INTERNAL ENERGY
Every object of matter, whether solid, liquid, or gas,
consists of atoms or molecules in rapid motion. The
kinetic energies of these particles constitute the internal
energy of the body of matter.
The temperature of the body is a measure of the average
kinetic energy of its particles.
Heat may be thought of as internal energy in transit.
When heat is added to a body, its internal energy
increases and its temperature rises; when heat is
removed from a body, its internal energy decreases and
its temperature falls.
TEMPERATURE
Temperature is familiar as the property of a body of
matter responsible for sensations of hot or cold to the
touch.
Temperature indicates the direction of internal energy
flow: When two objects are in contact, internal energy
goes from the one at higher temperature to the one at
lower temperature.
A thermometer is a device for measuring temperature.
Two things are necessary in constructing a
thermometer:
First we must have a property X that varies with
temperature T. The thermometric property should be one
that is easily measured like the expansion of a liquid, the
pressure in a gas, or the resistance of an electric circuit.
Other quantities which vary with temperature are
radiated energy, the color of emitted light, vapor
pressure etc.
The second requirement in constructing a thermometer
is the establishment of standard temperatures. This is
usually done by selecting lower and upper fixed points:
- The lower fixed point (ice point) is the temperature at
which water and ice coexist in thermal equilibrium under
a pressure of 1 atm.
- The upper fixed point (steam point) is the temperature
at which water and steam coexist in equilibrium under a
pressure of 1 atm.
TEMPERATURE SCALES
The Celsius (or centrigrade)
temperature scale assigns 0°C to
the freezing point of water and
100°C to its boiling point.
An absolute temperature scale
has its zero point in the absolute
zero of temperature. Lord Kelvin
devised this scale. The Kelvin (K)
has been adopted by the SI as the
base unit for temperature
measurement.
The Kelvin and the Celsius scales are related as follows:
TK = TC + 273
13.1 A mercury-in-glass thermometer may not be used at
temperatures below -40C. This is because mercury freezes at this
temperature.
a. What is the freezing point of mercury in the Kelvin scale?
b. What is the difference between this temperature and the freezing
point of water?
a. TK = TC + 273 = -40 + 273 = 233 K
b. 273 - 233 = 40 K
ΔT = Twater - THg = (0 - (-40)) = 40C
LINEAR EXPANSION
A change in any one dimension of a solid due to a
change in temperature is called linear expansion.
ΔL = α Lo ΔT
Units: m
where  is the coefficient of linear expansion (ºC-1)
13.2 An iron pipe is 300 m long at room temperature (20C). If the
pipe is to be used as a steam pipe, how much allowance must be
made for expansion, and what will the new length of the pipe be?
Lo = 300 m
to = 20C
tSTEAM = 100C
αFe = 1.2x10-5/C
ΔL = α Lo ΔT
= 1.2x10-5 (300) (100 - 20)
= 0.288 m
L = Lo + ΔL
= 300 + 0.288
= 300.29 m
AREA EXPANSION
The following formulas apply for area expansion:
ΔA = γ Ao ΔT
Units: m2
where γ is the coefficient of area expansion, γ = 2 α
13.3 A brass disk has a hole 80 mm in diameter punched in its
center at 20C. If the disk is placed in boiling water, what will be the
new area of the hole?
r = 80/2 = 40 mm
to = 20C, tf = 100C
αBRASS = 1.8x10-5/C
γ = 2α = 3.6x10-5/C
A = πr2
= π (40)2
= 5026.5 mm2
ΔA = γ Ao ΔT
= 3.6x10-5 (5026.5) (100 - 20)
= 14.5 mm2
A = Ao + ΔA
= 5026.5 + 14.5
= 5041 mm2
GAS LAWS AND KINETIC THEORY
Four measurable quantities of a sample: pressure,
volume, temperature and mass determine the state of a
given sample. In this unit we will study the thermal
behavior of gases.
IDEAL GASES
In a gas the individual molecules are so far apart that the
cohesive forces between them are usually very small.
Also, when a large quantity of gas is confined in a rather
small volume, the volume occupied by the molecules is
still a tiny fraction of the total volume.
An ideal gas is a gas whose behavior is completely
unaffected by cohesive forces or molar volumes. Of
course no real gas is ideal, but under ordinary
conditions of temperature and pressure, the behavior of
any gas conforms very closely to the behavior of an
ideal gas.
Boyle's Law
"If the temperature of a gas is held constant the volume
occupied by an enclosed gas is inversely proportional to
the pressure applied to it."
13.4 What volume of hydrogen gas at atmospheric pressure is
required to fill a 5.6x10-2 m3 tank under an absolute pressure of 1.7
x 106 Pa?
P1 = 101.3x103 Pa
P2 = 1.7x106 Pa
V2 = 5.6x10-2 m3
P1V1 = P2V2
6
2
PV
2 2
17
.
x
10
(
5
.
6
x
10
)
3
V1 
=
0.939
m

