chapter 18 the micro/macro connection - brian...

CHAPTER 18 – THE MICRO / MACRO CONNECTIONPHYS 212 S’14

CHAPTER 18THE MICRO/MACRO

CONNECTIONThis brief chapter revisits some of the concepts from

chapter 17, but from a different viewpoint. Here, we will look at the microscopic (i.e. molecular-level) properties of various materials, and consider how these are manifested

in macroscopic properties such as thermal energy and specific heat.

Slide 1


18.1 Molecular Speeds and Collisions

Slide 2

For the past couple of chapters, we have qualitatively considered gases as consisting of a large number of molecules that are moving –and occasionally colliding with each other and with the container walls. Now, let’s take a look at the properties of this motion – the speed of the molecules and the frequency of collisions – and see how these values are related to the state variables.If we could instantaneously measure and plot the speeds of every molecule in a sample of gas, it would look like the figure on the right.

STATISTICS ALERT

This is known as a Maxwell-Boltzmann distribution.



Slide 3

What the figure tells us is that the molecules have a distribution of speeds. At any given moment, a few are nearly stationary, and a few are moving at a couple of km/s. Somewhere in between, there is a most likely speed, as well as an average speed (these aren’t the same). The figure also tells us that as the gas temperature increases, the distribution widens and shifts toward faster speeds.



Slide 4

Mean Free PathAs we have seen, the trajectory of any one molecule through space is a complicated zigzag pattern, due to its frequent collisions. It will soon prove useful to have a definition of the average distance between collisions, called the “mean free path” (λ). We’ll skip the derivation (p. 504) and present the result:

λ = 1

4 2𝜋𝑁

𝑉𝑟2

Here, 𝑁 / 𝑉 is the number density of the molecules, while 𝑟 is the radius. This equation makes qualitative sense – collisions should occur more frequently with a high-density gas, and with one that contains fatter molecules.



Slide 5

Mean Free Path cont’If you’re not explicitly told the atomic or molecular radius, just use this approximation: 𝑟 = 0.5 × 10−10 m (monatomic) or 𝑟 = 1.0 ×10−10 m (diatomic) .

Pressure, Number Density, and TemperatureBuried in Example 18.1 is a useful relation. Starting from the ideal gas law in molar form: 𝑝𝑉 = 𝑁𝑘𝐵𝑇, we can write

𝑁

𝑉=

𝑝

𝑘𝐵𝑇(remember, 𝑘𝐵is the Boltzmann constant). Therefore, it’s possible to calculate mean free path using only pressure and temperature information.


Problem #1: Lottery Machine

Slide 6

A lottery machine uses blowing air to keep 2000 ping-pong balls bouncing around inside a 1.0 m X 1.0 m X 1.0 m box. The diameter of a ping-pong ball is 3.0 cm. What is the mean free path (in cm) between collisions? Is there any reason to suspect that our formula for λmight be slightly incorrect here?

Solution: in class

RDK EX 18.7


18.2 Pressure in a Gas

Slide 7

We’ve known for a few weeks now that a gas “has pressure” due to the large number of tiny collisions between its molecules and the container walls. It’s time to throw some math at this situation. Since we’re discussing collisions, we will be referring back to PHYS 211 quite often.The figure shows a single collision with a wall oriented normal to the 𝑥-axis. As a result of the collision, the 𝑥-component of the molecule’s velocity changes from +𝑣𝑥 to −𝑣𝑥 (the other two components of velocity don’t change). Using the impulse-momentum theorem,

𝐽𝑥,𝑚𝑤 = ∆𝑝𝑥 = 𝑚∆𝑣𝑥 = −2𝑚𝑣𝑥This represents the impulse on the molecule due to the wall. However, we’re interested in the impulse on the wall due to the molecule, which is equal to (see Newton’s 3rd law):

𝐽𝑥,𝑤𝑚 = −𝐽𝑥,𝑚𝑤 = +2𝑚𝑣𝑥



Slide 8

Next, we will assume that there are 𝑁coll collisions during a short time ∆𝑡. We’ll start by assuming that all of the colliding molecules have the same 𝑥-velocity (although we know this isn’t true, from just a few slides ago…bear with me here). The total impulse on the wall during ∆𝑡 will then be equal to

