Mon., 12/5 HW13: Tues. 12/6 T9 RE-T9; Lab Wed., 12/7 …bulldog2.redlands.edu/fac/eric_hill/Phys233/Lecture/Day 23 T8.pdf · 0 L x x 2L x 2 x x L 3 2 x L p x L x h 2 L x h 2 2 L x

Phys 233 Day 23, T8: Changes in Entropy 1

Thurs. 12/1 T8 Entr Change B3, S3 RE-T8; Lab: 2nd Draft L9

Mon., 12/5

Tues. 12/6

Wed., 12/7

Thurs. 12/8

T9 Heat Engines S2, S7

L12: Heat Engine (T9)

Review

HW13: T7: S.2, S.3, S.8; T8: S.5, S.8, S.9

RE-T9; Lab Notebook & Full Report 11 (no revisions)

PL 12; Quiz 10: T7, T8

Lab: 2nd Draft L10

Equipment

o Ppt.

Announcements

Time to look into REU’s

Calculating Entropy Changes

T8.1 The Entropy of a Monatomic Gas

T8.2 Entropy Depends on Volume

T8.3 A General Expression for Entropy

T8.4 Constant-Temperature Processes

T8.5 Handling Changing Temperatures

T8.6 Non-quasistatic Processes

T8.1 The Entropy of a Monatomic Gas

In the interest of avoiding higher math and brevity, the book takes a rather handwaving

approach to this. In fact, in class last time we actually took the first step of the more rigorous

approach – when we were working on the Boltzmann speed distribution, so I’ll sketch out that

approach. Hopefully this will provide some reinforcement of last time and some deeper

understanding. The only hitch is that, at one point, we’ll just have to trust the mathematicians –

for that, hopefully it looks plausible, but don’t feel like you have to completely ‘get’ it.

Multiplicity of a Monatomic Ideal Gas

We’ll start simple and consider a Monatomic Ideal Gas of one particle (I know, not very

gassy) and later we’ll think about how to extend our approach to handle an N-particle

gas.

The multiplicity is simply the number of microstates that are consistent with the

macrostate of having total energy U?

o Note: in a monatomic ideal gas, Kinetic energy is the only kind there is.

Well then, what defines a ‘microstate’ for a single gas atom?

Wave function

o We’ve essentially got a particle in a box. When you’re not looking, you

really can’t say where or if the particle is localized. As you remember from

Unit Q, you describe the objects with sine waves – everywhere in the box. So

we’re just left with the freedom of momenta. In this basis set, we just count

up the number of possible momenta.

o momentawave.#

1 how many different momenta are compatible with the given geometry of the box

and the energy U.

o First, deal with the constraint that the particle’s in the box.


o Remember how momentum and box size are related for a particle in a box.

Consider just one component of momentum

x

x

hp

Say the box is Lx long, then what wavelengths fit in the box?

So, the momenta compatible with the length are x

xxL

hnp

2

nx = 1,2,3,4,…

We could represent that graphically as

Now, if that were it, if the “box” were essentially one dimensional, since

2

22

2

82 x

xn

nmL

hn

m

pU

we’d say there was only one microstate compatible with a given energy. That would be

the end of the story: 11,1 D

How about in 2-D?

Other 2 dimensions

Similarly, the other two components of momentum that fit in the

box are

y

yyL

hnp

2 ny = 1,2,3,4,…

yL

h

2

yL

h

22

yL

h

23

We could plot out the possible states like this:

Lx 0

xx L2

2

2 xx

L

3

2 xx

L

px

xL

h

2 xL

h

22

xL

h

23

py


Each point where the lines intersect represents a possible momentum state, for

example,

yx L

h

L

hp

23,

22

Now, the multiplicity is, again, the number of micro states (momentum states) that are

compatible with having energy U where, again, m

pU

2

2

or mUp 2 where 22

yx ppp

So, essentially, we’re asking, how many states have the same magnitude of momentum. On our

plot, all points with the same magnitude of momentum lie on an arc of a circle.

px

xL

h

2 xL

h

22

xL

h

23

py

yL

h

2

yL

h

22

yL

h

23

px

xL

h

2 xL

h

22

xL

h

23

yL

h

2

yL

h

22

yL

h

23

mUp 2


So, that’s a question of how many grid points lie on the arc; in truth, very few lie exactly on the

arc; however, if we fudge it a little and ask, how many lie within some range of the arc, then

we’re really asking how many grid point are within the area of that shell.

