20.1

Chapter 20 The Kinetic Theory of Gases

Overview of where we are headed As I having been trying to emphasis, it is the First law of thermodynamics that sets up the structure to solve problems in thermodynamics. At this point, we have identified that the internal energy as Eint = Kmicro + Umicro where Kmicro is the total translational KE of the microstates in an ideal gas, which is monoatomic (as we move onto diatomic and polyatomic molecules, the KE will include rotational and vibrational modes as well). Furthermore, we have also identified that a measure of the internal energy of a monoatomic gas is the temperature that is proportional to the KE (T ∝ KE). Now we will derive and IGL using the Kinetic theory of Gases and extend the IGL using the very power Equipartition theorem from Statistical Physics.

Using the Kinetic Theory of Gases to Derive the IGL From a microscopic viewpoint, we use a model called the kinetic theory of gases to derive the IGL and more importantly show that the temperature and internal energy are a function of the average KE of the gas particles. What I find particularly of great interest here is by starting with a few basic assumptions, we can show that the microscopic behavior of molecules will produce a macroscopic effect. If you think about this, this is astounding!

To derive the IGL, we start by explaining in detail how pressure (p = F/A) is created by the gas. This implies the need to determine the force applied by a gas on the walls of the container. Outline of my approach to derive the microscopic pressure:

1. Calculate the average force applied by a gas on a container

2. Introduced the appropriate average speed – rms speed

3. Derive the microscopic pressure, where the IGL “falls out” of the calculation.

PhET IGL with one molecule in the box

Step 1: Calculate the average force applied by a single particle on a container wall A single gas particle only moving in the x-direction will collide with the walls elastically. This collision applies a pressure p = Favg/A on the walls. Using the impulse-momentum theorem, the wall applies an impulse on the gas particle and changes its momentum from –mvx to +mvx;

x xf x0 x x x xp p p p ( p ) 2p 2mv = − = − − = =

Interesting side note: the relationship ∆px is frame independent and is the basis for why temperature is frame independent.

Extending the number of gas particles, one assumes that all the N gas particles move with speed vx but only a fraction of this number (called Ncollision) is colliding with the walls at any one instant ∆t with the wall. The net impulse/force for one molecule colliding with the walls can be extend to Ncollision colliding gas particles:

x avg avg

single N collision

x avg xatom gas atoms

N2mv F t F 2

tp Fdt F t mv= = =

= ⎯⎯⎯→ ⎯⎯⎯⎯→

where the quantity Ncollision/∆t is the number of collisions per second. Determining Ncollision/∆t is the most challenging part of the derivation of the IGL. We first start by determining how far a gas particle travels during a time interval ∆t when moving at speed vx: ∆x = vx∆t. Looking at the box to the right, every one of the gas particles in the shaded region that is moving to the right will reach and collide with the wall during time ∆t. Gas particles outside this region will not reach

the wall during ∆t and will not collide. The shaded region tells us about the number of

20.2

molecules colliding with the right wall in a shaded volume A∆x. However, only half the gas particles are moving to the right so the number of collisions during ∆t is

( )

1

collision 2

rate of collisions collision1

x2 with right wall

number density volume spanned by moleculesN

of the gas colliding with the right wall

NN N N A x Av t

V 2V t 2V

=

= = =

⎯⎯⎯⎯⎯→

x

Av

The average force by N gas particles on the wall is

collision

x

2 2collision

avg x x x avgN N

Avt 2V

N2

t

N NF mv 2 Amv Amv F

2V V=

=

= = =

Notice that the expression Favg does not depend on any details of the particle collisions. We can now relax the assumption that all molecules have the same speed by replacing the squared velocity vx

2 with its average value. This is the average force applied by a gas on a container

2

avg x

1D motions and N molecules in a container

NF mv A

V=

where 2

xv is the quantity vx2 averaged over all the molecules in the container.

