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TopicsKinetic Molecular Theory and Its Applications

Kinetic Molecular Theory

The most immediately useful bit of information you can pull from the definition
of the kinetic molecular theory provided in the
summary is that the average kinetic
energy
of a gas is proportional to the absolute temperature.

âàùT

@@Equation @@ has a number of very serious implications. First of
all, any two gases at the same temperature will have the same kinetic energy.
Remember that kinetic energy E_{k} = 1/2mv^{2}, and that average kinetic energy
= 1/2m.

Here's where things get complicated. After some mathematical maneuvering we
find a more exact expression for the average velocity $\overline{v}$:

= =

k is the Boltzmann constant. Think of the Boltzmann constant as the gas constant R for individual molecules. Analogously, m is the mass per
molecule, just as M is the mass per mole. If you multiply k by Avogadro's
number, you'll get R.

Let's take a breather. In order to keep things simple, I have refrained from
including derivations. If you are at all mathematically inclined, however, I
suggest that you take a look in a good physics book (look under statistical
mechanics or ideal gases) at the derivations of and the other
equations I introduce. Alternatively, ask your instructor to show you. The
derivations can be painful, but they will prove to you that these equations have
meaning.

Let's get back into the fray. There are two other characterizations of v that
you should know: the most probable velocity $v_p$ and the root mean
square velocity $v_{\mbox{rms}}$. The most probable velocity is exactly
what it sounds like: the velocity at which the greatest number of molecules in a
gas travel. It can be expressed mathematically:

v_{p} = =

The root mean square velocity, which measures the typical velocity of molecules in a gas, is slightly tricky. To derive its value, find the square root of the mean of the squares of the average velocity. It is easier to understand
mathematically:

v_{rms}

=

=

=

Make sure that you see that v_{rms} = , NOT
v_{rms} = . The latter equation reduces to
v_{rms} = , which is not the case. v_{rms}
requires the mean of the squares of the velocities. Square the velocities
first, then take their mean.

When solving for these values of v, be sure to reduce all variables to
SI units. M is particularly insidious--it must be in kg/m^{3} if all the
other units are SI.

Maxwell-Boltzmann Speed Distributions

You'll often see the range of speeds plotted against the number of molecules on
a Maxwell-Boltzmann speed distribution. Plotting the values of
, v_{p}, and v_{rms}, we find that:

v_{p} < < v_{rms}

These three measures of v are not equal because the distribution is not
symmetrical about its peak. Such is the case because the lowest possible speed
is zero, while the highest is classically infinite. For this reason, the peak
of the distribution (v_{p}) will always be to the left of the average speed
(). The difference between v_{p} and v_{rms} is even
more exaggerated because it involves the mean of squares.

A Maxwell-Boltzmann speed distribution changes with temperature. As discussed
with the kinetic molecular theory, higher temperatures lead to higher
velocities. Thus the distribution of a gas at a hotter temperature will be
broader than it is at lower temperatures.

The total area underneath the Maxwell-Boltzmann speed distributions is equal to
the total number of molecules. If the area under the two curves is equal, than
the total number of molecules in each distribution are equal.

The Maxwell-Boltzmann speed distribution also depends on the molecular mass of
the gas. Heavier molecules have, on average, less kinetic energy at a given
temperature than light molecules. Thus the distribution of lighter molecules
like H_{2} is much broader and faster than the distribution of a heavier
molecule like O_{2}:

Diffusion and Mean Free Path

Diffusion is the spread of one substance through another. The fact that
molecules collide when they diffuse is the reason why it takes a considerable
amount of time for a gas to travel from one place to another. Think, for
example, of a smell released at one point in the room. Because gas molecules
move at such fast velocities, if there were no collisions the smell would fill
the room instantly.

The collision between gas molecules makes the calculation of the rate of
diffusion difficult. Instead, we'll focus on the mean free path. The mean
free path λ is the mean distance a molecule travels before it impacts
another molecule; given the huge number of collisions in a gas, the mean free
path is vastly smaller than any typical room or container. The mean free path
is calculated with the following formula:

λ =

N is the total number of molecules present. The rate of collisions is simply
v_{rms} divided by the mean free path:

rate of collision =

Effusion

Effusion is the rate at which a gas passes through a small hole into a
vacuum. The rate of effusion of a gas is directly proportional to
v_{rms}: