Chapter 14
The Ideal Gas Law and
Kinetic Theory of Gases
14.1 Molecular Mass, the Mole, and Avogadro’s Number
• The atomic mass scale facilitates comparison of the atomic
mass (protons, neutrons and electrons) of one atom with
another.
• The unit is called the atomic mass unit (symbol u). 1 u =
1.6605 x 10-27 kg.
• The atomic mass number given in the period table is a
weighted average of the mass of all isotopes of an element.
14.1 Molecular Mass, the Mole, and Avogadro’s Number
•One mole of a substance contains as many
particles as there are atoms in 12 grams of the
isotope carbon-12 (atomic mass =12 u).
•The number of atoms or molecules per mole
is a constant number known as Avogadro’s
number, (NA ).
NA = 6.022 x 1023 particles/mole of substance
N A  6.022 1023 mol 1
The number of moles, n, of a substance is:
N
n
NA
Where N is the total number of particles, and
NA is Avogadro’s Number.
14.1 Molecular Mass, the Mole, and Avogadro’s Number
The molar mass (M) of a substance is the mass
per mole in grams/mole. Note grams are not
SI units!
M ( g / mol)  (mparticle)( N A )
The number of moles, n, is then:
n(mol) 
mparticle( g )
M (g/mol)
Where
n = number of moles,
m = mass (in grams)
NA = Avagadro’s number
M = mass in grams per mole
14.1 Molecular Mass, the Mole, and Avogadro’s Number
• The mass per mole (M, in grams/mole) of a
substance has the same numerical value as the
atomic or molecular mass of the substance (in
atomic mass units).
• For example, the atomic mass of hydrogen is
1.00794 u (1 proton plus the occasional
neutron or two). Therefore, hydrogen has an
molar mass (Mhydrogen ) of 1.00794 g/mol.
Likewise, lithium has a molar mass of 6.941
g/mol, and magnesium has a molar mass of
24.305 g/mol.
• If SI units are necessary, change grams to kg.
EXAMPLE Moles of oxygen
How many moles
of oxygen gas, O2,
are there in 100 g
of the gas?
Assume that each
molecule of this
diatomic gas
consists of two O16 atoms.
EXAMPLE Moles of oxygen
QUESTION: How many moles of oxygen
gas, O2, are there in 100g?
Known:
Molecular mass of oxygen gas (O2) = 32u. Therefore molar
mass (M) = 32g
m = 100g
Find: n
What do we know? The number of moles in a sample equals the
mass of the sample divided by its molar mass.
n = m/M
EXAMPLE Moles of oxygen
QUESTION: How many moles of oxygen
gas, O2, are there in 100g?
Known:
Atomic mass of oxygen gas (O2) = 32u. Therefore molar mass
(M) = 32g
m = 100g
Find: n
Pertinent relationship: The number of moles in a sample equals
the mass of the sample divided by its molar mass.
n = m/M = 100 g/32g/mol = 3.13 moles
Moles, molar mass and particles
Argon gas has an atomic mass = 39.948 u, neon
has an atomic mass = 20.180 u, and helium has
an atomic mass = 4.0026 u.
You have a mole of each substance in 3
separate containers. Which container has the
greatest mass?
A. Argon
B. Helium
C. All have the same
D. No way to tell
Moles, molar mass and particles
Argon gas has an atomic mass = 39.948 u, neon
has an atomic mass = 20.180 u, helium has an
atomic mass = 4.0026 u.
You have a mole of each substance in 3
separate containers. Which container has the
most gas atoms?
A. Argon
B. Helium
C. All have the same
D. No way to tell
How many moles?
Consider a mixture of three different monatomic
gases: 1.10 g of argon (molar mass = 39.948
g/mol), 2.48 g of neon (molar mass = 20.180
g/mol), and 3.12 g of helium (molar mass =
4.0026 g/mol). How many moles of gas atoms
are in the container?
How many moles?
Consider a mixture of three different monatomic
gases: 1.10 g of argon (molar mass = 39.948
g/mol), 2.48 g of neon (molar mass = 20.180
g/mol), and 3.12 g of helium (molar mass =
4.0026 g/mol). How many moles of gas atoms
are in the container?
n = m/M (do separately for each)
na = .0275 mol, nn = .123 mol, nh = .779 mol
ntotal = .930 moles
EOC #8 percentages of atoms
Consider a mixture of three different monatomic
gases: 1.10 g of argon (molar mass = 39.948
g/mol), 2.48 g of neon (molar mass = 20.180
g/mol), and 3.12 g of helium (molar mass =
4.0026 g/mol). For this mixture, determine the
percentage of the total number of atoms that
corresponds to each of the components.
EOC #8 percentages of atoms
Consider a mixture of three different monatomic gases:
1.10 g of argon (molar mass = 39.948 g/mol), 2.48 g
of neon (molar mass = 20.180 g/mol), and 3.12 g of
helium (molar mass = 4.0026 g/mol). For this
mixture, determine the percentage of the total number
(N) of atoms that corresponds to each of the
components. Starting with argon:
fa = Na /Ntot but n = N/NA so use n (number of moles)
instead.
fa = na /ntot = (ma /Ma )/ntot and so forth. Multiply
answers by 100 to convert into percentages.
An ideal gas is an idealized
model for real gases with low
densities (i.e. the space that
the gas molecules occupy is
much smaller than the
container).
The pressure from the gas
comes from the force of gas
molecules when they collide
against the wall of the
container.
Properties of ideal gases were
determined experimentally
by investigators in the 17th19th centuries.
We can’t use:
Pb = Pt + ρgh because
a gas is not an
incompressible fluid,
and density is not
constant.
14.2 The Ideal Gas Law
Boyle’s Law (circa 1660s): It can be shown experimentally that, at constant
temperature, the pressure of an ideal gas is inversely proportional to the volume.
PV = constant, for a fixed number of moles at a given temperature.
P 1 V
This is the shape of the
graph for an inverse
relationship:
P = k(1/V) where k is a
constant (analogous to a
slope)
PV is equal to the same
value (k) anywhere on the
curve. This value changes
if the temperature changes.
Isotherm – curve or line of equal
temperature.
Gay-Lussac Law (circa 1700):
Experimentally, it can be
shown that at constant volume
the pressure of an ideal gas is
proportional to the temperature,
i.e. the relationship plots as a
line:
P T
or P = cT where c is a constant.
A temperature of absolute zero
theoretically corresponds to
where pressure is zero, i.e. the
molecules are not moving, so
they cannot hit the container.
Charles Law(circa 1780s)
Experimentally, it can be shown that at a constant pressure , the
volume of an ideal gas is proportional to the temperature:
V T
Or V = cT where c is a constant
http://www.grc.nasa.gov/WWW/K-12/airplane/aglussac.html
Avogadro’s Law (circa 1811)
• If two containers of
different ideal gases
are at the same
pressure, volume and
temperature, then
they contain the
same number of
particles, regardless
of the mass of those
particles.
Atomic mass = 2u
Atomic mass = 32u
The Ideal-Gas Law – A combination of Gay-Lussac,
Charles, Avogadro’s and Boyle’s Law
The pressure p, the volume V, the number of moles n and
the temperature T of an ideal gas are related by the idealgas law as follows:
where R is the universal gas constant, R = 8.31 J/mol K, and T
must be in Kelvin (Celsius + 273)
The ideal gas law may also be written as
where N is the number of molecules in the gas rather than the
number of moles n. The Boltzmann’s constant is kB = 1.38 ×
10−23 J/K, and is equal to R/NA.
The mass of a gas (!)
These 2 containers containing
H2 and O2 are at the same
pressure, volume and
temperature. Use the gas law
to determine which container
has greater mass:
A. O2
B. H2
C. Both have the same mass
D. Unable to determine
without numerical values
for the other variables
Atomic mass = 2u
Atomic mass = 32u
Calculating a gas temperature using ratios
A cylinder of gas is at 0◦ C. A piston
compresses the gas to half its original volume
and 3 times it’s original pressure. What is the
final temperature of the gas?
Calculating a gas temperature using ratios
A cylinder of gas is at 0◦ C. A piston compresses the gas to half its
original volume and 3 times it’s original pressure. What is the
final temperature of the gas?
T1 = (P1 V1 )/nR and T2 = (P2 V2 )/nR
T2 = (P2 V2 )/nR so T2 = T1 (P2 /P1 ) (V2 /V1 )
T1 = (P1 V1 )/nR
(P2 /P1 ) = 3, and (V2 /V1 ) = .5
T2 = 273 K (3) (.5) = 409 K
EOC #17 – party time
A clown at a birthday party has brought along a helium cylinder,
with which he intends to fill balloons. When full, each balloon
contains 0.036 m3 of helium at an absolute pressure of 1.20 x
105 Pa. The cylinder contains helium at an absolute pressure of
1.80 x 107 Pa and has a volume of 0.0031 m3. The temperature
of the helium in the tank and in the balloons is the same and
remains constant. What is the maximum number of balloons
that can be filled (rounded to a whole number)?
