1. Chapters 18, micro-macro connection
Lecture 25
Goals:
Chapters 18, micro-macro connection
Third test on Thursday at 7:15 pm.
1) For the last couple of weeks, all of our arguments have been macroscopic and today we will link those arguments to the microscopic world. For example, we have argued that a gas will contain a very large number of atoms or molecules that are moving around, colliding with each other and colliding with the walls of the container that they are in. In today’s lecture, we will try to understand how macroscopic quantities such as pressure and temperature relate to microscopic motion of particles. 1

2. 15 10 5 (m/s) Nitrogen molecules Percentage of near room temperature
Let’s first start by discussing the speed of particles that form a gas. The first question is, if I have a gas at a certain temperature T, how fast are the molecules moving? Are all the molecules moving with the same speed or is there a range of speeds?
This is a question that can be answered experimentally, by a variety of methods. This graph shows the results for nitrogen molecules near room temperature. The results are presented in the form of an histogram. The y-axis is the percentage of molecules, and the x-axis is the speed range. There will be very few molecules with a speed between 0-100m/s, similarly there will be few molecules with speeds exceeding 1000 m/s. The most likely speed range is between m/s. This is really fast by the way, exceeding 1000 miles per hour.
Now if you repeat this experiment, you will find exactly the same distribution. There is an important idea here. So a gas typically will contain on order 10^20 molecules or so, each molecule moving around randomly frequently colliding with other molecules and the wall of the container. So this is a very complicated system, yet there are average quantities that have very precise values. Here, for example, the percentage of molecules that have a speed range between m/s has a very precise, predictable value.
Now, when you change the temperature, this distribution changes. At higher temperatures, the distribution will shift to higher speeds, at lower temperatures, the distribution will shift to lower speeds. This is actually called Maxwell-Boltzmann distribution.
(m/s)
0-100

3. Atomic scale r r ≈1 angstrom=10-10 m
What is the typical size of an atom or a small molecule?
A) 10-6 m
B) m
C) m
r ≈1 angstrom=10-10 m r
Now, in these discussions, it is important to know the spatial scale of an atom, or a molecule. So, I would like to quickly ask this to you, what is the typical size of an atom or a small molecule?
The correct answer is meters, which is 1 angstrom, which is roughly what the Bohr radius is.
So, when you think of an ideal gas, the picture that you should have in mind is: you should think these billiard ball particles flying around randomly, each with a size of about 1 angstrom. There are no long-range interactions, however, there are frequent collisions, when the paths of two particles cross. These collisions change the speed of the particle and also the direction that it is moving. At atmospheric pressure at room temperature, each molecule goes through millions of collisions every second.

4. N/V: particles per unit volume
Mean free path
Average distance particle moves between collisions:
N/V: particles per unit volume
Now, in microscopic discussions, the idea of mean free path is very useful and is very commonly used. Mean free path is the average distance between collisions. So once again, molecules of the gas move freely until they encounter a collision, and mean free path is the average distance the molecule moves before it encounters another collision.
Mean free path is usually denoted by lambda. And it is given by this expression, it is not difficult to derive this expression, but we are not going to do that derivation here, but instead we will make qualitative arguments.
Here N/V is the particles per unit volume and r is the radius of the particle, so r^2 is the area
This expression is intuitive. Mean free path is inversely proportional to number of particles per unit volume. If there are more particles per unit volume, the collisions would be more frequent, and the average distance between collisions would drop. Similarly, if the area of the particles were larger, the collisions would be more frequent, and as a result mean free path would decrease.
In ultrahigh vacuum, MFP may be as large as thousands of kilometers
The mean free path at atmospheric pressure is:
λ=68 nm

5. Pressure in a gas
Now,let’s proceed with a discussion of pressure. So why does a gas have pressure? The pressure in a gas is due to collisions of the molecules with the walls of the container.
If we were to zoom in inside a gas container as shown here, what we would see is that molecules would be constantly bouncing of the walls of the container. Let’s consider a particular molecule with a x velocity component vx. If we consider the ideal case of perfectly elastic collision, then the molecule will rebound from the wall with the x-velocity component completely flipped.
Now, since the velocity of the molecule has changed, the momentum of the molecule has changed, and the momentum has changed due to the impulse experienced by the molecule from the wall. And this impulse will be to the left.
From newton’s third law, the wall gets an impulse equal and in the opposite direction to the right.
So for each molecular collision, the wall gets a tiny bit of kick to the right, a tiny bit of force to the right, and pressure is nothing but force per unit area, so that is what causes pressure.
The kick, the force for each collision is very tiny, but when you get on order 10^20 molecules colliding the wall every second then you get a macroscopic pressure.

