LECTURE 10- KINETIC THEORY 1Phys 124H- Honors Analytical Physics IBChapter 19Professor Noronha-Hostler

FIRST IMAGES OF A BLACK HOLE!

MAKE UP EXAM

Due next Friday BEFORE class.

Handwritten, carefully write out your steps (free body diagrams)

Make sure you write down the final answer

Put your NAME, ID on the test

Number according to the master exam

REVIEW HEAT CAPACITY

Q = CΔT ΔT =QC

Thermodynamics covers the exchange of energy from a system to the environment

p, V, TVariables used to describe a system:

So far we’ve mostly covered solids and liquid (and gases that act like fluids)

SPECIFIC HEAT CAPACITY DEMO

Wood = 1.76 J/g∘C

Metals < 1 J/g∘C

Styrofoam = 1.1 J/g∘CΔT =

Qcm

TODAY’S OBJECTIVES

Kinetic theory

Avogadro’s number

Ideal gas law

Distributions

PROPERTIES OF A GAS

How do we measure the amount of a gas?

NA = 6.02 ⋅ 1023 mol−1

A mole is the number of atoms in 12g of 12C

Number of atoms/molecules per moleAvogadro’s number

 All gases occupying the same volume under the same conditions of temperature and pressure contain the same number of atoms or

molecules

NUMBER OF MOLES

n =NNANumber of moles

Number of molecules

n =Msamp

M

n =Msamp

mNA

Mass of sample

Molar mass

Molecular mass

IDEAL GAS LAW (BARREL DEMO)

pV = nRT

Holds for any gas with a low density (assume molecules don’t interact)

Absolute pressure

Number of moles of gas

Gas constant R = 8.31J

mol ⋅ K

Temperature [K]

Under which of the following circumstances does a real gas behave like an ideal gas?

a) The gas particles move very slowly.

b) The gas particles do not collide with each other very often.

c) The interaction between the gas particles is negligible.

d) The interaction between the gas particles and the walls of the container is negligible.

e) There are only one kind of particles in the container.

Under which of the following circumstances does a real gas behave like an ideal gas?

a) The gas particles move very slowly.

b) The gas particles do not collide with each other very often.

c) The interaction between the gas particles is negligible.

d) The interaction between the gas particles and the walls of the container is negligible.

e) There are only one kind of particles in the container.

An ideal gas is enclosed within a container by a moveable piston. If the final temperature is two times the initial temperature and the volume is reduced to one-fourth of its initial value, what will the final pressure of the gas be relative to its initial pressure, P1?

A.)

B.)

C.)

D.)

E.)

8P1

4P1

2P1

P1

2P1

4

An ideal gas is enclosed within a container by a moveable piston. If the final temperature is two times the initial temperature and the volume is reduced to one-fourth of its initial value, what will the final pressure of the gas be relative to its initial pressure, P1?

A.)

B.)

C.)

D.)

E.)

8P1

4P1

2P1

P1

2P1

4

ANSWER

p1 =nRT1

V1

T2 = 2T1 V2 =V1

4Find p2 when and

p2V1

4= nR2T1

p2 =nR8T1

V1= 8p1

NUMBER OF MOLECULESDepending on the situation, sometimes it may be better to write in

terms of the number of molecules, N.

n =NNA

Recall

where the Boltzmann constant is k =RNA

= 1.38 ⋅ 10−23 J/K

pV = nRT pV = NRNA

T

pV = NkT

ISOTHERMAL EXPANSION

Remove lead shots Vi → Vf

T = constIsothermal

Isothermal expansion- volume increases while the temperature remains constant

p = nRT⏟=const

1V

Pressure is inversely related to the volume

WORK DONE BY IDEAL GAS

W = ∫Vf

Vi

pdVRecall from last week

Substituting in p = nRT1V

W = ∫Vf

Vi

nRT1V

dV = nRT∫Vf

Vi

1V

dV

W = nRT ln V |VfVi

= nRT (ln Vf − ln Vi) = nRT lnVf

Vi

ISOTHERMAL WORK

W = nRT lnVf

Vi

Expansion Vf > Vi

Compression Vf < Vi

ln [ > 1] → W > 0

ln [ < 1] → W < 0

If the system is expanding, the work done by the system is positive

If the system is compressing, the work done by the system is negative

WHICH P-V DIAGRAM DOES THE MOST WORK?

A.) A

B.) B

C.) C

D.) All equal

E.) not enough information

AB C

WHICH P-V DIAGRAM DOES THE MOST WORK?