𝐽wall = 2𝑁coll𝑚𝑣𝑥Now, you will may recall from PHYS 211 that this impulse is related to the average force of the collisions as

𝐽wall = 𝐹avg∆𝑡

Thus, the average force on the wall is

𝐹avg =2𝑁coll𝑚𝑣𝑥

∆𝑡But, 𝑁coll/∆𝑡 defines the net rate of collisions with the wall – as

derived on p. 505 of the text, this is equal to 1

2𝑁𝑉𝐴𝑣𝑥.



Slide 9

All together, the average force of the collisions is

𝐹avg = 21

2

𝑁

𝑉𝐴𝑣𝑥 𝑚𝑣𝑥 =

𝑁

𝑉𝑚𝑣𝑥

2𝐴

This equation is almost accurate. We’ll fix it in a moment.

The Root-Mean-Squared SpeedEarlier, we assumed that all molecules had the same 𝑣𝑥. That’s obviously not true…at any instant in time, half of the molecules (on average) have a positive 𝑣𝑥 while the rest have a negative 𝑣𝑥.Therefore, the average 𝑣𝑥 is zero!Remember, though, that the speed of a molecule (or anything at all, moving in 3 dimensions) is

𝑣 = 𝑣𝑥2 + 𝑣𝑦

2 + 𝑣𝑧2



Slide 10

We can thus define the average of the squared speed as

𝑣avg2 = 𝑣𝑥

2 + 𝑣𝑦2 + 𝑣𝑧

2avg

= 𝑣𝑥2

avg + 𝑣𝑦2

avg+ 𝑣𝑧

2avg

The square root of this is referred to as the root-mean-square (or rms) speed:

𝑣rms = 𝑣2 avg

Now, we must realize that there’s nothing particularly special about the 𝑥-axis. On average, we must have the same speed-squared in any direction:

𝑣𝑥2

avg = 𝑣𝑦2

avg= 𝑣𝑧

2avg

Therefore, 𝑣𝑥2

avg =𝑣rms2

3, and we can finally write

𝐹net =1

3

𝑁

𝑉𝑚𝑣rms

2 𝐴



Slide 11

FINALLY, we can find the pressure on the wall due to all of the molecular collisions:

𝑝 =𝐹

𝐴=1

3

𝑁

𝑉𝑚𝑣rms

2


Problem #2: Temperature & Pressure

Slide 12

The number density in a container of neon gas (atomic mass number = 20) is 5.00 x 1025 m-3. The atoms are moving with an rms speed of 660 m/s.What are the temperature and pressure inside the container?

Solution: in class

RDK EX 18.11


18.3 Temperature

Slide 13

An individual molecule in a gas has a mass m and (at one particular moment) a speed v. Its translational kinetic energy is, from PHYS 211,

𝜖 =1

2𝑚𝑣2

We know that each molecule in a sample of gas has a different speed, but there’s an average speed 𝑣rms, meaning that there’s an average translational kinetic energy,

𝜖avg =1

2𝑚𝑣rms

2

Combined with the gas pressure equation from the last section, we have

𝑝 =2

3

𝑁

𝑉𝜖avg


18.3 Temperature

Slide 14

Then, using the ideal gas law 𝑝𝑉 = 𝑁𝑘𝐵𝑇, we can write

𝜖avg =3

2𝑘𝐵𝑇

This tells us that a molecule’s average translational kinetic energy is proportional to the temperature of the gas. Notice that it does not depend on the molecule’s mass.

Since 𝜖avg =1

2𝑚𝑣rms

2 , we finally have an

expression for the rms speed:

𝑣rms =3𝑘𝐵𝑇

𝑚


Problem #3: Kinetic Energy & Temperature

Slide 15

If the average kinetic energy of the molecules in an ideal gas initially at 20°C is doubled, what is the final temperature of the gas?

A 10°C

B 40°C

C 313°C

D 586°C


Problem #4: Center of the Sun

Slide 16

What are the average kinetic energy and the rms speed of a proton at the center of the sun, where the temperature is 2.0 x 107 K?

