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PHY 171
Lecture 13
(February 15, 2012)
Kinetic Theory of Gases — Assumptions
• Assume that a gas is comprised of a large number N of molecules, each of mass m, moving in
random directions with a variety of speeds.
This assumption is well supported by experiment. Brownian motion proves the random motions
of molecules, and the fact that an entire dish of water does not evaporate at once indicates that
molecules must have a variety of speeds. Also, in your warm-up assignment, you read how
one could do an experiment in which gas molecules were passed through an evacuated tube
containing a “velocity selector” — a set of two circular plates with notches cut into them, so
that by varying the speed of the axle supporting these two plates, we could make the plates
rotate at just the right angular speed that only molecules with a specified velocity would be
fast enough to get through the notch in the second plate.
• On average, molecules in a gas remain far apart from one another, that is, the average separation
between molecules is much greater than the size of each molecule.
• Molecules obey the laws of classical mechanics, and interact only when they collide. In other
words, we ignore the weak attractive forces between molecules, because the potential energy
associated with these forces is small compared to the kinetic energy.
• Assume perfectly elastic collisions between molecules and of molecules with the walls of the
container.
• The time taken for a collision is much less than the time between collisions, that is, collisions
are considered to be instantaneous processes.
Qualitative explanation of pressure of a gas based on kinetic theory
The pressure of a gas arises from the collision of the molecules of the gas with the walls of the
container.
Qualitative explanation of Boyle’s Law
The above characterization helps us understand Boyle’s law. If we decrease the volume of a gas,
the molecules have less space to move around in, and therefore collide more frequently with the
walls of the container, thereby giving rise to higher pressure.
We will now derive quantitative expressions for pressure and temperature based on the kinetic
theory of gases.
PHY 171 (Winter 2012 )
Lecture 13
Derivation of Expression for Pressure from Kinetic Theory
Imagine that the molecules are in a rectangular vessel
whose walls have area A, and the length between opposite walls is l, as shown in the figure on the right.
Recall that the pressure exerted by the gas on the walls of
the container is due to the collision of the molecules with
the walls.
Now, focus on one molecule as it moves about inside the
box. Suppose this molecule has velocity v, which has components vx , vy , and vz in the x, y, and z directions respectively.
Let this molecule collide with the right wall along the x-axis, as shown in the figure above. Since
the collision is elastic, only the x-component of momentum will change. The initial momentum
is mvx directed along the positive direction of the x-axis. After colliding with the wall, the final
momentum of the molecule will be mvx , but along the negative direction of the x-axis, so we can
write the final momentum as −mvx . Therefore,
change in momentum, ∆(mv) = mvx − (−mvx ) = 2mvx
Now, since the length traveled along the x-axis is l — the length of the box, the time taken to go
from the right wall to the left wall and back to the right wall is
2l
∆t =
vx
The time ∆t between collisions is small, so the number of collisions per second is large. Therefore,
the average force averaged over many collisions is equal to the force during one collision divided
by the time between collisions. Now, recall that
dv
d
Force, F = ma = m = (mv)
dt
dt
and so, the force on the right wall
F =
2mvx
mvx2
∆(mv)
=
=
∆t
2l/vx
l
This is the force due to one molecule. Note that collisions with the top or sides do not affect vx ,
so we can ignore them in writing the above equation. Also, the molecule may collide with other
molecules on its way between walls, but this transfers the energy to the other molecules — when
averaged over many molecules (as we will do below), this will not affect the solution. Another
way of stating this is to say that for every molecule that collides with the right wall and exerts
a force mvx2 /l on it, there will be a counterpart that exerts the same force on the left wall, and
averaged over the large number of molecules, this situation is equivalent to our focus on the one
molecule moving back and forth between the left and right walls (along the x-axis).
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PHY 171 (Winter 2012 )
Lecture 13
Then, the force on the right wall due to all N molecules is
m 2
2
2
F =
vx1 + vx2
+ . . . + vxN
l
(L13.1)
where vx1 is the velocity of the first particle, vx2 is the velocity of the second particle, etc.
Now, the average value of the x-component of velocity is given by
2
2
2
2 vx1
+ vx2
+ . . . + vxN
vx =
N
(L13.2)
where hvx2 i denotes the average of vx2 .
Using eq. (2), we can write the above eq. (1) for force as
m F = N vx2
l
(L13.3)
But, from the Pythagorean theorem
2 2 2 2
v = vx + vy + vz
and since the velocities are assumed random, there is no reason why one component should be
preferred over the other, so that
2 2 2
vx = vy = vz
which gives
2
v = 3 vx2
Therefore, eq. (3) becomes
F =
m hv 2 i
N
l
3
(L13.4)
Eq. (4) is thus the expression for force acting perpendicular to any wall of the containing vessel.
To find the pressure on any wall, therefore, all we have to do is divide by the area of the wall. So,
the pressure is given by
F
1 Nm hv 2 i
P =
=
A
3
Al
In the denominator, Al = V , the volume of the container, so we get
P =
1 Nm hv 2 i
3
V
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(L13.5)
PHY 171 (Winter 2012 )
Lecture 13
Multiplying by the volume V on both sides, we can cast eq. (5) into a form that we can compare
with the ideal gas law. So
1
P V = Nm v 2
(L13.6)
3
Notice something interesting? Since the average kinetic energy is hKi =
in eq. (6) is simply 2 times the average kinetic energy hKi.
