# Chapter 10: Thermal Physics

Mechanics

Oct 27, 2013 (4 years and 8 months ago)

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Chapter 10: Thermal Physics

Temperature and the Zeroth law of Thermodynamics

We all know that a glass of ice water (ice and water) will eventually warm to room
temperature if left out, melting the ice in the process. And a cup of hot tea will eventually
co
ol to room temperature. But what is this thing we call "temperature"? I'm sure everyone
has a general idea of temperature, but for physics we need a precise definition.

Consider a cooler in which we place ice and a warm can of soda. The warm soda will
exc
hange energy (in the form of heat) with the ice, eventually cooling to the same
temperature as the ice. We haven't defined what heat is. For now let's just say that heat is
a form of energy which can be transmitted between things. Because the soda can
exch
ange heat with the ice, we say that they are in
thermal contact
. When they stop
exchanging heat energy, we say that they are in
thermal equilibrium
.

So what if we have 2 cans of soda, and we want to know if they are in thermal
equilibrium. How can we tell
? We would take a thermometer (a third object, C), and
place it in thermal contact with the first can of soda (A). After the thermometer and soda
have reached thermal equilibrium, we note the temperature. Then we place the
thermometer in thermal contact wi
th the second can of soda (B). Again, after reaching
thermal equilibrium, we note the temperature. If the two temperatures are the same, then
the two cans of soda are in thermal equilibrium.

This idea leads to the zeroth law of thermodynamics:

"If bodies
A and B are separately in thermal equilibrium with a third body, C, then A and
B will be in thermal equilibrium with each other if placed in thermal contact."

So finally, back to the initial question, what is temperature? "Temperature is the property
that

determines whether an object will be in thermal equilibrium with other objects. Two
objects in thermal equilibrium with each other are at the same temperature."

Thermometers and Temperature Scales

Thermometers are devices for measuring temperature. The c
ommon thermometers consist
of a quantity of mercury or alcohol (glycol) in a thin glass tube. As the temperature
varies, the fluid expands and contracts, and its height in the tube changes. A scale is
supplied to read off the height of the fluid directly i
n terms of temperature.

The simple thermometers described above have certain drawbacks. They don't operate
below the freezing point of the liquid, and thermometers based on different liquids may
not agree at all temperatures. Normally the thermometers are

calibrated for 0°C (32°F)
when ice and water are in equilibrium, and 100°C (212°F) when water and steam are in
equilibrium.

There are three temperature scales you need to be familiar with, the Celsius, Farenheit,
and Kelvin scales. You are probably famil
iar with the Celsius and Farenheit temperature
scales. The Kelvin scale (the SI standard) measures temperature relative to absolute zero.
Absolute zero is the lowest achievable temperature, corresponding to 0°K, or
-
273.15°C.
One kelvin unit equals on cels
ius unit
--

a change of temperature by 1°K equals a change
of temperature by 1°C. The following table summarizes the different scales.

Celsius

Farenheit

Kelvin

Steam point

100°

212°

373.15°

Ice point

32°

273.15°

Absolute zero

-
273.15°

-
459.67°

Conversion from T
C

T
C

T
F

= (9/5)T
C

+ 32

T
K

= T
C

+ 273.15

Thermal Expansion of Solids and Liquids

Common thermometers make use of the expansion of liquids with increasing
temperature. This phenomenon is called thermal expansion.

On a microscopic scale, th
ermal expansion occurs because the average separation of
atoms and molecules increases. The atoms and molecules in a substance are constantly
vibrating, but on average they remain a certain distance apart. If the temperature of the
substance increases, the
n the energy of the vibrations increases, meaning that the size
(amplitude) of the vibrations increases. The result is that the average separation between
atoms and molecules increases.

Empirically, we find that an object of initial length L
0

will undergo

a change of length,

L for a temperature change

T, given by:

L =

L
0

T.

The proportionality constant

is called the coefficient of linear expansion. The
coefficents of linear expansion for a variety of substances near room temperature are
listed in tab
le 10.1 of the text.

Exapmple: P10.3

The New River Gorge bridge in West Virginia is a 518 m long steel arch. How much will
its length change between temperature extremes
-
20°C and 35°C?

From table 10.1 we find the coefficent of linear expansion for steel
to be

= 11x10
-
6
/°C.
If L
0

= 518 m, and

T = 55°C, then

L =

L
0

T = (11x10
-
6
/°C)(518 m)(55°C) = 0.31 m = 31 cm.

Recall:

Thermometers and Temperature Scales

Thermometers are devices for measuring temperature. The common thermometers consist
of a quantity of m
ercury or alcohol (glycol) in a thin glass tube. As the temperature
varies, the fluid expands and contracts, and its height in the tube changes. A scale is
supplied to read off the height of the fluid directly in terms of temperature.

