Chapter 21
The Kinetic Theory of Gases
The description of the behavior of a gas in
terms of the macroscopic state variables P, V, and
T can be related to simple averages of microscopic
quantities such as the mass and speed of the
molecules in the gas. The resulting theory is called
the Kinetic Theory of Gases.
From the point of view of kinetic theory, a gas
consist of a large number of molecules making
elastic collisions with each other and with the walls
of container.
In the absence of external forces (we may
neglect gravity), there no preferred position of the
molecule in the container, and no preferred
directions for it’s velocity vector.
The molecules are separated, on the average,
by distances that are large compared with their
diameters, and they exert no forces on each other
except when they collide.
This final assumption is equivalent to
assuming a very low gas density, which is the
same as assuming that the gas is an ideal gas.
Because momentum is conserved, the
collisions the molecules make with each other
have no effect on the total momentum in any
directions – thus such collisions may be
neglected.
Molecular Model of an Ideal Gas
• The model shows that the pressure that a gas
exerts on the walls of its container is a
consequence of the collisions of the gas molecules
with the walls
• It is consistent with the macroscopic description
developed earlier
Assumptions for Ideal Gas Theory
•
The gas consist of a very large number of identical
molecules, each with mass m but with negligible size (this
assumption is approximately true when the distance between
the molecules is large, compared to the size).
The consequence → for negligible size molecules we
can neglect the intermolecular collisions.
•
The molecules don’t exert any action-at-distance forces
on each other. This means there are no potential energy
changes to be considered, so each molecules kinetic energy
remains unchanged. This assumption is fundamental to the
nature of an ideal gas.
Assumptions for Ideal Gas Theory
•
The molecules are moving in random
directions with a distribution of speeds that is
independent of direction.
•
Collisions with the container walls are elastic,
conserving the molecule’s energy and
momentum.
Pressure and Kinetic Energy
• Assume a container
is a cube
Edges are length d
• Look at the motion
of the molecule in
terms of its velocity
components
• Look at its
momentum and the
average force
Pressure and Kinetic Energy
• Since the collision is elastic, the y component of molecule’s velocity
remains unchanged, while the x component reverses sign. Thus the
molecule undergoes the momentum
change of magnitude 2mvx.
• After colliding with the right hand
wall, the x - component of
molecule’s velocity will not change
until it hits the left-hand wall and its
x - velocity will again reverses.
Δt = 2d / vx
Pressure and Kinetic Energy
• The average force, due to the
each molecule on the wall:
2 m vx
m v x2
p
Fi 


t (2d / v x )
d
• To get the total force on the
wall, we sum over all N
molecules. Dividing by the wall
area A then gives the force per
unit area, or pressure:
mv x2
2
2

F
m
v
m
v
F  i
d   x   x
P 

A
A
A
Ad
V
Pressure and Kinetic Energy
P
Since,
2
v
x
N
m v
V
2
x
mN

V
v
2
x
N
is just the average of the squares of x –
components of velocities:
mN 2
P
vx
V
Pressure and Kinetic Energy
Since, the molecules are moving in random directions the
2
2
2
v
v
vx ,
average quantities
y , and
z must be
equal and the average of the molecular speed



v v v v
2
and v2 = 3vx2,
pressure:
or
2
x
2
y
2
z
vx2 = v2/3 . Then the expression for
mN 2
P
v
3V
Pressure and Kinetic Energy
• The relationship
mN 2 can be written:
P
v
3V
2N
P 
3V

 1
 m v 
 2

___
2
• This tells us that pressure is proportional to the number
of molecules per unit volume (N/V) and to the average
translational kinetic energy of the molecules
Pressure and Kinetic Energy
• This equation also relates the macroscopic
quantity of pressure with a microscopic quantity
of the average value of the square of the
molecular speed
• One way to increase the pressure is to increase
the number of molecules per unit volume
• The pressure can also be increased by
increasing the speed (kinetic energy) of the
molecules
A 2.00-mol sample of oxygen gas is confined to a
5.00-L vessel at a pressure of 8.00 atm. Find the
average translational kinetic energy of an oxygen
molecule under these conditions.
A 2.00-mol sample of oxygen gas is confined to a
5.00-L vessel at a pressure of 8.00 atm. Find the
average translational kinetic energy of an oxygen
molecule under these conditions.
2

2N m v
P
3V  2 
K av
K av
m v2 3PV


w here N  nN A  2N A
2
2N
3PV


2 2N A 
 1.013  105 Pa atm  5.00  103 m 3 
2 2 m ol  6.02  1023 m olecules m ol
3 8.00 atm
K av  5.05  1021 J m olecule
Molecular Interpretation of Temperature
• We can take the pressure as it relates to the kinetic
energy and compare it to the pressure from the equation
of state for an ideal gas
2  N   1 2  Nk B T
P    m v  
3  V  2
V