Step 2: The Root-Mean Squared Velocity We need to be somewhat careful when averaging velocities. The velocity component vx has a sign. At any instant of time, half the molecules in a container are moving to the right (vx > 0) while the other half is moving to the left (vx < 0); thus the average velocity

is vx,avg = 0. If this weren’t true, the entire “stationary” container of colliding gas particles would not be in equilibrium and spontaneously move, which it doesn’t! So, it is meaningless to talk about the standard definition of the average. A new way of looking at “average speed” must be discussed. The problem arises with the negative sign of the velocity and must be eliminated. There is a type of average called the root-mean-square ≡ rms that does the following: (i) squares the velocities, (ii) take the average of the square of the velocities, and (ii) finally take the square root of this sum. That is, the root-mean-square speed vrms is (this is very similar to the stanard deviation before including the over counting)

2 2 2

1 2 N

rms

v v vv

N

+ + +=

In three-dimensions, the vrms is 2 2

2 2 2 2 1x Nx

rms x y z x

v vv v v v where v

N

+= + + =

Because the square root “undoes” the square, vrms must in some sense, give a type of average speed.

Example Suppose five gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s. Is the vrms the same as vavg? NO.

2 2 2 2 2

2

rms

500 600 700 800 900

5v v 714 m/s

+ + + += = =

Comparing this with the average,

avg

500 600 700 800 900

5v 700 m/s

+ + + += =

When comparing vrms and vavg, one sees that they are not the same and “roughly speaking”, vrms gives greater weight to the larger speeds than does vavg.

20.3

To extend the one-dimensional velocities to three-dimensions, note that there’s nothing special about the x-axis. The coordinate system is something that we impose on the problem, so on average, for randomly moving molecules (this assumption was first implemented by Maxwell)

2 2 2 2 2 2 21x y z rms x y z x x rms3

v v v v v v v 3v v v= = ⎯⎯→ = + + = ⎯⎯→ =

Making this replacement into the average force equation, the net average force on the wall of the container is

2 21x rms3

3D motions2 2

avg x avg rmsv v

1D motions 3D motions and N molecules

N NF mv A F mv A

V 3V== ⎯⎯⎯⎯⎯→ =

Step 3: Deriving the pressure and IGL Substituting the average force into the pressure definition, we get

avg 2

rms

N

3V

Fp mv

A= =

Rearranging this pressure equation, the IGL “pops into existence” from molecular magic:

( )2 22 1 2 1

rms translational trans rms3 2 3 2

IGL from molecular viewpoint

pV N mv NK pV where K mv= = =

Remarks

• The pressure is directly related to Krms = Ktrans of the molecules colliding with the container walls. This relationship relates the macroscopic pressure with the microscopic KE’s.

• In this derivation, did we ever mention collisions with the gas particles themselves? NO!

• There are two KEs terms identified here: i. Average randomized translational KE of a single molecule:

21

trans rms2K (single molecule) mv=

ii. Total translational KE in the whole gas is the total internal energy:

int micro microE K U= +

transNK=

• If the macro- and micro-forms of the IGL are to agree, then the temperature must be defined as

32

B trans trans trans B3 2

B

2nRT Nk

3pV T NK T K or K k T

k= == = =⎯⎯→

Important remark: We see that the (i) temperature is proportional to the average randomized translational KE per molecule, however, the more important results is that (ii) average translational KE per molecule depends only on the temperature, NOT on the pressure, volume, or type of molecule.

Deeper question on the Kinetic Theory of Gases Why does one have to consider collisions at all in the kinetic theory of Gases if ideal gases do not interact with each other (or ignored)? That is, in the kinetic theory of gases, it is assumed that the gas particles have no interactions except during collisions. However, in the subsequent derivation (see above on page 20.3) of

2 2rms int3 3

pV NK E= =

only the collisions of the gas particles with the container walls are considered, but not the collisions between gas particles. Why can we ignore them?

Answer

20.4

We can ignore the collisions between particles when deriving that equation because we are assuming that we are dealing with a rarified gas, i.e. the average distance between two molecules is large compared to their size. When calculating pressure, we are mainly concerned with the interaction with the walls, so this approximation is quite good; actually, an ideal gas is formally made of point particles, so the probability of a particle-particle collision would be rigorously zero in the ideal gas approximation.

But of course, to compute other quantities such as the mean free path we must consider the particles to be of finite size, i.e. we have to give up the ideal gas approximation. However, in such calculations, the ideal gas law (IGL) is used: this apparently makes no sense, because the IGL is derived under the assumption that we are dealing with point particles! The catch is that, even if the IGL is only rigorously valid for a gas of point particles (which by definition cannot collide with each other), it is a quite good approximation also for a gas of finite-size, but small particles: this is why we can use the IGL in the derivation of quantities such as the mean free path. And this is also why you can find in many books (or on Wikipedia) the somewhat misleading definition “An ideal gas is a theoretical gas composed of point particles that do not interact except when they collide elastically”.