P1
1013
. x103
Charles' Law
"The volume of a given amount of gas is directly
proportional to the absolute temperature when the
pressure is kept constant."
A graph of the temperature of a gas and its volume is
shown below. If the line is extrapolated to the left it will
intersect the x-axis at a particular value. This point is 273°C or 0°K. As matter cannot contract beyond a zero
volume or exert less than a zero pressure, it follows that 273°C has become known as absolute zero - the limit
beyond which temperature cannot be lowered.
13.5 A large balloon filled with air has a volume of 200 liters at 0C.
Calculate its volume at 57C if the pressure is unchanged.
V1 = 200 L
T1 = 273 K
T2 = 57 + 273 = 330 K
V1 V2

T1 T2
V1T2 200(330) = 241.75 L

V2 
273
T1
Gay-Lussac's Law
"At constant volume, the pressure of a gas is directly
proportional to the absolute temperature."
General Gas Law:
PV
PV
1 1
2 2

T1
T2
13.6 A tank with internal volume of 20 liters is filled with oxygen
under an absolute pressure of 6x106 Pa at 20C. The oxygen is to
be used in a high-flying aircraft, where the absolute pressure is
7x104 Pa and the temperature is -20C. What volume of oxygen can
be supplied by the tank under these conditions?
V1 = 20 L
P1 = 6x106 Pa
T1 = 20 + 273 = 293 K
P2 = 7x104 Pa
T2 = - 20 + 273 = 253 K
PV
PV
1 1
2 2

T1
T2
6
PV
T
6
x
10
(20)(253)
1 1 2

V2 
= 1480 L
4
7
x
10
(
293
)
P2 T1
If we consider the effect of a change in mass, the
equation becomes: PV
PV
1 1
2 2
m1T1

m2 T2
The mass is changed by injecting molecules at the left. The
density remains constant for constant pressure and temperature
and it increases as the volume is held fixed by the piston and the
temperature is fixed
13.7 The pressure on a helium storage tank reads 1.4x106Pa when
the temperature is 27C. The container develops a leak overnight,
and the pressure the next morning is found to be 1x106Pa at a
temperature of 17C. What percentage of the original mass of
helium remains inside of the container?
P2
P1 = 1.4x106 Pa
T1 = 27 + 273 = 300 K P1

P2 = 1x106 Pa
T2 = 17 + 273 = 290 K
m1T1 m2 T2
V1 = V2
fraction remaining =
m2
m1
m2 P2 T1 1x106 (300)


