Phys321 Fall07 Notes 01

8/6/2019 Phys321 Fall07 Notes 01

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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.

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Phys321 Fall07 Notes 01