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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.
Phys 233 Day 23, T8: Changes in Entropy 2
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
Phys 233 Day 23, T8: Changes in Entropy 3
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
Phys 233 Day 23, T8: Changes in Entropy 4
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
Phys 233 Day 23, T8: Changes in Entropy 5
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
Phys 233 Day 23, T8: Changes in Entropy 6
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
Phys 233 Day 23, T8: Changes in Entropy 7
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
Phys 233 Day 23, T8: Changes in Entropy 8
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
Phys 233 Day 23, T8: Changes in Entropy 9
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
Phys 233 Day 23, T8: Changes in Entropy 10
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
Phys 233 Day 23, T8: Changes in Entropy 11
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
Phys 233 Day 23, T8: Changes in Entropy 12
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
Phys 233 Day 23, T8: Changes in Entropy 13
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.
Phys 233 Day 23, T8: Changes in Entropy 14
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 )
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