That number is easily found by saying that, for every rectangle of ‘area’ A

h

L

h

L

hpp

yx

yx422

2

There’s one grid point, or there’s a ‘density of grid points’ of 2

41

h

App

pp

stateD yx

yx

p

Meanwhile, the ‘area’ in this momentum space that corresponds to having energy U is the area of

our quarter ring: pmUppA Uwithp 22

241

..

So, the multiplicity, i.e., number of states with the right amount of energy, is simply

pmUh

AAD UwithppD 2

2

42..2.1

3-D.

We did this for 1 particle in 2-D partly because it’s a lot easier to visualize and partly so we can

see the general trend as we now expand to 1 particle in 3-D and ultimately N particles in 3-D.

To go to 3-D we just need to add one more axis, and have a volume instead of an area in

‘momentum space.’

px

xL

h

2 xL

h

22

xL

h

23

yL

h

2

yL

h

22

yL

h

23

mUp 2

p


z

zzL

hnp

2 nz = 1,2,3,4,…

zL

h

2

zL

h

22

zL

h

23

Now, a particle can have any combination of allowed momentum

components and fit in the 3-D box. z

z

y

y

x

xL

hn

L

hn

L

hnp

2,

2,

2

for any nx, ny, nz.

This 3-D span of possibilities can be represented on a 3-D graph,

just like the 3-D possibilities of position in space. We call such a

plot a momentum space plot.

Every point the grid lines intersect is a momentum state that fits in

the volume.

Putting it another way, which will soon be useful, there is one state per

volume V

hV p

8

3

, or the density of states in momentum space is

3

81

h

V

VD

p

p

o Now deal with Energy constraint.

So, the particle’s state must fit in the box; it must also have energy U. For

our monatomic particle, we really mean it must have kinetic energy U.

That places another constraint on the momenta it can have, therefore what

states it can be in.

mUpppp zyx 22222

Back in ‘momentum space, this constraint prescribes a sphere of “radius”

mUp 2 (or rather, the 1/8th

of the sphere in the positive octant since

we’re talking standing waves).

pz

px

py

pz

pz=zL

h

2

py=yL

h

2

px=xL

h

2

px

py

px py

pz Vp=

V

h

L

h

L

h

L

h

zyx 8222

3

pz


In other words, there is one state per volume on the plot Vp.

Put two constraints together to find number of possible states – multiplicity

o So, how many states both satisfy the energy constraint and the volume

constraint, i.e., have energy U and momenta that fit in the volume, i.e., lie on

the shell and are grid points?

Of course, frightfully few of these points, if any will lie exactly on the shell, but if

we allow for an infinitesimal degree of uncertainty in the momenta (box walls

aren’t perfectly rigid, energy isn’t perfectly constant, etc.), then the shell has

thickness dp and volume mUdpdpmUdppVp 244812

81 . The 1/8

px

py

pz mUp 2

px

py

pz mUp 2


comes from only the positive octant of momentum space corresponding to unique

standing waves.

So the number of states compatible with both constraints is essentially the volume

of the shell (that has the appropriate energy) times the density of states (whose

momenta fit).

mUdph

V

h

VmUdp

V

stateVDV

p

ppp 331 881

Q - How does the multiplicity of 1 atom of helium with energy U in a container compare to that

of 1 atom of argon in an identical container with the same energy?

A. The helium atom has a greater entropy.

B. The argon atom has a greater entropy.

C. Both atoms have the same entropy.

D. There is not enough information to answer definitively.

The more massive Argon has the higher entropy.

Why? If we trace back through the derivation, more mass but same energy means greater

magnitude of momentum (p=sqrt (2mU)). Looking at the picture on the top of the page, that

larger momentum would define a larger spherical shell which thus intersects more grid points /

allowed momentum states. That’s a bit mathlandishly abstract perhaps, so think of this: each

allowed state that’s on a given shell has the same magnitude but a different direction for its

momentum. So, saying that a larger shell intersects more grid points is saying that there are

more allowed directions for the momentum.

What about N molecules?

px

py

pz mUp 2

dppVp

2

81 4


o Let’s do 2: Same basic program:

BpAp

BpApApAp

BpBpApApBA

VVh

V

countsover

VVDDcountsovercountsover

DVDV

countsover

..

