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Boltzmann Distribution Law
The motion of molecules is extremely chaotic
enormous rateAny individual molecule is colliding with others at an Typically at a rate of a billion times per second
We introduce the
(^) number density
(^) n V (^) ( E (^) )
This is called a distribution function
It is defined so that
(^) n V (^) ( E (^) ) (^) dE
(^) is the number of
molecules per unit volume with energy between
E
and
E
(^) dE
n From statistical mechanics, the number density is
V
E
n
0 e
(^) – E (^) / k B T
Boltzmann distribution law
divided byenergy state varies exponentially as the energyprobability of finding the molecule in a particularThe Boltzmann distribution law states that the
(^) k B T
equilibrium is shown at rightof gas molecules in thermalThe observed speed distribution
P ( v ) is called the
(^) Maxwell-
functionBoltzmann speed distribution
P ( v )
P ( v )
The peak shifts to the right as
T
(^) increases
with increasing temperatureThis shows that the average speed increases
possible speed is 0 and the highest is infinityThe asymmetric shape occurs because the lowest
N
v=NP
v )
P ( v ) is a probability distribution function, it gives
intervalthe fraction of molecules whose speeds lie in the
(^) dv
(^) centered on the speed
(^) v .
on the mass and on temperatureThe distribution of molecular speeds depends both
liquids than in gasesof gases even though the speeds are smaller inThe speed distribution for liquids is similar to that
where atoms can still move around.helium, which is known to be a “quantum solid”anymore, they vibrate. The only exception is solidIn solids, atoms do not have translational energy
P ( v ) dv
(^) = (^1)
(^0)
Example: What is the
(^) rms
(^) speed of hydrogen at
T
=300 K? How much slower are O
2 (^) molecules
compared to H
2 ?
M (H (^2) ) = 2.016 g/mole
v rms (^) = 1930 m/s at
(^) T (^) = 300 K
v rms (^) =
3 M RT
v rms (^) (O 2 )
v rms (^) (H 2 ) = M H 2
M O 2
v rms (^) (O 2 )
v rms
(^) (H
2 ) =
(^322) (^) =
4 1
O
2 (^) is 4 times slower than H
2
H (^2)
Kinetic Theory of the Gases, Part 2
The Mean Free Path
fashionmolecules in a randoma gas collides with otherA molecule moving through
referred to as aThis behavior is sometimes
(^) random-walk
process
The
(^) mean free path
decreasesmolecules per unit volume increases as the number of
12/08/
Lecture 6
The mean free path,
l , equals the average distance
v t traveled in a time interval
t divided by the
number of collisions that occur in that time interval:
collision frequency The number of collisions per unit time is the
collisionThe inverse of the collision frequency is the
(^) mean free time
l
v (^) t
(^) d 2 v (^) t
( ) n v = 1
(^) d (^2) N / V
f
d
2
vN / V
(units: particles s
)
of an ideal gaschange the temperatureSeveral processes can
Since
T
(^) is the same for
each process,
E
int (^) is also
the same
the work (1st law)the different paths, as isThe heat is different for
temperature isparticular change inassociated with aThe heat and work
(^) not
(^) unique
Molar Specific Heat
An ideal monatomic gas contains non-interacting atoms
increasing the translational kinetic energy of the gaswith a fixed volume, all of the energy goes intoWhen energy is added to a monatomic gas in a container There is no other way to store energy in such a gas
Therefore,
E
int (^) is a function of
T
(^) only:
E
int (^) = 3/
(^) nRT
The exact relationship depends on the type of gas
At constant volume,
Q
E
int (^) = (^) nC
V (^) T
onesThis applies to all ideal gases, not just monatomic
Solving for
C
V (^) gives
C
V (^) = 3/
R
(^) = 12.5 J/mol
(^). K^
For all monatomic gases
results for monatomic gasesThis is in good agreement with experimental
In a constant-pressure process,
E
int (^) = (^) Q
(^) + (^) W
(^) and
C
P (^) – (^) C
V (^) = (^) R
C This also applies to any ideal gas P (^) = 5/
R
(^) = 20.8 J/mol
(^). K^
We can also define
(^) C P
C V
= (^5) R / (^) 2
3 R / (^) 2 (^) =
3 5 (^) = (^1)
. 67
Molar Specific Heats of Other Materials
vibrational motions of the moleculesinclude contributions from the rotational andThe internal energy of more complex gases must
the thermal expansion is small andconstant pressure, very little work is done sinceIn the cases of solids and liquids heated at
C
P
and
C
V
are
approximately equal
Equipartition of Energy
must be taken into accountcontributions to internal energyWith complex molecules, other
center of masstranslational motion of theOne possible energy is the
various axes also contributesRotational motion about the around theWe can neglect the rotation
(^) y (^) axis in a
compared to thediatomic since it is negligible
(^) x
(^) and
(^) z (^) axes
Linear
molecule
freedom (each degree = 1/2The translational motion adds three degrees of
R
degrees of freedom (each degree = 1/2The rotational motion in a linear molecule adds two
R
more degree of freedom (each degree = 1The vibrational motion for a diatomic adds one
R
Therefore,
E
int (^) = 7/
(^) nRT
(^) and
C
V (^) = 7/
R
the molar specific heat is a function of temperatureThis is inconsistent with experimental results as
monatomic gasAt low temperatures, a diatomic gas acts like a C
V (^) = 3/
R
At about room temp. the value increases to
C
V (^) = 5/
R
not vibrational energyThis is consistent with adding rotational energy but
At high temperatures, the value increases to
C
V (^) = 7/
R
and translationalThis includes vibrational energy as well as rotational
are more complex (3For molecules with more than two atoms, the vibrations
N
atoms
-6(5) degrees of freedom)
The number of degrees of freedom becomes larger
the more “ways” there are to store energyThe more degrees of freedom available to a molecule,
This results in a higher molar specific heat