• Therefore, the temperature is a direct measure of the
average molecular kinetic energy
Molecular Interpretation of Temperature
• Simplifying the equation relating temperature and
kinetic energy gives
___
2
1
3
m v  k BT
2
2
• This can be applied to each direction,
1 ___2 1
m v x  k BT
2
2
with similar expressions for vy and vz
A Microscopic Description of Temperature
• Each translational degree of freedom
contributes an equal amount to the
energy of the gas
• A generalization of this result is called
the theorem of equipartition of energy
Theorem of Equipartition of Energy
• Each degree of freedom contributes
½kBT to the energy of a system, where
possible degrees of freedom in addition to
those associated with translation arise
from rotation and vibration of molecules
Total Kinetic Energy
• The total kinetic energy is just N times the kinetic energy
of each molecule
K tot trans
 1 ___2  3
3
 N  m v   Nk BT  nRT
2
2
 2
• If we have a gas with only translational energy, this is the
internal energy of the gas
• This tells us that the internal energy of an ideal gas
depends only on the temperature
Molar Specific Heat
• Several processes can
change the temperature of
an ideal gas
• Since ΔT is the same for
each process, ΔEint is also
the same
• The heat is different for the
different paths
• The heat associated with a
particular change in
temperature is not unique
Molar Specific Heat
• We define specific heats for two processes that
frequently occur:
– Changes with constant volume
– Changes with constant pressure
• Using the number of moles, n, we can define
molar specific heats for these processes
Molar Specific Heat
• Molar specific heats:
Q = nCV ΔT for constant-volume processes
Q = nCP ΔT for constant-pressure processes
• Q (under constant pressure) must account for both
the increase in internal energy and the transfer of
energy out of the system by work
• Q (under constant volume) account just for change
the internal energy
• Qconstant P > Qconstant V for given values of n and ΔT
Ideal Monatomic Gas
• A monatomic gas contains one atom per
molecule
• When energy is added to a monatomic gas in
a container with a fixed volume, all of the
energy goes into increasing the translational
kinetic energy of the gas
– There is no other way to store energy in
such a gas
Ideal Monatomic Gas
• Therefore,
E int  K tot
3
3
Nk B T  nRT
trans 
2
2
• ΔEint is a function of T only
• At constant volume,
Q = ΔEint = nCV ΔT
This applies to all ideal gases, not just monatomic ones
Monatomic Gases
• Solving
 E in t
3
 nC V  T  nR  T
2
for CV gives CV = 3/2 R = 12.5 J/mol . K for all
monatomic gases
• This is in good agreement with experimental
results for monatomic gases
Monatomic Gases
• In a constant-pressure process, ΔEint = Q + W and
 E in t  Q  W  n C P  T   P  V

• Change in internal energy depends only on
temperature for an ideal gas and therefore are the
same for the constant volume process and for
constant pressure process
nCV T  nC P T  nRT
CP – CV = R
Monatomic Gases
CP – CV = R
• This also applies to any ideal gas
CP = 5/2 R = 20.8 J/mol . K
Ratio of Molar Specific Heats
• We can also define
C P 5R / 2
g 

 1.67
CV 3 R / 2
• Theoretical values of CV , CP , and g are in
excellent agreement for monatomic gases
• But they are in serious disagreement with the
values for more complex molecules
– Not surprising since the analysis was for monatomic
gases
Sample Values of Molar Specific Heats
• A 1.00-mol sample of air (a diatomic ideal gas)
at 300 K, confined in a cylinder under a heavy
piston, occupies a volume of 5.00 L. Determine
the final volume of the gas after 4.40 kJ of
energy is transferred to the air by heat.
• A 1.00-mol sample of air (a diatomic ideal gas) at 300 K,
confined in a cylinder under a heavy piston, occupies a
volume of 5.00 L. Determine the final volume of the gas
after 4.40 kJ of energy is transferred to the air by heat.
nR T
The piston moves to keep pressure constant: V 
P
Q  nC P T  nCV  R  T
Q
Q
Q
2Q
T 



nCP n(CV  R)
 3R
 5nR
n
 R
 2

nR
nR  2Q  2Q 2 QV
V 
T 



P
P  5nR  5 P 5 nRT
• A 1.00-mol sample of air (a diatomic ideal gas) at 300 K,
confined in a cylinder under a heavy piston, occupies a
volume of 5.00 L. Determine the final volume of the gas
after 4.40 kJ of energy is transferred to the air by heat.
nR
nR  2Q  2Q 2 QV
V 
T 



