Kinetic Molecular
Theory & Gases
An Honors/AP Chemistry
Presentation
Kinetic Molecular Theory
Kinetic means motion
So the K.M.T. studies the
motions of molecules.
Solids - vibrate a little
Liquids - vibrate, rotate, and
translate (a little)
Gases - vibrate, rotate, and
translate (a lot)!
Basic Assumptions of KMT
 Gases consist of
large numbers of
molecules in
continuous
random motion.
 The volume of
the molecules is
negligible
compared to the
total volume.
 Intermolecular
interactions are
negligible.
 When collisions
occur, there is a
transfer of kinetic
energy, but no
loss of kinetic
energy.
 The average
kinetic energy is
proportional to
the absolute
temperature.
Gas Properties
Volume - amount of space (L or
mL)
Temperature - relative amount of
molecular motion (K)
Pressure - the amount of force
molecules exert over a given area
(atm, Torr, Pa, psi, mm Hg)
Moles - the number of molecules
(mol)
Temperature Conversions
C = 5/9(F-32)
F = 9/5C + 32
K = C + 273
So what is the absolute
temperature (K) of an
object at -40 oF?
Answer to Temperature
Conversion
-40 oF = -40 oC
-40 oC = 233 K or 230 K
Pressure Conversions
QuickTime™ and a
decompressor
are needed to see this picture.
1 atm = 760 mm Hg
= 760 Torr =
101,325 Pa = 14.7
psi
How many
atmospheres is 12.0
psi?
How many Torr is
1.25 atm?
How many Pascals is
720 mm Hg?
Answers to Pressure
Conversions
12.0 psi = .816 atm
1.25 atm =950. Torr
720 mm Hg = 96000 Pa
A Barometer
 A mercury barometer
measures air pressure by
allowing atmospheric
pressure to press on a
bath of mercury, forcing
mercury up a long tube.
The more pressure, the
higher the column of
mercury.
More on the barometer
 Although
American
meteorologists
will sometimes
measure the
height in inches,
typically this
pressure is
measured in
mm Hg.
 1 mm Hg = 1
Torr
S.T.P.
When making comparisons
we often use benchmarks or
standards to compare
against.
In chemistry Standard
Temperature is 0 oC (273K)
and Standard Pressure is 1
atm.
Boyle’s Law
If the amount and temperature
of the gas are held constant,
then the volume of a gas is
inversely proportional to the
pressure it exerts.
Mathematically this means that
the pressure times the volume
is a constant.
P*V = k
P1V1=P2V2
Boyle’s Law in Action
Sample Questions
The volume of a balloon is 852
cm3 when the air pressure is
1.00 atm. What is the volume if
the pressure drops to .750 atm?
A gas is trapped in a 2.20 liter
space beneath a piston exerting
25.0 psi. If the volume expands
to 2.75 L, what is the new
pressure?
The Answers are…
P1V1=P2V2
(1atm)(852cm3)= (.750atm)*V2
V2 =1140cm3
(25.0psi)(2.20L)=P2(2.75L)
P2 = 20.0 psi
Charles’ Law
If the amount and the pressure
of a gas are held constant, then
the volume of a gas is directly
proportional to its absolute
temperature.
Mathematically, this means that
the volume divided by the
temperature is a constant.
V/T = k
V1/T1=V2/T2
Charles Law in Action
Sample Questions
The volume of a balloon is 5.00 L
when the temperature is 20.0 oC.
If the air is heated to 40.0 oC,
what is the new volume?
3.00 L of air are held under a
piston at 0.00 oC. If the air is
allowed to expand at constant
pressure to 4.00 L, what is the
new Celsius temperature of the
gas?
The Answers Are…
V1/T1=V2/T2
5.00L/293K = V2/313K
V2=5.34L
273K/3.00L = T2/4.00L
T2=364K=91oC
The Gay-Lussac Law
 If the amount and volume of the gas
are held constant, then the pressure
exterted by the gas is directly
proportional to its absolute
temperature.
 Mathematically this means that the
pressure divided by the temperature
is a constant.
 P/T = k
 P1/T1=P2/T2
The Gay-Lussac Law in Action
Sample Questions
 A tank of oxygen is stored at 3.00
atm and -20 oC. If the tank is
accidentally heated to 80 oC, what is
the new pressure in the tank?
