Ch14. The Ideal Gas Law and Kinetic Theory
Molecular Mass, the Mole, and Avogadro's Number
To set up atomic mass scale, a reference value (along with a
unit) is chosen for one of the elements. The unit is called the
atomic mass unit (symbol: u). By international agreement, the
reference element is chosen to be the most abundant type or
isotope* of carbon, which is called carbon-12. Its atomic
mass * is defined to be exactly twelve atomic mass units, or
12 u. The relationship between the atomic mass unit and the
kilogram is
1
A portion of the periodic table showing the atomic number
and atomic mass of each element. In the periodic table it is
customary to omit the symbol “u” denoting the atomic mass
unit.
2
The molecular mass of a molecule is the sum of the atomic masses
of its atoms.
Macroscopic amounts of materials contain large numbers of
atoms or molecules. Even in a small volume of gas, 1 cm3, for
example, the number is enormous. It is convenient to express
such large numbers in terms of a single unit, the gram-mole, or
simply the mole (symbol: mol). One gram-mole of a substance
contains as many particles (atoms or molecules) as there are atoms
in 12 grams of the isotope carbon-12.
12 grams of carbon-12 contain 6.022 × 1023 atoms. The number
of atoms per mole is known as Avogadro’s number NA, after the
Italian scientist Amedeo Avogadro (1776–1856):
3
4
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).
5
Example 1. The Hope Diamond and
the Rosser Reeves Ruby
The Hope diamond (44.5 carats), which is almost pure carbon.
The Rosser Reeves ruby (138 carats), which 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
diamond and (b) the number of Al2O3 molecules in the ruby.
(a)
m = (44.5 carats)[(0.200 g)/(1 carat)] = 8.90 g
6
(b)
m = (138 carats)[(0.200 g)/(1 carat)] = 27.6 g.
.
Calculations like those in part (a) reveal that the Rosser
Reeves ruby contains 0.271 mol or
7
Check Your Understanding 1
A gas mixture contains equal masses of the
monatomic gases argon (atomic mass = 39.948 u) and
neon (atomic mass = 20.179 u). They are the only
gases in the mixture. Of the total number of atoms,
what percentage is neon?
0.664
8
The Ideal Gas Law
An ideal gas is an idealized model for real gases that have
sufficiently low densities.
9
The absolute pressure of an ideal gas is proportional to the number
of molecules or, equivalently, to the number of moles n of the gas (P
n). 
P
nT/V.
IDEAL GAS LAW
The absolute pressure P of an ideal gas is directly
proportional to the Kelvin temperature T and the
number of moles n of the gas and is inversely
proportional to the volume V of the gas: P = R(nT/V). In
other words,
where R is the universal gas constant and has the value of 8.31
J/(mol·K).
10
The constant term R/NA is referred to as Boltzmann’s
constant, in honor of the Austrian physicist Ludwig
Boltzmann (1844–1906), and is represented by the symbol k:
11
Example 2. Oxygen in the Lungs
In the lungs, the respiratory membrane separates tiny sacs of
air (absolute pressure = 1.00 × 105 Pa) from the blood in the
capillaries. These sacs are called alveoli, and it is from them that
oxygen enters the blood. 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 body temperature (310 K),
.
find the number of oxygen molecules in one of the sacs.
12
One mole of an ideal gas occupies a volume of 22.4 liters
at a temperature of 273 K (0 °C) and a pressure of one
atmosphere (1.013 × 105 Pa). These conditions of
temperature and pressure are known as standard
temperature and pressure (STP).
13
Conceptual Example 3.
Beer Bubbles on the Rise
The next time you get a chance, watch the bubbles rise in a
glass of beer. If you look carefully, you’ll see them grow in
size as they move upward, often doubling in volume by the
time they reach the surface. Why does a bubble grow as it
ascends?
The number of moles does increase as the bubble rises.
Each bubble acts as a nucleation site for CO2 molecules,
so as a bubble moves upward, it accumulates carbon
dioxide from the surrounding beer and grows larger.
14
Boyle’s law
A pressure-versus-volume
plot for a gas at a constant
temperature is called an
isotherm. For an ideal gas,
each isotherm is a plot of the
equation P = nRT/V =
constant/V.
15
Check Your Understanding 2
Consider equal masses of the three monatomic gases argon
(atomic mass = 39.948 u), krypton (atomic mass = 83.80 u),
and xenon (atomic mass = 131.29 u). The pressure and volume
of each is the same. Which gas has the greatest and which the
smallest temperature?
Xenon has the greatest and argon the
smallest temperature.
16
Example 4. Scuba Diving
In scuba diving, a greater water pressure acts on a diver at
greater depths. The air pressure inside the body cavities (e.g.,
lungs, sinuses) must be maintained at the same pressure as that
of the surrounding water; otherwise they would collapse. A
special valve automatically adjusts the pressure of the air
breathed from a scuba tank to ensure that the air pressure
equals the water pressure at all times. The scuba gear consists
of a 0.0150-m3 tank filled with compressed air at an absolute
pressure of 2.02 × 107 Pa. Assuming that air is consumed at a
rate of 0.0300 m3 per minute and that the temperature is the
same at all depths, determine how long the diver can stay
under seawater at a depth of (a) 10.0 m and (b) 30.0 m.
17
(a)
18
(b) The calculation here is like that in part (a).
Absolute water pressure is 4.02 × 105 Pa
Vf = 0.754 m3
19
Frenchman Jacques Charles (1746–1823) discovered that
at a constant pressure, the volume of a fixed mass (fixed
number of moles) of a low-density gas is directly
proportional to the Kelvin temperature (V T).
Charles’ law
20
Kinetic Theory of Gases
21
THE DISTRIBUTION OF
MOLECULAR SPEEDS
22
KINETIC THEORY
The pressure that a gas exerts is
caused by the collisions of its
molecules with the walls of the
container.
23
A gas particle is shown
colliding elastically with
the right wall of the
container and
rebounding from it.
24
25
26
Conceptual Example 5.
Does a Single Particle Have a Temperature?
Each particle in a gas has kinetic energy. Furthermore, the
equation