P
P  5nR  5 P 5 nRT
2
V 
5
(4.40  103 J )(5L)
 3.53L
J 

(1mol ) 8.314
(300 K )
mol  K 

V f  Vi  V  5.00L  3.53L  8.53L
Molar Specific Heats of Other Materials
• The internal energy of more complex gases
must include contributions from the rotational
and vibrational motions of the molecules
• In the cases of solids and liquids heated at
constant pressure, very little work is done since
the thermal expansion is small and CP and CV
are approximately equal
Adiabatic Processes for an Ideal Gas
• Assume an ideal gas is in an equilibrium state
and so PV = nRT is valid
• The pressure and volume of an ideal gas at any
time during an adiabatic process are related by
PV g = constant
• g = CP / CV is assumed to be constant during the
process
• All three variables in the ideal gas law (P, V, T )
can change during an adiabatic process
Equipartition of Energy
• With complex
molecules, other
contributions to internal
energy must be taken
into account
• One possible way to
energy change is the
translational motion of
the center of mass
Equipartition of Energy
• Rotational motion
about the various axes
also contributes
We can neglect the
rotation around the y
axis since it is negligible
compared to the x and z
axes
Equipartition of Energy
• The molecule can
also vibrate
• There is kinetic
energy and potential
energy associated
with the vibrations
Equipartition of Energy
• The translational motion adds three degrees of
freedom
• The rotational motion adds two degrees of
freedom
• The vibrational motion adds two more degrees
of freedom
• Therefore, Eint = 7/2 nRT and CV = 7/2 R
Agreement with Experiment
• Molar specific heat is a function of temperature
• At low temperatures, a diatomic gas acts like a
monatomic gas CV = 3/2 R
Agreement with Experiment
• At about room temperature, the value increases
to CV = 5/2 R
This is consistent with adding rotational energy but
not vibrational energy
• At high temperatures, the value increases
to CV = 7/2 R
This includes vibrational energy as well as rotational
and translational
Complex Molecules
• For molecules with more than two atoms, the
vibrations are more complex
• The number of degrees of freedom is larger
• The more degrees of freedom available to a
molecule, the more “ways” there are to store
energy
This results in a higher molar specific heat
Quantization of Energy
• To explain the results of the various molar
specific heats, we must use some quantum
mechanics
Classical mechanics is not sufficient
• In quantum mechanics, the energy is
proportional to the frequency of the wave
representing the particle
The energies of atoms and molecules are
quantized
Quantization of Energy
• This energy level diagram
shows the rotational and
vibrational states of a
diatomic molecule
• The lowest allowed state is
the ground state
Quantization of Energy
• The vibrational states are separated by larger
energy gaps than are rotational states
• At low temperatures, the energy gained during
collisions is generally not enough to raise it to
the first excited state of either rotation or
vibration
Quantization of Energy
• Even though rotation and vibration are
classically allowed, they do not occur
• As the temperature increases, the energy of the
molecules increases
• In some collisions, the molecules have enough
energy to excite to the first excited state
• As the temperature continues to increase, more
molecules are in excited states
Quantization of Energy
• At about room temperature, rotational energy is
contributing fully
• At about 1000 K, vibrational energy levels are
reached
• At about 10 000 K, vibration is contributing fully
to the internal energy
Molar Specific Heat of Solids
• Molar specific heats in solids also demonstrate a
marked temperature dependence
• Solids have molar specific heats that generally
decrease in a nonlinear manner with decreasing
temperature
• It approaches zero as the temperature
approaches absolute zero
DuLong-Petit Law
• At high temperatures, the molar specific heats
approach the value of 3R
This occurs above 300 K
• The molar specific heat of a solid at high
temperature can be explained by the equipartition
theorem
Each atom of the solid has six degrees of freedom
The internal energy is 3nRT and Cv = 3 R
Molar Specific Heat of Solids
• As T approaches 0,
the molar specific heat
approaches 0
• At high temperatures,
CV becomes a
constant at ~3R
Specific Heat and Molar Specific Heat of Some Solids and Liquids
Substance
Aluminum
Bismuth
c, kJ/kg·K
0.900
0.123
c*, J/mol·K
24.3
25.7
Copper
Gold
Ice (-100C)
0.386
0.126
2.05
24.5
25.6
36.9
Lead
Silver
Tungsten
0.128
0.233
0.134
26.4
24.9
24.8
Zink
Mercury
Water
0.387
0.140
4.18
25.2
28.3
75.2
The molar mass of copper is 63.5 g/mol. Use the
Dulong-Petit law to calculate the specific heat of
copper.
The molar mass of copper is 63.5 g/mol. Use the DulongPetit law to calculate the specific heat of copper.
The molar specific heat is the heat capacity per mole:
C mc
c  
 Mc
n
n
*
The Dulong-Petit law gives c* in terms of R:
c*= 3R
c
3(8 .314 J / m ol  K )
c

