Kinetic Molecular Theory (KMT)
Gases consist of very small particles (atoms or
molecules) which are separated by large
distances.
 Most of the volume occupied by a gas is
empty space.
Gas molecules move randomly at very high
speeds in all directions.
Pressure results from molecules colliding with
the walls of the container.
5-1
Kinetic Molecular Theory (KMT)
There are no intermolecular forces between
the gas molecules except when they collide.
All collisions between molecules are elastic
collisions.
 The internal energy of the system
remains the same.
The KE of molecules is dependent upon the
absolute temperature.
5-2
STP
Standard temperature and pressure,
 T = 273.15K ≈ 273K
 P = 1 atm = 760 mm Hg = 76 cm Hg
 1.0 mol of any gas = 22.4 L = 22.4 dm3
5-3
The Gaseous Phase
In the gaseous state, the molecules have
sufficient energy to overcome the
intermolecular forces that attract them to
each other.
Each molecule acts independent
of the others.
Gases have low densities
and completely fill the volume
that is available to them.
5-4
Units Used In Measuring Gases
Volume
mL, L, cm3, dm3, m3
Temperature
Must use K in gas laws.
Pressure
atm, torr, kPa, Pa
Moles
Amounts measured in moles.
5-5
What Makes a Gas an Ideal Gas?
Any ideal gas assumes the:
 volume that the gas molecules
themselves occupy is neglected.
 gas molecules do not attract each other.
Ideal gases are approximated when the
intermolecular attractions are very small.
5-6
Low pressures and high temperatures are the
conditions by which a gas approaches the
ideal.
 Low pressures enable gas molecules to
spread very far apart.
 High temperatures represent a large
amount of kinetic energy which
overcomes attractive forces.
Small gas molecules such as hydrogen with
only dispersion forces best approximate the
ideal.
5-7
Deviations from the ideal gas law occur most
at low temperatures and high pressures.
The ideal gas law (IGL) is given by
PV = nRT
L•atm ,
where the gas constant R = 0.0821
mol•K
P is the pressure in atm, V is the volume in L,
n is the number of moles, and T is the Kelvin
or absolute temperature.
5-8
Ideal Gas Law Problems
What is the pressure exerted by 31.0 moles of
carbon dioxide that is contained in a 110. liter
tank at 28°C?
n = 31.0 moles V = 110. L
R = 0.0821
T = 28°C = 301 K
L•atm
mol•K
5-9
PV = nRT
P =
L•atm
31.0 mol × 0.0821
mol•K
× 301 K
110. L
P = 6.96 atm
5 - 10
A gas with a mass of 3.47 g occupies 282 mL
at 23.0°C and 743 mm Hg. Determine the
molecular mass of the gas.
m = 3.47 g
V = 282 mL
T = 295 K
L•atm
P = 743 mm Hg R = 0.0821
mol•K
m
PV = nRT
n=
MM
mRT
MM =
PV
5 - 11
MM =
L•atm
3.47 g × 0.0821
mol•K
× 295 K
743 mm × 1 atm × 282 mL × 1 L
760 mm
103 mL
MM = 305 g•mol-1
5 - 12
Charles’s Law
Charles’s law says that at constant
pressure, the volume of a gas is directly
proportional to the absolute temperature
(K).
Remember the gas laws require
temperatures to be expressed in Kelvin to
avoid dividing by zero.
5 - 13
Two quantities, V and T, are directly
proportional if by whatever factor T changes,
V changes by the same factor.
A graph of V vs T yields a straight line
through the origin.
The following graph looks to be an exception
but it is not because the temperature is
expressed in °C.
5 - 14
The graph shows that when the line is
extrapolated to -273.15°C, you expect the
volume to be zero.
At 0 K, a gas is predicted to have a zero
volume but all gases either liquefy or solidify
before reaching that temperature.
5 - 15
.
-273.15 °C = 0 K (absolute zero)
5 - 16
Deriving Charles’s Law
Charles’s law can be derived from the ideal
gas law.
Assume the same gas has existed at two
different temperatures while the pressure
and the number of moles of gas remained
constant.