Hint: There is the same amount of moles of gas (n) in x balloons
as there is in the whole cylinder.
Moveable Piston
A frictionless gas-filled cylinder
(V = πr2 h) is fitted with a
movable piston, as the drawing
shows. The height h is 0.110 m
when the temperature is 271 K
and increases as the temperature
increases. What is the value of h
when the temperature reaches 315
K? Another ratio problem?
PV = nRT
P(πr2 h) = nRT
(πr2 h) = nRT/P
πr2 h2 =(nRT2 )/P2
πr2 h1 =(nRT1 )/P1
And if we divide out quantities
that don’t change
h2 = (T2/P2 )
h1 =(T1/P1)
Too many unknowns?
Moveable Piston
FBD of piston and mass
PgasA
PaA
Fg
ΣF = 0
Pgas A = PatmA + mg
Pgas = Patm + mg/A
Where cross-sectional area of cap
Use Newton’s Law to find
an expression for pressure
and see if pressure depends
on volume, temperature or
number of moles if the
container is capped by a
moveable piston.
Moveable Piston
FBD of piston and mass
PgasA
PaA
Fg
ΣF = 0
Pgas A = PatmA + mg
Pgas = Patm + mg/A
Where cross-sectional area of cap
Pgas = Patm + mg/A
If T changes does P change?
If V changes does P change?
If n changes does P change?
NO! P is constant for a
moveable piston
Moveable Piston
A frictionless gas-filled
cylinder (V = πr2 h) is fitted
with a movable piston, as
the drawing shows. The
height h is 0.110 m when the
temperature is 271 K and
increases as the temperature
increases. What is the value
of h when the temperature
reaches 315 K? Another
ratio problem?
P(πr2 h) = nRT
(πr2 h) = nRT/P
πr2 h2 =(nRT2 )/P2
πr2 h1 =(nRT1 )/P1
And if we divide out quantities
that don’t change
h2 = T2
h1 = T1
Two different gases at the same temperature are in
cylinders with identical moveable pistons. Which of the
following is different for the gases (besides volume)?
a. Pressure
b. Number of moles
c. Both pressure and number
of moles
d. Either pressure and/or
number of moles, but it is
impossible to determine
which without more
information.
The diving bell (#24)
A primitive diving bell consists of a cylindrical tank
with one end open and one end closed. The tank is
lowered into a freshwater lake, open end downward.
Water rises into the tank, compressing the trapped air,
whose temperature remains constant during the
descent. The tank is brought to a halt when the
distance between the surface of the water in the tank
and the surface of the lake is 40.0 m. Atmospheric
pressure at the surface of the lake is 1.01 x 105 Pa.
Find the fraction of the tank's volume that is filled
with air.
Another Planet – Number density
On a hypothetical planet, the atmospheric pressure is 3.0
x 106 Pa, and the temperature is 900 K. On the earth's
surface the atmospheric pressure is 1.0 x 105 Pa, while
the surface temperature can reach 320 K. Researchers
can determine how “thick” the atmosphere of this
planet is compared to that of Earth, by comparing their
number densities (N/V) . Number density is the
number of particles per unit volume (units of m-3 )
Find the ratio:
(N/V)Planet / (N/V)Earth.
It is necessary to use the “particle version” of the Ideal
Gas Law: PV = NkT.
PV diagrams
• Each point on the
graph represents
unique values of
V,P.
• If the number of
moles are known, T
can be determined
with the ideal gas
law.
PV Diagram
Rank the
temperature
of one mole
of an ideal gas
on different
points on the
PV diagram
What is the ratio
Tf /Ti for this
process?
A.
B.
C. 4
D. 2
E. 1 (no change)
Kinetic Gas Theory
• The pressure of the gas comes from the
force due to the collisions with the
container wall, divided by the surface
area.
• Collisions with the container wall do
not change the speed of the ideal gas
(only direction) molecules, but collisions
with other gas molecules change both
speed and direction.
Speed distribution in an ideal gas
• A fixed amount of an ideal gas in a container has gas
molecules moving at a number of speeds.
• If the temperature increases, the speeds increase.
• Note that the range of speeds (width) increases as well.
Maxwell distribution curves for O2
Speed distribution in an ideal gas
http://www.chm.davidson.edu/ChemistryApplets/KineticMolecularTheory
/Maxwell.html
Maxwell distribution curves for O2
Kinetic Gas Theory
The molecules of an ideal gas travel at many different
speeds. Use vrms as an “average speed”.