 A piston is trapped in place at a
temperature of 25 oC and a pressure
of 112 kPa. At what celcius
temperature is the pressure 102 kPa?
The Answers are…
P1/T1=P2/T2
(3atm)/(253 K)= P2/ (353 K)
P2 =4.19 atm
(298 K)/(112 kPa)=T2/(102kPa)
T2 = 271 K = -2 oC
Avogadro’s Law
If the temperature and the
pressure of a gas are held
constant, then the volume of a
gas is directly proportional to
the amount of gas.
Mathematically, this means that
the volume divided by the # of
moles is a constant.
V/n = k
or
V/m = k
V1/n1=V2/n2 or V1/m1=V2/m2
Avogadro’s Law in Action
Sample Questions
 The volume of a balloon is 5.00 L
when there is .250 mol of air. If
1.25 mol of air is added to the
balloon, what is the new volume?
 3.00 L of air has a mass of about
4.00 grams. If more air is added so
that the volume is now 24.0 L, what
is the mass of the air now?
The Answers Are…
V1/n1=V2/n2 or V1/m1=V2/m2
5.00L/.250 mol = V2/1.50 mol
V2=30.0 L
4.00g/3.00L = m2/24.0L
m2=32.0 g
The Combined Gas Law
This law combines the inverse
proportion of Boyle’s Law with
the direct proportions of
Charles’, Gay-Lussac’s, and
Avogadro’s Laws.
P1V1/(n1T1) = P2V2/(n2T2)
or
P1V1/T1 = P2V2/T2
Four Gas Laws in One
The combined gas law could be
used in place of any of the
previous 4 gas laws.
For example, in Boyle’s Law, we
assume that the amount and
temperature are constant. So if
we cross them off of the
combined gas law:
P1V1/(n1T1) = P2V2/(n2T2)
P1V1 = P2V2
Another Example
A sample of hydrogen has a
volume of 12.8 liters at 104
oF and 2.40 atm. What is
the volume at STP?
The answer is:
P1V1/(n1T1) = P2V2/(n2T2)
P1=2.40atm,V1=12.8L,
T1=104oF=40oC=313K,
T2=273K, P2=1atm, n1=n2
(2.4atm)(12.8L)/(313K) =
(1atm)V2/(273K)
V2 = 26.8 L
The Ideal Gas Law
If, P1V1/(n1T1) = P2V2/(n2T2)
Then PV/(nT) = constant
That constant is R, the ideal
gas law constant.
R = .0821 L*atm/(mol*K)
R = 8.314 J/(mol*K)
So, PV=nRT
But what about…
Since n = m/M, we can
substitute into PV = nRT
and get
PVM = mRT
Since D = m/V, we can
substitute in again and get
PM = DRT
So which one is it?
Like a good carpenter, it is good
to have many tools so that you
can choose the right tool for the
right job.
If I am solving a gas problem
with density, I use PM = DRT.
If I am solving a gas problem
with moles, I use PV = nRT.
If I am solving a gas problem
with mass, I use PVM = mRT.
Such as….
Under what pressure would
oxygen have a density of 8.00
g/L at 300 K?
PM = DRT
P(32 g/mol) = (8 g/L)(.0821
latm/molK)(300 K)
P = 6.16 atm
An Important Number
What is the volume of 1 mole of a
gas at STP?
PV = nRT V = nRT/P
V =
(1mol)*(.0821Latm/molK)(273K)/
(1atm)
V = 22.4 L
This is called the standard molar
volume of an ideal gas.
Gas Stoichiometry
We had said that stoichiometry
implied a ratio of molecules, or
moles. Up until now we only
used mole ratios.
However Avogadro said that the
volume is directly proportional
to the number of molecules.
This means that we can do
stoichiometry with volume or
moles.
Example 1 of Gas
Stoichiometry
What volume of hydrogen is
needed to synthesize 6.00
liters of ammonia?
N2 (g) + 3 H2 (g) --> 2 NH3 (g)
6.00 L H2 x (2 NH3/3 H2) =
4.00 L NH3
Example 2 of Gas
Stoichiometry
What mass of nitrogen is needed to
synthesize 20.0 L of ammonia at 1.50
atm and 25 oC?
