• 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  nRT NkBT
P    mv  

3 V  2
V
V

• Temperature is a direct measure of the average
molecular kinetic energy
___
1
3
2
mo v  kBT
2
2
The Kelvin Temperature of
an ideal gas is a measure of
the average translational
kinetic energy per particle:
K tot trans
 1 ___2  3
3
3
 N  m v   NkBT  nRT  PV
2
2
2
 2
k =1.38 x 10-23 J/K Boltzmann’s Constant
Root-mean-square speed:
vrms
3kT
 v 
m
2
Total Kinetic Energy
• The total kinetic energy is just N times the kinetic
energy of each molecule
 1 ___2  3
3
K tot trans  N  m v   NkBT  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
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.
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 )278K

(32u )(1.66 x1027 kg / u )
 466m / s
Is this fast?
YES!
Speed of
sound:
343m/s!
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  4 N 

2

k
T
B


3/2
v 2e  mv
Ludwig Boltzmann
1844 – 1906
Austrian physicist
Contributed to
Kinetic Theory of Gases
Electromagnetism
Thermodynamics
Pioneer in statistical mechanics
2
/ 2 kBT
QuickCheck 18.3
A rigid container holds both hydrogen gas (H2)
and nitrogen gas (N2) at 100C. Which statement
describes their rms speeds?
A. vrms of H2 < vrms of N2.
B. vrms of H2 = vrms of N2.
C. vrms of H2 > vrms of N2.
© 2013 Pearson Education, Inc.
Slide 18-34
QuickCheck 18.3
A rigid container holds both hydrogen gas (H2)
and nitrogen gas (N2) at 100C. Which statement
describes their rms speeds?
A. vrms of H2 < vrms of N2.
B. vrms of H2 = vrms of N2.
C. vrms of H2 > vrms of N2.
© 2013 Pearson Education, Inc.
Slide 18-35
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
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
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
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 Heats
Isobaric requires MORE HEAT than Isochoric for the
same change in Temperature!!!!
The total change in thermal
energy for ANY PROCESS,
due to work and heat, is:
This applies to all ideal gases, not
just monatomic ones! WOW!
CP and CV
Note that for all
ideal gases:
where
R = 8.31 J/mol K is
the universal gas
constant.
Slide 17-80
Important Concepts
Agreement with Experiment: Diatomic Hydrogen
acts like a monatomic gas at low temperature!
•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.
Quantization of Energy.
•To explain the results of the various molar
specific heats, we must use some quantum
mechanics.
– Classical mechanics is not sufficient
•This energy level diagram shows the rotational
and vibrational states of a diatomic molecule.
•The lowest allowed state is the ground state.
•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.
Section 21.4
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
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
Adiabatic Processes
Adiabatic Processes: Q=0
 An adiabatic process is one for which:
Ti Vig-1 = Tf Vfg-1
where:
 Adiabats are steeper than
hyperbolic isotherms because
only work is being done to
change the Temperature. The
temperature falls during an
adiabatic expansion, and rises
during an adiabatic
compression.
Slide 17-88
A 4.00-L sample of a nitrogen gas confined to a
cylinder, is carried through a closed cycle. The
gas is initially at 1.00 atm and at 300 K. First, its
pressure is tripled under constant volume. Then,
it expands adiabatically to its original pressure.
Finally, the gas is compressed isobarically to its
original volume. (a) Draw a PV diagram of this
cycle. (b) Find the number of moles of the gas. (c)
Find the volumes and temperatures at the end of
each process (d) Find the Work and heat for each
process. (e) What was the net work done on the
gas for this cycle?
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 g = constant
g = 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, DEint = 0 and
the initial and final states are the
same and no change in temperature is
expected.
– No change in temperature is expected
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