= 0.738
6
14
. x10 (290)
m1 PT
1 2
= 73.8% He remains
THE IDEAL GAS LAW
PV
PV
1 1
 2 2
n1T1 n2 T2
PV = nRT where R is the universal gas
constant
R = 8.31 J/mol K = 0.0821 L atm/mol K = 1.99
cal/mol K
STP conditions:
At a temperature of 273 K and
a pressure of 1 atm, 1 mol of any gas
occupies a volume of 22.4 liters.
13.8 How many grams of oxygen will occupy a volume of 1600
liters at a pressure of 2 atm and a temperature of 190C?
V = 1600 L
P = 2 atm
T = 190 + 273 = 463 K
Moxygen = 2(16) = 32 g/mol
R = 0.0821 L atm/mol K
PV = nRT
PV
(2)(1600)

n
= 84.18 mol
RT (0.0821)(463)
m = nM
= 84.18 mol (32 g/mol)
= 2694 g
AVOGADRO'S NUMBER AND THE
IDEAL GAS
Avogadro’s hypothesis formulated in
1811 states that “equal volumes of gas
at the same pressure and temperature
contain equal numbers of molecules.”
The number of molecules per mole is
known as Avogadro’s number (NA).
NA = 6.02x1023 molecules.
The total number of molecules in a gas (N) equals the
product of the number of moles of gas (n) and the
number of molecules per mole (NA); thus N = n NA and
the ideal gas equation can be written as follows:
P V = n R T = (N/NA) R T
Where: (N/NA) R = k and k = 1.38 x l0-23 J/K.
k is known as Boltzmann’s constant.
POSTULATES OF THE KINETIC THEORY OF GASES
- A gas consists of a large number of molecules moving
in random directions with a variety of speeds.
- The average distance between any two molecules in a
gas is large compared to the size of an individual
molecule.
- The molecules obey the laws of classical mechanics
and are presumed to interact with one another only
when they collide.
- Collisions between molecules or between a molecule
and the walls of the container are perfectly elastic.
The pressure exerted by an enclosed gas may be
rewritten in light of these postulates and is given by the
equation:
1
2
PV  Nmv
3
where N is the total number of molecules in the gas, m is
the mass of an individual molecule, and v2 is the average
value of the squares of the velocities of the molecules of
the gas.
ROOT-MEAN SQUARE VELOCITY (vrms)
The root-mean-square velocity is often confused with
average velocity. The root-mean-square velocity refers
to the square root of the average of the squares of the
magnitudes of the velocities of the molecules in a gas.
ej  cv  v
vrms  v
2
1/ 2
2
1
2
2
h
 v ... / N
2
3
1/ 2
The average or the mean speed is equal to the sum of
the speeds of the molecules divided by the number of
molecules.
b
g
v  v1  v2  v3 ... / N
KINETIC ENERGY AND ABSOLUTE TEMPERATURE
From the kinetic theory, PV 
1
Nmv 2
3
and from the ideal gas law, P V = n R T
The average kinetic energy of a molecule in a gas is
directly proportional to the absolute temperature of the
gas:
3
-23 J/K
k
=
1.38
x
l0
KE  kT
2
13.8 A 5.00 L vessel contains 0.010 kg of an ideal gas at 100C and
a pressure of 2 atm.
a. How many molecules are in the vessel?
V=5L
T = 100 + 273 = 373 K
P = 2 atm
R = 0.0821 L atm/mol K
PV = nRT
PV
(2)(5)
n

= 0.33 mol
RT (0.0821)(373)
N = 0.33 mol (6.02x1023 molecules/mol)
= 2x1023 molecules
b. Determine the root-mean-square speed of the molecules in the
gas.
m = 0.01 kg
k = 1.38 x l0-23 J/K
0.010kg
-26 kg/molecule
m = mass/molecule =
=
5x10
2 x1023 molecules
3
KE  kT
2
1 2 3 2
mv  kT
2
2
3(138
. x1023 )(373)
= 555.7m/s
v
26
5x10
3kT
v
m
13.9 What is the root-mean-square speed of a nitrogen molecule
at a temperature of 300 K? The mass of a nitrogen molecule is
4.65 x10-26 kg.
T = 300 K
m = 4.65x10-26 kg
3kT
v
m
3(138
. x1023 )(300)
= 517 m/s
v
26
4.65x10