Temperature and Kinetic Theory of Gases slides

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Temperature and theTemperature and the
Kinetic Theory of Gases
Physics Rapid Learning Series
www.RapidLearningCenter.com/
© Rapid Learning Inc. All rights reserved.
Wayne Huang, Ph.D.
Keith Duda, M.Ed.
Peddi Prasad, Ph.D.
Gary Zhou, Ph.D.
Michelle Wedemeyer, Ph.D.
Sarah Hedges, Ph.D.

Learning Objectives
How gases cause pressure
The relationship between
By viewing this tutorial you will learn…
The relationship between
temperature and kinetic energy
The relationships between
common temperature scales
Kinetic Molecular Theory
How gas properties relate to
each other
3/72
How to apply several gas laws
The difference between ideal
and real gases
Diffusion and effusion
Concept Map
Physics
Studies
Previous content
New content
Matter and EnergyMatter and Energy
Gases
One state is
Volume
Pressure
MolecularMolecular
Rates of EffusionRates of Effusion
and Diffusion
4/72
TemperatureTemperature
# of Moles
Molar Mass
and Density
Molar Mass
and Density
Speeds
Gas Laws
Have properties
Related to each other with

Average KineticAverage Kinetic
Energy and
Temperature
5/72
Definition - Kinetic Energy
Kinetic Energy (KE) - the energy due
to motion of an object.j
m = mass of the object
v = speed of the object
2
mv
2
1
KE =
6/72
p j
Thus, the kinetic energy of an object is
proportional to the square of its speed.

Definition - Temperature
Temperature – Proportional to the
average kinetic energy of the molecules.
Energy due to motion
(Related to how fast the
molecules are moving)
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As temperature
increases
Molecular
motion
increases
Temperature and Kinetic Energy
Increased average kinetic energy
Temperature is
Decreased
Temperature is
Increased
Loss of heat
Gain of heat
energy
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Decreased average kinetic energy
Loss of heat
energy

Definition - Average Kinetic Energy
Average Kinetic Energy (KEAVE) - The
energy due to motion of an object.
3
R = universal gas constant = 8.31 (Joules/Kelvins *mole)
T = temperature in Kelvins
RT
2
3
KE =
9/72
Thus the temperature of an object is proportional
to the average kinetic energy of its molecules. As
the gas heats up, its molecules oscillate at a
faster rate.
Calculating Average Kinetic Energy
3
Temperature is defined as proportional to average
kinetic energy…how do you calculate it?
Avg. KE = Average Kinetic Energy (in J, Joules)
RT
2
3
KEAvg. = R = Gas constant (use 8.31 J/K mol)
T = Temperature (in Kelvin)
Find the average kinetic energy of a sample of O2 at 28°CExample:
( )3
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Avg. KE = ? J
R = 8.31 J/K mol
T = 28°C + 273 = 301 K
Avg. KE = 3752 J/mole
( ) 301K
moleK
J8.31
2
3
KEAvg. ×
×
=

Average Translational Kinetic Energy
The average translational kinetic energy Kave is
related to the temperature T by the relationship:
3
kT
2
3
Kave =
AN
Rk =
k, a constant is the ratio of the universal gas
constant to Avogadro’s number, NA.
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AN
T
N
R
2
3
K
A
ave =
Thus, the average translational kinetic energy per
molecule is given by:
Kelvins, Celsius, and Fahrenheit
There are three commonly used units for
temperature.
Kelvin (K) Scale. It’s also referred to as the
absolute scale, 0 K is the temperature at
which molecules of an object have a kinetic
energy of 0.
Celsius (C) Scale. This scale is based on the
boiling and freezing points of water It is
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boiling and freezing points of water. It is
commonly used in science.
Fahrenheit (F) Scale. This is the scale used
on weather channels and is frequently used
in the United States.