The simple thermomet
ers described above have certain drawbacks. They don't operate
below the freezing point of the liquid, and thermometers based on different liquids may
not agree at all temperatures. Normally the thermometers are calibrated for 0°C (32°F)
when ice and water

are in equilibrium, and 100°C (212°F) when water and steam are in
equilibrium.

There are three temperature scales you need to be familiar with, the Celsius, Fahrenheit,
and Kelvin scales. You are probably familiar with the Celsius and Fahrenheit temperat
ure
scales. The Kelvin scale (the SI standard) measures temperature relative to absolute zero.
Absolute zero is the lowest achievable temperature, corresponding to 0°K, or
-
273.15°C.
One Kelvin unit equals on Celsius unit
--

a change of temperature by 1°K
equals a change

of temperature by 1°C. The following table summarizes the different scales.

Celsius

Fahrenheit

Kelvin

Steam point

100°

212°

373.15°

Ice point

32°

273.15°

Absolute zero

-
273.15°

-
459.67°

Conversion from T
C

T
C

T
F

= (9/5)T
C

+ 32

T
K

= T
C

+ 273.15

Thermal Expansion of Solids and Liquids

Common thermometers make use of the expansion of liquids with increasing
temperature. This phenomenon is called thermal expansion.

On a microscopic scale, thermal expansion occurs because the average

separation of
atoms and molecules increases. The atoms and molecules in a substance are constantly
vibrating, but on average they remain a certain distance apart. If the temperature of the
substance increases, then the energy of the vibrations increases,
meaning that the size
(amplitude) of the vibrations increases. The result is that the average separation between
atoms and molecules increases.

Empirically, we find that an object of initial length L
0

will undergo a change of length,

L for a temperature change

T, given by:

L =

L
0

T.

The proportionality constant

is called the coefficient of linear expansion. The
coefficients of linear expansion for a variety of substances near room temperature are
listed in table 10.1 of the text.

Example
: P10.3

The New River Gorge bridge in West Virginia is a 518 m long steel arch. How much will
its length change between temperature extremes
-
20°C and 35°C?

From table 10.1 we find the coefficient of linear expansion for steel to be

= 11x10
-
6
/°C
.
If L
0

= 518 m, and

T = 55°C, then

L =

L
0

T = (11x10
-
6
/°C)(518 m)(55°C) = 0.31 m = 31 cm.

Similar calculations can be done for area and volume expansion, for example,
determining the change in area of a metal plate, or the change in volume of a slab o
f
concrete when it is heated or cooled. The resulting equations are:

A =

A
0

T, and

V =

V
0

T

where

is the coefficient of surface expansion and equals 2

, and

is the coefficient of
volume expansion and equals 3

.

Macroscopic Description of an Ideal
Gas

This long title is basically a way to introduce the ideal gas law into a Physics course. It is
found that under many conditions, the pressure P, volume V, temperature T, and moles n,
of a gas are related by:

PV = nRT

where R is the universal gas consta
nt, R = 8.31 J/mol K in SI units. Recall that in SI
units, pressure is measured in Pa = N/m², volume in m³, and temperature in °K. If the
pressure is given in atmospheres (atm) and the volume in liters (L), then you can use the
value of R = 0.0821 L atm/mo
l K.

You may recall the fact that 1 mol of gas at STP (standard temperature and pressure
occupies a volume of 22.4L. This result can be obtained by using the above value of R,
letting P = 1atm, T = 0°C = 273°K, and n = 1mol. V = nRT/P = (1mol)(0.0821 L at
m/mol
K)(273°K)/(1atm) = 22.4 L.

Example: P10.21

One mole of gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is
heated at constant volume until the pressure triples, what is the final temperature? (b) If
the gas is heated so t
hat both the pressure and volume are doubled, what is the final
temperature?

(a) Since the number of moles and the volume are held constant, we have

P
0
/T
0

= nR/V = P
f
T
f
.

Since P
f

= 3P
0

and we're given that T
0

= 300°K, then

T
f

= (P
f
/P
0
)T
0

= (3)(300°K) = 900
°K = 627°C.

NOTICE that if the temperature is left in Celsius, then the result would be 81°C which is
WRONG.

(b) Now only the number of moles remain constant, so

P
0
V
0
/T
0

= nR = P
f
V
f
/T
f
. This time we're told that P
f

= 2P
0

and V
f

= 2V
0

so

T
f

= T
0
(P
f
/P
0
)(V
f
/
V
0
) = (300°K)(2)(2) = 1200°K = 927°C.

