• Temperature ~ Average KE of each particle
• Particles have different speeds
• Gas Particles are in constant RANDOM motion
• Average KE of each particle is: 3/2 kT
• Pressure is due to momentum transfer
Speed ‘Distribution’ at
CONSTANT Temperature
is given by the
Maxwell Boltzmann
Speed Distribution
Mean Free Path
 A single molecule follows a
zig-zag path through a gas
as it collides with other
molecules.
 The average distance
between the collisions is
called the mean free path:
 (N/V) is the number density of the gas in m−3.
 r is the the radius of the molecules when modeled as
hard spheres; for many common gases r ≈ 10−10 m.
© 2013 Pearson Education, Inc.
Slide 18-20
QuickCheck 18.1
The temperature of a rigid container of oxygen gas
(O2) is lowered from 300°C to 0°C. As a result, the
mean free path of oxygen molecules
A. Increases.
B. Is unchanged.
C. Decreases.
© 2013 Pearson Education, Inc.
Slide 18-21
QuickCheck 18.1
The temperature of a rigid container of oxygen gas
(O2) is lowered from 300°C to 0°C. As a result, the
mean free path of oxygen molecules
A. Increases.
B. Is unchanged.
C. Decreases.
© 2013 Pearson Education, Inc.
λ depends only on N/V, not T.
Slide 18-22
Pressure and Kinetic Energy
•
•
•
•
•
•
Assume a container is a cube with
edges d.
Look at the motion of the molecule in
terms of its velocity components and
momentum and the average force
Pressure is proportional to the number
of molecules per unit volume (N/V) and
to the average translational kinetic
energy of the molecules.
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
___
2  N  1
2 
P =    mo v 
3  V  2

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 
=
= NkBT
P
v  nRT
 m=


3 V  2

• Temperature is a direct measure of the average
molecular kinetic energy
Molecular Interpretation of
Temperature
• Simplifying the equation relating
temperature and kinetic energy gives
___
1
3
2
mo v = kBT
2
2
• This can be applied to each direction,
1 ___2 1
m v x = kBT
2
2
– with similar expressions for vy and vz
Total Kinetic Energy
• The total kinetic energy is just N times the kinetic
energy of each molecule
 1 ___2  3
3
K
=
N =
mv  =
NkBT
nRT
tot trans
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
Kinetic Theory Problem
A 5.00-L vessel contains nitrogen gas at
27.0°C and 3.00 atm. Find (a) the total
translational kinetic energy of the gas
molecules and (b) the average kinetic energy
per molecule.
Hot Question
Suppose you apply a flame to 1 liter of water for a certain
time and its temperature rises by 10 degrees C. If you apply
the same flame for the same time to 2 liters of water, by how
much will its temperature rise?
a) 1 degree
b) 5 degrees
c) 10 degrees
d) zero degrees
Ludwig Boltzmann or Dean Gooch?
• 1844 – 1906
• Austrian physicist
• Contributed to
– Kinetic Theory of Gases
– Electromagnetism
– Thermodynamics
• Pioneer in statistical
mechanics
Distribution of Molecular Speeds
•
•
•
The observed speed distribution of gas
molecules in thermal equilibrium is
shown at right
NV is called the Maxwell-Boltzmann
speed distribution function
mo is the mass of a gas molecule, kB is
Boltzmann’s constant and T is the
absolute temperature
 mo 
NV = 4π N 