Furthermore, there is a more fundamental problem with the usual derivation of pV=²/₃Eint. The mass of the particles appears in the derivation until the very end when it cancels out exactly (i.e., if all energies are equal via the equipartition theorem). Since they are not equal, the derivation just inserts “mean energy” instead of “energy” and it’s all done. Now it is obvious, that a gas of identical particles will have a mean energy, and since the contribution to pp is proportional to the energy, the derivation is perfectly accurate. What is not obvious is, why the mean energies of sets of different particles should coincide. If they don't, then the whole derivation gets much more complicated (the collisions with the walls and between particles are not “elastic” any more), and collisions matter. But an elastic collision between identical particles is very simple: they swap their momenta. In other words, the particle just gets exchanged, and continues as if there had been no collision.

This last piece is complicated and subtle but extremely interesting in learning the flaws of the IGL. But this is what makes physics so beautiful – physics is about the discussions of the details, not the regurgitation of information and numbers. If you get anything from me, I hope you at least take that away from me.

The kinetic-molecular model contains a hidden assumption about the temperature of the container walls. What is this assumption? What would happen if this assumption were not valid? Answer: The assumption of elastic collisions with the walls is equivalent to assuming that the walls are at the same temperature as the gas. If the walls are hotter or colder than the gas, the gas atoms will gain or lose energy when they collide with the walls.

Collisions Between Molecules and the Mean Free Path We have ignored the possibility that two gas molecules might collide. If they are really zero-volume points, they never collide. But consider a more realistic model in which the molecules are rigid spheres with radius r. Suppose a nonzero volume gas particle is zipping along in a straight line, the molecule will follow a convoluted zig-zag path in which it frequently collides with other molecules (see image on the right). What is the average distance between collisions – this is called the mean free path? Without going into the mathematical details, the answer is

2

the average distanced traveled between collisions

1mean free path

4 2 r (N/V) =

20.5

The mean free path is inversely proportional to the number of molecules per unit volume

(N/V) and inversely proportional to the cross-sectional area r2 of a molecule; the more molecules there are and the larger the size of a molecule, the shorter the mean distance between collisions. Note that the mean free path distance does not depend on the speed of the molecule.

Interpretation a. If you try to walk through a crowd, your mean free path—the distance you can travel

on average without running into another person—depends on how large the people are (their cross-sectional area A) and how closely they are spaced (number density N/V). Larger people that are densely packed will be harder to move through and therefore, ones mean free path will be shorter.

A familiar molecular example of the mean free path occurs when someone opens a bottle of strong perfume a few feet away from you. If molecular speeds are hundreds of meters per second, you might expect to smell the perfume almost instantly. But that isn’t what happens. As you know, it takes many seconds for the molecules to diffuse across the room.

b. Let’s analyze this mean free path distance with the following question: if the temperature is increased at constant pressure, how should the mean free path behave as? Using the IGL, the mean free path can be written in terms of pressure:

B

B

2 2N/V p/k T

k T1

4 2 r N/V 4 2 r p=

= = =

If the temperature is increased at constant pressure, the gas expands, the average

distance between molecules increase and increases (less likely to collide with another gas particle). On the other hand, if the pressure is increase at constant

temperature, the gas compresses (pV = constant) and decreases (more likely to collide with another gas particle).

Molecular Speeds From the temperature relationship, we find that the rms speed is

== ⎯⎯→ = B2 31rms B rms2 2

molecule molar

3k T 3RT

m Mmv k T v

The Maxwell-Boltzmann speed distribution f(vx) in three dimensions is a probability distribution curve describing particles speeds in idealized gases that changes with temperature.

2B2

B

mv /2k T

3/2m

e2 k T

f(v)dv 4 v dv−

=

Higher temperatures lead to higher velocities but in terms of the distribution function, the distribution of a gas at a hotter temperature will be broader than it is at lower temperatures. That is, a broad number of particles will have higher speeds.

At a given temperature T, gas particles of different masses will have the same Krms but different vrms speeds. For example, the speed distribution curve below shows that the lower helium mass particles (MHe = 4.0 g/mol) will have, on the average, higher speeds than the higher masses argon or xenon (MXe = 131.3 g/mol).