Once again, forgetting to use the kelvin scale yields an incorrect result.

N
A

= 6.02 x 10
23

This is the number of molecules in one mole of gas (or any substance). If we have n
mo
les of gas, then the number of molecules is N = nN
A
, or n = N/N
A
. Thus we can
rewrite the ideal gas law as:

PV = (N/N
A
)RT = N(R/N
A
)T = Nk
B
T

where k
B

is Boltzmann's constant. The value of k
B

is

k
B

= R/N
A

= (8.31 J/mol K)/(6.02 x 10
23
/mol) = 1.38 x 10
-
23

J/K
.

The Kinetic Theory of Gases

We can relate the properties of an ideal gas to the mechanics that we studied in earlier
chapters. To do this, we must picture the gas as a collection of molecules moving about
inside a container. To be specific, we make the
following assumptions (from the text):

1.

The number of molecules is large, and the average separation between them is
large compared with their dimensions. This means that the molecules occupy a
negligible volume in the container.

2.

The molecules obey Newton
's laws of motion, but as a whole they move
randomly.

3.

The molecules undergo elastic collisions with each other and with the walls of the
container. Thus in the collisions kinetic energy is constant.

4.

The forces between molecules are negligible except duri
ng a collision.

5.

The gas under consideration is a pure substance; that is all molecules are identical.

In this model, the pressure of the gas is due to the molecules hitting the walls of the
container. If there were only a few molecules, then this would b
e more like banging. But
because there are many molecules (of order 10
23
), the collisions of molecules with the
walls is constant and very uniform across the wall. Imagine the difference between
holding an umbrella in a slow rain versus a downpour and I th
ink you've got the picture.

A calculation of the average force on a wall yields the following expression for the
pressure in terms of the rms velocity of the molecules:

P = (2/3)(N/V)(
\
frac12;mv²). The pressure is proportional to the number of molecules p
er
unit volume (the number density of molecules) and the average kinetic energy of the
molecules, KE
avg

= ½mv²
avg
. This is truly an amazing result. It relates the mechanics of
microscopic particles (their average kinetic energy) to a macroscopic property (
pressure).

In this model, the temperature of a gas is related to the average kinetic energy of the
molecules of the gas. To see this, rewrite the above expression as:

PV = (2/3)N(½mv²
avg
).

Since the ideal gas law tells us that PV=nRT
--

the same PV
--

the
n we can equate nRT =
(2/3)N(½mv²
avg
). Use the relations n = N/N
A

and R/N
A

= k
B

to get:

T = (2/3k
B
)(½mv²
avg
)

or, in the more standard form:

KE
avg

= ½mv²
avg

= (3/2)k
B
T

This is another amazing result. It relates the average kinetic energy of a molecule of ga
s
to the temperature of the gas as a whole!

Finally, a comment about the quantity v²
avg
. This is the average of the square of the
velocities. Often we want to speak of an average velocity (not squared). But the average
velocity is zero, because there are
as many molecules moving to the right (positive) as to
the left (negative) on average, and they cancel to give an
average velocity

of zero. So
instead we use the rms velocity. The letters rms stand for root
-
mean
-
square, which
translates to the square root
of the average of the square of the velocities. The rms
velocity is

v
rms

= Sqrt{v²
avg
} = Sqrt{3k
B
T/m} = Sqrt{3RT/M}

where M is the molar mass of the molecule.

Example: P10.38

A cylinder contains a mixture of helium and argon gas in equilibrium at a tempe
rature of
150°C. (a) What is the average kinetic energy of each type of molecule? (b) What is the
rms speed of each type of molecule?

(a) KE
avg

= (3/2)k
B
T = (3/2)(1.38x10
-
23

J/K)(150+273°K) = 876x10
-
23

J = 8.76x10
-
21

J.
Since the kinetic energy doesn't dep
end on the type of atom or molecule (
i.e.

no mass
dependence), the result is the same for both gases.

(b) v
rms

= Sqrt{3RT/M}

This does depend on the molecular mass. M
He

= 4.0x10
-
3

kg/mol and M
Ar

= 39.948x10
-
3

kg/mol. The resulting rms speeds are:

v
rms He

= Sqrt{3(8.31 J/mol K)(423K)/(4.0x10
-
3

kg/mol) = 1620 m/s

v
rms Ar

= Sqrt{3(8.31 J/mol K)(423K)/(40.x10
-
3

kg/mol) = 513 m/s.

