physics 2a chapter 13

• Temperature ~ Average KE of each particle

• Particles have different speeds

• Gas Particles are in constant RANDOM motion

• Average KE of each particle is: 3/2 kT

• Pressure is due to momentum transfer

Speed ‘Distribution’ at

CONSTANT Temperature

is given by the

Maxwell Boltzmann

Speed Distribution

© 2013 Pearson Education, Inc.

A single molecule follows a

zig-zag path through a gas

as it collides with other

molecules.

The average distance

between the collisions is

called the mean free path:

(N/V) is the number density of the gas in m−3.

r is the the radius of the molecules when modeled as

hard spheres; for many common gases r ≈ 10−10 m.

Mean Free Path

Slide 18-20


The temperature of a rigid container of oxygen gas

(O2) is lowered from 300C to 0C. As a result, the

mean free path of oxygen molecules

A. Increases.

B. Is unchanged.

C. Decreases.

QuickCheck 18.1

Slide 18-21


The temperature of a rigid container of oxygen gas

(O2) is lowered from 300C to 0C. As a result, the

mean free path of oxygen molecules

A. Increases.

B. Is unchanged.

C. Decreases.

QuickCheck 18.1

Slide 18-22

λ depends only on N/V, not T.

Pressure and Kinetic Energy

• Assume a container is a cube with edges d.

• Look at the motion of the molecule in terms of its velocity components and momentum and the average force

• Pressure is proportional to the number of molecules per unit volume (N/V) and to the average translational kinetic energy of the molecules.

• This equation also relates the macroscopic quantity of pressure with a microscopic quantity of the average value of the square of the molecular speed

• One way to increase the pressure is to increase the number of molecules per unit volume

• The pressure can also be increased by increasing the speed (kinetic energy) of the molecules

___22 1

3 2o

NP m v

V

Molecular Interpretation of

Temperature• We can take the pressure as it relates to the kinetic

energy and compare it to the pressure from the

equation of state for an ideal gas

• Temperature is a direct measure of the average

molecular kinetic energy

___2 B2 1

3 2

N nRT Nk TP mv

V V V

___2

B

1 3

2 2om v k T

2 3rms

kTv v

m Root-mean-square speed:

The Kelvin Temperature of

an ideal gas is a measure of

the average translational

kinetic energy per particle:

k =1.38 x 10-23 J/K Boltzmann’s Constant

___2

tot trans B

1 3 3 3

2 2 2 2K N mv Nk T nRT PV

Total Kinetic Energy

• The total kinetic energy is just N times the kinetic

energy of each molecule

• If we have a gas with only translational energy, this is

the internal energy of the gas

• This tells us that the internal energy of an ideal gas

depends only on the temperature

___2

tot trans B

1 3 3

2 2 2K N mv Nk T nRT

Kinetic Theory Problem

A 5.00-L vessel contains nitrogen gas at

27.0C and 3.00 atm. Find (a) the total

translational kinetic energy of the gas

molecules and (b) the average kinetic energy

per molecule.

Kinetic Theory ProblemCalculate the RMS speed of an oxygen molecule

in the air if the temperature is 5.00 °C.

The mass of an oxygen molecule is 32.00 u

(k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)

3rms

kTv

m What is m?

m is the mass of one

oxygen molecule in kg.

What is u?

How do we get the mass in kg?

Kinetic Theory ProblemCalculate the RMS speed of an oxygen molecule

in the air if the temperature is 5.00 °C.

The mass of an oxygen molecule is 32.00 u

(k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)

3rms

kTv

m

23

27

3(1.38 10 / )278

(32 )(1.66 10 / )

x J K K

u x kg u

466 /m s

What is m?m is the mass of one

oxygen molecule.

Is this fast? YES!Speed of

sound:

343m/s!

Distribution of Molecular Speeds• The observed speed distribution of gas

molecules in thermal equilibrium is

shown at right

• NV is called the Maxwell-Boltzmann

speed distribution function

• mo is the mass of a gas molecule, kB is

Boltzmann’s constant and T is the

absolute temperature

Ludwig Boltzmann

1844 – 1906

Austrian physicist

Contributed to

Kinetic Theory of Gases

Electromagnetism

Thermodynamics

Pioneer in statistical mechanics

2

3 / 2

/ 22

B

42

Bmv k ToV

mN N v e

k T


A rigid container holds both hydrogen gas (H2)

and nitrogen gas (N2) at 100C. Which statement

describes their rms speeds?

A. vrms of H2 < vrms of N2.

B. vrms of H2 = vrms of N2.

C. vrms of H2 > vrms of N2.

QuickCheck 18.3

Slide 18-34


A rigid container holds both hydrogen gas (H2)

and nitrogen gas (N2) at 100C. Which statement

describes their rms speeds?

A. vrms of H2 < vrms of N2.

B. vrms of H2 = vrms of N2.

C. vrms of H2 > vrms of N2.

QuickCheck 18.3

Slide 18-35

More Kinetic Theory Problems

A gas molecule with a molecular mass of 32.0 u has a speed of 325 m/s. What is the temperature of the gas molecule?

A) 72.0 K B) 136 K C) 305 K D) 459 K

E) A temperature cannot be assigned to a single molecule.