20.6

Let’s calculate the rms speed of a single molecule of H2 and N2 at room temperature (20oC). The rms speed values at these temperatures are

rms 2

3RT 3(8.31 J/ mol

Mv (H ) ==

K )(293 K

3

)

(2.02 10 kg/ mol−

)1906 m/s=

Or we can use the kB form of vrms:

23

B

rms 2 27

3k T 3(1.38 1

6

0 )(293 K)

m 1. 7 10 k )g(2

J/Kv (H ) 1906 m/s

−

−

== =

rms 2 rms 2v (H ) 1900 m/s 4250 mph and v (N ) 511 m/s 1150 mph= =

Lower mass gas particles move faster, whereas higher mass ones move slower even though they have the same rms translational KE. H2 is the most plentiful element in the universe, yet our atmosphere is mostly N2 and O2. Why doesn’t it contain H2?

Answer: It is believed that during the early stages of the earth’s evolution, there was lots of H2. However, it escaped into outer space. One might ask how can H2 escape if the escape velocity of the earth is 11,200 m/s? The atmosphere early in the life of earth was composed of three gas particles, and their rms speeds are

rms 2 rms 2 rms 2v (O ) 478 m/s v (N ) 511 m/s v (H ) 1902 m/s= = =

Although vrms(H2) is only 17% of the escape velocity, some molecules move at speeds many times faster according to the Maxwell-Boltzmann speed distribution curve (look at the tail end of the curve), and therefore, can have speeds that exceed the escape velocity. In a time of about five billion years since the formation of the earth, there has been enough time for essentially all the hydrogen to escape that was formed in the beginnings of the earth.

Example: Evaporation of water at room temperature (needs better explanation!!)

Explain how water in a tea saucer is evaporated at room temperature.

I think of temperature as the average KE of the water molecules. While the average molecule doesn't have enough energy to break the inter-molecular bonds, a non-average molecule does (i.e., Maxwell-Boltzmann Distribution function guarantees this). Water is a liquid because the dipole attraction between polar water molecules makes them stick together. At STP (acting somewhat like a vice), you need a comparatively large temperature of 100°C (translating to a high average energy distributed among the microscopic degrees of freedom, most relevantly the kinetic ones) for water molecules to break free in bulk, creating bubbles of water vapor within the liquid. However, at the surface of the liquid, lone molecules may end up getting enough KE to break free due to the random nature of molecular motion at basically any temperature. On the flip side, water molecules in the atmosphere may enter the liquid at the surface as well, which is measured by equilibrium vapor pressure.

Example 20.1

20.7

On earth, STP is based on the average atmospheric pressure at the surface and on a phase change of water that occurs at an easily produced temperature, being only slightly cooler than the average air temperature. The atmosphere of Venus is almost entirely carbon dioxide (CO2), the pressure at the surface is a staggering 93 atm, and the average temperature is 470°C. Venusian scientists, if they existed, would certainly use the surface pressure as part of their definition of STP (93 atm, 470oC). What are (a) the rms speed and (b) the mean free path of carbon dioxide molecules at Venusian STP? (c) Is the mean path distance shorter or longer? Explain your reasoning with both a physical and mathematic answer. The radius of a CO2 molecule is 116 pm.

Solution For starters, carbon dioxide is a “linear molecule” (all three atoms are in a co-linear line). The C=O bond is 116 pm (1.16×10-10m), so the entire molecule is 232 pm. The mass of a CO2 molecule is

2

27 26

CO C O pm m 2m (12 2 16)m 44 1 661 10 kg 7 31 10 kg

− −= + = + = =

The Venusian atmosphere has the following temperature and pressure: 450oC = 743 K

and 93 atm = 9.42×106 Pa. a. The rms speed of a CO2 molecule is

23

B

rms rms26

3k T 3(1 38 10 J/K)(743 K)v 649 m/s v (Venus)

m 7 31 10 kg

−

−

= = = =

How does this compare to the vrms value of CO2 in our atmosphere?

23

B

rms rms26

3k T 3(1 38 10 J/K)(293 K)v 407 m/s v (Earth)

m 7 31 10 kg

−

−

= = = =

The ratio of these speeds is

rms

rms

v (Venus) 649 m/s1.60 faster

v (Earth) 407 m/s= =

b. Because the high air pressure on the Venusian surface, one expects that the mean-free path of the CO2 molecules to be shorter (λ ∝ 1/p), so, they will interact much more.