Recall

For an ideal gas: PV = nRT

Using a microscopic model of a gas, one finds that the pressure of the gas is
related to the a
verage kinetic energy of a molecule of gas:

P = (2/3)(N/V)(½mv²
avg
)

Chapter 11: Heat

The Mechanical Equivalent of Heat

As
eluded

to earlier, heat is a form of energy. Carefully constructed experiments have
shown tha
t when mechanical energy is "lost" due to non
-
conservative forces, that it turns
into heat. By expanding our concept of energy to include heat
--

along with kinetic
energy of translation and rotation, gravitational potential energy, and elastic potential
e
nergy
--

we regain conservation of energy in a more general form.

Heat energy is (still) commonly measured in calories. One calorie is the energy needed to
raise the temperature of 1 gram of water from 14.5°C to 15.5°C. To convert to joules, the
SI unit o
f energy, 1 cal = 4.186 J. The Calories (big c) used for food equal 1000 calories
(little c).

Specific Heat

If we add heat to something, generally its temperature increases. The specific heat is a
measure of the relation between heat energy and temperatur
e. Since the amount of heat
added must also depend on how much of the substance there is, we also include the mass
in the relation:

c = Q/m

T.

Table 11.1 contains a list of specific heats.

Conservation of Energy: Calorimetry

First, we want to consider problems where heat energy is transferred between objects. A
controlled system, that is allows energy to be transferred between well defined o
bjects, is
called a calorimeter. We will consider cases where heat leaves one substance and goes to
a second. What is being transferred is heat energy, Q = mc

T. The temperatures, masses
and specific heats of the substances may be different, but the heat t
hat leaves on
substance got to another:

Q
out

= Q
in

For now, make both energies positive, that is, make

T positive. The following examples
illustrate the general procedure.

Example: P11.11

Lead pellets, each of mass 1.00 g, are heated to 200°C. How many p
to 500 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C?
Neglect any heat transfer to or from the container.

The amount of heat lost by the lead pellets when cooled to 25°C must equal the heat
needed to

raise the temperature of water to 25°C. First, how much heat must be added to
the water?

Q
w

= c
w
m
w
(T
f

-

T
w
)

where:

c
w

= specific heat of water = 4186 J/kg°C

m
w

= mass of water = 0.50 kg

T
f

= final temperature = 25°C

T
w

= initial water temperature = 20
°C

T
Pb

= initial lead temperature = 200°C

c
Pb

= specific heat of lead = 128 J/kg°C

m
p

= mass of a lead pellet = 0.001 kg

N = number of lead pellets needed.

The heat lost by N lead pellets is:

Q
Pb

= Nm
p
c
Pb
(T
Pb

-

T
f
)

Set the two heats equal to each othe
r and solve for N:

N = (c
w
/c
Pb
)(m
w
/m
p
){(T
f

-

T
w
)/(T
Pb

-

T
f
) = (4186/128)(0.50/0.001){(25
-
20)/(200
-
25)} =
467.

Example: P11.16

A 100 g aluminum calorimeter contains 250 g of water. The two are in thermal
equilibrium at 10°C. Two metallic blocks are placed
in the water. One is a 50 g piece of
copper at 80°C. The other sample has a mass of 70 g and is originally at a temperature of
100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific
heat of the unknown second sample.

The

heat lost by the metallic blocks must equal the heat gained by the water and
aluminum calorimeter.

Q
Al+w

= m
Al
c
Al
(T
f

-

T
Al
) + m
w
c
w
(T
f

-

T
w
) = (m
Al
c
Al

+ m
w
c
w
)(T
f

-

T
Al+w
).

Q
blocks

= m
Cu
c
Cu
(T
Cu

-

T
f
) + m
x
c
x
(T
x

-

T
f
)

Again, equate the two heats and this time

solve for c
x
, the specific heat of the unknown
sample. All other quantities are known.

c
x

= {(m
Al
c
Al

+ m
w
c
w
)(T
f

-

T
Al+w
)
-

m
Cu
c
Cu
(T
Cu

-

T
f
)}/(m
x
(T
x

-

T
f
)) =
{((0.10kg)(900J/kg°C) + (0.25kg)(4186J/kg°C))(20
-
10°C)
-

(0.05kg)(387J/kg°C)(80
-
20°C)}/(0.07kg)(10
0
-
20°C) =1800 J/kg°C.

Latent Heat and Phase Changes

When a substance undergoes a
change of phase
, heat is absorbed or given off
without

a
change of temperature. Familiar phase changes are from solid to liquid (ice melting to
water), the reverse of liqui
d to solid (water freezing to ice), liquid to gas (water boiling
away to steam) or its reverse of gas to liquid (steam condensing to water). The more rare
phase change from solid to gas is possible, as are a number of more specialized cases
such as:

from
(ferro)magnetic to non
-
magnetic (paramagnetic)

from superconducting to normal

Phase changes generally occur because the internal structure of a substance is changed.
We characterize the energy absorbed of given off by the
latent heat
, L, so that Q = mL i
s
the amount of heat for a mass m of the substance. When the phase change is between
solid and liquid, use the latent heat of fusion, L
f
, and when the change is between liquid
and gas, use the latent heat of vaporization, L
v
. Table 11.2 is a list of latent

heats for a
variety of substances.