1
m hv 2 i, the term m hv 2 i
2
So, eq. (6) can be written in the form
PV =
i 2
1 h
N 2 hKi = N hKi
3
3
Compare eq. (7) to the ideal gas law P V = NkB T ; we can then write it as NkB T =
Canceling N from both sides, we get
hKi =
3
kB T
2
(L13.7)
2
N hKi.
3
(L13.8)
where kB is the Boltzmann constant, and T is the absolute temperature (in Kelvins). In words,
eq. (8) says that
the average kinetic energy of molecules in an ideal gas is directly proportional to the absolute
temperature.
Therefore, the higher the temperature, the faster the molecules are moving on the average. Check
out the java applet at mc2.cchem.berkeley.edu/Java/molecules for a demonstration — decrease the temperature of the (red) molecules on the left to 1 K, and compare their motion to that
of the (blue) molecules on the right.
Note that even though we derived this result by neglecting collisions between particles, our results
will hold even when we consider collisions. As noted above, our assumption of elastic collisions between gas molecules guarantees that there will always be a molecule that collides with momentum
mvx with the right wall corresponding to one that bounced off the left wall with this momentum.
Moreover, another assumption of the kinetic theory is that the time spent during collisions is
negligible compared to the time between collisions. Therefore, our neglect of collisions is meant
merely to simplify the math. In the same manner, we could have chosen a container of any shape,
but chose a cube to keep the math simple. Finally, although we have calculated the pressure on
one wall only, it follows from Pascal’s law (from the chapter on Fluids) that the pressure is the
same on all the walls, and at every point in the interior.
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PHY 171 (Winter 2012 )
Lecture 13
Connecting the average kinetic energy to temperature allows us to understand a host of processes
from the microscopic point of view. Let us consider two examples: evaporation and boiling of a
liquid. Note that the assumptions of kinetic theory relate strictly to gases, but we will see below
how they can qualitatively explain phenomena for liquids and solids as well.
Evaporation
Evaporation is a phenomenon in which molecules escape from the surface of a liquid, thereby
converting parts of the liquid into vapor. If you leave water out in a small dish, some of it will
evaporate no matter what the temperature. The higher the temperature, the greater is the amount
evaporated.
This is easy to explain from the microscopic point of view, in which random collisions give some
of the molecules enough kinetic energy to be able to escape from the surface. You get more
evaporation at higher temperatures, because higher temperatures imply that on the average, there
are more molecules that will be able to gain enough kinetic energy to escape the surface.
Boiling
As a liquid is heated, all the molecules get more energy, and more and more molecules start
acquiring enough energy to escape as vapor. At any temperature, therefore, we can define a
vapor pressure, which can be compared to the atmospheric pressure. The boiling point of a liquid
can then be defined as the temperature at which the vapor pressure of the liquid is equal to the
atmospheric pressure, because this means that the majority of water molecules have enough energy
to escape the liquid.
In fact, this explains why bubbles form in boiling but not in evaporation. Since the vapor pressure
is lower than atmospheric pressure for evaporation, the pressure inside the liquid is equal to
atmospheric pressure plus liquid pressure, and bubbles cannot form. On the other hand, since the
vapor pressure is equal to atmosphere pressure for boiling, bubbles of vapor form throughout the
liquid.
Moreover, when you go up a mountain where the atmospheric pressure is lower or decrease air
pressure in some other way (e.g., using a vacuum pump), the liquid molecules can attain enough
energy at a lower temperature to make the vapor pressure equal to this lower atmospheric pressure.
For example, as you decrease the pressure, the boiling point of water will decrease. At one
atmospheric pressure (sea level), the boiling point of water is 100◦ C. At 0.5 bar (7 psi), the
boiling point of water is 80◦ C. As you approach very low pressures (0.03 bar = 0.5 psi), you can
make water boil at 26.5◦ C (almost at room temperature). On the other hand, if you increase the
atmospheric pressure, the water molecules will need to go to a higher temperature for them to
gain enough energy to make their vapor pressure equal to or greater than the higher atmospheric
pressure. This is the principle used in a pressure cooker. (Remember also that boiling water is
convenient for cooking because as long as the water is boiling, you have a constant temperature.
This is true whether you are boiling faster or slower; this relates to how quickly the phase change
is happening, but not to temperature change; it remains at the boiling point until all the water
has converted to 100◦ C).
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PHY 171 (Winter 2012 )
Lecture 13
Another interesting point to note is that the boiling point can also be changed by adding impurities.
Generally, the presence of such impurities in water increases the boiling point of water. This
is usually because the impurities dilute the concentration of water (i.e., the number of water
molecules per unit volume decreases), so there are fewer molecules that can change to vapor at
any given temperature. As a result, one would have to go to a higher temperature to achieve the
same vapor pressure, i.e., the boiling point would be higher. For example, concentrated sugarwater solutions used for making candies and caramel boil at temperatures in excess of 150◦ C
(Source: Tufts School of Engineering course lecture notes).
On the other hand, adding impurities to water will lower its freezing point. One way to think
about this is that the presence of the impurities makes it harder for the nucleation process — the
gathering of minute solid particles of ice around which more molecules can arrange to form the
lattice of solid ice. That is why salt is put on roads before a snowstorm.
Additional material covered in this lecture has been included in the notes for Lecture
14 to preserve continuity.
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