Lecture #6

Physics 7ACassandra Paul

Summer Session II 2008

Today…

• Finish: Intro to Particle Model of Energy• Particle Model of Bond Energy• Particle Model of Thermal Energy• Solid and liquid relations• Modes and equipartition

Last time:

PEpair-wise

System: Two Particles, one bondInitial: v=0, r=1.12σFinal: v~0 r=3σ

Wait! We don’t have an equation for PE pair-wise!

It’s ok, we have something better… a graph!

Work

ΔPE = Work

PEf – PEi = Work

0ε – (-1ε) = Work Work = 1ε

EnergyAdded

i

f

Is the energy added in our example the same as the ±ΔEbond needed for a phase

change?

A. Yes, and this is always the case

B. No never, this energy added is equal to the total energy of the system.

C. Yes, but only when the system consists of only two particles and one bond.

EnergyAdded

i

fi

f

EnergyRemoved

What is ΔEbond at the microscopic level?

A. The total amount of energy it takes to break (or form) one bond in a system.

B. The total amount of energy released when one bond in a system breaks (or forms).

C. The average amount of energy it takes to break (or form) all of the bonds in a system

D. The total amount of energy it takes to break (or form) all of the bonds in a system.

E. The total amount of energy released when all bonds in a system are broken (or formed).

So what happens when we have more particles?

We will get there, but first…

Particle Model of Bond Energy

A tool for exploring the energy associated with breaking bonds at

the microscopic level

Lets go back to our definition of ΔEbond

at the microscopic level:

In the Particle Model of Bond Energy:ΔEbond is equal to the total amount of energy it

takes to break (or form) all of the bonds in a system.

Is there such a thing as instantaneous Ebond?

Ebond + Ethermal = Etot ????

Yes! We need a definition for instantaneous Eb…

The particle is at rest at equilibrium, what is it’s total energy? What form(s) is

it in?

r0

Etot = -1ε =Ebond

= Ebond + Ethermal

Ebond is equal to the potential energy of the system when the particle is at rest at equilibrium.

KE + PE = Etot

0 + -1ε = Etot = -1ε + 0

Note: don’t think is proof Ebond = PE we will talk about this soon...

Now back to the question of Ebond when you have more than two

particles…

Let’s start a little easier than a 18 particle system.

What is the total bond energy of this system?

A. ~1εB. ~2εC. ~-1εD. ~-2εE. ~-3ε

r0r0

Are these two particles bonded?

We need to draw to scale to answer:

1σ 1σ 1σ

How far apart are the two on the outside?

~2σ (2.24σ to be exact)

1.21σ 1.21σ

1σ 1σ 1σ

2.24σ

The Ebond of the system is:

-1ε + -1ε +.03ε=2.03ε

So what is our definition of instantaneous Ebond?

• Ebond is the total amount of potential energy a system of particles possesses when the particles are at rest.

• Ebond = Σall pairs (PEpair-wise) EXACT DEFINITION

But what about when we have too many particles to count?

1σ 1σ 1σ

1.21σ 1.21σ

We don’t want to spend all day counting, so we need to develop an approximation

Closest Atomic Packing

The red particle has 6 nearest-neighbors in the same plane, three more on top and thenthree more on the bottom for a total of 12 nearest-neighbors. If you add any more to the system, they are no longer nearest-neighbors. (They are NEXT-nearest-neighbors.)

Here’s what it looks like when there areall packed together.

Developing an approximation:

How many nearest neighbors does every particle have?

Condtions for our approximation:1.We only want to consider nearest neighbor bonds

Ebond = Σn-n bonds (-ε)

2.We don’t want to have to count

Ebond = (tot # n-n bonds) (-ε)

12 bonds associated with every particle (for close packing, other packings have different #’s)But we know there are two particles associated with every bondSo we must divide by 2 in order to get the total number of NN bonds

Start with: Ebond = Σall pairs (PEpair-wise)

Ebond = n-n/2 (tot # of particles) (-ε) Ebond = 6*(tot # of particles)(-ε)

What if you were given this 2-D packing?

How many nearest neighbors does each atom have?

A. 9 nearest neighborsB. 8 nearest neighborsC. 2 nearest neighborsD. 4 nearest neighbors

What if you were given this 2-D packing?

How many nearest neighbors does each atom have?

D. 4 nearest neighbors

If we had 1 mole of this substance what would be the value of Ebond?

Ebond = n-n/2 (tot # of particles) (-ε)

A. -1.204x1024εB. -2.408x1024εC. 1.204x1024εD. 2.408x1024εE. -4.816x1024ε

Hint, how many particles are in a mole?

Ok ready to start another model?

We’ve talked about everything except Eth at the microscopic level… so guess what we’re going to cover next?