If in a calorimetry problem one of the substances undergoes a phase change, then the
calculation of the heat lost of gained must be modified to include the energy involved in
the phase change.

Example: P11.20

A 50 g ice

cube at 0°C is heated until 45 g has become water at 100°C and 5.0 g has
become steam at 100°C. How much heat was added to accomplish this?

First, what are the steps required to go from the ice cube initial state to the water plus
steam final state?

1.

Melt

the ice into water

2.

Heat the water from 0°C to 100°C

3.

Turn 5.0g of the 100°C water into steam.

Let's calculate the heat required for each step, and then sum them to get the total heat
required.

1.

The heat required to melt 50.0g of ice is Q
1

= mL
f

= (0.05k
g)(3.33x10
5
J/kg) =
1.7x10
4
J.

2.

The heat required to raise the temperature of 50g of water from 0°C to 100°C is
Q
2

= mc

T = (0.05kg)(4186J/kg°C)(100°C) = 21000 J = 2.1x10
4
J.

3.

The heat required to vaporize 5.0g of water is Q
3

= mL
v

= (0.005kg)(2.26x10
6
) =
1.1
x10
4
J.

The total heat required is Q
tot

= Q
1

+ Q
2

+ Q
3

= (1.7 + 2.1 + 1.1)x10
4
J = 4.9x10
4
J.

Heat Transfer by Conduction, Convection, and Radiation

There are three ways to transfer heat: conduction, convection, and radiation.

Conduction occurs when heat p
ropagates from a source to a destination through
an intervening material. For instance, water is heated by placing it in a container
and placing the container of a stove. Heat is transferred through the pot to the
water.

In convection, heat is transferred

by currents in a gas or liquid. For example,
bread is baked in an oven by the heated air in the oven. In a traditional oven, the
convection is natural; in a convection oven, the air is forced to circulate by a fan,
and this is called forced convection.

R
on,

heat transfer occurs through the emission and absorption of electro
-
magnetic radiation, in particular, infrared radiation. This is the manner in which
food is broiled or grilled. A microwave oven also uses radiation, but instead of
infrared, micr

We begin by looking at conduction.

Conduction

In conduction, an amount of energy, Q, is transported through a material in a given
amount of time

t. The
heat transfer rate
, H, is defined as H=Q/

t.

Now imagine heat passing throu
gh a block of material with area A and thickness L, one
side of the block at temperature T
1

and the other side at temperature T
2
. It is found that
the heat transfer rate is proportional to the area and the temperature difference, and
inversely proportional

to the thickness.

H = Q/

t = kA(T
2

-

T
1
)/L

where k is a constant for the material called the thermal conductivity. Table 11.3 lists
values of thermal conductivity, k, for a number of materials.

Convection

We will say little about convection. Not that this isn't an important proc
ess
--

convection
is responsible for ocean currents (the Gulf stream, the Alaskan current, etc.), and is a
major factor in weather.

All objects at a finite temperature emit energy as electro
-
s in the visible band; most heat transfer occurs through
radiation in the infrared band. The heating element of a toaster glows red, and light bulbs
are hot.

The rate at which an object emits radiation is given by Stefan's law:

P =

AeT
4

where

P is the power radiated in Watts,

is a constant equal to 6.0x10
-
8
W/m
2
K
4
,

A is the surface of the object in square meters,

e is the emissivity, a material dependent constant between 0 and 1, and

T is the object's temperature in kelvins

W
hen an object is surrounded by other objects at finite temperature, then it not only emits
radiation, but will also absorb radiation from the surroundings. The net rate at which
energy is gained or lost is the difference between the rate of emission, and t
he rate of
absorption:

P
net

=

Ae(T
4
-
T
0
4
)

where T
0

is the temperature of the surroundings.

Example: P11.37

A sphere that is to be considered as a perfect black
-
and is at 200°C in a room where the temperature is 22°C. Calculate the net rate at which

A blackbody is an ideal absorber of radiation and has an emissivity of 1.0. Use P
net

=

Ae(T
4
-
T
0
4
). The area of a sphere is 4

r
2
, therefore:

P
net

= (6.0x10
-
8
W/m
2
K
4
)4

(0.060 m)
2
(1.0)((473°K)
4

-

(295°K)
4
) = 110 W.