Intuitively, a rough estimate gives pVenus ≈ 100pEarth and the mean free path should be 1/100 shorter than that on Earth if pressure was the only variable (not ture). Let’s see if this is true.

B Venus

Venus 2 22 Venus Venus

B Venus

23

1 2 60

k T1 1

p4 2 r N/V 4 2 r p4 2 r

k T

(1.38 10 J/K)(743 K) 14.3 nm

4 2 (1.16 10 m) (9.42 10 Pa)

−

−

= = =( )

= = =

What is the mean free distance on earth for CO2? 23

B Earth

Earth 2 1 2 5

Earth

0

k T (1.38 10 J/K)(293 K)52.6 nm

4 2 r p 4 2 (1.16 10 m) (1.01 10 Pa)

−

−

= = = =

Therefore, the mean free distance between CO2 molecules on Earth vs. Venus is

Earth

Venus

52.6 nm3.70

13.9 nm= =

That is, a CO2 molecule on Earth travels 3.8 times farther before a collision. Stated differently, since the atomic diameter of carbon dioxide is 2.32×10-9 m, then the mean path difference between the two planets is

20.8

10 10

9 9

Venus Earth

1.12 1 22.7 23 atomic diameter atomic diameters2.6 10 m 52.6 10 m

2.32 10 m 2.32 10 m

− −

− −

= =

Conceptual Questions

Why is there a minimum possible temperature (absolute zero) but no absolute maximum temperature?

Answer: make remarks about classical theory without quantum and special theory of relativity.

Conceptual Questions

Question: What is boiling? The difference between boiling and bubbling Language dilemma: boiling is different than bubbling in this example

Answer: According to the Maxwell-Boltzmann’s speed distribution, there is a range of speeds that water molecules can have in, say, a glass of water. Something special happens at the air-water boundary: the faster molecules have sufficient KE to break away from the water and form a vapor pressure just above the water surface, as shown in the image to the right. These escaping molecules are kept in place within the vapor barrier because atmospheric pressure is higher than the vapor pressure (patm > pvapor). But there are also faster moving water molecules deeper within the glass that phase transition (just like at the surface) and form micro-sized bubbles with an internal pressure pbubble inside the bubble. The difference between what happens at the air-water boundary is that the pressure at the surface of the water is lower compared to the pressure at, say, the bottom of the glass (pglass > pbubble). Remember that pglass = patm + ρgh. So, every time a bubble is created at the bottom of the glass no substantial sized bubble is created in ordinary (everyday) situations. There are two key points here; (i) atmospheric pressure is the cause for not having a bubbling effect in room temperature water. Reducing atmospheric pressure allows the faster moving molecules to cause a bubbling effect – the phase transition between water into water vapor, where the bubbling pressure is compared to the pressure of the surrounding water (pglass ≈ pbubble). When this occurs, then the glass of water would suddenly start bubbling at room temperatures. (ii) The bubbling effect occurs not because there is heat transfer that increases the internal energy of water but because of the very low atmospheric pressure. There is NO heat transfer occurring here! If one was to suddenly reduce atmospheric pressures to zero, then all of the water would spontaneously bubble.

Note that this is quite different from the process of boiling water in a pot. When boiling water in a pot, there is a constant heat transfer from the stove to the water such that it increases the overall temperature of the water. Although the boiling of water needs deeper concepts that I am not willing to take at the moment, boiling water requires nucleation sites. However, normal boiling of water increases the KE of the water molecules at the bottom of the pot (pbubble), which in turn, form bubbles because pbubble > ppot. As these bubbles raise to the surface, there is a release of tapped gases in the bubbles that we normally call boiling. With constant bubbles raising up along with convection, water is said to be boiling and raise the temperature to “100oC” (remember from lab that a pot of boiling water will never reach 100oC because of evaporation or heat transfer).