2

3

....

....

2

8

.

1

.

1

..

Over counts: Interchange particles and you get the same picture, i.e., the

labels of “particle A” and “particle B” aren’t actually pasted on the real

particles. So counted twice as many possibilities as there really are. = 2

Energy constraint: U1 + U2 = U

mUppppppp zyxzyx 222

2.

2

2.

2

2.

2

1.

2

1.

2

1.

So it’s best to call "" &... BApBpAp VVV which is the “volume” of a

6-D shell of constant p and thickness dp; i.e., it’s the surface area

of that 6-D shell times thickness dp.

In general, the hyper-area of a shell of constant radius p in d

dimensions is 1

2

2/

!1

2 d

d

d

p

o While I for one can’t easily visualize such a thing and must

take the mathemeticians’ word for it, I can see that it works

for the two cases I am familiar with: if d =2 we get the old

2 p – circumference of a circle and if d = 3 we get 4 p2,

though that involves knowing how to handle (½!) .

Since we’re dealing with standing waves so all components of p

must be positive, we require only the positive segment,

1

2

2/

!1

2

2

1 d

d

dd

p

So, the “volume” is dppVd

d

dd

BAp

1

2

2/

21

&.!1

2 here d = 6 (6

components, 6 ‘dimensions’)

o Generalize to N particles, each with 3 components of momentum

Overcounts = N! (same as when we were choosing which coin to call

‘first’ ‘second’, … in the two-state coin flip system)

23

13

3

12

3

22 2

3

/)(

!!

),( NN

NN

N mUVNfpN

mU

h

V

NVU

In the last step, all the constants and terms that only depend upon

N are grouped to define the factor f(N). The book’s expression is a

good approximation to this.

I know, that looks pretty ugly, but the idea is simple: given the discrete

momenta states that an individual particle can occupy, that we’ve got N

particles and only U energy to around, how many different ways can we

distribute the particles through the states?

As for Entropy then,

23 /)(ln),,(ln),,( NN

N mUVNfkNVUkNVUS


T8T.2 - How does the entropy of 1 mol of helium gas in a container at room temperature

compare to 1 mol of argon gas in an identical container at the same temperature?

A. The helium gas has a greater entropy.

B. The argon gas has a greater entropy.

C. Both gases have the same entropy.

D. There is not enough information to answer definitively.

This is essentially the same question as before. If U = 3/2 NkT, then same temperature

means same internal energy. With that in mind, changing the particle mass just changes m, and

you can see that the gas of more massive particles has the greater entropy.

T8.2 Entropy Depends on Volume

The text particularly notes that entropy depends upon the volume and why that is. In this

derivation, the volume dependence enters because increasing the volume / the lengths of the

sides increases the density of states / shinks our grid in momentum space – for the same total

energy more momentum states exist. The book says that the reason for the volume dependence

is not because there are more locations that the atoms can occupy. That is true… for a basis set

of energy eigen states. In point of fact, one can choose to cast the problem in terms of position

eigen states rather than energy eigen states, and then it would be quite true that increasing

volume would increase the number of such position states.

T8.3 A General Expression for Entropy Changes

The book’s argument here is fairly qualitative; that’s to avoid some multi-variable calculus.

You’re all taking a multivariable calculus class, so I’ll go the other rout.

You should have met in your math class the idea that, if you have a function that depends upon

multiple variables, then there are multiple ways to effect a small change in the function’s value:

Mathematical generality: for function F(x,y,z,…),

...,...),,(,,,

dzz

Fdy

y

Fdx

x

FzyxdF

zyzxzy

Look at our expression for Entropy as a mathematical function. While we derived it for the Ideal

Gas, the notion that it might generally depend upon such parameters as U, V, and N doesn’t seem

farfetched; indeed, it may well depend upon some other parameters too, just as a stand, say

parameter X. Thus you can effect a change in the entropy by changing any one of these

parameters:

dXX

SdN

N

SdV

V

SdU

U

SXNVUdS

NVUXVUXNUXNV ,,,,,,,,

...),...,,(

Now, as you already know,

U

S

T

1

So the first term can be rephrased giving


dXX

SdN

N

SdV

V

SdU

TXNVUdS ...),...,,(

1

What about the other terms?