Starting with Newton’s 2nd Law (what else?) it can be
shown that: PV = 2/3 N (1/2 mvrms 2 )
Where:
• N = number of molecules in the container
• m = mass of one molecule of the ideal gas (really small number)
• vrms = rms speed,
• P = pressure,
• V = volume
• 1/2 mvrms 2 is the kinetic energy = K of the “ average
molecule” in the container
Kinetic Gas Theory
Newton’s Law predicts:
PV = 2/3 N (1/2 mv 2rms )
Experimental evidence (ideal gas law) states:
PV = NkT (molecular version)
The N term cancels out and leaves:
KEav/molecule = (3/2)kT
This shows that:
• The temperature of an ideal gas is determined only by
its temperature
• The product PV is an energy term: PV = (2/3N)KE
Example root mean squared speed of a cesium atom
What is the rms speed for cesium atoms at a temperature of 1 x 10-6 K (a
very low temperature!)
14.3 Kinetic Theory of Gases
Example Total microscopic kinetic energy
What is the total microscopic kinetic energy of a mole of an ideal gas at 0˚ C
and standard atmospheric pressure ?
Which system has the largest average
translational kinetic energy per
molecule?
A. 2 mol of He at p = 2 atm, T = 300 K
B. 2 mol of N2 at p = 0.5 atm, T = 450 K
C. 1 mol of He at p = 1 atm, T = 300 K
D. 1 mol of N2 at p = 0.5 atm, T = 600 K
E. 1 mol of Ar at p = 0.5 atm, T = 450 K
Microscopic Kinetic energy of a volume of gas
Ktot = 3/2 N k T
where N is the number of molecules of gas in the
volume: Nk = nR
where n is the number of moles of gas.
The total internal energy, U, of a gas, consists of
the kinetic energy AND the potential energy of
its molecular bonds
In a monatomic gas (molecules composed of a
single atom), there are no molecular bonds:
Total internal energy U = 3/2 N k T
Uint of a monatomic ideal gas
In a monatomic ideal gas, all energy is kinetic. The
average kinetic energy per molecule is:
2
KE  12 mvrms
 32 kT
U  N 32 kT  32 nRT
Monatomic gases include helium, neon, argon, krypton,
xenon and radon. Watch out Superman!
Example Problem using number density
• A cylinder of monatomic gas has a pressure
that is twice as great as standard amospheric
pressure and a number density (N/V) of 4.2 x
1025 m-3. The rms speed of the atoms is 660
m/s. Identify the gas.
EOC # 37
Initially, the translational rms speed of a
molecule of an ideal gas is 578 m/s. The
pressure and volume of this gas are kept
constant, while the number of molecules is
doubled. What is the final translational rms
speed of the molecules?
EOC #38
Two gas cylinders are identical. One contains the
monatomic gas neon (Ne), and the other
contains an equal mass of the monatomic gas
xenon (Xe). The pressures in the cylinders are
the same, but the temperatures are different.
Determine the ratio KEAvg,xenon / KEAvg,neon of
the average kinetic energy of a xenon atom to
the average kinetic energy of a neon atom.
EOC #42
Helium (He), a monatomic gas, fills a 0.009 m3
container. The pressure of the gas is 5.60 x 105
Pa. How long would a 0.25 horsepower engine
have to run (1 hp = 746 W) to produce an
amount of energy equal to the internal energy
of this gas?
EOC #42
Helium (He), a monatomic gas, fills a 0.009 m3
container. The pressure of the gas is 5.60 x 105
Pa. How long would a 0.25 horsepower engine
have to run (1 hp = 746 W) to produce an
amount of energy equal to the internal energy
of this gas?
Ans: 40.5 seconds
EOC #9
A cylindrical glass of water (H2O) has a radius
of 3.75 cm and a height of 10.00 cm. The
density of water is 1.00 g/cm3. How many
moles of water molecules are contained in the
glass?
Fixed Piston cap (isochoric)
You have two containers of equal volume (in this case the
piston is not free to move, it is fixed). One is full of helium
gas. The other holds an equal mass of nitrogen gas. Both
gases have the same pressure. How does the temperature of
the helium compare to the temperature of the nitrogen?
Mass is measured in grams or kg.
A. Thelium > Tnitrogen
B. Thelium < Tnitrogen
C. Thelium = Tnitrogen
He
N2