Definition- Kelvin Scale
Kelvin (K) – Temperature scale with
b l tan absolute zero
Temperatures cannot fall below an absolute zero
A temperature scale with absolute zero is needed in Gas
Law calculations because you can’t have negative
pressures or volumes.
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K273C =+
Absolute vs. Relative Scales
Kelvin (K)
0K is the temperature at which molecules of an object have zero
kinetic energy (zero motion)
When in doubt use Kelvins for thermodynamics calculations.
Celsius (C)
Water freezes at 0 °C
Water boils at 100 °C
T (Celsius) = T (Kelvins) - 273.15
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Fahrenheit (F)
Water freezes at 32 °F
Water boils at 212 °F
T (Fahrenheit) = (9/5) * T (Celsius) + 32

Temperature Scales
Molecules are
completely still
(kinetic energy = 0
Increasing
molecular motion
(kinetic energy)
0 Kelvins Water freezes
273.15 K
0 °C
32 °F
Water Boils
373.15 K
100 °C
212 °F
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Decreasing temperature Increasing temperature
Decreasing kinetic energy
Loss of heat energy
Increasing kinetic energy
Gain of heat energy
32 F 212 F
Note - Absolute Temperature
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Question: Temperature and Kinetic Energy
If the temperature of an ice cube is -8 °C, what is the
average kinetic energy (in Joules) of the water
molecules?
0 Joules because the water
molecules are crystallized as ice
and therefore not moving.
0 Joules because the
temperature is below 0 K
Pick the best answer:
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temperature is below 0 K.
Greater than 0 Joules because
the temperature is above
absolute zero or 0 K.
Answer: Temperature and Kinetic Energy
If the temperature of an ice cube is -8 °C, what is the
average kinetic energy (in Joules) of the water
molecules?
0 Joules because the water
molecules are crystalized as ice
and therefore not moving.
0 Joules because the
Pick the best answer:
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Greater than 0 Joules because
the temperature is above
absolute zero or 0 K

Note - Temperature Scales
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Gas BehaviorGas Behavior
and Kinetic
Molecular Theory
20/72

Definition- Theory and KMT
Theory – An attempt to explain why or
h b h i ti thhow behavior or properties are as they
are. It’s based on empirical evidence.
Kinetic Molecular Theory (KMT) – An
attempt to e plain gas beha ior
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attempt to explain gas behavior
based upon the motion of molecules.
Assumptions of the KMT
All gases are made of atoms or molecules.
Gas particles are in constant, rapid, random
motion
1
2 motion.
The temperature of a gas is proportional to the
average kinetic energy of the particles.
Gas particles are not attracted nor repelled from
one another.
3
4
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All gas particle collisions are perfectly elastic (no
kinetic energy is lost to other forms).
The volume of gas particles is so small compared
to the space between the particles, that the
volume of the particle itself is insignificant.
5
6

KMT and Gas Behavior
The Kinetic MolecularThe Kinetic Molecular
Theory and its
assumptions can be
used to explain gas
behavior.
23/72
Definition - Pressure
Pressure – Force of gas
particles running into a
surface.
24/72

Pressure and Number of Molecules
A b f C lli i P
If pressure is molecular collisions with the
container…
As number of
molecules
increases,
there are more
molecules to
collide with
the wall
Collisions
between
molecules and
the wall
increase
Pressure
increases
25/72
As # of molecules increases, pressure increases.
Pressure (P) and # of molecules (n) are directly
proportional (∝)
nP ∝
Pressure and Volume
A l C lli i P
If pressure is molecular collisions with the
container…
As volume
increases,
molecules can
travel farther
before hitting
the wall
Collisions
between
molecules and
the wall
decrease
Pressure
decreases
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As volume increases, pressure decreases.
Pressure and volume are inversely proportional.
V
1
P ∝

Pressure and Temperature
If temperature is related to molecular motion…
and pressure is molecular collisions with the
container…
As temperature
increases,
molecular
motion
increases
Collisions
between
molecules and
the wall
increase
Pressure
increases
27/72
As temperature increases, pressure increases.
Pressure and temperature are directly proportional.
TP ∝
Pressure Inside and
Outside a Container
28/72

Definition- Atmospheric Pressure
Atmospheric Pressure Pressure dueAtmospheric Pressure – Pressure due
to the layers of air in the atmosphere.
Less layers of
air
Lower
atmosphericClimb in
altitude
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air pressurealtitude
As altitude increases, atmospheric pressure decreases.
Pressure In Versus Out
A container will expand or contract until the
pressure inside = atmospheric pressure outside
Expansion will lower the internal pressure
Example: A bag of chips is bagged at sea level. What happens if
the bag is then brought up to the top of a mountain.
Contraction will raise the internal pressure
(Volume and pressure are inversely related)
The internal pressure is from low
altitude (high pressure)
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The internal pressure is higher than the external pressure.
The bag will expand in order to reduce the internal pressure.
( g p )
The external pressure is high
altitude (low pressure).Higher
pressure
Lower
pressure
Lower
pressure

When Expansion Isn’t Possible
Example: An aerosol can is left in a car trunk in the summer. What
happens?
Rigid containers cannot expand.
Can
Explodes!
happens?
The temperature inside the can
begins to rise.
As temperature increases,
pressure increases.
Higher
pressure
Lower
pressure
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The internal pressure is higher than the external pressure.
The can is rigid—it cannot expand, it explodes!
Soft containers or “movable pistons” can expand and contract.
Rigid containers cannot.
Gas Laws
32/72

General Strategy for Gas Law Problems
Id tif titi b th i it1
The following steps are a general way to approach these
problems.
Identify quantities by their units.
Make a list of known and unknown quantities
in symbolic form.
Look at the list and choose the gas law that
relates all the quantities together.
Pl titi i d l
1
2
3
4
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Plug quantities in and solve.4
Pressure Units
Several units are used when describing pressure.
Unit Symbol
atmospheres atm
Pascals, kiloPascals
millimeters of mercury
pounds per square inch
Pa, kPa
mm Hg
psi
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1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

Definition- STP Conditions
Standard Temperat re and Press reStandard Temperature and Pressure
(STP) – 1 atm (or the equivalent in
another unit) and 0°C (273 K).
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Problems often use “STP” to indicate quantities…
Don’t forget this “hidden” information when
making your list!
KMT and Gas Laws
The Gas Laws are
experimental
observations of gas
behavior that the Kinetic
Molecular Theory
explains.
36/72
explains.

“Before” and “After” in Gas Laws
This section has 4 gas laws which have “before”
and “after” conditions.
For example:
2
2
1
1
n
P
n
P
=
Where P1 and n1 are pressure and # of moles “before”
37/72
and P2 and n2 are pressure and # of
moles “after”
Both sides of the equation are talking about the same
sample of gas—with the “1” variables before a change,
and the “2” variables after the change
Avogadro’s Law
Avogadro’s Law relates # of particles (moles) and
pressure.
Where Temperature and Pressure are held constant
V = Volume
n = # of moles of gas
2
2
1
1
n
V
n
V
=
Example: A sample with 0.15 moles of gas has a volume of 2.5 L.
What is the volume if the sample is increased to 0.55
moles?
The two volume units must match!
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moles?
n1 = 0.15 moles
V1 = 2.5 L
n2 = 0.55 moles
V2 = ? L
0.55mole
V
0.15mole
2.5L 2
=
2V
0.15mole
2.5L0.55mole
=
×
V2 = 9.2 L

Boyles’ Law
Boyles’ Law relates pressure and volume.
Where temperature and # of molecules
are held constant
P = pressure
V = volume2211 VPVP =
The two pressure units must match and
the two volume units must match!
Example: A gas sample is 1.05 atm when 2.5 L. What volume is it
if the pressure is changed to 745 mm Hg?
P it d t t h t
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P1 = 1.05 atm
V1 = 2.5 L
P2 = 745 mm Hg
V2 = ? L
V2 = 2.7 L
Pressure units need to match—convert one:
=0.980 atm
2V0.980atm2.5L1.05atm ×=×
2V
0.980atm
2.5L1.05atm
=
×
745 mm Hg
= ______ atm
mm Hg
atm1
760
0.980
Charles’ Law
Charles’ Law relates temperature and pressure.
Where pressure and # of molecules are
held constant
21 VV V = Volume
T = Temperature
2
2
1
1
T
V
T
V
=
The two volume units must match and
temperature must be in Kelvin!
Example: What is the final volume if a 10.5 L sample of gas is changed
from 25°C to 50°C?
T t d t b i K l i !
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V1 = 10.5 L
T1 = 25°C
V2 = ? L
T2 = 50°C
V2 = 11.4 L
Temperature needs to be in Kelvin!
= 298 K
= 323 K
25°C + 273 = 298 K
50°C + 273 = 323 K
323K
V
298K
10.5L 2
=
2V
298K
10.5L323K
=
×

Combined Gas Law
The combined gas law assumes that nothing is
held constant.
P = Pressure
VPVP
Each “pair” of units
V = Volume
n = # of moles
T = Temperature
22
22
11
11
Tn
VP
Tn
VP
=
Each pair of units
must match and
temperature must be in
Kelvin!
Example: What is the final volume if a 0.125 mole sample of gas at 1.7
atm, 1.5 L and 298 K is changed to STP and particles are
added to make 0.225 moles?
P1 = 1.7 atm
V1 = 1.5 L
STP is standard temperature (273 K) and pressure (1 atm)
41/72
n1 = 0.125 mole
T1 = 298 K
P2 = 1.0 atm
V2 = ? L
n2 = 0.225 mole
T2 = 273 K V2 = 4.2 L
273K0.225mole
V1.0atm
298K0.125mole
1.5L1.7atm 2
×
×
=
×
×
2V
298K0.125mole1.0atm
1.5L1.7atm273K0.225mole
=
××
×××
Why You Only Really Need 1 out of the 4 Laws!
2211 VPVP
=
The combined gas law can be used for all “before”
and “after” gas law problems!
2211 TnTn
22
12
11
11
Tn
VP
Tn
VP
=
For example, if volume is held constant, then
and the combined gas law becomes:
21 VV =
42/72
When two variables on opposites sides are the same, they
cancel out and the rest of the equation can be used.
22
2
11
1
Tn
P
Tn
P
=

“Transforming” the Combined Gas Law
Watch as variables are held constant and the
combined gas law “becomes” the other 3 laws.
VPVP
22
22
11
11
Tn
VP
Tn
VP
=Hold pressure and
temperature constant
Avogadro’s Law
22
22
11
11
Tn
VP
Tn
VP
=Hold moles and
temperature constant
Boyles’ Law
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22
22
11
11
Tn
VP
Tn
VP
=Hold pressure and
moles constant
Charles’ Law
How to Memorize What’s Held Constant
How do you know what to hold constant for each
law?
Hold Pressure and Temperature constantAvogadro’s Law
Hold moles and Temperature constantBoyles’ Law
Avogadro was a Professor at Turin University (Italy)
The last letter of his first name Robert is T
44/72
Hold Pressure and moles constantCharles’ Law
The last letter of his first name, Robert, is T
Charles was from Paris

Example of Using only the Combined Law
Example: What is the final pressure if a 15.5 L sample of gas at 755
mm Hg and 298 K is changed to STP?
P 755 H
“moles” is not mentioned in the problem—therefore
P1 = 755 mm Hg
V1 = 15.5 L
T1 = 298 K
P2 = 1.0 atm
V2 = ? L
T2 = 273 K
p
it is being held constant.
It is not needed in the combined law formula.
Pressure units must match!
1 atm = 760 mm Hg
= 760 mm Hg
22
22
11
11
Tn
VP
Tn
VP
=
45/72
V2 = 14.1 L
273K
VHg760mm
298K
15.5LHg755mm 2×
=
×
2V
298KHg760mm
15.5LHg755mm273K
=
×
××
Ideal Gas Law
46/72