= Ebond + Ethermal KE + PE = Etot

Particle Model of Thermal Energy

A tool for exploring the energy associated with the movement and

potential movement of particles at the microscopic level

What is thermal energy at the particle level?

• Bond Energy is that which is associated with the PE of the particles when they are at rest.

What is thermal energy at the particle level?

• Bond Energy is that which is associated with the PE of the particles when they are at rest.

• Thermal Energy is that which is associated with the oscillations (or translational motion) of the particle.

So can we say that PE = Ebond and KE = Ethermal?

NO!!!

In DL You should have derived:

• For Solids and Liquids:PE = Ebond + ½ Ethermal

KE = ½ Ethermal

Why is Ethermal split between PE and KE?Think about a mass spring, in order to make the

spring oscillate faster through equilibrium, we must stretch the spring further from equilibrium, thus increasing the PE as well.

In DL You should have derived:

• For Solids and Liquids:PE = Ebond + ½ Ethermal

KE = ½ Ethermal

Do these equations hold for gases?Lets look at monatomic gases…

AtomAtom

Hmmm, no spring.

MONOTOMIC gases

What MUST be equal to zero?

A.PEB.KEC.Ebond

D.Ethermal

E.PE and Ebond

AtomAtom

Hmmm, no spring.

MONOTOMIC gases

What MUST be equal to zero?

D. PE and Ebond

AtomAtom

Hmmm, no spring.

= Ebond + Ethermal KE + PE = Etot

KE = Ethermal = EtotFor gases:

This brings us to MODES

Mode: A ‘way’ for a particle to store energy.

Gases have different ‘ways’ to have energy than liquids and gases!

Each mode contains (½ kb T) of energywhere kb is Boltzmann’s constant: kb = 1.38x10-23 J/K,and T = Temperature in Kelvin

But more on this value later……

3 KEtranslational modes

Modes of an atom in monoatomic gasModes of an atom in monoatomic gas

Every atom can move in three directions

0 PE modes

GasNo bonds, i.e. no springs

3 KEtranslational modes

Modes of an atom in solid/liquidModes of an atom in solid/liquid

Every atom can move in three directions

Plus 3 potential energy along

three directions

3 PE modes

So solids and Liquids have 6 modes total!

Cassandra don’t solids and each have liquids have 12 nearest neighbors and thus 12 springs, and so if each spring has a KE and PE mode, aren’t there 24 modes total!?

Cassandra don’t solids and each have liquids have 12 nearest neighbors and thus 12 springs, and so if each spring has a KE and PE mode, aren’t there 24 modes total!?

This is tricky! Yes they each have 12 BONDS but they can only move in 3 DIMENTIONS. (We live in 3-D not 12-D) So the while the particle can move diagonally, this is really only a combination of say to the right, up, and out therefore, the number of modes are DIFFERENT than the number of bonds.

In DL you will figure out how to count modes for diatomic gases too…

But there is one more part about the Particle model of bond energy that we have not talked about yet…

Equipartition of EnergyEquipartition of EnergyIn thermal equilibrium, EIn thermal equilibrium, Ethermal thermal is shared equally among all the is shared equally among all the “active” modes available to the particle. In other words, each “active” modes available to the particle. In other words, each “active” mode has the same amount of energy given by :“active” mode has the same amount of energy given by :

EEthermal per modethermal per mode = (1/2) k = (1/2) kBBTT

Liquids and Solids Gas

Let’s calculate the Thermal Energy of a mole of monatomic gas at 300K….

• EEthermal per modethermal per mode = (1/2) k = (1/2) kBBTT

• = ½ (1.38x10-23J/K)(300K) • = ½ (1.38x10-23J/K)(300K)(3)(6.02x1023)• = 3.74 kJ

(# of modes per particle) (# of particles)

What about the Total KE for a monatomic gas?

How does the KE compare to the Ethermal of a monatomic gas?

A. KE>EthermalB. KE<EthermalC. KE=EthermalD. Depends on the SubstanceE. Impossible to tell

Monatomic gases (only)

Etot = KE +PE =Ebond +Ethermal

Etot = KE =Ethermal

Same question as before!

Quiz Monday

I will send out an email saying what you should know, no later than Thursday afternoon.

Have a good weekend!

DL sections

• Swapno: 11:00AM Everson Section 1• Amandeep: 11:00AM Roesller Section 2• Yi: 1:40PM Everson Section 3• Chun-Yen: 1:40PM Roesller Section 4

Introduction to the Particle Model Introduction to the Particle Model Potential Energy between two atomsPotential Energy between two atoms

separation

Flattening: atoms have negligible forcesat large separation.

Repulsive: Atoms push apart as they get too close

r

PE

Distance between the atoms