establishes the relationship
between the average kinetic energy per particle and the
temperature of an ideal gas. Is it valid, then, to conclude that a
single particle has a temperature?
A single gas particle does
not have a temperature.
27
Check Your Understanding 3
The pressure of a monatomic ideal gas is doubled, while its
volume is reduced by a factor of four. What is the ratio of the
new rms speed of the atoms to the initial rms speed?
vrms ,new
vrms ,initial
 0.707
28
Example 6.
The Speed of Molecules in Air
Air is primarily a mixture of nitrogen N2 (molecular mass = 28.0
u) and oxygen O2 (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.
29
30
THE INTERNAL ENERGY OF A
MONATOMIC IDEAL GAS
The internal energy of a substance is the sum of the various kinds of
energy that the atoms or molecules of the substance possess. A
monatomic ideal gas is composed of single atoms. These atoms are
assumed to be so small that the mass is concentrated at a point, with
the result that the moment of inertia I about the center of mass is
negligible.
31
Diffusion
The process in which molecules move from a region of higher
concentration to one of lower concentration is called diffusion.
The host medium, such as the air or water, is referred to as
the solvent, while the diffusing substance, like the perfume
molecules, is known as the solute. Relatively speaking,
diffusion is a slow process, even in a gas.
32
Conceptual Example 7.
Why Diffusion Is Relatively Slow
In Example 6 we have seen that
a gas molecule has a
translational rms speed of
hundreds of meters per second
at room temperature. At such a
speed, a molecule could travel
across an ordinary room in just
a fraction of a second. Yet, it
often takes several seconds, and
sometimes minutes, for the
fragrance of a perfume to reach
the other side of a room. Why
does it take so long?
33
When a perfume molecule diffuses through air, it
makes millions of collisions each second with air
molecules. The speed and direction of motion change
abruptly as a result of each collision. Between
collisions, the perfume molecule moves in a straight
line at a constant speed. Although a perfume
molecule does move very fast between collisions, it
wanders only slowly away from the bottle because of
the zigzag path resulting from the collisions. It would
take a long time for a molecule to diffuse in this
manner across a room. Usually, however, convection
currents are present and carry the fragrance across
the room in a matter of seconds or minutes.
34
Using diffusion, a transdermal patch delivers a drug directly into
the skin, where it enters blood vessels. The backing contains the
drug within the reservoir, and the control membrane limits the rate
of diffusion into the skin. Another way to control the diffusion is to
adjust the concentration of the drug in the reservoir by dissolving it
in a neutral material.
35
(a) Solute diffuses through the channel from the region of higher
concentration to the region of lower concentration. (b) Heat is
conducted along a bar whose ends are maintained at different
temperatures.
36
FICK’S LAW OF DIFFUSION
The mass m of solute that diffuses in a time t through a
solvent contained in a channel of length L and crosssectional area A is
where  C is the concentration difference between the ends of
the channel and D is the diffusion constant.
SI Unit for the Diffusion Constant: m2/s
37
Example 8.
Water Given Off by Plant Leaves
38
Large amounts of water can be given off by plants. It has been
estimated, for instance, that a single sunflower plant can lose
up to a pint of water a day during the growing season. Figure
14.17 shows a cross-sectional view of a leaf. Inside the leaf,
water passes from the liquid phase to the vapor phase at the
walls of the mesophyll cells. The water vapor then diffuses
through the intercellular air spaces and eventually exits the leaf
through small openings, called stomatal pores. The diffusion
constant for water vapor in air is D = 2.4 × 10–5 m2/s. A
stomatal pore has a cross-sectional area of about A = 8.0 × 10–
11 m2 and a length of about L = 2.5 × 10–5 m. The
concentration of water vapor on the interior side of a pore is
roughly C2 = 0.022 kg/m3, while that on the outside is
approximately C1 = 0.011 kg/m3. Determine the mass of water
vapor that passes through a stomatal pore in one hour.
39
40
Check Your Understanding 4
The same solute is diffusing through the same solvent in each
case referred to in the table below, which gives the length and
cross-sectional area of the diffusion channel. In each case, the
concentration difference between the ends of the diffusion
channel is the same. Rank the diffusion rates (in kg/s) in
descending order (largest first).
Length
Cross-Sectional
Area
(a)
½L
A
(b)
L
½A
c, b, a
(c)
L
2A
41
Concepts & Calculations Example 9.
The Ideal Gas Law and Springs
42
There are three identical chambers containing a piston and a
spring whose spring constant is k = 5.8 × 104 N/m. The
chamber in part a is completely evacuated, and the piston just
touches its left end. In this position, the spring is unstrained. In
part b of the drawing, 0.75 mol of ideal gas 1 is introduced into
the chamber, and the spring compresses by x1 = 15 cm. In part c,
0.75 mol of ideal gas 2 is introduced into the chamber, and the
spring compresses by x2 = 24 cm. Find the temperature of each
gas.
P = F/A
F = kx
P = kx/A
43
44
Concepts & Calculations Example 10.
Hydrogen Atoms in Outer Space
In outer space the density of matter is extremely low, about one
atom per cm3. The matter is mainly hydrogen atoms (m = 1.67
× 10–27 kg) whose rms speed is 260 m/s. A cubical box, 2.0 m on
a side, is placed in outer space, and the hydrogen atoms are
allowed to enter. (a) What is the magnitude of the force that the
atoms exert on one wall of the box? (b) Determine the pressure
that the atoms exert. (c) Does outer space have a temperature,
and, if so, what is it?
45
(a)
46
(b)
(c)
47
Problem 22
REASONING AND SOLUTION If the pressure at the surface is P1
and the pressure at a depth h is P2, we have that P2 = P1 +  gh. We
also know that P1V1 = P2V2. Then,
V1 P2 P1  gh
gh