6. Consider a gas with all molecules traveling at a speed vx hitting a wall.
If (N/V) increases by a factor of 2, the pressure would:
A) decrease
B) increase x2
C) increase x4
If m increases by a factor of 2, the pressure would:
A) decrease
B) increase x2
C) increase x4
If vx increases by a factor of 2, the pressure would:
A) decrease
B) increase x2
C) increase x4

7. (vx2)avg= (vy2)avg= (vz2)avg
P=(N/V)mvx2
Because we have a distribution of speeds:
P=(N/V)m(vx2)avg
For a uniform, isotropic system:
(vx2)avg= (vy2)avg= (vz2)avg
Root-mean-square speed:
(v2)avg=(vx2)avg+(vy2)avg+(vz2)avg=Vrms2

8. Microscopic calculation of pressure
P=(N/V)m(vx2)avg
=(1/3) (N/V)mvrms2
PV = (1/3) Nmvrms2
We finally arrive at a pretty remarkable result. The pressure of a gas, which is a macroscopic quantity, can be calculated from purely microscopic physics. (N/V) is the number of particles per second and vrms is basically a measure of the average speed of the particles.
We can rewrite this in a form that resembles the ideal gas law, PV=NkT.

9. Micro-macro connection
PV = (1/3) Nmvrms2
PV = NkBT (ideal gas law)
kBT =(1/3) mvrms2
The average translational kinetic energy is:
We will now make the micro-macro connection by tying this result to the ideal gas law. This is the result that we have just derived. Now ideal gas law tells us that, PV=NkT where k is the Boltzmann’s constant, which is the gas constant divided by Avagadro’s number.
Now, equating these right-hand-sides to each other, we see that kT=(1/3) m v^2.
This finally gives temperature a very precise meaning. So far, we have always been a little vague about the temperature. We argued that temperature is a measure of thermal energy of a system but we had not precisely defined what temperature means. So temperature is a direct measure of how fast the particles are moving. For a given temperature, and given the mass of the molecules, we can exactly calculate the rms speed of the molecules that we expect. .
So now, you know how to calculate roughly the average speed for nitrogen molecules at room temperature. You know how to calculate that 600 m/s number that we discussed in the beginning of lecture.
We can also relate temperature to average translational kinetic energy:
εavg=(1/2) mvrms2
εavg=(3/2) kBT

10. The average kinetic energy of the molecules of an ideal gas at 10°C has the value K1. At what temperature T1 (in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K1?
(A) T1 = 20°C
(B) T1 = 293°C
(C) T1 = 100°C
Suppose that at some temperature we have oxygen molecules moving around at an average speed of 500 m/s. What would be the average speed of hydrogen molecules at the same temperature?
(A) 100 m/s
(B) 250 m/s
(C) 500 m/s
(D) 1000 m/s
(E) 2000 m/s
Solve these problems on the board.
mH vH^2 = mO vO^2

11. Equipartition theorem
Things are more complicated when energy can be stored in other degrees of freedom of the system.
monatomic gas: translation
solids: translation+potential energy
diatomic molecules: translation+vibrations+rotations

12. Equipartition theorem
The thermal energy is equally divided among all possible energy modes (degrees of freedom). The average thermal energy is (1/2)kBT for each degree of freedom.
εavg=(3/2) kBT (monatomic gas)
εavg=(6/2) kBT (solids)
εavg=(5/2) kBT (diatomic molecules)
Note that if we have N particles:
Eth=(3/2)N kBT =(3/2)nRT (monatomic gas)
Eth=(6/2)N kBT =(6/2)nRT (solids)
Eth=(5/2)N kBT =(5/2)nRT (diatomic molecules)

13. Specific heat Eth=(6/2)N kBT =(6/2)nRT (solid) ΔEth=(6/2)nRΔT=nCΔT
Molar specific heats can be directly inferred from the thermal energy.
Eth=(6/2)N kBT =(6/2)nRT (solid)
ΔEth=(6/2)nRΔT=nCΔT
C=3R (solid)
The specific heat for a diatomic gas will be larger than the specific heat of a monatomic gas:
Cdiatomic=Cmonatomic+R

14. Entropy
A perfume bottle breaks in the corner of a room. After some time, what would you expect? A) B)

15. very unlikely probability=(1/2)N
The probability for each particle to be on the left half is ½.
probability=(1/2)N

16. Second Law of thermodynamics
The entropy of an isolated system never decreases. It can only increase, or in equilibrium, remain constant.
The laws of probability dictate that a system will evolve towards the most probable and most random macroscopic state
Isolated is very important.
When you have a hot system in contact with a cold system, this is also the reason why heat always flows from the hot to the cold.