Temperature ~ Average KE of all particles

Molecular Interpretation of

Temperature

• Simplifying the equation relating

temperature and kinetic energy gives

• This can be applied to each direction,

– with similar expressions for vy and vz

___2

B

1 3

2 2om v k T

___2

B

1 1

2 2xmv k T

Equipartition of Energy• Each translational degree of freedom contributes an

equal amount to the energy of the gas

– In general, a degree of freedom refers to an independent means by which a molecule can possess energy

• Each degree of freedom contributes ½kBT to the

energy of a system, where possible degrees of

freedom are those associated with translation,

rotation and vibration of molecules

Monatomic and Diatomic GasesThe thermal energy of a monatomic gas of N atoms is

A diatomic gas has more thermal energy than a monatomic

gas at the same temperature because the molecules have

rotational as well as translational kinetic energy.

Molar Specific HeatsIsobaric requires MORE HEAT than Isochoric for the

same change in Temperature!!!!

The total change in thermal

energy for ANY PROCESS,

due to work and heat, is:

This applies to all ideal gases, not

just monatomic ones! WOW!

CP and CV Note that for all ideal gases:

whereR = 8.31 J/mol K is the universal gas constant.

Slide 17-80

Important Concepts

Agreement with Experiment: Diatomic Hydrogen

acts like a monatomic gas at low temperature!

•Molar specific heat is a function of

temperature.

•At low temperatures, a diatomic gas

acts like a monatomic gas.

– CV = 3/2 R

•At about room temperature, the value

increases to CV = 5/2 R.

– This is consistent with adding

rotational energy but not

vibrational energy.

•At high temperatures, the value

increases to CV = 7/2 R.

– This includes vibrational

energy as well as rotational

and translational.

Quantization of Energy.

•To explain the results of the various molar

specific heats, we must use some quantum

mechanics.

– Classical mechanics is not sufficient

•This energy level diagram shows the rotational

and vibrational states of a diatomic molecule.

•The lowest allowed state is the ground state.

•The vibrational states are separated by larger

energy gaps than are rotational states.

•At low temperatures, the energy gained during

collisions is generally not enough to raise it to the

first excited state of either rotation or vibration.

Section 21.4

In a constant-volume process, 209 J of

energy is transferred by heat to 1.00 mol

of an ideal monatomic gas initially at

300 K. Find (a) the increase in internal

energy of the gas, (b) the work done on

it, and (c) its final temperature

What additional kind of energy makes CV larger

for a diatomic than for a monatomic gas?

A. Charismatic energy.

B. Translational energy.

C. Heat energy.

D. Rotational energy.

E. Solar energy.

Reading Question 18.2

Slide 18-12

What additional kind of energy makes CV larger

for a diatomic than for a monatomic gas?

A. Charismatic energy.

B. Translational energy.

C. Heat energy.

D. Rotational energy.

E. Solar energy.


Slide 18-13

An adiabatic process is one for which:

where:

Adiabats are steeper than

hyperbolic isotherms because

only work is being done to

change the Temperature. The

temperature falls during an

adiabatic expansion, and rises

during an adiabatic

compression.

Adiabatic Processes

Slide 17-88

Adiabatic Processes: Q=0

Ti Vig-1 = Tf Vf

g-1

A 4.00-L sample of a nitrogen gas confined to a

cylinder, is carried through a closed cycle. The

gas is initially at 1.00 atm and at 300 K. First, its

pressure is tripled under constant volume. Then,

it expands adiabatically to its original pressure.

Finally, the gas is compressed isobarically to its

original volume. (a) Draw a PV diagram of this

cycle. (b) Find the number of moles of the gas. (c)

Find the volumes and temperatures at the end of

each process (d) Find the Work and heat for each

process. (e) What was the net work done on the

gas for this cycle?

Adiabatic Processes for an

Ideal Gas• An adiabatic process is one in which no energy is

transferred by heat between a system and its surroundings (think styrofoam cup)

• Assume an ideal gas is in an equilibrium state and so PV = nRT is valid

• The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV g = constant

g = CP / CV is assumed to be constant

All three variables in the ideal gas law (P, V, T ) can change during an adiabatic process

Special Case: Adiabatic Free

Expansion

• This is an example of adiabatic free expansion

• The process is adiabatic because it takes place in an insulated container

• Because the gas expands into a vacuum, it does not apply a force on a piston and W = 0

• Since Q = 0 and W = 0, DEint = 0 and the initial and final states are the same and no change in temperature is expected. – No change in temperature is expected

The second law of thermodynamics says that

A. The entropy of an isolated system never

decreases.

B. Heat never flows spontaneously from

cold to hot.

C. The total thermal energy of an isolated

system is constant.

D. Both A and B.

E. Both A and C.


Slide 18-15

A. Both microscopic and macroscopic processes

are reversible.

B. Both microscopic and macroscopic processes

are irreversible.

C. Microscopic processes are reversible and

macroscopic processes are irreversible.

D. Microscopic processes are irreversible and

macroscopic processes are reversible.

In general,


Slide 18-16

Systems A and B are both monatomic gases. At this instant,

A. TA > TB.

B. TA = TB.

C. TA < TB.

D. There’s not enough information to compare their

temperatures.

QuickCheck 18.6

Slide 18-51

Systems A and B are both monatomic gases. At this instant,

A. TA > TB.

B. TA = TB.

C. TA < TB.

D. There’s not enough information to compare their

temperatures.

QuickCheck 18.6

Slide 18-52

A has the larger average energy per atom.

physics 2a chapter 13

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