Definition - Ideal Gas
Ideal Gas – All of the assumptions of
the Kinetic Molecular Theory (KMT)
lidare valid.
Ideal Gas Law – Describes properties
of a gas under a set of conditions.
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nRTPV =
This law does not have “before” and “after”—
there is no change in conditions taking place.
Definition- Gas Constant (R)
nRTPV =
Gas Constant (R) – Constant equal to
the ratio of P×V to n×T for a gas.
Values for R
kPaL×
Use this one
when the P unit
is “kPa”
48/72
8.31
0.0821
Kmole
kPaL
×
×
Kmole
atmL
×
×
is kPa
Use this one
when the P unit
is “atm”

Memorizing the Ideal Gas Law
nRTPV =
Phony Vampires are not Real Things
49/72
Ideal Gas Law Example
An example of the Ideal Gas Law:
P = Pressure
V = Volume
Choose your “R” based
upon your “P” units.
nRTPV = n = # of moles
R = Gas constant
T = Temperature
upon your P units.
T must be in Kelvin!
nRTPV =
What is the pressure (in atm) of a gas if it is 2.75 L, has 0.25
moles and is 325 K?
Example:
P = ?
Choose the “0.0821” for “R” since the problem asks for
“ ”
50/72
( ) 325K
Kmole
atmL0.08210.25moles2.75LP ×
×
××=×
P ?
V = 2.75 L
n = 0.25 moles
T = 325 K
Phydrogen = 2.43 atm
“atm”
R = 0.0821 (L×atm) / (mol×K)
( )
2.75L
325K
Kmole
atmL0.08210.25moles
P
×
×
××
=

Definition - Molar Mass
Molar mass (MM) – Mass (m) per
moles (n) of a substancemoles (n) of a substance.
n
m
MM =
Therefore:
m
51/72
Therefore:
MM
n =
Ideal Gas Law and Molar Mass
The Ideal Gas Law is often used to determine molar
mass.
nRTPV
m
n =and RT
m
PVnRTPV = MM
n =and RT
MM
PV =
A gas is collected. The mass is 2.889 g, the volume is 0.936
L, the temperature is 304 K and the pressure is 98.0 kPa.
Find the molar mass
Example:
P = 98.0 kPa
Choose the “8.31” for “R” since the problem uses “kPa”
( )kPaL2 889g
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P 98.0 kPa
V = 0.936 L
m = 2.889 g
T = 304 K
MM = ? g/mole
MM = 79.6 g/moleR = 8.31 (L×kPa) / (mol×K)
( ) 304K
Kmole
kPaL8.31
MM
2.889g
0.936L98.0kPa ×
×
×=×
( ) 304K
Kmole
kPaL8.31
0.936L98.0kPa
2.889g
MM ×
×
×
×
=

Definition - Density
D i R i f lDensity – Ratio of mass to volume
for a sample.
V
m
D =
53/72
Ideal Gas Law and Density
Using the density equation with the Ideal Gas Law:
V
m
D =andRT
MM
m
PV =
MM
RT
V
m
P =
MM
RT
DP =
A gas is collected. The density is 3.09 g/L, the volume is
0.936 L, the temperature is 304 K and the pressure is 98.0
kPa. Find the molar mass
Example:
54/72
P = 98.0 kPa
V = 0.936 L
D = 3.09 g/L
T = 304 K
MM = ? g/mole
MM = 79.6 g/mole
Choose the “8.31” for “R” since the problem uses “kPa”
R = 8.31 (L×kPa) / (mol×K)
( )
MM
304K
Kmole
kPaL8.31
L
g3.0998.0kPa
×
×
×
=
( )
98.0kPa
304K
Kmole
kPaL8.31
L
g3.09MM
×
×
×
=

Real Gases
55/72
Real Gases
Real Gas – Two of the assumptions of the
Kinetic Molecular Theory are not valid.
Wrong Assumption 1: Gas particles are not
attracted nor repelled from one another
Wrong Assumption 2: The volume of gas particles
Reality: Gas particles do have attractions and
repulsions towards one another.
56/72
Wrong Assumption 2: The volume of gas particles
is so small compared to the space between the
particles, that the volume of the particle itself is
insignificant
Reality: Gas particles do take up space—thereby
reducing the space available for other particles to be.