The book gives a hint by observing that, for a monatomic ideal gas undergoing an adiabatic

process

ConstTV 32/ so ConstVT 23/

But also

TU

So

ConstVU 23/

Which means that N

VUNfkNVUS 23/ln),,( remains unchanged under such a process.

So, then maybe we should expect S to depend on just Q rather than all of dU.

Here’s a way of getting more rigorous:

Now, there’s actually a mathematical identity that goes

xyz

y

z

z

x

y

x

it can be seen graphically. Imagine multiplying both sides of the

equation by y.

zz

xx

yy

z

z

xy

y

x

y

xyz

From the left-hand side, you get a corresponding change in x, x, and

these two together move you along vector L from the origin to the

point indicated. Meanwhile, on the right, you’d first move along

vector R1, in the y-z plane, and the corresponding z times the dx/dz

derivative would move you along vector R2, arriving at the same point.

So, we rephrase

T

PP

TV

U

U

S

V

S 1

Then,

dXX

SdN

N

S

T

PdV

T

dUXNVUdS ...),...,,(

y

z

x

L R1

R2


Of course, PdV = -w, and q = dU – w. Note: I’m using lower case w and q so the notation

reminds us we’re talking differentially little amounts of work and heating, just as dU is a

differentially small change in energy while U is a big change.

dXX

SdN

N

S

T

qXNVUdS

dXX

SdN

N

S

T

wdUXNVUdS

...),...,,(

...),...,,(

I’m leaving the other two terms so you can easily see the limitations to the book’s T8.17 - they

vanish only if N is not changing and nothing else that effects entropy is changing.

Notationally Challenged: Now, there’s something a little awkward about the notation – in this

derivation we’ve been considering a differentially small change in entropy, dS, so the Q that’s

featured in this equation is just for a differentially small step (over which the temperature can be

treated as constant.) In point of fact, the equation can stand as it is, but it would be nice to build

into its notation something that explicitly reminds us that it’s valid only on the differential scale.

There are different ways of trying to make that fact explicit in the notation – rather than using Q

you could use q, or some people (such as Moore) write dQ, but that would make a mathematician

cringe since “dQ” should mean “differential change in heat”, but we just mean “differentially

small heating”, so a popular compromise is to write dQ.

T8.4 Constant-Temperature Processes

T

qXNVUdS ),...,,(

Even when N and other parameters are being held constant, T generally is not over a finite

process, so this relation generally needs to be integrated and T allowed to vary appropriately.

However, there are three common cases in which T does remain constant.

1) A reservoir is involved

a. You’re considering the entropy change of a reservoir as energy flows into it, but

not enough to significantly change T

b. You’re considering the entropy change of a system that’s in contact with a

reservoir, so its temperature is getting held constant.

2) A phase transition. In essence, this is making or breaking bonds and so putting energy

and freedoms into play that didn’t used to be thermally accessible or taking them out of

play.

T8S.6 – Imagine that 22 g of helium gas in a cylinder expands quasistatically while in contact

with a reservoir at a temperature of 25°C and does 85 J of work on its surroundings in the

process. What is the entropy change of the gas? Of the reservoir?

T

QS , where T= 25°C+273K=298K; WUQ but 0TU , so

JJWQ 855


K

J

T

QS

298

85

Meanwhile, if the reservoir is having the same amount of work done it and we say that its

temperature is essentially the same constant, then it must have the opposite heating and so equal

and opposite change in entropy.

T8S.7 – Imagine that you allow 0.40 mol of nitrogen to expand from a volume of 0.005 m3 to a

volume of

0.015 m3 while the gas is held at a constant temperature of 304 K. By how much does the

entropy of the gas change?

T

QS , where T= 304K; WUQ but 0TU , so

i

f

V

VNkTWQ ln for an

isothermal expansion. So,

5

15ln/1038.110022.640.0ln

ln2323 KJ

mole

moleculesmole

V

VNk

T

V

VNkT

Si

fi

f

T8B.8 - A puddle containing 0.80 kg of water at 0°C freezes on a cold night, becoming ice at

0°C. What is the entropy change of the water that is now ice? (The latent heat of freezing water

is 333 kJ/kg.)

KC

kgkJkg

T

mL

T

QS

2730

/33380.0

T8T.4 - In a tub 100 kg of water at 30°C absorbs 100 J from an electric heater. We can consider

the water to be a reservoir in this process (True or False).

