Equilibrium Between Particles - Lecture Slides | PHYS 213, Study notes of Physics

Material Type: Notes; Professor: Thaler; Class: Univ Physics: Thermal Physics; Subject: Physics; University: University of Illinois - Urbana-Champaign; Term: Fall 2008;

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Physics 213: Lecture 11, Pg 1
213
213
Misc.
Misc.
Notes
Notes
The end is near – don’t get behind
All EXcuses must be taken to 231 Loomis
before noon, Tuesday, April 24.
The PHYS 213 (combined) final exam time
is 8-10 am Monday, Dec. 15. The conflict
exam is 8-10 am Wed., Dec. 17.
The deadline for changing your final exam
time is 10pm, Tuesday, April 22.
HW 5 is due Tuesday, at 8 am.
HW 6 is due next Saturday, at 8 am.
Course Survey = 2 bonus points
(accessible at top of HW6)
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Physics 213: Lecture 11, Pg 1

Misc.

Misc.

Notes

Notes

The end is near – don’t get behind

All EXcuses must be taken to 231 Loomisbefore noon, Tuesday, April 24.

The PHYS 213 (combined) final exam timeis 8-10 am Monday, Dec. 15. The conflictexam is 8-10 am Wed., Dec. 17.

The deadline for changing your final examtime is 10pm, Tuesday, April 22.

HW 5 is due Tuesday, at 8 am.

HW 6 is due next Saturday

, at 8 am.

Course Survey = 2 bonus points(accessible at top of HW6)

Physics 213: Lecture 11, Pg 2

Lecture 11 Lecture 11

Equilibrium Between Particles Equilibrium Between Particles

Agenda for today Agenda for today

Free Energy and Chemical Potential

Simple defects in solids

Ideal gases, revisited

Reference for this Lecture:

Elements

Ch 11

Reference for Lecture 12:

Elements

Ch 12

Physics 213: Lecture 11, Pg 4

The path ahead... The path ahead...



We have thus far studied systems in which volumes can beexchanged

this led to mechanical equilibrium and equal

pressures in the two volumes.



We then studied systems in which energy can be exchanged

this

led to thermal equilibrium and equal temperatures of the systems.



Now we consider systems in which

particles

can be exchanged, e.g.,

Particles can move from place to place

Particles can combine (chemistry, vacancy-interstitialrecombination, electron-hole recombination, nuclear reactions…)into new types.



This will lead to “chemical equilibrium”, in which the “chemicalpotential” is equalized, the free energy of the system is minimized,and the

total

entropy is maximized (which after all is the most

probably state of affairs...)



And lots and lots of applications...

We start with a concrete example...

Physics 213: Lecture 11, Pg 5

Simple defects in solids (1)

Simple defects in solids (1)



Atoms can pop out of position and sit in interstitial sites.

These defects (and others) play a crucial role in thethermal weakening of materials, etc.



In thermal equilibrium, how many are there?

(determines how hot is safe for some structure to get)



We calculate the number N

I

of interstitials to minimize F

Let’s say that an atom can either go to the surface, making a‘normal’ site, or sit at an interstitial site, with extra energy

I

For big crystals, the surface energy and entropy changes arenegligible. so the changes to F from making an interstitial comefrom its own extra energy and entropy.

Interstitial

Physics 213: Lecture 11, Pg 7

ACT 1

ACT 1

Solution

Solution

As we let the temperature of the solid

0, what

fraction of the atoms will sit at the interstitial sites?

a) noneb) halfc) all of them

Because it costs energy to sit at the interstitialsites (if energy were lowered, then the crystalstructure itself would rearrange itself tominimize the energy), at some sufficiently lowtemperature the entropy gain by sitting at thesite (more places to sit) will dominated by theenergy cost

the free energy will be

minimized by “staying at home”.We might expect a Boltzmann-like dependenceon temperature...

Physics 213: Lecture 11, Pg 8

Defects in Solids (2)Defects in Solids (2)

Say we have

N atoms, N possible interstitial sites



We want to know

N

I

, the number of interstitials at temperature T.



method: minimize F(N

I

), Helmholtz free energy for N

i

interstitials

» Call F(0)=0 for convenience» Assume the vibrational entropy is not much changed by

making an interstitial but U is increased by an amount

Ι

!

(

)

(

)

(

)

ln

!(

)!

I

I

I

I

I

I

I

N

F N

U N

TS N

N

kT

N

N

N

=

=

∆ −

# ways to put N

I

identical

particles in N bins.