Useful data: the mass of a proton is 1.67 × 10−27kg, 𝑘𝐵 = 1.38 × 10−23 J/K

Solution: in class

RDK EX 18.25


18.4 Thermal Energy and Specific Heat

Slide 17

In this section, we will examine a rather elegant connection between thermal energy and specific heat. The thermal energy of a system of molecules (as opposed to a single molecule) can be expressed as

𝐸th = 𝐾micro + 𝑈micro

where 𝐾micro is the kinetic energy of the moving molecules and 𝑈micro is the potential energy associated with stretching or compressing molecular bonds.In a monatomic gas, there are no bonds, and thus there is no potential energy (gravity is too weak to count). Furthermore, the only kinetic energy is translational. Thus, for a gas consisting of 𝑁 atoms,

𝐸th = 𝜖1 + 𝜖2 +⋯+ 𝜖𝑁 = 𝑁𝜖avg =3

2𝑁𝑘𝐵𝑇 =

3

2𝑛𝑅𝑇



Slide 18

This result indicates that thermal energy is directly proportional to temperature. We’ve been using this result for a while (mostly to justify our first-law bar charts), but now we finally have a proof.

A change in thermal energy is therefore related to a change in

temperature as ∆𝐸th =3

2𝑛𝑅∆𝑇. But we already know from the last

chapter that ∆𝐸th = 𝑛𝐶𝑉∆𝑇. Combined, this suggests what we already found, in Table 17.4 of the text:

𝐶𝑉 =3

2𝑅 = 12.5

J

mol ∙ K



Slide 19

The degrees of freedom of a system represents the number of distinct and independent modes of energy storage. For example, a single atom in a monatomic gas has three DOF – kinetic energy based on motion along the 𝑥-, 𝑦-, and 𝑧-axes:

𝜖 =1

2𝑚𝑣2 =

1

2𝑚𝑣𝑥

2 +1

2𝑚𝑣𝑦

2 +1

2𝑚𝑣𝑧

2 = 𝜖𝑥 + 𝜖𝑦 + 𝜖𝑧

More complex molecules have additional DOF. For instance, a diatomic molecule has both kinetic and potential energy associated with the vibration of the molecular bond, as well as rotational kinetic energy associated with a possible end-over-end tumbling of the molecule.

3 DOF



Slide 20

The equipartition theorem is a powerful concept in statistical physics whose proof is beyond the scope of this course. However, we can state its key point:The thermal energy of a system of particles is equally divided among all the possible degrees of freedom. For a system of Nparticles at temperature T, the energy stored in each DOF is 𝟏

𝟐𝑵𝒌𝑩𝑻, or in terms of moles,

𝟏

𝟐𝒏𝑹𝑻.

For example, a monatomic gas has 3 DOF, and thus 𝐸th =3

2𝑁𝑘𝐵𝑇



Slide 21

Can the equipartition theorem help us make some sense of the specific heat of a solid? According to the simplified diagram below, each atom in a solid should have 6 DOF, associated with kinetic and potential energy along all three axes. This suggests that

𝐸th = 3𝑁𝑘𝐵𝑇 = 3𝑛𝑅𝑇

As we did for the monatomic gas, we compare the change in thermal energy:

∆𝐸th = 3𝑛𝑅∆𝑇to the definition of molar specific heat:

∆𝐸th = 𝑛𝐶∆𝑇

This suggests that 𝐶 = 3𝑅 = 25.0 J/mol∙K. This is close to the values shown for elemental solids in Table 17.2. It’s not exact for reasons that will be mentioned in class.



Slide 22

For molecules of a diatomic gas, the situation is a bit more complex. Quantum effects come into play, which render some of the energy modes “inaccessible” at room temperature. There’s a good explanation on pp. 512-513 of the text, but the details aren’t important. All we really want is the result, which is that at “commonly-used temperature” (i.e. not overly hot or cold), diatomic gases have

𝐸𝑡ℎ =5

2𝑁𝑘𝐵𝑇 =

5

2𝑛𝑅𝑇

and

𝐶𝑉 =5

2𝑅 = 20.8

J

mol ∙ K


Problem #5: Thermal Energy of Aluminum

Slide 23

What is the thermal energy of 100 cm3 of aluminum at 100°C?