(For water, c = 4186 J/K/kg.)

So, does the water’s temperature change much?

KKkgJkg

J

mc

QTTmcUQ

4186

1

/4186100

100 that’s not much compared with

its initial 30°C, so “TRUE”, it can be considered a reservoir.

T8.5 Handling Changing Temperatures (but ~constant specific heat)

Let’s say the only way we’re changing a system’s energy is through heating, under that

condition,

mcdTdUq

Then the temperature is most certainly changing, but in a smooth way


if

final

initial

final

initial

TTmcT

mcdT

T

qS

T

mcdT

T

qXNVUdS

/ln

),...,,(

T8T.5 - 100 g of water is slowly heated from 5 to 25°C. What is its change in entropy?

(For water, c = 4186 J/K/kg.)

A. 582 J/K

B. 29.2 J/K

C. 139 J/K

D. 6.95 J/K

E. 0

F. Other (specify)

T8.6 Non-quasistatic Processes

If I asked you which resulted in the greatest change in gravitational potential – my taking the

vernercular up the side of Mt. San Jancinto or my hiking up the mountain, you’d say ‘they’re the

same’ because the change in potential energy doesn’t depend on how you went about changing it,

it doesn’t depend on the path you took, just the two end points.

So, if you had to do the math to compute the change in gravitational potential energy, you’d

integrate of the path that was the simplest, regardless of whether it was the path actually taken or

not.

Similarly, the change in entropy for a system doesn’t depend on how you got it between one

volume and another, one energy and another, one number of particles and another, it just depends

on what the initial and final volume, energy, and particle count were.

Again, if you had to do the math to compute the change in entropy, you’d integrate over the

process that was mathematically simplest, regardless of whether it was the process actually

employed or not.

T8T.9 - A sample of helium gas with an initial temperature Ti in an insulated cylinder with initial

volume Vi is suddenly and violently compressed to one-half of its original volume. The absolute

temperature of the gas after the compression is observed to have doubled. (Note: In a

quasistatic adiabatic compression of a monatomic gas, TV⅔ =constant.) A suitable replacement

for this process would be

A. Slow isothermal compression from Vi to ½ Vi

B. Slow adiabatic compression from Vi to ½ Vi

C. Heating the gas to double its temperature followed by slow adiabatic compression

from Vi to ½ Vi

D. Slow adiabatic compression from Vi to ½ Vi followed by heating to bring the

temperature to 2Ti

E. Slow adiabatic compression from Vi to ½ Vi followed by cooling to bring the

temperature to 2Ti

F. None of the above.


Both D and E seem plausible, but the question is whether we’ll need to cool down after the

compression or warm up. To determine this, we can invoke that 3/2

3/23/2

f

iifffii

V

VTTVTVT so if we cut the volume in half, we will more than double the

temperature: 223/2

iif TTT , and thus we’ll need to cool down, answer E.

T8T.8 - A stone of mass m is dropped from a height h into a bucket of water. What would be a

suitable replacement process for this non-quasistatic process?

A. Lower the stone gently into the water.

B. Lower the stone gently in the water, and then heat the water until it has mgh more

energy that it had before.

C. Lower the stone gently in the water, and then stir the water gently to give it mgh

more energy.

D. Raise the water up to height h, gently put the stone in it, and then lower both back

to the ground.

E. Either (B) or (C)

F. None of the above.

Answer B. we can lower the rock into the water and then heat the water to the tune of Q=mgh,

so we’ll have T

mghS if we assume that T hardly changes. If we allow that T will change

during the process, then we can use that the water’s specific heat is constant so TcmQ w such

that w

wcm

mghTmghTcmQ and so

iw

w

i

w

i

iw

i

f

w

T

T

w

Tcm

mghcm

T

Tcm

T

TTcm

T

Tcm

T

dTmcS

f

i

1ln1lnlnln

(note form math land: 1ln if 1 . So, if 1iwTcm

mgh, that approximation will hold

and we’ll be back at the constant T approximation: T

mghS )

Documents

Mon., 12/5 HW13: Tues. 12/6 T9 RE-T9; Lab Wed., 12/7 …bulldog2.redlands.edu/fac/eric_hill/Phys233/Lecture/Day 23 T8.pdf · 0 L x x 2L x 2 x x L 3 2 x L p x L x h 2 L x h 2 2 L x