LECTURE 11- KINETIC THEORY 2Phys 124H- Honors Analytical Physics IBChapter 19Professor Noronha-Hostler

TODAY’S OBJECTIVES

Kinetic theory

Distributions

Mean free path

WHY THEN DOES SMELL TAKE TIME TO SPREAD?

Repeated collisions of particles prevents smell to spread as quick as the rms

COLLISIONS WITH PARTICLES

TRANSLATIONAL KINETIC ENERGY

Kavg = ( 12

mv2)avg

=12

m (v2)avg

=12

mv2rms, vrms =

3RTM

substitute in

=12

m3RTM

, NA =Mm

substitute in

=3RT2NA

, substitute in k =RNA

TRANSLATIONAL KINETIC ENERGY

Kavg =32

kT

At a fixed temperature T, all types of gas have the same average translational kinetic energy

Two identical, sealed containers have the same volume. Both containers are filled with the same number of moles of gas at the same temperature and pressure. One of the containers is filled with helium gas and the other is filled with neon gas. Which one of the following statements concerning this situation is true?

a.) The speed of each of the helium atoms is the same value, but this speed is different than that of the neon atoms.

b)

c) The pressure within the container of helium is less than the pressure in the container of neon.

d) The internal energy of the neon gas is greater than the internal energy of the helium gas.

e) The rms speed of the neon atoms is less than that of the helium atoms

KNeavg > KHe

avg

Two identical, sealed containers have the same volume. Both containers are filled with the same number of moles of gas at the same temperature and pressure. One of the containers is filled with helium gas and the other is filled with neon gas. Which one of the following statements concerning this situation is true?

a.) The speed of each of the helium atoms is the same value, but this speed is different than that of the neon atoms.

b)

c) The pressure within the container of helium is less than the pressure in the container of neon.

d) The internal energy of the neon gas is greater than the internal energy of the helium gas.

e) The rms speed of the neon atoms is less than that of the helium atoms

KNeavg > KHe

avg

ANSWER

vrms =3RTM

MNe > MHe

vHerms > vNe

rms

A monatomic gas is stored in a container with a constant volume. When the temperature of the gas is T, the rms speed of the atoms is . What is the rms speed when the gas temperature is increased to 3T?

vrms

A.)

B.)

C.)

D.)

E.)

vrms

9vrms

33vrms

3vrms

9vrms

A monatomic gas is stored in a container with a constant volume. When the temperature of the gas is T, the rms speed of the atoms is . What is the rms speed when the gas temperature is increased to 3T?

vrms

A.)

B.)

C.)

D.)

E.)

vrms

9vrms

33vrms

3vrms

9vrms

NEXT WEEK

Distributions

Types of expansions

Degrees of freedom

MEAN FREE PATHHow long does a

molecule travel before colliding?

λ

λ =1

2πd2N/V

m m

d

d d

ASSUMPTIONS FOR λ

Assume molecule has and all other molecules at rest

Assume molecules are sphere with diameter d

Assume collisions collisions occur when molecules are with d of each other

v = const

DENSITY AND MEAN FREE PATH

λ =1

2πd2N/VRecall ρ = N/V

λ =1

2πd2ρ

As ⇑ ρ, ⇓ λ

As ⇑ d, ⇓ λ

(more collisions)

(bigger molecules)

DISTRIBUTIONS OF MOLECULAR SPEEDS

What fraction of molecules have

What fraction of molecules have

v > vrms?

v > 2vrms?

MAXWELL’S SPEED DISTRIBUTION

P(v) = 4π ( M2πRT )

3/2

v2e− Mv22RTProbability distribution

∫∞

0P(v)dv = 1

This means the probability to find a molecule traveling at a certain speed

If you integrate over all velocities, the probability to find the particle has to

be one.