Useful data: the density of aluminum is 2700 kg/m3, and its atomic mass number is 27

Solution: in class

RDK EX 18.31


18.5 Thermal Interactions and Heat

Slide 24

For the purposes of solving problems in chapter 17, we assumed that two systems that are brought into contact will eventually attain the same temperature – a condition called thermal equilibrium. Let’s see how this occurs, on the microscopic level.

The hypothetical setup that we use in this analysis is shown in the figure. The systems are separated by a “thin, stiff barrier”… this prevents atoms from moving between systems and ensures that the volumes don’t change, but it does permit atoms to collide at the boundary as if the barrier wasn’t present. There are 𝑁1 atoms in system 1, 𝑁2 atoms in system 2, and 𝑁 =𝑁1 +𝑁2 total atoms.



Slide 25

The initial temperatures of the two interacting systems are 𝑇1𝑖 and 𝑇2𝑖, and therefore their initial thermal energies are

𝐸1𝑖 =3

2𝑁𝑘𝐵𝑇1𝑖 =

3

2𝑛𝑅𝑇1𝑖 and 𝐸2𝑖 =

3

2𝑁𝑘𝐵𝑇2𝑖 =

3

2𝑛𝑅𝑇2𝑖

(we’re assuming monatomic gases…for diatomic, use 5/2).

The total energy of the two systems is 𝐸tot = 𝐸1𝑖 + 𝐸2𝑖. Since the two systems are isolated from the environment, this total energy cannot change, although the energy in each individual system can change.

As the figure shows, when a “fast” atom collides elastically with a “slow” atom, the former loses energy while the latter gains energy. Since the average speed is higher on the hotter side, this agrees with our reasoning that the hot side cools while the cold side warms.



Slide 26

So, how do the systems know to stop exchanging thermal energy once they’re at the same temperature? Well, they don’t. Even after the temperatures equilibrate, we will continue to encounter collisions between faster atoms and slower atoms. However, at this point, the faster atom is equally likely to be on either side of the barrier. As such, the net transfer of thermal energy is zero.Overall, thermal equilibrium can be expressed by the formula

𝜖1 avg = 𝜖2 avg = 𝜖tot avg

That is to say, the average translational kinetic energy of an atom in system 1 is equal to the average translational kinetic energy of an atom in system 2 (which therefore equals the average TKE of any atom in the total system).

Of course, since 𝜖avg =3

2𝑘𝐵𝑇, we can also say that at thermal

equilibrium, 𝑇1𝑓 = 𝑇2𝑓. This is exactly what we expected.



Slide 27

Does this result also indicate that the final thermal energies 𝐸1𝑓and

𝐸2𝑓 are equal? Let’s see…

The total thermal energy at equilibrium is 𝐸tot = 𝑁𝜖avg. *

The equilibrium condition is 𝜖1 avg = 𝜖2 avg = 𝜖tot avg.

Combined, we have𝐸1𝑓𝑁1

=𝐸2𝑓𝑁2

=𝐸tot

𝑁1 +𝑁2Rearranging, we have

𝐸1𝑓 =𝑁1

𝑁1 + 𝑁2𝐸tot =

𝑛1𝑛1 + 𝑛2

𝐸tot

𝐸2𝑓 =𝑁2

𝑁1 + 𝑁2𝐸tot =

𝑛2𝑛1 + 𝑛2

𝐸tot

* Knight uses the subscript “th” here, just above Eq. 18.41. It’s not necessarily wrong, it’s just inconsistent with the rest of his notation on this page.



Slide 28

This result (presented in terms of number of atoms and number of moles) indicates that the final thermal energies of each system are not the same, unless we have the unlikely case that the systems contain the same number of atoms.