(single occupancy)

Need energy

ΙΙΙΙ

for each defect

Physics 213: Lecture 11, Pg 10

Defects (3)

Defects (3)

To find N

I

we

minimize F

set the ‘derivative’ to zero:

0

(ln(

)

ln(

))

ln(

)

I

I

I

I

I

I

I

N

dF

kT

N

N

N

kT

dN

N

N

=

= ∆ +

= ∆

So:

ln(

)

I

I

I

I

kT

I

I

N

N

e

N

N

kT

N

N

= −

=

Usually, N

I

<< N, so:

I

I

kT

N

e

N

=

This looks like a simple Boltzmann expression.

Exponential increaseas T increases

Physics 213: Lecture 11, Pg 11

Defects (4)

Defects (4)

Atoms can also go out to the surface and leave vacancies.

Following exactly the same math:

V

V

kT

V

N

e

N

N

=

Usually, N

V

<< N, so:

V

V

kT

N

e

N

=

vacancy

Again, surface energy/entropychanges in a BIG crystal are negligible,unlike in this little picture.

Physics 213: Lecture 11, Pg 13

ACT 2

ACT 2

Solution

Solution

We just saw that fraction of vacancies is given byAssume the binding energy of the atom to it’s ‘home site’

is 0.5 eV. If we want to keep the fraction of vacanciesless than 1%, what is the maximum temperature weshould heat the material to?

a) 100

C

b) 1000

C

c) 10,

C

V

V

kT

N

e

N

=

5

ln(0.01)

0.5eV

4.6(8.

10

eV/K

1264 K

1

000 C

V

V

kT

V

e

kT

kT

T

=

= −

=

=

=

×

=



Physics 213: Lecture 11, Pg 14

Particle equilibrium: chemical potential Particle equilibrium: chemical potential

μ μ

μ μ μμμμ



These solid defects provide examples in which some

particle

number

can vary. We want a convenient way to treat such

problems—we invent a new property,

chemical potential (

, which

simplifies the business of maximizing total S via minimizing F insuch problems (just as “T” simplified the description of how todistribute energy to maximize total S). Using

μμμμ

lets us treat

problems where we can’t quite list all the states.



Roughly speaking,

μμμμ

is the amount the (free) energy of a system

changes if we add one particle.

Volume exchange

p

1

= p

2

mechanical equilibrium

Energy exchange

T

1

= T

2

thermal equilibrium

Particle exchange

μ μ

μ μ

1

=

μμμμ

2

chemical equilibrium

Physics 213: Lecture 11, Pg 16

Defects in solids and

Defects in solids and



Now let’s look at a solid which doesn’t have time for defects todiffuse to or from the surface (a common case)



Now making an interstitial leaves behind a vacancy, i.e., dN

V

= dN

I

If we

start

with a perfect crystal:

N

V

= N

I

Note this is different than our first example—exchanging particlebetween two volumes— there dN

1

dN

2

Therefore

μ

I

μ

V

= 0

Interstitialvacancy

V

V

I

I

I

I

I

I

V

dF

dF

dF

dF

dF

dN

dN

dN

dN

dN

Physics 213: Lecture 11, Pg 17

Defects in solids and

Defects in solids and



In equilibrium:

I

V

, so:

(

)

2

2

2

ln(

)

ln(

)

0

2

ln(

)

0 since here

(

)

if

then

I

V

T

T

V

I

I

V

I

V

V

I

V

V

I

V

V

kT

kT

T

I

V

V

kT

V

V

I

N

N

kT

kT

N

N

N

N

N

kT

N

N

N

N

N

e

e

N

N

N

N

N

N

N e

∆ +∆

∆ +

=

∆ + ∆

=

=

=

=

≡ ∆ + ∆

=

=

This looks almost like Boltzmann, except for

kT,

since

both

defects make entropy.

Physics 213: Lecture 11, Pg 19

ACT 3

ACT 3

We just saw that the product of interstitials andvacancy occupancies is given byAssume we start with

N

I

N = N

V

N

. If we double

N

I

, what will be the new

N

V

N

a) 10

b) 10

c) 10

d) 2 x 10

e) 10

2

T

I

V

kT

N N

e

N

=

Physics 213: Lecture 11, Pg 20

ACT 3

ACT 3

Solution

Solution

We just saw that the product of interstitials andvacancy occupancies is given byAssume we start with

N

I

N = N

V

N

. If we double

N

I

, what will be the new

N

V

N

a) 10

b) 10

c) 10

d) 2 x 10

e) 10

2

T

I

V

kT

N N

e

N

=

When we add more interstitial atoms,

some of these ‘fill in’ the openvacancies. Since the product isconstant, if one doubles the othermust halve.