DISTRIBUTION OF SPEEDS

vavg

vrms

vmode

DISTRIBUTION OF HEIGHT (NBA)

DISTRIBUTION OF INCOME

SPEED DISTRIBUTION

Closed containers A and B both contain helium gas at the same temperature. There are n atoms in container A and 2n atoms in container B. At time t = 0 s, all of the helium atoms have the same kinetic energy. The atoms have collisions with each other and with the walls of the container. After a long time has passed, which of the following statements will be true?

a) The atoms in both containers have the same kinetic energies they had at time t = 0 s.

b) Both containers have a wide range of speeds, but the distributions of speeds are the same for both A and B.

c) Average kinetic energy for atoms in container B is greater than A.

d) Average kinetic energy for atoms in container A is greater than B.

e) Both containers have a wide range of speeds, but the distributions of speeds has a greater range for container B than that for container A

Closed containers A and B both contain helium gas at the same temperature. There are n atoms in container A and 2n atoms in container B. At time t = 0 s, all of the helium atoms have the same kinetic energy. The atoms have collisions with each other and with the walls of the container. After a long time has passed, which of the following statements will be true?

a) The atoms in both containers have the same kinetic energies they had at time t = 0 s.

b) Both containers have a wide range of speeds, but the distributions of speeds are the same for both A and B.

c) Average kinetic energy for atoms in container B is greater than A.

d) Average kinetic energy for atoms in container A is greater than B.

e) Both containers have a wide range of speeds, but the distributions of speeds has a greater range for container B than that for container A

ANSWER

Kavg =32

kT

vrms =3RTM

M =Msamp

n

vrms =3nRTMsamp

Kinetic energy must be the same at the same T

The rms speeds are not the same. But the distribution is the same.

HOW TO USE A DISTRIBUTION

P(v) = 4π ( M2πRT )

3/2

v2e− Mv22RT∫

∞

0P(v)dv = 1We know

What fraction of molecules have v > vrms?

∫∞

vrms

P(v)dv = 1

What fraction of molecules have v1 < v < v2?

∫v2

v1

P(v)dv = 1

EXAMPLE

Given a simpler distribution P(v) = A ⋅ e−Av

Is this normalized? i. e. ∫∞

0P(v)dv = 1

∫∞

0A ⋅ e−Avdv = A∫

∞

0e−Avdv

= A−1A

e−Av |∞0 = − e−Av |∞

0 = − 0 + 1 = 1

WHAT FRACTION ARE 2 TIMES THE RMS?

∫∞

2vrms

P(v)dv = ∫∞

2vrms

A ⋅ e−Avdv

We just found that the generic form of this integral is

∫ A ⋅ e−Avdv = − e−Av

putting in the new bounds we find

−e−Av |∞2vrms

= 0 + e−2Avrms = e−2Avrms

AVERAGE VS. RMS

In order to get the average

vavg = ∫∞

0vP(v)dv

Add in velocity to get the average

vrms = ∫∞

0v2P(v)dv

In order to get the root mean squared

Add in velocity^2 to get the rms

These are known as moments of the

distribution. The average is the first moment and the root mean squared is the second moment.

INTEGRATION BY PARTS

∫ udx = ux − ∫ xdu

∫ Av ⋅ e−Avdv = A ∫ v ⋅ e−Avdv

integrate this

∫ v ⋅ e−Avdv = v−1A

e−Av − ∫−1A

e−Avdv

Integration by parts

= −vA

e−Av +1A ∫ e−Avdv = −

vA

e−Av −1

A2e−Av

FINDING THE AVERAGE

∫ Av ⋅ e−Avdv = A∫ v ⋅ e−Avdv = A (−vA

e−Av −1

A2e−Av)

vavg = − (v +1A ) e−Av

vrms = ∫∞

0v2P(v)dv Something to try at

home

FOR THE MAXWELL’S DISTRIBUTION

vavg = ∫∞

0vP(v)dv

vrms = ∫∞

0v2P(v)dv

vavg =8RTπm

vrms =3RTM

MODE OF THE DISTRIBUTION

vmode

mode=maximum of the distribution

dP(v)dv

= 0

After some math

vp =2RTM

Most probable speed from Maxwell’s speed

distribution

MOLECULES IN THE TAIL

v ≫ vrms

Particles in the tail of the distribution

Rain. High speed tail of the distribution molecules evaporate -> clouds

INTERNAL ENERGY OF A GAS

Eint = NKavgMonatomic ideal gas N = nNARecall

Kavg =32

kTEint = nNAKavg Recall

Eint = nNA32

kT R = kNARecall

Eint =32

nRTAlso only dependent

on T

The assumption here is that E only depends on the translational KE and not also rotational (needed for molecules)