Problem #6: Thermal Interaction

Slide 29

Systems A and B are interactingthermally. At the particularmoment in time shown here…

A 𝑇𝐴 > 𝑇𝐵

B 𝑇𝐴 = 𝑇𝐵

C 𝑇𝐴 < 𝑇𝐵

D Both systems are about to collapse into an all-consuming fireball

RDK STT 18.5


18.6 Irreversible Processes & 2nd Law

Slide 30

A common approach to solving problems in all branches of physics is to apply conservation of energy. However, for the situation described in the last section, this approach doesn’t really help. In an individual collision, it’s perfectly possible for energy to be transferred from the colder system to the hotter system while conserving energy. Yet, on a macroscopic scale, we don’t observe this happening. Thermal energy always moves from hot to cold…it’s an irreversible process.



Slide 31

Since macroscopic phenomena come about due to microscopic events, how are the former irreversible while the latter are reversible?The text provides a good analogy here. Consider two boxes which initially contain a different number of balls (𝑁1𝑖 > 𝑁2𝑖). Once every second, a ball is chosen completely at random, and moved to the other box. This process is reversible – an individual ball is equally capable of moving in either direction. However, since we’re selecting balls at random, we are always more likely to be selecting a ball from the box that currently contains more of

them. Therefore, on average, the balls are more likely to move from box 1 to box 2 than the other way around. This system will stabilize when 𝑁1 ≈𝑁2.



Slide 32

But since the individual selection of each ball is random, isn’t it at least possible for 𝑁1to increase while 𝑁2 decreases? Well, it is. But it’s exceedingly improbable. Look at the case of coin-flipping:• If you flip 2 coins, then there’s a 1 in 2 chance that they both

come up the same, and a 1 in 2 chance that the number of heads and tails are the same.

• If you flip 4 coins, there’s a 1 in 8 (12.5%) chance that they’ll all come up the same, but a 3 in 8 (37.5%) chance that you’ll get equal numbers of heads and tails.

• For 6 coins, there’s a 3.12% chance that they’ll all be the same, but a 31.25% chance of getting equal numbers of heads and tails.

• As the number of coins increases to Avogadro’s number, the odds of 𝑁heads ≈ 𝑁tails becomes overwhelmingly more favorable than the odds of having a significant difference between 𝑁heads and 𝑁tails.



Slide 33

Order, Disorder, and EntropyThe figure shows various arrangements of atoms in a container. At top, they are arranged in an ordered “lattice”, while at the bottom,

they are arranged completely at random. The top arrangement is exceedingly unlikely to have spontaneously formed, whereas the bottom arrangement is much less likely. If we had some way of creating the top arrangement, it would evolve into something resembling the bottom arrangement after a certain amount of time.



Slide 34

Order, Disorder, and EntropyThe degree of disorder is a state variable called the entropy of the system. Our textbook implies that it is difficult to calculate entropy, and this is true. However, we can present (without derivation) a very useful equation for one particular case…the change in a system’s entropy (∆𝑺) when an amount of heat Q is added at a constant temperature is

∆𝑆 =𝑄

𝑇If the temperature changes during the process, calculating the change in entropy does indeed become too difficult.From the above equation, we see that entropy must have SI units of J/K (just like Boltzmann’s constant 𝑘𝐵…that’s not a coincidence).



Slide 35

The 2nd Law of ThermodynamicsThere are numerous ways to state the 2nd law. The formal statementis this:

The entropy of an isolated system (or group of systems) never decreases. The entropy either increases until the system reaches equilibrium, or, if the system began in equilibrium, the entropy stays the same.

It’s important to remember the “isolated” part. Humans are very good at creating order out of disorder, but this always requires external sources of energy. However, an isolated system never spontaneously generates order out of randomness.



Slide 36

The 2nd Law of Thermodynamics cont’Based on our coin-flipping analogy, we can make the following informal statements of the 2nd law:

When two systems at different temperatures interact, heat energy is transferred spontaneously from the hotter to the colder system, never from the colder to the hotter.

and

The time direction in which the entropy of an isolated macroscopic system increases is “the future”.

That’s heavy, and not in a 𝑊 = 𝑚𝑔 sense. We’ll discuss it further in class.


Problem #7: Isothermal Compression

Slide 37

An ideal gas is compressed isothermally. The change in entropy of the gas for this process is

A positive

B negative

C zero

D impossible to determine

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