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Chapter 14
The Ideal Gas Law and
Kinetic Theory
14.1 Molecular Mass, the Mole, and Avogadro’s Number
To facilitate comparison of the mass of one atom with another, a mass scale
know as the atomic mass scale has been established.
kg106605.1u 1 27−=
The atomic mass is given in atomic
mass units. For example, a Li atom
has a mass of 6.941u.
The unit is called the atomic mass unit (symbol u). The reference
element is chosen to be the most abundant isotope of carbon, which is
called carbon-12. Carbon-12 has exactly 12 atomic mass units.
14.1 Molecular Mass, the Mole, and Avogadro’s Number
One gram-mole of a substance contains as
many particles as there are atoms in 12
grams of the isotope carbon-12.
123 mol10022.6 −=AN
AN
Nn =
number of
moles
number of
atoms
The number of atoms per mole is
known as Avogadro’s number, NA.
14.1 Molecular Mass, the Mole, and Avogadro’s Number
ANm
Nmn
particle
particle=
The mass per mole (in g/mol) of a substance
has the same numerical value as the atomic or
molecular mass of the substance (in atomic
mass units).
AN
Nn =
moleper Mass
mn =
For example Hydrogen has an atomic
mass of 1.00794 g/mol, while the mass
of a single hydrogen atom is 1.00794 u.
14.1 Molecular Mass, the Mole, and Avogadro’s Number
Example: The Hope Diamond and the Rosser Reeves Ruby
The Hope diamond (44.5 carats) is almost pure carbon. The Rosser
Reeves ruby (138 carats) is primarily aluminum oxide (Al2O3). One
carat is equivalent to a mass of 0.200 g. Determine
(a) the number of carbon atoms in the Hope diamond and
(b) the number of Al2O3 molecules in the ruby.
(a)
AnNN =
( ) ( ) ( ) molg12
carat 1g 200.0carats 5.44=
moleper Mass
mn =
0.74 moles
( )( )123 mol10022.6mol 74.0 −=N
atoms1046.4 23=N
14.1 Molecular Mass, the Mole, and Avogadro’s Number
(b)
moles 271.0=n
atoms1063.1 23
moleper Mass
mn =
AnNN =
( ) ( ) ( )
( ) ( )
molg96.101
carat 1g 200.0carats 138
99.15398.262
+
=
the number of Al2O3 molecules in the ruby.
( )( )123 mol10022.6mol 271.0 −== AnNN
14.2 The Ideal Gas Law
The condition of low density means that the
molecules are so far apart that they do not
interact except during collisions, which are
effectively elastic.
TP
At constant volume the pressure
is proportional to the temperature.
An ideal gas is an idealized model for real
gases that have sufficiently low densities.
• No interactions
• All collisions are elastic
14.2 The Ideal Gas Law
At constant temperature, the pressure is
inversely proportional to the volume.
VP 1
The pressure is also proportional
to the amount of gas.
nP
V
nTP
14.2 The Ideal Gas Law
The absolute pressure of an ideal gas is directly proportional to the Kelvin
temperature and the number of moles of the gas and is inversely proportional
to the volume of the gas.
V
nRTP =
nRTPV =
( )KmolJ31.8 =R
THE IDEAL GAS LAW
V
nTP
14.2 The Ideal Gas Law
nRTPV =
AN
Nn =
( )KJ1038.1
mol106.022
KmolJ31.8 23
123
−
−=
==
AN
Rk
TN
RNPV
A
=
NkTPV =Boltzmann Constant
nRTPV = NkTPV =OR
No. of moles No. of particles
14.2 The Ideal Gas Law
Example: Oxygen in the Lungs
In the lungs, the respiratory membrane separates tiny sacs of air
(pressure 1.00x105Pa) from the blood in the capillaries. These sacs
are called alveoli. The average radius of the alveoli is 0.125 mm, and
the air inside contains 14% oxygen. Assuming that the air behaves as
an ideal gas at 310 K (body temperature), find the number of oxygen
molecules in one of
these sacs.NkTPV =
air of molecules109.1 14
kT
PVN =
( ) ( ) ( )( )K 310KJ1038.1
m10125.0Pa1000.123
33
345
−
−
=
N
? sac onein oxygen of molecules ofNumber =
( ) ( )14.0109.1 14 molecules107.2 13
14.2 The Ideal Gas Law
Consider a sample of an ideal gas that is taken from an initial to a final
state, with the amount of the gas remaining constant.
nRTPV =
i
ii
f
ff
T
VP
T
VP=
constant == nRT
PV
Constant T, constant n: iiff VPVP = Boyle’s law
Constant P, constant n:
i
i
f
f
T
V
T
V= Charles’ law
14.3 Kinetic Theory of Gases
• As a result, the atoms and molecules
have different speeds.