Real Gas Law
The Real Gas Law takes into account the
deviations from the Kinetic Molecular Theory.
nRTPV =Ideal Gas Law nRTPV =
( ) nRTnbV
V
an
P 2
2
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
Ideal Gas Law
Real Gas Law
Also called “van der Waals equation”
Take into account the
57/72
Take into account the
change in pressure due
to particle attractions
and repulsions
Takes into account the
space the particles
take up
“a” and “b” are constants that you look up for each gas!
Real Gas Law Example
Example: At what temperature would a 0.75 mole sample of CO2 be
2.75 L at 3.45 atm?
( ) nRTnbV
V
an
P 2
2
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
2.75 L at 3.45 atm?
(van der Waals constants for CO2: a = 3.59 L2atm/mol2
b = 0.0427 L/molP = 3.45 atm
V = 2.75 L
n = 0.75 mole
T = ? K
a = 3.59 L2atm/mol2
b = 0.0427 L/mol
Choose the “0.0821” for “R” since the problem uses
“atm”
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T = 164 K
( ) ( ) ( ) T
Kmol
atmL0.08210.75mol
mol
L0.04270.75mol2.75L
(2.75L)
mol
atmL3.59(0.75mol)
3.45atm 2
2
22
×
×
××=×−
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛ ×
+
R = 0.0821 (L×atm) / (mol×K)

Distribution of MolecularDistribution of Molecular
Speeds and Mean Free
Path
59/72
Definition - Mean Free Path
Mean Free Path - (λ). The mean free
path is the average distance traveled by
an atom or molecule before it collidesan atom or molecule before it collides
with another atom or molecule.
V
Nπd2
1
λ 2
=
d = molecular diameter
N/V = number of molecules
per unit volume
60/72
Thus, the number of collisions a molecule undergoes
increases as the number of molecules in a container
increases and as the diameter of the molecules
increases. In a reaction mixture, as the number of
collisions increases the reaction rate increases.

Definition- Vrms
Root-Mean-Square Speed (vrms) –
One measure of the average speed of
l l imolecules in a gas.
M
3RT
vrms =
R = 8.31 J/mol*k; T
= temperature [K];
M = molar mass
[kg/mol]
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[kg/mol]
Thus Vrms increases with temperature and decreases
as the molecules increase in size.
61/43
61/43
Molecular Speed
The pressure P exerted by n moles of a gas may be
determined from the root-mean-square speed vrms
of its molecules.
P = pressure; n = moles of gas; V = volume; M = molar mass
3V
nMv
P
2
rms
=
62/72
Note that the pressure increases with increasing
average speed and molar mass.
Note that pressure decreases with increasing
volume.

Maxwell Speed Distribution
Vave is just the average speed. Some molecules
move much faster and some molecules move much
slower.
The probability that a molecule in a gas at
temperature T is moving at a given speed is shown
by a probability function known as the Maxwell
distribution.
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Diagram- Maxwell’s Distribution
Maxwell's Speed Distribution Law
2
2.5
^-3m/s)
0
0.5
1
1.5
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
Molecular Speed (m/s)
Probability(10^
0 deg C
200 deg C
N t th t i th f 0 C th d ith th
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Note that in the curve for 0 °C, the speed with the
maximum probability is around 350 m/s.
For 200 °C, the speed with the maximum probability is
higher, around 500 m/s.