5 - 17
Applying the IGL at the two different
temperatures yields:
P1V1 = nRT1
P2V2 = nRT2
Dividing one equation by the other yields:
P1V1
P2V2
V1
T1
=
nRT1
nRT2
V2
=
T2
=>
P1V1
P2V2
=
nRT1
nRT2
(Charles’s Law)
5 - 18
Charles’s Law Problem
A sample of oxygen has a volume of 350 mL at
60.0°C. What will be the volume of oxygen at
standard temperature if the pressure remains
constant?
V1 = 350 mL
T1 = 60.0°C = 333 K
P1 = P2
V2 = ?
T2 = 0.0°C = 273 K
P2 = P1
5 - 19
.
V1
T1
V2
=
T2
V1
V2 =
× T2
T1
V2 = 350 mL × 273 K = 290 mL O2
333 K
5 - 20
Boyle’s Law
Boyle’s law says that at a constant
temperature for a fixed quantity of gas, the
volume of the gas is inversely proportional
to the pressure.
Two quantities, P and V, are inversely
proportional because by increasing either
variable by a factor causes a decrease in
the other variable that was the inverse of
the first factor.
5 - 21
A graph of V vs P yields a hyperbola.
When the pressure is doubled (2 ×), the volume
is halved (1/2).
5 - 22
A graph of V vs 1/P yields a straight line.
5 - 23
Deriving Boyle’s Law
Boyle’s law can be derived from the ideal
gas law.
Assume the same number of moles of gas has
remained at the same temperature while the
pressure and volume have varied.
5 - 24
Applying the IGL at the two different
pressure yields:
P1V1 = nRT1
P2V2 = nRT2
Dividing one equation by the other yields:
P1V1
P2V2
=
nRT1
nRT2
P1V1 = P2 V2
=>
P1V1
P2V2
=
nRT1
nRT2
(Boyle’s Law)
5 - 25
Boyle’s Law Problem
A gas tank holds 2750 L of nitrogen at
920 mm Hg. If the temperature remains
constant, what will be the volume of nitrogen
at standard pressure?
P1 = 920 mm
V1 = 2750 L
T1 = T 2
P2 = 760 mm Hg
V2 = ?
T2 = T1
5 - 26
.
P1V1 = P2V2
P1V1
V2 =
P2
V2 = 920 mm × 2750 L = 3300 L N2
760 mm
5 - 27
Avogadro’s Hypothesis and Law
Avogadro’s hypothesis says that equal
volumes of gases at the same temperature
and pressure contain equal numbers of
molecules.
Alternatively, 1.00 mole of any gas at STP
(0°C, 1.00 atm) occupies a volume of
22.4 L.
5 - 28
Avogadro’s law says that the volume of a gas
maintained at constant temperature and
pressure is directly proportional to the
number of moles of gas.
5 - 29
Deriving Avogadro’s Law
Avogadro’s law can be derived from the ideal
gas law.
Assume the pressure and temperature
remain constant.
5 - 30
Applying the IGL at the two different
volumes yields:
P1V1 = n1RT1
P2V2 = n2RT2
Dividing one equation by the other yields:
P1V1
P2V2
=
V1
V2
=
n1
n2
n1RT1
n2RT2
=>
P1V1
P2V2
=
n1RT1
n2RT2
(Avogadro’s Law)
5 - 31
General Gas Law
The general gas law (combined gas law) is
used when you are not restricting the IGL to
only two variables (P, V or V, T). In most
instances, two of the three physical
properties of a gas are subject to change.
The general or combined gas law can be
derived from the ideal gas law.
5 - 32
Assume the same gas has existed at two
different temperatures while only the number
of moles of gas have remained constant.
Mr. Bean has run out of gas!
