis commonly used to identify phase transitions
Latent heat is defined asQ
Lm
= ,
i.e. the heat needed to transform the system from one phase
to another phase
In order thatL is well defined, the convention is that
1. P is constant (usually 1 atm)2. no other work is done besides the mechanical work of
constant pressure expansion or compression
For example
Ice melting 80 cal gL = Water boiling 540 cal gL =
Compare to
Water (0C 100C) 100 cal gQ
m=
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Enthalpy
Define Enthalpy (H): work needed to createroom for the system
H U PV = +
internal energy
Enthalpy is useful for considering energy changes under constant
pressure conditions. Under normal conditions, P is usually
constant (e.g. 1 atm).
If a system is changed from one state to another, due to externalwork or heat, then the external energy input will serve to
1. increase the internal energy by dU, and2. do work against (atmospheric) pressure, if the volume
changes as well, by PdV
P
P
P P P
volume expands (dV)
P
P
P
PdVis associated
with this partPP
P
P
P
energy input
(heat or work)system
external pressure (P)
dH dU PdV = +
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Enthalpy continued
P
P
P P P
volume expands (dV)
P
P
P
PdVis associated
with this partPP
P
P
P
energy input
(heat or work)system
external pressure (P)
dH dU PdV = +
Recall the 1st Law
( ) other
dU Q W
Q PdV W
= +
= + +
where we have usedany other form of work
done on the system
(besides volumechange; e.g. electrical,
chemical)
otherW PdV W = +
total workwork done by
compression
Therefore
other
dH dU PdV Q W = + = +
This means that the enthalpy change is caused only by heat and
other forms of work, not compression-expansion work. If
, then dH 0otherW = Q=
Recall (from pg. 20 of these notes or eq. 1.45 in the text book)
( )PP PP
U PdV U PdV H CT T
+ = = T=
or
P
P
HC
T
=
= enthalpy capacity
which is similar to
V
V
UCT
= =
energy capacity
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Example: the enthalpy change associated with boiling water
It is customary for chemists to be concerned with of chemical
reactions because there is usually a substantial change of volume
and the reaction takes place in a laboratory at constant(atmospheric) pressure.
H
Boiling water at 100C, 1 atm
Heat Source
H
steam (water vapor)
at 1 atm1 mole ~ 18 g
40660 JH = for 1 mole (18 g)
Note 18 2260 J is the Latent Heat
for the transformation
QH Lm
= =
The enthalpy change for this reaction can be written
( )steam water steam steam water
H U P V
V V V V V V
= +
= ~ >>
Hence
steamP V PV
Treat steam as an ideal gas
( ) ( )8.31 J K 373 K 3100 JPV RT = = =
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This means that out of the 40660 JH Q = = ,
31008%
40660 of the heat energy is used to push the
atmosphere away to make room for the water
vapor.
40660 310092%
40660
of the heat energy is used to break the
chemical bonds between the H2O
molecules in liquid so that the escape
into a gaseous form.
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Another Example: Enthalpy of Formation
{2 2
liquidgas (1.5 mole total)
1H O H
2 H+
14243
2O
2 2
1H O (ga
2+ s) has more internal energy than (liquid) water, so
when hydrogen burns to form water, it releases 286 kJ/mole of
water produced.286 kJH =
Enthalpy
of Formation This means that enthalpy decreases,heat is released (exothermic)
Like the previous example, the volume of the liquid is negligible
compare to the volume of the gas.
Therefore
gas H U P V U PV = + = +
Formation of 1 mole of water requires 1 mole of H2 and 0.5 mole
of O2, which means a total of 1.5 mole of gas. The work done on
the gas when it collapses to form water is
( )( ) ( )-1 -1
1.5 8.31 J mol K 300 K4 kJ
gasPV nRT =
= ~
Therefore, of the 286 kJH Q = = produced, 4 kJ comes fromthe work done and the rest (282 kJ) comes from the change of
chemical bonds.