A.) A

B.) B

C.) C

D.) All equal

E.) not enough information

AB C

WHAT IF THE TEMPERATURE CHANGES?

W = ∫Vf

Vi

nRT1V

dV

It’s no longer possible to pull out T.

Special case 1: Constant volume (isochoric)

dV = 0 W = 0

Special case 2: Constant pressure (isobaric) W = ∫Vf

Vi

pdV

ISOBARIC- CONSTANT PRESSURE

W = ∫Vf

Vi

pdV

W = p∫Vf

Vi

dV

W = pV |VfVi

= p(Vf − Vi)

W = pΔV

KINETIC THEORY (DEMO)

mv

L

L

n Moles

We assume the density is low enough that particles only

collide with the walls

How do we describe the motion of the particles inside the box?

KINETIC THEORY

mv

L

L

n M

For a collision with the wall the particle’s change in momentum is

Δpx = pxf − pxi

= mvx − (−mvx) = 2mvx

Particles continually hit the wall, time to travel back and forth between walls Δt =

2Lvx

Fx =Δpx

Δt=

2mvx

2L/vx

F =mvx

LForce of an individual particle colliding against the wall

momentum

PRESSURE OF ALL PARTICLES

Recall the pressure is

p =Fx

A then for this box of length L p =Fx

L2

Counting up contributions from all the particles in the box in the x direction

ptot =Fx,tot

L2=

mv21,x /L + mv2

2,x /L + … + mv2N,x /L

L2

ptot =m (v2

1,x + v22,x + … + v2

N,x)L3

PRESSURE CONTINUED

px =m (v2

1,x + v22,x + … + v2

N,x)L3

V = L3

px =mV (v2

1,x + v22,x + … + v2

N,x)(v2

x )avg ≡1N ∑

N

v2x

px =mN(v2

x )avg

v

ROOT MEAN SQUARED

p =mN(v2

x )avg

V=

mN(v2)avg

3V

Root Mean Squared vrms ≡ (v2)avg v2rms = (v2)avg

For all three dimensions v2 = 3v2x

v2x = v2/3

p =mNv2

rms

3V

MACROSCOPIC VS. MICROSCOPIC

p =mNv2

rms

3VPressure is macroscopic, can easily measure the pressure of a tube of gas

Volume is macroscopic

p =mNv2

rms

3VThe molecular mass, number of molecules,

and root mean squared of the particle velocity are all microscopic quantities. We need special instruments to measure them!

How can we write the pressure in terms of the molar mass, M?

A.)

B.)

C.)

D.)

E.)

p =MNv2

rms

3V

p =nMv2

rms

3V

p = nMv2rms

p =nMv2

rms

3Vmp =

nMNAv2rms

3V

WRITE IN TERMS OF MOLAR MASSNumber of moles Number of moleculesn =

NNA

M = mNAMolar Mass

p =mNv2

rms

3V=

mnNAv2rms

3V

p =nMv2

rms

3V

VOLUME DEPENDENCE

p =nMv2

rms

3V⇓ V , ⇑ p

BOYLE’S LAW

pV=nRT

If the temperature is held constant, then the right hand side is constant.

Then at T=const, if the pressure or volume is changed

p1V1 = p2V2

SPEED OF AN IDEAL GAS

p =nMv2

rms

3VpV = nRT

Ideal Gas LawPressure from kinetic theory

pV =13

nMv2rmsRearranging

Substitute

13

nMv2rms = nRT

v2rms =

3RTM

vrms =3RTM

MOLECULE ROOT MEAN SQUARED SPEEDS

vrms =3RTM

The root mean squared of particles depends only on the temperature and

their molar mass

Particles move extremely quickly! Faster than the speed of sound (v=343 m/s)

v < vrms

AVERAGE VS. RMS

vrms ≥ vavg

This always holds for the root mean squared over the average.Given the set of velocities: v={4,5,10,1,13}

vavg =4 + 5 + 10 + 1 + 13

5vrms =

42 + 52 + 102 + 12 + 132

5

= 6.6 = 7.9

WHAT IS THE AVERAGE AND RMS OF v = {0.1, 39.5, 0.02, 0.03, 0.05, 0.001}

A.)

B.)

C.)

D.)

E.)

vavg = 6.6

vavg = 10

vavg = 1.6

vavg = 16

vavg = 1.1

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

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

ISOCHORIC EXPANSION

MOLECULES

PHASES

DEGREES OF FREEDOM

ISOBARIC EXPANSION

ADIABATIC