ISOCHORIC EXPANSION dV = 0

MOLAR SPECIFIC HEAT AT CONSTANT VOLUME

Q = nCvΔTHeat vs. T

Molar specific heat at a constant volumeTo relate to the internal energy, 1st law of thermodynamics

ΔEint = Q − W

Eint =32

nRT

ΔEint = nCvΔT − W

W = 0Isochoric expansion

ΔEint = nCvΔT

CONTINUED

ΔEint =32

nRΔTΔEint = nCvΔT Recall

32

nRΔT = nCvΔT

Cv =32

R

Cv = 12.5J

mol ⋅ K

 monatomic gas in a constant-volume process

OTHER SPECIFIC HEATS

Eint = nCVT

A change in the internal energy Eint of a confined ideal gas depends only on the change in the temperature, not on

what type of process produces the change.

CHANGE IN INTERNAL ENERGY

ΔEint = ΔnCVT

Regardless of the process, all have the same change in the internal energy

ISOBARIC EXPANSION dp = 0

Q = nCpΔT

Molar specific heat at a constant pressure

ΔEint = Q − W

What is work for an isobaric expansion?

W = pΔVFrom last week

MOLAR SPECIFIC HEAT AT CONSTANT PRESSURE

Q = nCpΔTΔEint = Q − W

W = pΔV = nRΔT

substituting in

ΔEint = nCpΔT − W ΔEint = nCvΔTrecall

nCvΔT = nCpΔT − WIdeal gas law

substituting in

nCvΔT = nCpΔT − nRΔT

Cv = Cp − R

SPECIFIC HEAT

Cv = Cp − R

Specific heat at constant volume vs. constant pressure

Cp > CV

Q = nCΔTRecalling

For a constant volume it’s easier to change the temperature than for a constant pressure!

In other words for p=const, you lose energy to work. For V=const you don’t lose energy to work.

TYPES OF MOTION OF MOLECULES

translational motions (moving left-right and up-down)rotational motions (spinning about an axis like a top)oscillatory motions (atoms oscillating towards/aways from one another

DEGREES OF FREEDOMdegrees of freedom (f), independent ways in which molecules can store energy.

Each such degree of freedom has associated with it an average energy

Kmolecule =12

kT Kmol =12

RT

Avg. energy per molecule Avg. energy per mole

Ktot =12

nRT

SPECIFIC HEAT & DEGREES OF FREEDOM

For 3 spacial dimensions (translational degrees of freedom)

Eint =32

nRT

Generic f

Then, the specific heat is Cv =32

R →f2

R

Eint =32

nRT →f2

nRT

Cp = CV + RSince

Cp =f2

R + R

PHASES OF MATTER AND DEGREES OF FREEDOM

DEGREES OF FREEDOM

ADIABATIC (Q=0)

Example sound waves through air

(very quick or well-insulated)

DURING AN ADIABATIC PROCESS

pVγ = const where γ =Cp

CV

Before and after an adiabatic process

piVγi = pfV

γf

Substituting in the ideal gas law p =nRT

V

nR⏟const

TVγ−1 = const TVγ−1 = const

TiVγ−1i = TfV

γ−1f

Summary of processes

FREE EXPANSION

Free expansion is a different type of adiabatic process (previous equations don’t apply)

Ti = Tf

Then for an ideal class if there is no change in the temperature

piVi = pfVf

NEXT WEEK

Last regular class of the semester (review follows)

Second law of thermodynamics

Refrigerators

Carnot cycle

Microscopic view