The Kinetic Theory of Gases
1. Deriving a Gas Law
c
Imagine a rectangular box with sides a, b, and c, containing
only one molecule. The molecule has a mass m and a velocity
⃑ . Collisions between the molecule and the walls of the box
are elastic, and the walls of the box are immovable. Thus, the
molecule rebounds from the wall with the same speed it had
before the collision.
We wish to find the average force exerted by the wall on the
molecule over a period of time. We calculate this by using the
impulse given to the molecule at each collision:
⃑
a
m
𝑣⃑𝑥
b
⃑⃑
⃑
( )
Now we need the values of m∆v and ∆t in terms of m and ⃑ . Because the molecule rebounds
at the same speed, but in the opposite direction, its change in velocity is –2 ⃑ . Thus:
⃑
(
⃑ )
⃑
Since we want the average force, rather than the instantaneous force, we will take Δt to be the
time between successive collisions instead of the time during which the force is acting. This may
seem odd, but it works out fine when there are 1023 atoms having trillions of collisions per
nanosecond per square millimeter. So, ∆t, is the time that it takes for the molecule to travel the
length of the box and return again to the same wall. This is equal to the distance travelled, 2c,
divided by the speed, ⃑ . Combining these formulae, we arrive at:
(
⁄ )
The force on the wall is equal to the force on the molecule, but in the opposite direction, so we
drop the negative sign.
We can now find the pressure on the wall, since pressure is simply force per unit of area.
⁄
,
This can be rearranged to form:
where V is the volume of the box.
.
Of course, any real box of gas will contain more than one molecule. We can take this into
account by multiplying the force calculated for one molecule by the number of molecules in the
box, N. Since not all of the molecules will have the same velocity in the x direction, we will use
the average squared velocity in our equation. [Note that "the average squared velocity" is not the
( ) ]:
same as "the average velocity, squared". That is ( )
(
)
The Kinetic Theory of Gases
The last adjustment that we need to make is to switch from using only the velocity in the x
direction to using simply the velocity.
By using three-dimensional Cartesian coordinates, any
velocity can be divided into components in the x, y, and z
𝑣𝑦
directions. Furthermore, by using the Pythagorean
theorem It can be shown that:
𝑣
for any velocity.
Since this is true for each particular velocity, it must
also be true for the averages of the squared velocities:
(
)
(
)
(
(
)
𝑣𝑥
𝑣𝑧
)
This equation is particularly helpful because the coordinates x, y, and z, have been chosen
arbitrarily. That is, there can be no preferred direction. Thus:
(
)
(
(
)
)
(
)
(
)
(
)
(
)
By substituting this into equation 1, we arrive at:
(
)
(
[
or,
)
]
[
]
This is a very significant result. Consider how it compares to the perfect gas law arrived at
experimentally.
Perfect Gas Law (experimental):
PV = NkT,
[
Gas Law from Kinetic Theory:
where T is the temperature in Kelvin
degrees, and k is Boltzman’s constant.
]
Combining theory with experiment, we arrive at the conclusion that the temperature of a gas is
directly proportional to the average kinetic energy of its molecules.That is,
[
[
]
]
The establishment of this proportion is one of the most important successes of the kinetic
theory of matter. It explains the relationship between mechanical energy and internal energy. It
is the beginning of an explanation of the differences in diffusion rates of different gases, the
differences in the speed of sound in different gases, and many other disparate phenomena.
The Kinetic Theory of Gases
2. Average Velocity of Gas Molecules
Even though we can now find the average kinetic energy of molecules in a gas, there is no easy
way to calculate the mean speed of molecules in a gas. Finding this value requires a knowledge
of how the kinetic energy is distributed among the molecules. The characteristics of this
distribution, called the Maxwell-Boltzman distribution, are well understood (by someone), but
using them to calculate the mean speed of gas molecules is difficult, time-consuming, and not
particularly instructive.
Instead, we will use a value that is simple to calculate to represent an "average" speed of
molecules in a gas. This value is the square root of the mean value of the squared speeds of the
molecules. We arrive at this value as follows.
[
(
)
(
]
)
√(
)
√
This value, √( ) , is called the "root-mean-square speed" or "rms speed" and written
"vrms". It is different from the mean speed because the "average squared velocity" is not the same
as the "average velocity, squared". That is ( )
(
) .
Rather than finding the rms speed of molecules using equation 2, we will confine ourselves to
problems involving ratios between speeds of molecules in gases that are in thermal equilibrium
(that is, they are at the same temperature).
Example: A gas sample consists of a mixture of carbon dioxide (MM = 44 grams/mole) and
helium (MM = 4 grams/mole). If the rms speed of the CO2 molecules is 4.23 x 102 m/s, what
is the rms speed of the helium atoms?
Solution: Because the two gases are in the same container, they must be in thermal equilibrium.
That is, their molecules have the same average kinetic energy.
(
Thus,
√(
P.S.
)
(
)
)
√
(
( )
√
( )
√
(
)
)
= 1.40 x 103 m/s.
(
)