Chapter 12: Th
e Laws of Thermodynamics

Heat and Internal Energy

The first law of thermodynamics concerns the internal energy of a system, and the ways
that energy can be changed. First, the internal energy of something is the energy
contained within its atoms and molecu
les. This does NOT include the kinetic energy or
potential energy of the object or substance itself. We are now thinking about the stored in
the constituent atoms and molecules of a "system".

For example, 100g of ice and 100g of hot water sitting on a tab
le have the same
gravitational potential energy, mgh, but they have different temperatures, and different
amounts of internal energy. The H
2
O molecules in the ice have less energy on average
than the molecules in the hot water.

The internal energy of a sy
stem can be changed by:

doing work on or the system or letting the system do work on the outside world.

The first law of thermodynamics says that energy is conserved if we properly account for
the internal energy of the system
, and the transfer of heat or work that change the internal
energy.

If U stands for the internal energy, Q for the heat

to the system, and W for the
work done
BY

the system, then the first law is expressed by:

U = U
f

-

U
i

= Q
-

W.

to the system, the internal energy increases. If the system does work, it
must lose a corresponding amount of internal energy.

Example: P12.1

The only form of energy possessed by "molecules" of a monatomic ideal gas is
translational kinetic energy. Using
the results from the discussion of kinetic theory in
section 10.6, show that the internal energy of a monatomic ideal gas at pressure P and
occupying volume V may be written as U = (3/2)PV.

First, examples of monatomic gases are helium, neon, and argon. Mo
natomic gases are
composed of single atoms
--

the atoms do not bind together into molecules. So the phrase
"molecules of a monatomic ideal gas" is a bit awkward from my point of view. For our
purposes, you can replace that phrase with "atoms of a noble gas
".

Again, for our purposes, the atoms of a noble gas act like point masses. They can move
about, bounce off the container and each other, but there's nothing to rotate, and no
internal vibrations in which to store energy. Thus, all the internal energy is t
ranslational
kinetic energy, ½mv
2
avg
. And in chapter 10 we learned that the average kinetic energy of
an atom in a gas is KE
avg

= ½mv
2
avg

= (3/2)k
B
T. The internal energy of a gas is the sum of
the energy of all the constituents:

U =

KE
i

= N KE
avg

= nN
A
KE
a
vg

= (3/2)n(N
A
k
B
)T.

Recall that k
B

= R/N
A
, so the quantity in parenthesis is R.

Making use of the ideal gas law, PV = nRT, we find:

U = (3/2)nRT = (3/2)PV.

This result is useful in several of the homework problems.

Work and Heat

Work Done by a Gas, PV di
agrams

A gas at pressure P, changing volume by an amount

V, does work W=P

V. If we make
a plot with volume on the horizontal axis and pressure on the vertical axis, then we see
that the work done
by the gas in an expansion

is equal to the area under the curve from
the point at P
i
, V
i

to P
f
, V
f
. Such a plot is c
alled a PV diagram.

If the pressure is not constant while the volume changes, the work can still be calculated
as the area under the PV curve. To do this we must know the shape of the PV curve, that
is, we must know how the pressure and volume changed in
going from the initial point to
the final point.

Example: P12.6

Sketch a PV diagram of the following process. (a) A gas expands at constant pressure P
1

from volume V
1

to volume V
2
. It is then kept at constant volume while the pressure is
reduced to P
2
. (b
) A gas is reduced in pressure from P
1

to P
2

while its volume is held
constant at V
1
. It is then expanded at constant pressure P
2

to a final volume, V
2
. (c) In
which of the processes is more work done? Why?

Sketch (a) and (b). (c) Since P
1
>P
2
, the area un
der the PV curve for (b) is less than (a),
and the work done in process (b) is less than (a).

Example: P12.5

A gas expands from I to F along the three paths indicated in the figure. Calculate the
work done by the gas along paths (a) IAF, (b) IF, and (c) I
BF.

In each case, the work done by the gas is equal to the area under the PV curve. To get the
work in joules, the pressure must be expressed in pascals = N/m² and the volume in m³.
The conversion factors are 1 atm = 1.01x10
5
Pa, and 1 liter = 10
-
3
m³. Ther
efore P
i

=
4.04x10
5
Pa, P
f

= 1.01x10
5
Pa, V
i

= 2.0x10
-
3
m³, and V
f

= 4.0x10
-
3
m³.

(a) The area under the curve IAF is W
a

= P
i
(V
f

-

V
i
) = 808 J.

(b) Recall that the area of a triangle is ½base x height. The area under the curve IF is W
b

= ½(P
f

-

P
i
)(V
f

-

V
i
)
+ P
f
(V
f

-

V
i
) = ½(3.03x10
5
Pa)(2.0x10
-
3
m³) + (1.01x10
5
Pa)(2.0x10
-
3
m³) = 505 J.