Diffusion and Effusion
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Definition- Diffusion and Effusion
Diffusion – A gas spreads throughout
a space.
Effusion A gas escapes through a
Perfume is sprayed in one corner of the
room and a person on the other side smells
it after a moment.
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Effusion – A gas escapes through a
tiny hole.
Air leaks out of a balloon overnight and is
flat the next day.

Diffusion, Effusion and Mass
Temperature is A heavier object Heavy molecules
The mass of a particle affects the rate of diffusion
and effusion.
Temperature is
proportional
to average
kinetic energy
A heavier object
with the same
kinetic energy
as a lighter
object moves
slower than
the lighter
object
Heavy molecules
move slower
than smaller
molecules
67/72
Diffusion: If molecules move slower, it will take them longer to
reach the other side of the room.
Effusion: If molecules move slower, it will take them longer to
find the hole to escape through.
Both rates of diffusion and effusion are inversely proportional
to molecular mass.
Effusion and Graham’s Law
Effusion rates are related by Graham’s Law.
r1 = Rate of Effusion for molecule 1
r2 = Rate of Effusion for molecule 2
MM M l l f l l 1
21
MM
MMr
=
MM1 = Molecular mass for molecule 1
12 MMr
Example: A gas molecule effuses 0.355 times as fast as O2. What is
the molecular mass of the molecule?
Molecule 1 = O2
Molecule 2 = unknown molecule
If the unknown molecule is 0.355
times as fast as O2,
then make the rate of O2 = 1 and the
e32.00g/mol
MM
0.355
1 2
=
68/72
r1 = 1
r2 = 0.355
MM1 = 32.00 g/mole
MM2 = ? g/mole
then make the rate of O2 1 and the
rate of unknown = 0.355
Molecular mass of O2:
O 2 × 16.00 = 32.00 g/mole
e32.00g/mol
MM
0.355
1 2
2
=⎟
⎠
⎞
⎜
⎝
⎛
( ) 2
2
MM
0.355
1
e32.00g/mol =⎟
⎠
⎞
⎜
⎝
⎛
×
MM = 254 g/mole

Diffusion and Graham’s Law
Distances traveled during diffusion:
d1 = Distance traveled for molecule 1
d2 = Distance traveled for molecule 2
MM M l l f l l 1
21
MM
MM
d
d
=
12 MMd
Example: A gas molecule is 4 times as heavy as O2. How far does it
travel in the time that oxygen travels 0.25 m?
Molecule 1 = unknown molecule
Molecule 2 = O2 le128.00g/mo
e32.00g/mol
0.25m
d1
=
69/72
d1 = ?
d2 = 0.25 m
MM1 = 4×32.00 g/mole = 128 g/mole
MM2 = 32.00 g/mole
Molecular mass of O2:
O 2 × 16.00 = 32.00
g/mole
4
1
0.25m
d1
=
0.25m
2
1
d1 ×=
d1 = 0.125 m
Real gases do
not use 2 of the
Real gases do
not use 2 of the
Temperature is
proportional to
Temperature is
proportional to
Rates of Effusion
and Diffusion are
inversely
Rates of Effusion
and Diffusion are
inversely
Learning Summary
assumptions of
the KMT
assumptions of
the KMT
average kinetic
energy.
average kinetic
energy.
inversely
proportional to
molecular mass
inversely
proportional to
molecular mass
Several Gas LawsSeveral Gas Laws
70/72
Ideal gases follow the
assumption of the
Kinetic Molecular
Theory (KMT)
Ideal gases follow the
assumption of the
Kinetic Molecular
Theory (KMT)
Several Gas Laws
are used to
determine
properties under a
set of conditions
Several Gas Laws
are used to
determine
properties under a
set of conditions

Congratulations
You have successfully completed
the tutorial
Temperature and
The Kinetic TheoryThe Kinetic Theory
Wh t’ N t
Chemistry :: Biology :: Physics :: Math
What’s Next …
Step 1: Concepts – Core Tutorial (Just Completed)
Step 2: Practice – Interactive Problem Drill
Step 3: Recap – Super Review Cheat Sheet
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