DEMO Liquid Nitrogen or water, bell jar, and air pump

Reduce pressure to show that LN2 freezes or water “boils” according to IGL

Internal Energy Function is process independent

The derivation of the IGL lead to the internal energy function. Since changes in internal energy are path/process independent, this function applies to any situation, regardless of which process it is. Using the first law, all heatings can determine via Q = ∆Eint − W, we

20.9

first will calculate the (i) heatings for isochoric and isobaric, and follow up with (ii) the work for an adiabatic process. The internal energy function is

= +int micro microE K U = = =3 3

trans B2 2

per molecule per mole

NK Nk T nRT

Clearly, the internal energy depends only on the temperature, not on its pressure or volume:

= = 3 3int B2 2

E T TnR Nk Ideal gas monatomic gas

A summary of the heat capacities at constant volume and pressure show that

3 5

V P P V2 2 P VC R 12.5 J/mol K C nR 20.8 J/mol K C C R C /C= = = = + =

and all ideal gas (monatomic) thermodynamics processes show that

Process IGL condition Work W Heat Q Eint

Isothermic p1V1 = p2V2 p1V1ln(V2/V1) = p1V1ln(p1/p2) W = −Q 0

Isobaric T1 /V1= T2/V2 −pV = −nR∆T nCP∆T nCV∆T

Isochoric T1 /p1 = T2/p2 0 nCV∆T nCV∆T

Adiabatic p1V1 = p2V2

− nCV∆T 0 −nCV∆T

No assumptions have been made about the gas type other than being a monatomic gas (point-like) and therefore, the molar heat capacities are gas-type independent for monatomic gases.

PhET States of Matter – heat from solid→liquid→gas

How do the predicted gas heat capacities values compare with experimentally measured values? The table below that summarizes the details.

The theoretical predictions are highlighted in green while experimental measurements are in white. The predictions for monatomic gases hold quite well for most monatomic gases. However, the IGL predictions failed miserably as soon as one analyzes gases that are not monatomic but diatomic or polyatomic gases:

20.10

Extension of the IGL to nonzero volume molecules This tells us that our point-molecule model is not good enough and we need a more sophisticated model. The way this model is extend is using the very power equipartition theory. A first attempt to describe the details of heat capacity is the number of ways (degree of freedom ≡ DOF) a material has the ability to store energy.

i. What do we really mean by a monatomic gas? Since nonzero volume gas molecules only interact during collisions, when a collision occurs it results in a point-molecule having three freedoms to move in the translational directions (vx, vy, vz) after a collision. The total number of degrees of freedom is 3 = DOF. This 3 shows up in the change in internal energy as

3int 2

T monatomic gas)E nR ( =

ii. One problem with point molecules is they cannot rotate because they’re not extend objects. One can extend the IGL assumptions to incorporate the simplest model of a diatomic gas (envisioned them as two point molecules kissing). In addition to moving with translational motion, they can also rotate about their center of mass locations. Such molecules now have two additional freedoms to move with two different rotations (φ, θ) after a collision. So, the total number of degrees of freedom is 3 + 2 = 5 = DOF.

iii. A polyatomic gas can be envisioned as two-point masses with an interaction force between the atoms (little elastic dumbbell; bar = interaction). Such molecules now have one possible additional freedom with vibrational modes in the molecules. So, the total number of degrees of freedom is 3 + 2 + 1 = 6 = DOF.

PhET States of Matter – heat from solid→liquid→gas

• Neon is monatomic and 3 freedoms to move (vx, vy, vz) via collisions

• Oxygen-2 is diatomic with 5 freedoms to move (vx, vy, vz, φ, θ) via collisions

• Water is polyatomic with 6 freedoms to move (vx, vy, vz, φ, θ, f) via collisions. Use the Interaction Potential part of the PhET applet to explain this.

What is happening physically? When heating a monatomic gas at constant volume, all of the transferred energy increases only the random translational KE (can’t go anywhere else), which in turn increases the temperature and internal energy quickly. However, when heating diatomic or polyatomic gases, part of the energy has to supplied to the other DOF and these gases reach a lower temperature and take longer to reach higher temperatures too. To reach the same temperature of the monatomic gas, additional heating must be supplied to simultaneously increase all of the DOF, which including rotational and vibrational energies. Polyatomic gases will have larger molar heat capacities because they have more ways to store internal energy compare to diatomic and monatomic gas as the table of heat capacity values clearly shows. Said differently, with more degrees of freedom, there are more places for the energy to go. Therefore, as the number of degrees of

20.11

freedom increase, there is less energy in each “type of motion” which is a smaller temperature.

Silly analogy: A bucket of water spread into a number of glasses. Water = energy, number of glasses = DOF, amount of water in one glass = temperature.