k
T
π
2
B


3/2
2
v e
− mv 2 / 2 kBT
Molecular Speeds and
Collisions
Speed Summary
• Root mean square speed
=
v rms
kT
3kBT
=
1.73 B
mo
mo
• The average speed is somewhat lower than
the rms speed
8k T
kT
=
v avg
B
=
1.60 B
mo
π mo
• The most probable speed, vmp is the speed at
which the distribution curve reaches a peak
• vrms > vavg > vmp
=
v mp
2kBT
kT
=
1.41 B
m
m
Some Example vrms Values
At a given temperature, lighter molecules move faster, on
the average, than heavier molecules
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
The Kelvin Temperature of
an ideal gas is a measure of
the average translational
kinetic energy per particle:
3 / 2kT
= KE
= 1/ 2mv
2
rms
k =1.38 x 10-23 J/K Boltzmann’s Constant
Root-mean-square speed:
v=
rms
=
v
2
3kT
m
Kinetic Theory Problem
Calculate the RMS speed of an oxygen molecule
in the air if the temperature is 5.00 °C.
The mass of an oxygen molecule is 32.00 u
(k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)
vrms
3kT
=
m
What is m?
m is the mass of one
oxygen molecule in kg.
What is u?
How do we get the mass in kg?
Kinetic Theory Problem
Calculate the RMS speed of an oxygen molecule
in the air if the temperature is 5.00 °C.
The mass of an oxygen molecule is 32.00 u
(k = 1.3 8x 10 -23 J/K, u = 1.66 x 10 -27 kg)
vrms
3kT
=
m
What is m?
m is the mass of one
oxygen molecule.
−23
3(1.38 x10 J / K )278 K
=
(32u )(1.66 x10−27 kg / u )
= 466m / s
Is this fast?
YES!
Speed of
sound:
343m/s!
A cylinder contains a mixture of helium
and argon gas in equilibrium at 150°C.
(a) What is the average kinetic energy
for each type of gas molecule?
(b) What is the root-mean-square speed
of each type of molecule?
More Kinetic Theory Problems
A gas molecule with a molecular mass of 32.0 u has a
speed of 325 m/s. What is the temperature of the gas
molecule?
A) 72.0 K
B) 136 K
C) 305 K
D) 459 K
E) A temperature cannot be assigned to a single
molecule.
Temperature ~ Average KE of all particles
Equipartition of Energy
•
•
•
•
•
•
Each translational degree of freedom contributes an
equal amount to the energy of the gas
– In general, a degree of freedom refers to an
independent means by which a molecule can
possess energy
Each degree of freedom contributes ½kBT to the
energy of a system, where possible degrees of
freedom are those associated with translation,
rotation and vibration of molecules
With complex molecules, other contributions to
internal energy must be taken into account
One possible energy is the translational motion of
the center of mass
Rotational motion about the various axes also
contributes
There is kinetic energy and potential energy
associated with the vibrations
Monatomic and Diatomic Gases
The thermal energy of a monatomic gas of N atoms is
A diatomic gas has more thermal energy than a monatomic
gas at the same temperature because the molecules have
rotational as well as translational kinetic energy.
Molar Specific Heat
• We define specific heats for two processes
that frequently occur:
– Changes with constant pressure
– Changes with constant volume
• Using the number of moles, n, we can
define molar specific heats for these
processes
• Molar specific heats:
– Q = nCV DT for constant-volume processes
– Q = nCP DT for constant-pressure processes
Ideal Monatomic Gas
• Therefore, ∆Eint = 3/2 nRT
− ∆E is a function of T only
• In general, the internal energy of an ideal
gas is a function of T only
– The exact relationship depends on the type of
gas
• At constant volume, Q = ∆Eint = nCV ∆T
– This applies to all ideal gases, not just
monatomic ones
Ratio of Molar Specific Heats
• We can also define the ratio of molar specific heats
CP 5R / 2
=
γ =
= 1.67
CV 3R / 2
• Theoretical values of CV , CP , and γ 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
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
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
Sample Values of Molar Specific
Heats
In a constant-volume process, 209 J of
energy is transferred by heat to 1.00 mol
of an ideal monatomic gas initially at
300 K. Find (a) the increase in internal
energy of the gas, (b) the work done on
it, and (c) its final temperature
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
• An adiabatic process is one in which no energy is
transferred by heat between a system and its
surroundings (think styrofoam cup)
• 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 γ = constant
γ = CP / CV is assumed to be constant
All three variables in the ideal gas law (P, V, T ) can
change during an adiabatic process
Special Case: Adiabatic Free
Expansion
• This is an example of adiabatic free
expansion
• The process is adiabatic because it
takes place in an insulated container
• Because the gas expands into a
vacuum, it does not apply a force on
a piston and W = 0
• Since Q = 0 and W = 0, ∆Eint = 0 and
the initial and final states are the
same and no change in temperature is
expected.
– No change in temperature is expected
Adiabatic Process
• The PV diagram shows
an adiabatic expansion
of an ideal gas
• The temperature of the
gas decreases
– Tf < Ti in this process
• For this process
Pi Viγ = Pf Vfγ and
Ti Viγ-1 = Tf Vfγ-1
A 2.00-mol sample of a diatomic ideal
gas expands slowly and adiabatically
from a pressure of 5.00 atm and a
volume of 12.0 L to a final volume of
30.0 L.
(a) What is the final pressure of the gas?
(b) What are the initial and final
temperatures?
(c) Find Q, W, and ∆Eint.
Important Concepts
QuickCheck 17.7
Three possible processes A, B,
and C take a gas from state i to
state f. For which process is the
heat transfer the largest?
A.
B.
C.
D.
Process A.
Process B.
Process C.
The heat is the same for all three.
Slide 17-58
QuickCheck 17.7
Three possible processes A, B,
and C take a gas from state i to
state f. For which process is the
heat transfer the largest?
A.
B.
C.
D.
Process A.
Process B.
Process C.
The heat is the same for all three.
Same for all three
∆Eth = W + Q
Most negative for A ... ... so Q must be most positive.
Slide 17-59
Reading Question 18.2
What additional kind of energy makes CV larger
for a diatomic than for a monatomic gas?
A.
B.
C.
D.
E.
Charismatic energy.
Translational energy.
Heat energy.
Rotational energy.
Solar energy.
Slide 18-12
Reading Question 18.2
What additional kind of energy makes CV larger
for a diatomic than for a monatomic gas?
A.
B.
C.
D.
E.
Charismatic energy.
Translational energy.
Heat energy.
Rotational energy.
Solar energy.
Slide 18-13
Reading Question 18.3
The second law of thermodynamics says that
A. The entropy of an isolated system never
decreases.
B. Heat never flows spontaneously from
cold to hot.
C. The total thermal energy of an isolated
system is constant.
D. Both A and B.
E. Both A and C.
Slide 18-15
Reading Question 18.4
In general,
A. Both microscopic and macroscopic processes
are reversible.
B. Both microscopic and macroscopic processes
are irreversible.
C. Microscopic processes are reversible and
macroscopic processes are irreversible.
D. Microscopic processes are irreversible and
macroscopic processes are reversible.
Slide 18-16
QuickCheck 18.6
Systems A and B are both monatomic gases. At this instant,
A.
TA > TB.
B.
TA = TB.
C.
TA < TB.
D.
There’s not enough information to compare their
temperatures.
Slide 18-51
QuickCheck 18.6
Systems A and B are both monatomic gases. At this instant,
A.
TA > TB.
B.
TA = TB.
C.
TA < TB.
D.
There’s not enough information to compare their
temperatures.
A has the larger average energy per atom.
Slide 18-52
QuickCheck 17.13
A gas in a container expands
rapidly, pushing the piston out.
The temperature of the gas
A.
B.
C.
D.
Rises.
Is unchanged.
Falls.
Can’t say without knowing more.
Slide 17-89
QuickCheck 17.13
A gas in a container expands
rapidly, pushing the piston out.
The temperature of the gas
A.
B.
C.
D.
Rises.
Is unchanged.
Falls.
Can’t say without knowing more.
Slide 17-90