• 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

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

B

2 1

3 2

NP mv nRT Nk T

V

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

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.

Hot Question Suppose you apply a flame to 1 liter of water for a certain

time and its temperature rises by 10 degrees C. If you apply

the same flame for the same time to 2 liters of water, by how

much will its temperature rise?

a) 1 degree b) 5 degrees c) 10 degrees d) zero degrees

Ludwig Boltzmann or Dean Gooch?

• 1844 – 1906

• Austrian physicist

• Contributed to

– Kinetic Theory of Gases

– Electromagnetism

– Thermodynamics

• Pioneer in statistical

mechanics

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

2

3 / 2

/ 22

B

42

Bmv k ToV

mN N v e

k T

Molecular Speeds and

Collisions

Speed Summary

• Root mean square speed

• The average speed is somewhat lower than the rms speed

• The most probable speed, vmp is the speed at which the distribution curve reaches a peak

• vrms > vavg > vmp

B Brms

31.73

o o

k T k Tv

m m

B Bmp

21.41

k T k Tv

m m

B B

avg

81.60

o o

k T k Tv

m m

Some Example vrms Values

At a given temperature, lighter molecules move faster, on

the average, than heavier molecules

Speed Distribution • The peak shifts to the right

as T increases – This shows that the average

speed increases with increasing temperature

• The asymmetric shape occurs because the lowest possible speed is 0 and the highest is infinity

23/ 2 1/ 2 rmskT KE mv

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

Kinetic Theory Problem Calculate 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 Problem Calculate 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!

A cylinder contains a mixture of helium

and argon gas in equilibrium at 150°C.

(a) What is the average kinetic energy

for each type of gas molecule?

(b) What is the root-mean-square speed

of each type of molecule?

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

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

• With complex molecules, other contributions to

internal energy must be taken into account

• One possible energy is the translational motion of

the center of mass

• Rotational motion about the various axes also

contributes

• There is kinetic energy and potential energy

associated with the vibrations

Monatomic and Diatomic Gases The 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 Heat

• We define specific heats for two processes

that frequently occur:

– Changes with constant pressure

– Changes with constant volume

• Using the number of moles, n, we can

define molar specific heats for these

processes

• Molar specific heats:

– Q = nCV DT for constant-volume processes

– Q = nCP DT for constant-pressure processes

Ideal Monatomic Gas

• Therefore, DEint = 3/2 nRT

DE is a function of T only

• In general, the internal energy of an ideal

gas is a function of T only

– The exact relationship depends on the type of

gas

• At constant volume, Q = DEint = nCV DT

– This applies to all ideal gases, not just

monatomic ones

Ratio of Molar Specific Heats

• We can also define the ratio of molar specific heats

• Theoretical values of CV , CP , and g are in excellent agreement for monatomic gases

• But they are in serious disagreement with the values for more complex molecules

– Not surprising since the analysis was for monatomic gases

5 / 21.67

3 / 2P

V

C R

C Rg

Agreement with Experiment • 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

Sample Values of Molar Specific

Heats

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

Molar Specific Heats of Other

Materials

• The internal energy of more complex gases

must include contributions from the

rotational and vibrational motions of the

molecules

• In the cases of solids and liquids heated at

constant pressure, very little work is done,

since the thermal expansion is small, and CP

and CV are approximately equal

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

Adiabatic Process

• The PV diagram shows

an adiabatic expansion

of an ideal gas

• The temperature of the

gas decreases

– Tf < Ti in this process

• For this process

Pi Vig = Pf Vf

g and

Ti Vig-1 = Tf Vf

g-1

A 2.00-mol sample of a diatomic ideal

gas expands slowly and adiabatically

from a pressure of 5.00 atm and a

volume of 12.0 L to a final volume of

30.0 L.

(a) What is the final pressure of the gas?

(b) What are the initial and final

temperatures?

(c) Find Q, W, and DEint.

Cyclic Processes

• A cyclic process is one that starts and ends in the same state

– On a PV diagram, a cyclic process appears as a closed curve

• DEint = 0, Q = -W

• In a cyclic process, the net work done on the system per cycle equals the area enclosed by the path representing the process on a PV diagram