N2 (g) + 3 H2 (g) --> 2 NH3 (g)
20.0 L NH3 x (1 N2/2 NH3) = 10.0 L N2
PVM = mRT
(1.5 atm)(10 L)(28 g/mol) =
m(.0821Latm/molK)(298K)
m = 17.2 g N2
Dalton’s Law
When we talk about air pressure,
we need to understand that air is
not oxygen.
Air is a solution of nitrogen
(78.09%), oxygen (20.95%),
argon (.93%), and CO2 (.03%).
So when we talk about air
pressure, which gas are we
talking about?
ALL OF THEM!
Dalton’s Law of Partial Pressures
states that the total pressure of
a system is equal to the sum of
the partial (or individual)
pressures of each component.
Ptotal = P1 + P2 + … Px
So if air pressure is 1 atm, then
we can assume that the N2 is
.78 atm, the O2 is .21 atm, and
the Ar is about .01 atm.
A Corollary
If we extend Boyle’s Law
and Avogadro’s Law, we
could infer that, at constant
temperature and volume,
the pressure of a gas is
directly proportional to its
pressure.
P1/Ptotal = n1/ntotal
An important example
QuickTime™ and a
decompressor
are needed to see this picture.
 A sample of CaCO3
is heated, releasing
CO2, which is
collected over water
(a typical practice).
The pressure in the collection bottle is the sum
of the pressure of the CO2 plus the pressure of the
water vapor (since some water always
evaporates).
Ptotal = PCO2 + PH2O
Sample Water Vapor
Pressures
Water Vapor Pressure Table
Temperature
Pre ssure
(¡C)
0.0
5.0
10.0
12.5
15.0
15.5
16.0
16.5
17.0
17.5
18.0
18.5
19.9
(mmHg)
4.6
6.5
9.2
10.9
12.8
13.2
13.6
14.1
14.5
15.0
15.5
16.0
16.5
Temperature
Pre ssure
(¡C)
19.5
20.0
20.5
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.5
25.0
26.0
(mmHg)
17.0
17.5
18.1
18.6
19.2
19.8
20.4
21.1
21.7
22.4
23.1
23.8
25.2
Temperature
Pre ssure
(¡C)
27.0
28.0
29.0
30.0
35.0
40.0
50.0
60.0
70.0
80.0
90.0
95.0
100.0
(mmHg)
26.7
28.3
30.0
31.8
42.2
55.3
92.5
149.4
233.7
355.1
525.8
633.9
760.0
So in our example
If a total pressure of 365
Torr is collected at 25 oC in
a 100 ml collection bottle:
What is the partial pressure
of CO2?
What mass of CaCO3
decomposed?
Here’s how it works
Ptotal = PCO2 + PH2O
365 Torr = PCO2 + 23.8 Torr
PCO2 = 341.2 Torr = .449 atm
 PVM = mRT
(.449 atm)(.100 L)(44.0 g/mol)
= m(.0821Latm/molK)(298K)
m = .0807 g CO2
Corollary Problem
A gas collection bottle
contains .25 mol of He, .50
mol Ar, and .75 mol of Ne.
If the partial pressure of
Helium is 200 Torr:
What is the total pressure in
the system?
What are the partial
pressures of Ne and Ar?
The answers are…
nHe = .25 mol, nAr = .50 mol, nNe = .75
mol, PHe = 200 Torr.
ntotal =1.50 mol
Ptotal/Phe = ntotal/nHe
Ptotal/200Torr = 1.50 mol/.25 mol
Ptotal = 1200 Torr
PAr/Ptotal = nAr/ntotal
Par/1200 = .50 mol/1.50 mol
PAr = 400 Torr
PNe = 1200 Torr - 400 Torr - 200 Torr
Pne = 600 Torr
Temperature and Kinetic
Energy
Earlier, I stated that
temperature is a relative
measure of molecular
motion.
By definition, Kinetic energy
is a measure of the energy
of motion.
Pretty similar right?
Yes they are
KEav = 3/2*R*T
The average kinetic energy
depends only on the absolute
temperature.
R, the Ideal Gas Law Constant,
should be 8.314 J/molK, since
we will want the energy in the
proper SI unit of Joules.
A Thought Question
Which of the following ideal
gases would have the
largest average kinetic
energy at 25oC? He, N2, CO,
or H2
They are all the same!