(c) The area under the curve IBF is W
c

= P
f
(V
f

-

V
i
) = 202 J.

Example: P12.17

A gas is enclosed in a container fitted with a piston of cross
-
sectional area 0.150 m². The
pressu
re of the gas is maintained at 6000 Pa as the piston moves inward 20.0 cm. (a)
Calculate the work done by the gas. (b) If the internal energy of the gas decreases by 8.00
J, find the amount of heat removed from the system during the compression.

(a) Since

the volume of the gas decreases, work is done
on

the gas, or the work done
by

the gas is negative. The amount of work done is W = P

V = PA

s = (6000 Pa)(0.150
m²)(
-
0.200 m) =
-
180 J.

(b) While the gas is compressed, heat is added or removed such that the

pressure remains
constant. Using the first law,

U = Q
-

W, or Q =

U + W =
-
8.00J + (
-
180 J) =
-
188 J.
Remember that Q is defined as the heat added to the system, a negative Q implying that
heat is removed. Therefore the amount of heat removed is 188 J.

Heat Engines and the Second Law of Thermodynamics

One of the earliest and most important applications of thermodynamics is the
understanding of engines. For our purposes, a heat engine is a system that goes through
changes in pressure, volume, and tempera
ture, in a cyclic fashion. That is, the system
repeatedly follows the same changes in P, V, and T, returning to its starting conditions to
begin another cycle.

By far the most common engines involve a system consisting of a gas, so this is what we
will co
ncentrate on. We can represent the state of the gas on a PV diagram. (Since, for a
given quantity of an ideal gas, T = PV/nR, specifying PV fixes T.) A cyclic process traces
out a closed curve on the PV diagram. (diagram) The net work done by the engine pe
r
cycle equals the area enclosed by the curve,

W
cycle

= Area enclosed in PV curve.

A heat engine can be represented schematically as a system connected to a hot reservoir
and a cold reservoir. The system absorbs an amount of heat Q
h

from the hot reservoir
,
gives off an amount of heat Q
c

to the cold reservoir, and produces an amount of work W
on each cycle. Since the thermodynamic process of the engine is cyclic, it returns to its
starting condition after each cycle. Being in the same condition means that t
he internal
energy of the system is the same, or

U
cycle

= 0 for a complete cycle.

The first law tells us that

U
cycle

= Q
cycle

-

W
cycle
, or, since

U
cycle

= 0, Q
cycle

= Q
h

-

Q
c

=
W
cycle
. This is just conservation of energy. The system takes in an amount of heat energy
Q
h
, and gives off an amount of hea
t energy Q
c

and work W
cycle
.

The efficiency, e, of the engine is defined as the fraction of the heat energy taken in that
is converted to work:

e = W/Q
h

= (Q
h

-

Q
c
)/Q
h

= 1
-

Q
h
/Q
c
. If the engine converts all the input energy to heat
(W = Q
h
) then the effi
ciency is 100%. If no work is produced then the efficiency is 0.

The Second Law of Thermodynamics

As applied to heat engines, the second law of thermodynamics says that the amount of
work produced by a heat engine is always less than the amount of input e
nergy, or, it is
impossible for an engine to be 100% efficient. Mathematically, W < Q
h
, never equal to.

Recall:

U = Internal energy of a system

Q = Heat energy absorbed by the system

W = Work done by the system

The First Law of Thermodynamics:

U = Q
-

W

Heat Engines and the Second Law of Thermodynamics

One of the earliest and most important applications of thermodynamics is the
understanding of engines. For our purposes, a heat engine is a system that goes through
changes in pressure, volume, and te
mperature, in a cyclic fashion. That is, the system
repeatedly follows the same changes in P, V, and T, returning to its starting conditions to
begin another cycle.

By far the most common engines involve a system consisting of a gas, so this is what we
wi
ll concentrate on. We can represent the state of the gas on a PV diagram. (Since, for a
given quantity of an ideal gas, T = PV/nR, specifying PV fixes T.) A cyclic process traces
out a closed curve on the PV diagram. (diagram) The net work done by the engi
ne per
cycle equals the area enclosed by the curve,

W
cycle

= Area enclosed in PV curve.

A heat engine can be represented schematically as a system connected to a hot reservoir
and a cold reservoir. The system absorbs an amount of heat Q
h

from the hot rese
rvoir,
gives off an amount of heat Q
c

to the cold reservoir, and produces an amount of work W
on each cycle. Since the thermodynamic process of the engine is cyclic, it returns to its
starting condition after each cycle. Being in the same condition means t
hat the internal
energy of the system is the same, or

U
cycle

= 0 for a complete cycle.