5 - 33
Applying the IGL at two different pressures,
temperatures, and volumes yields:
P1V1 = nRT1
P2V2 = nRT2
Dividing one equation by the other yields:
P1V1
P2V2
=
nRT1
nRT2
P1V1
P2 V2
=
T1
T2
=>
P1V1
P2V2
=
nRT1
nRT2
(General Gas Law)
5 - 34
General Gas Law Problem
What volume will 4.50 L of SO2 at -27°C and
600. mm Hg occupy if the temperature
changes to 40.°C and the pressure to
700. mm Hg?
P1 = 600. mm Hg
V1 = 4.50 L
T1 = -27°C = 246 K
P2 = 700. mm
V2 = ?
T2 = 40.°C = 313 K
5 - 35
P1V1
=
T1
P2V2
T2
600. mm × 4.50 L × 313 K
V2 =
246 K × 700. mm
V2 = 4.91 L SO2
5 - 36
Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases
equals the sum of the pressures that each
gas would exert if it was by itself.
Each gas of behaves independently of the
others.
PT = P1 + P2 + … Pn
where n = total number of gases
5 - 37
To determine the pressure of one of the
gases present in the mixture, the mole
fraction is used.
Mole fraction, X, is a dimensionless number
(no unit) which expresses the ratio of the
number of moles of a particular component to
the total number present in the mixture.
n1
P1 = X1 × PT and X1 =
n2
5 - 38
Dalton’s Law of Partial Pressures
A 195 cm3 sample of helium is collected over
water at 16°C and 97.5 kPa. What volume
would the dry gas occupy at 117 kPa and
28°C?
P1 = 97.5 kPa
V1 = 195 cm3
T1 = 16°C = 289 K
P2 = 117 kPa
V2 = ?
T2 = 28°C = 301 K
5 - 39
PT = PHe + PW
PHe = PT – PW
PHe = 97.5 kPa – 1.81 kPa = 95.7 kPa
PW is the vapor pressure of water at 16°C
which can be found in a table of vapor
pressures of water.
It is important to note that the total pressure,
97.5 kPa, is the sum of the water vapor
pressure and the pressure of helium.
5 - 40
P1V1
=
T1
P2V2
T2
97.5 kPa × 195 cm3 × 301 K
V2 =
289 K × 117 kPa
V2 = 1.70 × 102 cm3 He
5 - 41
Molar Volume Not At STP
What is the density of argon at 37°C and
103.7 kPa?
P1 = 101.3 kPa
V1 = 22.4 dm3
T1 = 0°C = 273 K
P1V1
T1
=
P2 = 103.7 kPa
V2 = ?
T2 = 28°C = 310 K
P2V2
T2
5 - 42
.
V2 =
101.3 kPa × 22.4 dm3 × 310 K
273 K × 103.7 kPa
V2 = 24.8 dm3
M
=
D
V
39.95 g
3
=
D =
1.62
g/dm
24.6 dm3
5 - 43
Gas Stoichiometry
What mass of water is produced when
625 dm3 of hydrogen gas measured at 28°C
and 97.3 kPa is burnt in the presence of
oxygen as shown in the following reaction.
2H2(g) + O2(g) → 2H2O(g)
P1 = 97.3 kPa
V1 = 625 dm3
T1 = 27°C = 301 K
P2 = 101.3 kPa
V2 = ?
T2 = 0°C = 273 K
5 - 44
.
P1V1
P2V2
=
T1
T2
V2 =
97.3 kPa × 625 dm3 × 273 K
301 K × 101.3 kPa
V2 = 544 dm3 H2
STP, standard temperature and pressure, is
101.3 kPa and 0°C.
5 - 45
.
m = 544
×
dm3
H2 ×
18.02 g H2O
2 mol H2O
1 mol H2
×
3
22.4 dm H2
2 mol H2O
2 mol H2
= 438 g H2O
Remember that Avogadro’s hypothesis states
that 1 mol of any gas occupies a volume of
22.4 L (or 22.4 dm3) at STP.
This is the reason that the volume of the
hydrogen had to be determined at 0°C before
doing the stoichiometry.
5 - 46
More Gas Stoichiometry
If 528 g of carbon disulfide react with oxygen,
what volume of sulfur dioxide will be formed
at 29°C and 98.2 kPa?
CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)
m = 528 g CS2
5 - 47
V = 528 g CS2 ×
×
22.4 L SO2
1 mol SO2
P1 = 101.3 kPa
V1 = 311 L SO2
T1 = 0°C = 273 K
1 mol CS2
×
76.13 g CS2
2 mol SO2
1 mol CS2
= 311 L SO2
P2 = 98.2 kPa
V2 = ?
T2 = 29°C = 302 K
5 - 48
P1V1
P2V2
=
T1
T2
V2 =
101.3 kPa × 311 L × 302 K
273 K × 98.2 kPa
V2 = 355 L SO2
Starting with 528 g CS2, you determine the
volume of SO2 produced at 0°C, but the problem
asked for the volume at 29°C.
5 - 49
Effusion and Diffusion
According to Kinetic Molecular Theory (KMT),
the average kinetic energy of gas molecules
is given by
KE = ½mv2
Because the KE determines the temperature
of a substance, the molecules of hydrogen
gas have the same KE as those of Kr at the
same temperature.
5 - 50
A consequence of this is that the H2
molecules must be moving faster than those
of Kr to compensate for their smaller mass.
Two different types of molecular motion are
diffusion and effusion.
 Diffusion is the spreading out of one
substance throughout a second
substance. An example of this is
opening a bottle of ammonia (NH3).
5 - 51
 Effusion is the escape of gas molecules
through a tiny hole into evacuated
space.
To quantitatively compare gases, Graham’s
law is used. Graham’s law is given by:
r1
r2 =
MM2
MM1
where r1 and r2 are the rates of effusion and
the MM’s are the molar masses.
5 - 52
Graham’s law states that the rate of effusion
of gas molecules is inversely proportional to
the square root of their masses.
This means that a lighter gas such as H2
effuses quicker than a more massive gas like
Ar.
5 - 53
Graham’s Law Problem
Compare the rates of effusion of hydrogen
and krypton.
r1 = H2
MM1 = 2.02 g•mol-1
r1
r2 =
MM2
MM1
=
r2 = Kr
MM2 = 83.80 g•mol-1
83.80 g•mol-1
2.02 g•mol-1
= 6.44
H2 effuses 6.44 times faster than Kr.
5 - 54
Real Gases
Ideal gases require two conditions:
 The gas molecules themselves do not
occupy any volume.
 There are no intermolecular attractions.
Real gases start to approach these ideal
conditions at low pressures and high
temperatures.
5 - 55
Real gases contradict these conditions
because:
 the volume available to the molecules
making up a real gas is less than the
volume of the container because the
molecules take up some space.
 the pressure exerted on the sides of the
container is diminished because the
intermolecular attractions result in fewer
collisions with the sides of the container.
5 - 56
van der Waal’s Equation
(Pobs + a(n/V)2) × (V – nb) = nRT
The first term involving the Pobs is the corrected
pressure factor. This correction is necessary
because of the intermolecular attractions. As a
result of these attractions, the observed
pressure will be smaller than predicted from
the ideal gas law.
5 - 57
The second term, V – nb, is the corrected
volume which will be smaller than the
container volume because the gas molecules
themselves have volume.
 There will be less free space for the
molecules to move about.
The constants a and b are called empirical
constants because their value is determined
by fitting the equation to the experimental
results.
5 - 58
The values for a and b are different for
different molecules and can be found in a
table of physical constants.
5 - 59
van der Waals Problem
Calculate the pressure that H2O will exert at
50°C if 1.00 mol occupies 32.0 L assuming
H2O obeys the van der Waals equation. The
constants for H2O are:
L2•atm
a = 5.46
mol2
T = 50°C = 323K
b = 0.0305
V = 32.0 L
L
mol
n = 1.00 mol
5 - 60
(Pobs + a(n/V)2) × (V – nb) = nRT
.
2•atm
L
2)
(1.00
mol/32.0
L)
(Pobs + 5.46
×
mol2
× 32.0 L – 0.0305 L
mol
= 0.0821 L•atm × 323 K = 0.829 atm
mol•K
5 - 61