Equipartition of Energy and Degrees of Freedom How do we know how much energy is associated with each DOF for complex molecules? It is the principle of equipartition of energy that tells us how much energy each DOF gets. Although it is beyond the scope of this course to derive this, we will summarize its results now.

Principle of Equipartition of Energy Each degree of freedom (linear, angular or vibrational) has, on average, an associated KE per molecule of ½kBT.

1. Monatomic gas: DOF = 3 Collisions of gas molecules cause point molecules to change their translational motion in three independent velocity directions: vx, vy, vz. These three DOF implies its internal energy is

3V 21 1

int B B V2 2 5p V3 DOF 2

C RE N DOF k T N 1 1 1 k T nC T

C C R R

= = = + + = ⎯⎯→

= + =

2. Diatomic gas: DOF = 3 + 2 = 5 A diatomic molecule has two additional degrees of freedom for the two possible axes of rotations:

5V 21 1

int B B V2 2 72 DOF p V 2

C RE N DOF k T N 3 1 1 k T nC T

C C R R

= = = + + = ⎯⎯→

= + =

3. Polyatomic Molecules: DOF = 3 + 2 + 1 + … = 6 The large values of CV or some polyatomic molecules show contributions of vibrational energy, to the already other degrees of freedom. In addition, if a molecule is not in a straight-line, the number of degrees of freedom for rotations increases from 2 to 3.

V5 1int B B V2 2

pone vibrational dof

24.9 J/mol K

33.24 J/mol K

C 3RE N k T k T nC T

C 4R

= = = + = ⎯⎯→

= =

V1 1

int B B V2 2

p V1 or more DOF

C 3RE N DOF k T N 5 1 k T nC T

C C R 4R

== = + + =

= + =

⎯⎯→

Water is more complicated and has more DOF (see website link). When these values are compared with the table, it agrees well with what is observed in the lab.

20.12

Vibrational motion can also contribute to the heat capacities of gases. Molecular bonds are not rigid; they can stretch and bend, and the resulting vibrations lead to additional degrees of freedom and additional energies. For most diatomic gases, however, vibrational motion does not contribute appreciably to heat capacity. The reason for this is a little subtle and involves some concepts of quantum mechanics. Briefly, vibrational energy can change only in finite steps. If the energy change of the first step is much larger than the energy possessed by most molecules, then nearly all the molecules remain in the minimum-energy state of motion. In that case, changing the temperature does not change their average vibrational energy appreciably, and the vibrational degrees of freedom are said to be “frozen out.” In more complex molecules the gaps between permitted energy levels are sometime much smaller, and then vibration does contribute. The rotational energy of a molecules also changes by finite steps, but they are usually much smaller; the “freezing out” of rotational degrees of freedom occurs only in rare instances, such as for the hydrogen molecule below about 100 K.

Conceptual Questions

• Dielectric heating and microwaves; polar vs. nonpolar molecules.

• Why do some microwaved foods cool faster?

• Plastic is an organic molecule with symmetry? Question do high vs. low quality plastics get hot? How about quartz vs. glass?

Conceptual Questions

Suppose that you want to heat a gas so that its temperature will be as high as possible. Would you heat it under conditions of constant pressure or constant volume? Why?

Answer: The temperature T is related to its internal energy by Eint = 3/2∙nRT and the 1st

law is Eint = Q + W. It is desired to heat a gas so that its temperature will be as high as possible. If the process occurs at constant pressure, so that the volume of the gas increases, the gas does work. The available heat is used to do work and to increase the internal energy of the gas. On the other hand, if the process is carried out at constant volume, the work done is zero, and all of the heat increases the internal energy of the gas. Since the internal energy increases by a greater amount when the process occurs at constant volume, the temperature increase is greatest under conditions of constant volume. Therefore, if it is desired to heat a gas so that its temperature will be as high as possible, you should heat it under conditions of constant volume.

Example 20.2

The rms speed of the molecules in 1.0 g of hydrogen gas (H2) is 1800 m/s. a. What is the total translational kinetic energy of the gas molecules? b. What is the internal energy of the gas? c. 500 J of work are done to compress the gas while, in the same process, 1200 J of

heat energy are transferred from the gas to the environment. Afterward, what is the rms speed of the molecules?

