Thermal Physics Tutorial 3 - Molecular Kinetic Theory
The model, which we will study in this tutorial, tells us how gases behave. The topic fits quite well with mechanics because you need to understand about collisions to make sense of the Kinetic Theory model. It may be worth revising these ideas in Units 2 and 4.
In mechanics we have been looked at big things, like tennis balls, cars, people, to name but a few. This can be referred to as the macroscopic level. In this piece of work we will go down to the molecular level, where things are very small. We call this the microscopic level.
A Scottish botanist, Robert
Brown, first suggested that gases were in a state of random motion. He observed
that pollen particles seem to jiggle about for no apparent reason. We can see
the same by observing smoke particles under a microscope.
The random jerky movement of the particles is called Brownian
motion and is due to the continuous and random bombardment by air molecules.
Pressure is exerted by the
impact of molecules on the sides of a container. We can use the idea to derive
an expression for gas pressure. We must make the following assumptions:
The gas consists of a large number of identical
molecules, such that the statistical treatments are meaningful.
All collisions between
molecules and the walls of the container are perfectly elastic.
Intermolecular forces are negligible, as is gravity.
Molecules move in straight
lines and at constant speed
between collisions.
Collision
times are negligible
compared with time between collisions.
Volume
of the gas molecules is negligible compared
to that of the gas.
Newton’s laws of motion are applicable.
Health Warning: The derivation of the kinetic theory equation is rather tedious but it's on the syllabus.
Here we go:
Stage 1 Momentum Change
Consider a large box, side
l, full of molecules:
Now consider one molecule of
mass
m
Þ
its momentum in the
x direction is
mc
At face 1 there is a perfectly
elastic collision so that the momentum is reversed to
-mc_{x}.
This results in a momentum
change of:
mc_{x}
- (-mc_{x}) = 2mc_{x}
After colliding with face 1, the molecule travels a distance 2l before it collides again with face 1 again. This takes a time t, which is given by:
t = 2l/c_{x}
Stage 2 Impulse
So that we can get the
pressure, we need to know the force. This
can be worked out by Newton II, considering the impulse,
which is the change of momentum.
impulse
=
FDt
Rearranging gives us:
F =
change in momentum
Dt
Time interval is the time taken for the molecule to move up the box to the far end, bounce off and come back again. Time = distance / speed = 2l / c_{x}
Þ
F
= 2mc_{x}
= m(c_{x})^{2}
2l/c_{x} l
Stage 3 Working out the pressure
Since pressure = force/area, we can work out the pressure on face 1.
The
area of face
1 =
l^{2}.
p = F = mc_{x}^{2}/l = mc_{x}^{2}
A l^{2}
l^{3}
Suppose we have N
molecules of gas. The pressure will be:
p = m ( c_{x}_{1}^{2}
+
c_{x}_{2}^{2}
+ c_{x}_{3}^{2}....+
c_{xN}^{2})
= m(N_{c}_{x}^{2})
l^{3} l^{3}
Now l^{3} is the volume of the gas, so we can rewrite the equation:
Stage 4 The Statistics
The
term
<c_{x}
^{2}>
("cx-squared bar") is called the mean
square speed of the molecules in the x
direction. The bar is written over
the
c^{2}, as in the yellow
box. (Notice the different way of writing it in the text. You may
see this in some books. It is also almost impossible to write c-bar in
this web-page editor!) However the molecules are moving randomly in
the container and very few would be moving exactly parallel to the x-axis. However we can
consider each molecule’s velocity to be the resultant of three components,
c_{x},
c_{y},
and
c_{z}.
As in two dimensions, the three components can be combined by Pythagoras to give the resultant velocity.
Similarly we can combine the mean square velocities:
Since there are a large number
of molecules we can assume that there are equal numbers moving in each of the
co-ordinate directions:
So we can write:
So our final equation becomes:
Stage 5 Density
However we can go further.
Since
Nm is the total mass of
the gas and
V
is the volume, we can
express the pressure in terms of the density.
density = mass
=
Nm
volume
V
This gives us the equation:
Be careful of the difference between the square of the mean
(<c>)^{2} ["c-bar squared"] and the mean square
<c^{2}>
["c-squared bar"]. Consider
1 + 2 + 3 = 6. The average is 2 and
the square of the mean is 4. (1^{2}
+ 2^{2} + 3^{2}
) ÷ 3 = (1 + 4 + 9) ÷ 3 = 14 ÷ 3 = 4.67, so there is quite a difference.
The most probable speed is that
at which the greatest number of molecules are moving. The mean
speed is the average value of all the speeds. The root
mean square speed is the square root of the mean square speed of the
molecules. They are slightly but significantly different.
Question 1 |
The observed speed of ten particles at a particular instant are shown in the table:
(a) What is the most probable speed? (b) What is the mean speed? (c) What is the rms speed of the molecules? |
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Question 2 |
Nitrogen gas is kept in a closed container at a temperature of 27 ^{o}C and a pressure of 1.0 × 10^{5} Pa. The density of nitrogen is 1.25 kg m^{-3}. Calculate: (a) the rms speed of the molecules; (b) the temperature at which the molecules travel twice as fast. |
The internal energy of a material is the sum of its potential and kinetic
energies. In an ideal gas the
molecules are so far apart that intermolecular forces can be ignored.
Therefore there is no potential energy.
So all the energy is kinetic. The
total kinetic energy is shared randomly
throughout the molecules in the gas. Therefore
there is a random distribution of the speeds of the molecules, which we can show
on a graph.
If no energy is being transferred as heat between an object and its surroundings, we say that it is in thermal equilibrium. If a gas is in thermal equilibrium and it is not being compressed or expanded, the average kinetic energy will remain constant, so will the temperature.
We have seen that:
pV = nRT
pV = 1/3 Nmc^{2}
It does not take a genius to see that:
nRT = 1/3 Nmc^{2}
We also know that kinetic energy,
E_{k} = 1/2 mv^{2}
(or
1/2 mc^{2}).
We can now rewrite our
expression as (since 2/3 ×1/2 = 1/3):
nRT = 2/3 N[1/2mc^{2}]
We can rearrange this to write:
1/2mc^{2} = 3/2nRT
= 3/2RT
N N/n
Since N is the total number of molecules and
n
is the number of moles,
N/n
will always be
N_{A},
which is Avogadro's number, 6.02 ×10^{23}.
Þ
1/2mc^{2} = 3/2RT
N_{A}
Now
R/N_{A} is called
Boltzmann's constant and is given the
code
k.
We can easily work out the value for
k.
k = 8.31
= 1.38 × 10^{-23} J K^{-1}
6.02
× 10^{23}
So the equation now becomes:
The translational kinetic energy of an ideal gas molecule is NOT dependent on what the gas is. It is only dependent on the temperature. So the kinetic energy of helium atoms is the same as the kinetic energy of xenon atoms at a given temperature.
Remember the bear trap. Temperature must be in Kelvin.
Question 3 |
A cylinder of volume 0.25 m^{3} contains nitrogen gas at a temperature of 17 ^{o}C and a pressure of 1.0 × 10^{5} Pa. Molar mass of nitrogen = 0.028 kg mol^{-1}. Calculate: a) The number of moles of gas in the cylinder b) The rms speed of the gas molecules at 17 ^{o}C c) The average translational KE of a nitrogen molecule; d) The total kinetic energy of the gas in the cylinder. |