intE Q WD

Isothermal Process • At right is a PV diagram of an isothermal

expansion

• The curve is a hyperbola

• The curve is called an isotherm

• The curve of the PV diagram

indicates PV = constant

– The equation of a hyperbola

• Because it is an ideal gas and the

process is quasi-static,

PV = nRT and

f f f

i i i

V V V

V V V

nRT dVW P dV dV nRT

V V

ln i

f

VW nRT

V

Isothermal Process

• An isothermal process is one that occurs at

a constant temperature

• Since there is no change in temperature,

DEint = 0

• Therefore, Q = - W

• Any energy that enters the system by heat

must leave the system by work

intE Q WD f

i

V

VW P dV PV nRT

Isobaric Processes

• An isobaric process is one that occurs at a

constant pressure

• The values of the heat and the work are

generally both nonzero

• The work done is W = P (Vf – Vi) where P

is the constant pressure

intE Q WD f

i

V

VW P dV PV nRT

Isovolumetric Processes

• An isovolumetric process is one in which there is

no change in the volume

• Since the volume does not change, W = 0

• From the first law, DEint = Q

• If energy is added by heat to a system kept at

constant volume, all of the transferred energy

remains in the system as an increase in its internal

energy

intE Q WD f

i

V

VW P dV PV nRT

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

Thermo Processes • Adiabatic

– No heat exchanged

– Q = 0 and DEint = W

• Isobaric

– Constant pressure

– W = P (Vf – Vi) and DEint = Q + W

• Isovolumetric

– Constant Volume

– W = 0 and DEint = Q

• Isothermal

– Constant temperature

DEint = 0 and Q = -W

intE Q WD

ln i

f

VW nRT

V

A gas is taken through the cyclic process as shown.

(a) Find the net energy transferred to the system by heat during one complete cycle. (b) What If? If the cycle is reversed—that is, the process follows the path ACBA—what is the net energy input per cycle by heat?

intE Q WD f

i

V

VW P dV

A sample of an ideal gas goes through the process as shown. From A to B, the process is adiabatic; from B to C, it is isobaric with 100 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kJ of energy leaving the system by heat. Determine the difference in internal energy E(B) – E(A).

intE Q WD f

i

V

VW P dV PV nRT

Important Concepts

Heat Engine

• A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work

• A heat engine carries some working substance through a cyclical process

• The working substance absorbs energy by heat from a high temperature energy reservoir (Qh)

• Work is done by the engine (Weng)

• Energy is expelled as heat to a lower temperature reservoir (Qc)

DEint = 0 for the entire cycle

eng

h

We

Q

DEint = 0 for the entire cycle

Thermal Efficiency of a Heat

Engine

• Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature

• We can think of the efficiency as the ratio of what you gain to what you give

eng1h c c

h h h

W Q Q Qe

Q Q Q

DEint = 0 for the entire cycle

Rank in order, from largest to smallest, the work Wout

performed by these four heat engines.

A. Wb > Wa > Wc > Wd

B. Wb > Wa > Wb > Wc

C. Wb > Wa > Wb = Wc

D. Wd > Wa = Wb > Wc

E. Wd > Wa > Wb > Wc

eng1h c c

h h h

W Q Q Qe

Q Q Q

Rank in order, from largest to smallest, the work Wout

performed by these four heat engines.

A. Wb > Wa > Wc > Wd

B. Wb > Wa > Wb > Wc

C. Wb > Wa > Wb = Wc

D. Wd > Wa = Wb > Wc

E. Wd > Wa > Wb > Wc

eng1h c c

h h h

W Q Q Qe

Q Q Q

Perfect Heat Engine

• No energy is expelled to the cold reservoir

• It takes in some amount of energy and does an equal amount of work

• e = 100%

• It is an impossible engine

Could this heat

engine be built?

A. Yes.

B. No.

C. It’s impossible to tell without knowing

what kind of cycle it uses.

e

H C

H

W Q Q

W

Q

and 1c c c

c

h h h

Q T Te

Q T T

Could this heat

engine be built?

A. Yes.

B. No.

C. It’s impossible to tell without knowing

what kind of cycle it uses.

e

H C

H

W Q Q

W

Q

and 1c c c

c

h h h

Q T Te

Q T T

Analyze this engine to determine (a) the net work done per cycle, (b) the engine’s thermal efficiency and (c) the engine’s power output if it runs at 600 rpm. Assume the gas is monatomic and follows the ideal-gas process above.