Solution

20.13

We are given vrms = 1800 m/s. For 1.0 g of molecular hydrogen gas

23 231mol moleculesN 1 0 g 0 5 mol 6 02 10 3 01 10 molecules

2 0 g mol

= = =

The mass of one hydrogen molecule (H2) is twice that of a proton: 2∙1.661×10-27 kg =

3.322×10-27 kg. a. The total kinetic energy of the gas (i.e., internal energy) is

2 23 27 21 1translation rms rms2 2

translation

K NK N mv (3 01 10 ) (3 322 10 kg)(1800 m/s)

1 6 kJ K

−= = =

= =

a. From the kinetic theory of gases, the temperature is given by

272 21

int rms2 23

B B

2 2 3 322 10 kgT E mv (1800 m/s)

3k 3k 3(1 38 10 J/K)

260 K T

−

−

= = =

= =

For diatomic gases the internal energy is

5 5int int2 2

E nRT (0 50 mol)(8 31J/mol K)(260 K) 2 7 kJ E= = = =

b. The system starts at state-1 given by Eint,1 = 2700 J, and after transferring Q + W = −1200 + 500 = −700 J, the new internal state-2 is Eint,2 = Eint,1 Q + W = 2000 J, a decrease in internal energy. This allows one to determine the temperature of state-2 and therefore, the vrms for state-2. From the Eint,2, the temperature of state-2 is

2 int

2 2T E (2000 J) 192 5 K

5nR 5(0 50 mol)(8 31J/mol K)= = =

The new rms speed is

23

B 2rms,2 rms,227

3k T 3(1 38 10 J/K)(192 5 K)v 1550 m/s v

m 3 322 10 kg

−

−

= = =

Example 20.3

A 1.0 kg ball is at rest on the floor in a 2.0 m × 2.0 m × 2.0 m room of air at STP. Air is 80% nitrogen (N2) and 20% oxygen (O2) by volume. a. What is the internal energy of the air in the room? b. What fraction of the internal energy would have to be conveyed to the ball for it to be

spontaneously launched to a height of 1.0 m? c. By how much would the air temperature have to decrease to launch the ball? d. Your answer to part c is so small as to be unnoticeable, yet this event never

happens. Why not?

Solution a. The internal energy of the air is a combination of both N2 and O2:

2 2 2 2

5 5 5int int,N int,O N B O B total B2 2 2

E E E N k T N k T N k T= + = + =

where Ntotal is the total number of molecules. The identity of the molecules makes no difference since both are diatomic and ideal gases. We first have to determine the number of molecules in the room, which is given by the IGL:

26

total 23

B

pV (101,300 Pa)(2 m 2 m 2 m)N 2 15 10

k T (1 38 10 J/K)(273 K)−

= = =

The internal energy is then

20.14

26 235 5int total B2 2

6 6

int

E N k T (2 15 10 )(1 38 10 J/K)(273 K)

2 03 10 J 2 0 10 J E

−= =

= =

This is a significant amount of internal energy in the air. b. A 1 kg ball launched from rest to a height of 1 m will have gravitational potential

energy ∆Ug = mg∆y = 9.81 J. If some of the room’s internal energy was to be transferred to the ball spontaneously, what fraction of internal energy of the room is required to spontaneously launch the ball to a height of 1 m? Looking at the ratio of these two energies, we get

6

6

9 8 J4 8 10 0.00048%

2 03 10 J

−= ⎯⎯→

c. To determine the change in temperature of the room after siphoning out a tiny portion of internal energy, the fractional change of the temperature can be determined using the chain rule via

approximate

int int

int int

dT dTdT dE T E

dE dE= ⎯⎯⎯⎯⎯→ =

where, of course, the fractional change in internal energy is ∆Eint = −9.81 J (there is a minus sign since it is a decrease in internal energy). From the equation of the internal energy of a diatomic gas, we calculate

( )int 5 5total B total B2 2 5

int total B2

dE d 1 1N k T N k

dt dt dE /dt N k= = ⎯⎯→ =

Now, the change in temperature ∆T is written down as

int int

26 235 5int total B2 2

E E 9 81JT 0 0013 K T

dE /dT N k (2 15 10 )(1 38 10 J/K)−

− = = = = − =

Is this detectable? Relative to room temperature, the fractional ratio is

60 0013 KT 4.0 10

293 K

− = =

Simple answer is NO!