• The particles are in constant,
random motion, colliding with
each other and with the walls of
the container.
• Each collision changes the
particle’s speed.
14.3 Kinetic Theory of Gases
KINETIC THEORY
( ) ( )L
mv
vL
mvmv2
2 force Average
−=
+−−=
= maF
( )t
mv
t
vmF
=
=
𝑎 =∆𝑣
∆𝑡
collisions successivebetween Time
momentum Initial-momentum Final force Average =
=∆𝑝
∆𝑡
14.3 Kinetic Theory of Gases
L
mvF
2
=
For a single molecule, the magnitude of average force is:
For N molecules, the average force is:
=
L
vmNF
2
3 root-mean-square
speed
===
3
2
2 3 L
vmN
L
F
A
FP
volume
Since 𝑃 =𝐹
𝐴and A = L2
14.3 Kinetic Theory of Gases
=
V
vmNP
2
3
( ) ( )2
21
322
31
rmsrms mvNmvNPV ==
NkT KE
kTmvrms 232
21KE ==
=
3
2
3 L
vmNP
𝑁𝑘𝑇 =2
3𝑁(𝐾𝐸) 𝐾𝐸 =
3
2𝑘𝑇
14.3 Kinetic Theory of Gases
kTmvrms 232
21KE ==
THE INTERNAL ENERGY OF A MONATOMIC IDEAL GAS
nRTkTNU23
23 ==
14.3 Kinetic Theory of Gases
Example: The Speed of Molecules in Air
Air is primarily a mixture of nitrogen N2 molecules (molecular mass
28.0 u) and oxygen O2 molecules (molecular mass 32.0 u). Assume
that each behaves as an ideal gas and determine the rms speeds
of the nitrogen and oxygen molecules when the temperature of the air
is 293 K.
kTmvrms 232
21 = m
kTvrms
3=
For Nitrogen
123 mol106.022
molg0.28−
=m
sm511=rmsv
kg1065.4g1065.4 2623 −− ==
m
kTvrms
3=
( )( )kg1065.4
K293KJ1038.1326
23
−
−
=
For Oxygen sm478=rmsv
Problem: 3
Since the sample has a mass of
135 g, the mass per mole is135 g
5.00 mol= 27.0 g/mol
The periodic table of the elements ➔
.
A mass of 135 g of an element is known to contain 30.1 x 1023 atoms.
What is the unknown element?
Aluminum
𝑛 =𝑁
𝑁𝐴=
30.1 × 1023
6.022 × 10235 moles
Atomic mass of the substance is 27.0 u
Problem: 16
Therefore, the mass m of helium in the blimp is,
A Goodyear blimp typically contains 5400 m3 of helium (He) at an absolute
pressure of 1.10 x 105 Pa. The temperature of the helium is 280 K. What is
the mass (in kg) of the helium in the blimp?
𝑛 =𝑃𝑉
𝑅𝑇
PV = nRT
=(1.1 × 105𝑃𝑎)(5400 𝑚3)
8.31 𝐽/(𝑚𝑜𝑙. 𝐾)(280 𝐾) 2.55 × 105 𝑚𝑜𝑙𝑒𝑠
From Periodic table →1 mole of Helium is 4.0026 g
𝑚 = (2.55 × 105𝑚𝑜𝑙)(4.0026𝑔
𝑚𝑜𝑙)
= 1.0 × 106𝑔
1.0 × 103 𝑘𝑔
Problem: 39
ionosphere earth's surface3T T=
( )
( )
ionosphere
rms ionosphere ionosphere
earth's surfaceearth's surfacerms earth's surface
3
3 1.733
kTv Tm
TkTv
m
= = = =
Dividing the first equation by the
second and using the fact that
An oxygen molecule is moving near the earth's surface. Another
oxygen molecule is moving in the ionosphere (the uppermost
part of the earth's atmosphere) where the Kelvin temperature is
3.00 times greater. Determine the ratio of the translational rms
speed in the ionosphere to that near the earth's surface.
kTmvrms 232
21 =
m
kTv
ionosphere
ionosphererms
3)( =
and
m
kTv earth
earthrms
3)( =
For Practice
Ch. 14
FOC 1, 3, 6, 7 & 8.
Problems: 10, 18 & 57.