The first law tells us that

U
cycle

= Q
cycle

-

W
cycle
, or, since

U
cycle

= 0, Q
cycle

= Q
h

-

Q
c

=
W
cycle
. This is just conservation of energy. The system takes in an amount of heat energy
Q
h
, and gives off an amount of hea
t energy Q
c

and work W
cycle
.

The efficiency, e, of the engine is defined as the fraction of the heat energy taken in that
is converted to work:

e = W/Q
h

= (Q
h

-

Q
c
)/Q
h

= 1
-

Q
h
/Q
c
. If the engine converts all the input energy to heat
(W = Q
h
) then the effi
ciency is 100%. If no work is produced then the efficiency is 0.

The Second Law of Thermodynamics

As applied to heat engines, the second law of thermodynamics says that the amount of
work produced by a heat engine is always less than the amount of input
energy, or, it is
impossible for an engine to be 100% efficient. Mathematically, W < Q
h
.

The Carnot Engine

The Carnot engine is a theoretical device that is useful because it was shown to be the
most efficient engine operating between a hot and cold rese
rvoirs at temperatures T
h

and
T
c
, allowed by the laws of Physics. From a study of the Carnot engine, we can deduce the
maximum

efficiency for an engine, only knowing the hot and cold temperatures at which
heat is transferred. The result is:

e
c

= 1
-

(T
c
/T
h
)

where, of course, the temperatures must be in kelvin.

Notice that for the Carnot efficiency to equal 1, the cold temperature must be zero kelvin
--

absolute zero. Since it is impossible to get anything to a temperature of absolute zero, it
is impossible

to make an engine that is 100% efficient. Because most exhaust goes to the
atmosphere at a temperature of about 300K, the usual route to higher efficiency is to raise
T
h

for the engine. But this runs into practical difficulties such as the ability of mate
rials to
stand up to large temperature swings.

Example: P12.26

A heat engine performs 200 J of work in each cycle and has an efficiency of 30%. For
each cycle of operation, (a) how much heat is absorbed and (b) how much heat is
expelled?

Using e = W/Q
h
,

we deduce that the heat absorbed is

Q
h

= W/e = 200J/0.3 = 667 J.

And since W = Q
h

-

Q
c
, we find that

Q
c

= Q
h

-

W = 667 J
-

200 J = 467 J.

Example: P12.29

In one cycle, a heat engine absorbs 500 J from the high temperature reservoir and expels
300 J to a

low temperature reservoir. If the efficiency of this engine is 60% of the
efficiency of a Carnot engine, what is the ratio of the low temperature to the high
temperature in the Carnot engine?

First, the efficiency of this engine is e = 1
-

(Q
c
/Q
h
) = 1
-

(300/500) = 0.40 = 40%. If this
is 60% of the Carnot efficiency, then e = 60% e
c
, or e
c

= e/0.6 = 0.4/0.6 = 0.67 = 67%.
Finally, to get the ratio of the low to high temperatures, use the relation

e
c

= 1
-

(T
c
/T
h
), so that

(T
c
/T
h
) = 1
-

e = 1
-

0.67 = 0.33.

Entropy

Another result from the study of heat engines is the concept of entropy. Entropy is a
measure of the amount of
disorder

in a system. For example, the internal structure of ice
is more ordered than that of water, which is again more ordered than
steam. The entropy
of ice is less than the entropy of water is less than the entropy of steam (for a fixed
number of H
2
O molecules).

Entropy is represented by the letter S. In our discussions, we will only deal with changes
in entropy, in the same way tha
t we only deal with changes in internal energy. "The
change in entropy,

S, between two equilibrium states is given by the heat transferred,

Q, divided by the absolute temperature, T, of the system in this interval":

S =

Q/T.

We can restate the second
law of thermodynamics in terms of entropy:
the entropy of
the universe increases in all natural processes.

This does not mean that entropy is
always increasing everywhere. The entropy of a small system can be decreased, but the
second law says that there w
ill be a corresponding increase in the entropy outside of the
system which exceeds the decrease in entropy of the system.

Example: Melting a Piece of Lead

Calculate the change in entropy when 300 g of lead melts at 327?C (600?K). Lead has a
latent heat o
f fusion of 2.45x10
4
J/kg.

Let's determine the amount of heat absorbed to melt the lead, then use

S =

Q/T to
determine the change in entropy.

Q = mL
f

= (0.30kg)(2.45x10
4
J/kg) = 7.35x10
3
J.

Thus

S = (7.35x10
3
J)/(600?K) = 12.3J/K.