 1
V2 P1
P1
P1
V1
(1.000  103 kg/m 3 )(9.80 m /s2 )(0.200 m )
 1
 1.02
5
V2
(1.01 10 Pa)
48
Problem 41
REASONING AND SOLUTION
a. As stated, the time required for the first solute molecule to
2
traverse a channel of length L is t  L /(2D) . Therefore, for
water vapor in air at 293 K, where the diffusion constant is
D=2.4*10-5m2/s , the time t required for the first water molecule
to travel L  0.010 m is

L
0.01m
t

 2.1s
5 2
2D 2 2.4 10 m / s
2
2


49
b. If a water molecule were traveling at the translational rms speed
for water, the time t it would take to travel the distance L  0.010 m
would be given by t  L / vrms, where, according to Equation 14.6
1
( KE  mv2rms ), v
. Before we can use the last
2
rms  2KE/ m
expression for the translation rms speed vrms, we must determine the
mass m of a water molecule and the average translational kinetic
energy KE
Using the periodic table on the inside of the text’s back cover,
we find that the molecular mass of a water molecule is
2(1.00794 u)  15.9994 u  18.0153 u
Mass of two
hydrogen atoms
Mass of one
oxygen atom
50
3
18.015310 kg / m ol
 26
m

2
.
99

10
kg
23
1
6.02210 m ol
The average translational kinetic energy of water molecules at 293
K is, according to Equation 14.6,
KE 

3
3
 23
–23J
1
.
38

10
kT

(1.3810
2
2
v rms 
2KE
m


 21 –

/J/K)
K 293
K

6
.
07

10
293 K = 6.0710J

–21
 21
22 6.07
10
6.07  10 JJ
 26
–26
2.99  10 kg
2.99 10
kg
  637 m / s
 637 m/s
51
Thus, the time t required for a water molecule to travel the
distance at this speed is
t
L
v rms
0.010 m
5
–5

 11.6

10
ss
.
6

10
637 m /s
c. In part (a), when a water molecule diffuses through air, it makes
millions of collisions each second with air molecules. The speed
and direction changes abruptly as a result of each collision.
Between collisions, the water molecules move in a straight line at
constant speed. Although a water molecule does move very
quickly between collisions, it wanders only very slowly in a zigzag
path from one end of the channel to the other. In contrast, a water
molecule traveling unobstructed at its translational rms speed [as
in part (b)], will have a larger displacement over a much shorter
time. Therefore, the answer to part (a) is much longer than the
52
answer to part (b)
Problem 42
REASONING AND SOLUTION Fick’s law of diffusion gives
5
2
4
2
2
3
L D A C  4.2  10 m / s  4.0  10 m  3.5  10 kg/m 
3
v 


7.0

10
m/s
8
t
m
8.4  10 kg
53
Problem 44
REASONING AND SOLUTION
a. The average concentration is Cav = (1/2) (C1 + C2) = (1/2)C2 =
m/V = m/(AL), so that C2 = 2m/(AL). Fick's law then becomes m
= DAC2t/L = DA(2m/AL)t/L = 2Dmt/L2. Solving for t yields
t  L /  2D 
2
b. Substituting into this expression yields
t = (2.5 * 10–2 m)2/[2(1.0 * 10–5 m2/s)] = 31 s
54