Gasoline Engine

• In a gasoline engine, six processes occur

during each cycle

• For a given cycle, the piston moves up and

down twice

• This represents a four-stroke cycle

• The processes in the cycle can be

approximated by the Otto cycle

The Conventional Gasoline

Engine

Gasoline Engine – Intake Stroke

• During the intake stroke,

the piston moves

downward

• A gaseous mixture of air

and fuel is drawn into the

cylinder

• Energy enters the system

as potential energy in the

fuel

Gasoline Engine – Compression

Stroke

• The piston moves upward

• The air-fuel mixture is

compressed adiabatically

• The temperature increases

• The work done on the gas is

positive and equal to the

negative area under the curve

Gasoline Engine – Spark

• Combustion occurs when the

spark plug fires

• This is not one of the strokes

of the engine

• It occurs very quickly while

the piston is at its highest

position

• Conversion from potential

energy of the fuel to internal

energy

Gasoline Engine – Power Stroke

• In the power stroke, the gas expands adiabatically

• This causes a temperature drop

• Work is done by the gas

• The work is equal to the area under the curve

Gasoline Engine – Valve Opens

An exhaust valve opens as the piston reaches

its bottom position

• The pressure drops suddenly

• The volume is approximately constant

– So no work is done

• Energy begins to be expelled from the

interior of the cylinder

Gasoline Engine – Exhaust

Stroke

• In the exhaust stroke, the

piston moves upward

while the exhaust valve

remains open

• Residual gases are

expelled to the

atmosphere

• The volume decreases

Otto Cycle

The Otto cycle

approximates the

processes occurring in an

internal combustion

engine

Otto Cycle Efficiency

• If the air-fuel mixture is assumed to be an

ideal gas, then the efficiency of the Otto cycle

is

g is the ratio of the molar specific heats

• V1 / V2 is called the compression ratio

1

1 2

11e

V Vg

Otto Cycle Efficiency, cont

• Typical values:

– Compression ratio of 8

g = 1.4

– e = 56%

• Efficiencies of real engines are 15% to 20%

– Mainly due to friction, energy transfer by conduction, incomplete combustion of the air-fuel mixture

g

D34. The compression ratio of an Otto cycle, as shown in Figure 22.13, is VA/VB = 8.00. At the beginning A of the compression process, 500 cm3 of gas is at 100 kPa and 20.0C. At the beginning of the adiabatic expansion the temperature is TC = 750C. Model the working fluid as an ideal gas with Eint = nCVT = 2.50nRT and

= 1.40. (a) Fill in this table to follow the states of the gas:

T (K) P (kPa) V (cm3) Eint

A 293 100 500

B

C 1 023

D

A

(b) Fill in this table to follow the processes:

Q (input) W(output) Eint

AB

BC

CD

DA

ABCDA

Heat Pumps and Refrigerators

• Heat engines can run in reverse

– This is not a natural direction of energy transfer

– Must put some energy into a device to do this

– Devices that do this are called heat pumps or

refrigerators

• Examples

– A refrigerator is a common type of heat pump

– An air conditioner is another example of a heat pump

Coefficient of Performance

• The effectiveness of a heat pump is

described by a number called the

coefficient of performance (COP)

• In heating mode, the COP is the ratio of the

heat transferred in to the work required

energy transferred at high tempCOP =

work done by heat pump

hQ

W

COP, Heating Mode

• COP is similar to efficiency

• Qh is typically higher than W

– Values of COP are generally greater than 1

– It is possible for them to be less than 1

• We would like the COP to be as high as

possible

COP, Cooling Mode

• In cooling mode, you “gain” energy from a

cold temperature reservoir

• A good refrigerator should have a high COP

– Typical values are 5 or 6

COP cQ

W

Carnot Engine – Carnot Cycle

A heat engine operating in an ideal, reversible cycle (now called a

Carnot cycle) between two reservoirs is the most efficient engine

possible. This sets an upper limit on the efficiencies of all other engines

Carnot Cycle, PV Diagram

• The work done by the

engine is shown by the

area enclosed by the

curve, Weng

• The net work is equal

to |Qh| – |Qc|

DEint = 0 for the entire

cycle

Efficiency of a Carnot Engine

• Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs

• Temperatures must be in Kelvins

• All Carnot engines operating between the same two temperatures will have the same efficiency

and 1c c c

c

h h h

Q T Te

Q T T

Carnot’s Theorem

• No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs

– All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle

– The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc.

Notes About Carnot Efficiency

• Efficiency is 0 if Th = Tc

• Efficiency is 100% only if Tc = 0 K

– Such reservoirs are not available

– Efficiency is always less than 100%

• The efficiency increases as Tc is lowered and as Th

is raised

• In most practical cases, Tc is near room

temperature, 300 K

– So generally Th is raised to increase efficiency

Carnot Cycle in Reverse

• Theoretically, a Carnot-cycle heat engine

can run in reverse

• This would constitute the most effective

heat pump available

• This would determine the maximum

possible COPs for a given combination of

hot and cold reservoirs

Carnot Heat Pump COPs

• In heating mode:

• In cooling mode:

C

h h

h c

Q TCOP

W T T

c cC

h c

Q TCOP

W T T

26. A heat pump, shown in Figure P22.26, is

essentially an air conditioner installed backward. It

extracts energy from colder air outside and deposits it in

a warmer room. Suppose that the ratio of the actual

energy entering the room to the work done by the

device’s motor is 10.0% of the theoretical maximum

ratio. Determine the energy entering the room per joule

of work done by the motor, given that the inside

temperature is 20.0°C and the outside temperature is –

5.00°C.

Carnot cycle

10.100 0.100

Carnot efficiency

h hQ Q

W W

293 K0.100 0.100 1.17

293 K 268 K

h h

h c

Q T

W T T

1.17 joules of energy enter the room by heat for each joule of work done.

Can this refrigerator be built?

c cC

h c

Q TCOP

W T T

COP

H C

c

W Q Q

Q

W