Since Keav = 3/2*R*T, the
mass does not make a
difference (ideally).
KE =
3/2*(8.314J/molK)*(298K)
KE = 3716 J/mol
Speed vs Kinetic Energy
In physics, you learned that KE =
1/2*m*v2. The velocity, v,
describes the speed of an object
in a specific direction. If the
mass, m, is measured in kg and
the velocity is measured in m/s,
then the kinetic energy would
be measured in Joules.
Physics to Chemistry
Rewriting the physics version,
we could say that
v=√(2*KE/m).
In chemistry, the Kinetic energy
is measured in J/mol, so the
mass would have to be
measured in Kg/mol which is
essentially molar mass.
Root Mean Square
Speed
In Chemistry, we are not
worried about velocities in
multiple directions. We
want an average speed
independent of direction.
We call this Vrms - the root
mean square speed.
Vrms = √(3RT/M)
A Thought Question
Revisited
Which of the following ideal
gases would have the
largest root mean square
speed at 25oC? He, N2, CO,
or H2
This Time They Are
Different
 Vrms = √(3RT/M)
 For He, Vrms = √(3RT/M) =
√(3*8.314J/molK*298K)/4g/mol = 1363 m/s
 For N2, Vrms = √(3RT/M) =
√(3*8.314J/molK*298K)/28g/mol = 515 m/s
 Since the molar mass is the same for N2 and CO,
their Vrms would be the same, 515 m/s.
 For H2, Vrms = √(3RT/M) =
√(3*8.314J/molK*298K)/2g/mol = 1928 m/s
 Because H2 is the lightest, it moves the fastest.
And this leads us to…
QuickTime™ and a
decompressor
are needed to see this picture.
Rate1*√M1 = Rate2*√M2
Graham’s Law
The rate of effusion (or
diffusion) is inversely
proportional to the
square root of the
molar mass.
Effusion is the process
of a gas escaping from
one container through
a small opening.
Diffusion is the process
of a gas spreading out
in a large container.
Rate vs. Speed
QuickTime™ and a
decompressor
are needed to see this picture.
 When we say rate, we are
talking about an amount of
gas (moles, grams, or even
liters) per unit of time.
 This is not the same as
speed which is distance
over time.
 However, the main idea is
the same; lighter gases
move/effuse faster.
For example
Under a given set
of conditions,
oxygen diffuses at
10 L/hr. A different
gas diffuses at 20
L/hr under the
same conditions.
What is the molar
mass of this gas?
2 ways to solve this
By the equation:
Qui ckT ime™ and a
dec ompress or
are needed to s ee this pic ture.
 Rate1*√M1 = Rate2*√M2
 10 L/hr(√32g/mol) = 40 L/hr *√M2
 M2 = 2 g/mol
By Logic:
 If the rate of the unknown gas is 4
times faster, it must be 42, or 16,
times lighter.
 32 g/mol divided by 16 is 2 g/mol.
Real vs. Ideal
At the start of the presentation,
we talked about the major
assumptions of the Kinetic
Molecular Theory.
If a gas obeys the KMT, it is
ideal.
If it doesn’t obey the KMT, it is
real.
So What does that Mean?
The molecules of an ideal gas
do not interact with one
another, except to collide
elastically.
The molecules of a real gas
will interact, to some degree.
Since no gases are always
ideal, the trick is to make a
real gas behave ideally.
Real Gases Behaving Ideally
If we don’t want the
molecules attracting or
repelling one another, the
first issue is to use a
nonpolar gas.
If we use smaller amounts
of the gas, there are less
chances of them interacting.
Real Gases Behaving Ideally
If we put the gas in larger
volumes, the molecules will not
interact as much.
Likewise, if we keep the gas
under low pressure , the
molecules will not interact as
much.
This could also be stated by
having molecules that have low
densities.
Real Gases Behaving Ideally
The smaller the molecules, the
less likely they are to interact.
Lastly, at higher temperatures
the molecules are moving too
fast to actually interact with one
another - they are more likely to
collide elastically.
Phase Diagrams
A phase
diagram
shows how
the different
states of
matter exist
based on the
pressure and
temperature.
A Typical Phase Diagram
Water is not Typical…
…and Helium is weird!