 0 .392 J / g  K  0 .392 k J / k g  K
M
63 .5 g / m ol
*
This differs from the measured value of 0.386 kJ/kg·K given
in Table by less than 2%
The specific heat of a certain metal is measured to
be 1.02 kJ/kg·K. (a) Calculate the molar mass of
this metal, assuming that the metal obeys the
Dulong-Petit law. (b) What is the metal?
The specific heat of a certain metal is measured to be 1.02
kJ/kg·K. (a) Calculate the molar mass of this metal, assuming
that the metal obeys the Dulong-Petit law. (b) What is the
metal?
(a)
c*  Mc
c *  3R
Mc  3R
3R 3(8.314 J / mol  K )
M

 0.02445mol / kg  24.45mol / g
3
c
1.02 10 J / kg  K
(b) The metal must be magnesium, which has a molar
mass of 24.31 g/mol
Boltzmann Distribution Law
• The motion of molecules is extremely chaotic
• Any individual molecule is colliding with others at
an enormous rate
Typically at a rate of a billion times per second
• We add the number density nV (E )
This is called a distribution function
It is defined so that nV (E ) dE is the number of
molecules per unit volume with energy between E
and E + dE
Number Density and Boltzmann Distribution
Law
•
From statistical mechanics, the number density is
nV ( E )  n0 e
•
•
E

k BT
This equation is known as the Boltzmann
Distribution Law
It states that the probability of finding the
molecule in a particular energy state varies
exponentially as the energy (½mv2) divided by
k BT
Distribution of Molecular Speeds
• The observed speed
distribution of gas
molecules in thermal
equilibrium is shown in
figure. The number of
molecules having speeds
in the range v to v+dv is
equal to the area of the
shaded rectangle, Nvdv.
The function Nv
approaches zero as v
approaches infinity.
Distribution of Molecular Speeds
• As indicated in figure, the
average speed is somewhat
lower than the rms speed.
The most probable speed
vmp is the speed at which
the distribution curve
reaches a peak.
vrms 
vavg 
8k BT
vmp 
2k BT
k BT
 1.41
m0
m0
m0

3k BT
k BT
 1.73
m0
m0
2
vavg
k BT
 1.6
m0
v r m s  v a vg  v m p
Distribution Function
• The fundamental expression that describes the
distribution of speeds in N gas molecules is
 m 
N V  4 N 

 2 k B T 
3/2
2  mv 2 / 2 k B T
v e
m is the mass of a gas molecule, kB is
Boltzmann’s constant and T is the absolute
temperature
Speed Distribution
• The peak shifts to the right as T increases
This shows that the average speed increases with
increasing temperature
• The asymmetric shape occurs because the lowest
possible speed is 0 and the highest is infinity
Speed Distribution
• The distribution of molecular speeds depends
both on the mass and on temperature
• The speed distribution for liquids is similar to that
of gases
Evaporation
• Some molecules in the liquid are more energetic
than others
• Some of the faster moving molecules penetrate
the surface and leave the liquid
This occurs even before the boiling point is reached
• The molecules that escape are those that have
enough energy to overcome the attractive forces
of the molecules in the liquid phase
• The molecules left behind have lower kinetic
energies
Therefore, evaporation is a cooling process
Mean Free Path
• The molecules move with constant speed
along straight lines between collisions
• The average distance between collisions is
called the mean free path
• The path of an individual molecule is random
The motion is not confined to the plane of the
paper
Mean Free Path
• A molecule moving through a
gas collides with other
molecules in a random
fashion
• This behavior is sometimes
referred to as a random-walk
process
• The mean free path
increases as the number of
molecules per unit volume
decreases
Mean Free Path
• The mean free path is related to the
diameter of the molecules and the density of
the gas
• We assume that the molecules are spheres of
diameter d
• No two molecules will collide unless their
paths are less than a distance d apart as the
molecules approach each other
Mean Free Path
• The mean free path, ℓ, equals the average
distance vavgΔt traveled in a time interval Δt
divided by the number of collisions that occur in
that time interval:
v t
1


2
2
 d v  t  nV  d nV
Collision Frequency
• The number of collisions per unit time is the
collision frequency:
ƒ   d vn V
2
• The inverse of the collision frequency is the
collision mean free time