1 Kinetic Theory of an Ideal Gas
1.1
Pressure in an ideal gas
Assume that we have N gas molecules (atoms) of mass m in a container of volume V: We
will also assume that these molecules are noninteracting in the sense that they experience no
long range interactions with each other. However, they do randomly scatter off each other
elastically. We will also assume that there is no pressure gradient within our volume (i.e.
our container is not accelerating or via the equivalence principle not in a gravitational eld)
so that these molecules have a uniform spatial distribution as well as an isotropic velocity
distribution within the box. That is, all possible directions are equally likely. Note that
these are the basic assumptions for an ideal gas and provide an excellent approximation, for
example, of a low density helium gas.
For these molecules, we will assume that there is a probability distribution, P (vx ) ;
describing the probability of an individual molecule having a velocity in the range
vx ! vx + vx : If we normalize this probability distribution to 1; then the likelihood
of nding some number of molecules in our volume within the range vx ! vx + vx is
simply
Z vx + vx
N (vx ) = N
P (vx ) dvx = N P (vx ) vx :
(1)
vx
If we de ne the range vx to be too small, then the total number of molecules in this range
will become vanishing small. So for the time being we will de ne it to be small enough to
accurately de ne the velocity and yet when multiplied by the total number of molecules
in our volume, N , to allow for a signi cant number, N (vx ), to fall within this range of
velocities.
As an aside, the velocity probability distribution for the gas must be spatially independent
to allow for a uniform density. Additionally it must only depend on the magnitude of the
velocity, v, as that leads to an isotropic velocity distribution. This means that the probability
distribution for the x component of the velocities must be of the form
Z Z
P (vx ) =
P (v) dvy dvz :
(2)
So a subtlety that might have been overlooked in writting down equation (1) is that it had
been implicity assumed that all velocities in the y or z direction are allowed. As such all
possible velocity components in the y and z directions for the probability distribution have
been integrated over and we are only restricting the range of velocities in the x direction.
Now consider the molecules that are within a short distance, `; of the wall of our
container. We will de ne this distance to be less than the mean free path between scattering
events. This will allow us to imagine them traveling with a uniform velocity toward the
wall followed by a re ection off the wall unimpeded by scattering with another molecule.
Such a scattering event would likely have the effect of changing the momentum of this
molecule before it re ects off of the wall, hence impacting the resultant pressure. As we
shall see this distance cancels out during our calculation, but for conceptual reasons we will
always consider it to be less than a mean free path. Adjacent to the wall we will consider a
small volume of our gas given by = A` where A is some small surface area of the wall.
1
Additionally we will de ne the x direction as normal to this section of our wall. In this
geometry, a single molecule bouncing off the wall will experience a change in momentum
of
px = 2mvx :
(3)
Since there is a uniform density throughout the gas, the total number of molecules in the
volume is given by
N
n=
:
(4)
V
So that the number of molecules in our volume within a range of velocities vx ! vx + vx
is given by
N
P (vx ) vx :
(5)
n (vx ) = nP (vx ) vx =
V
In a time given by t = `=vx ; all of these these molecules will undergo the same change
in momentum, px = 2mvx ; via an elastic collision with the wall. Thus, the total rate of
change in momentum for these molecules is
2mvx2
px
N
= n (vx )
(6)
= 2mvx2 AP (vx ) vx :
t
`
V
This is effectively the force of the wall on these particular molecules. From Newton's third
law, for every action there is an equal and opposite reaction, the force of these molecules
upon the wall is simply the negative of this expression. Now the pressure upon this small
area by these select molecules is found by dividing by the area, A, or
N
P = 2mvx2 P (vx ) vx :
(7)
V
Here we have denoted P as the pressure originating only from the molecules within the
velocity range vx ! vx + vx : To determine the total pressure we need to sum over all
of the molecules taking into account the probability distribution P (vx ) : Let us do this by
taking the limit as vx ! dvx and performing the integral over the distribution. This leads
to
Z
Z
N 1 2
P = dP = 2m
vx P (vx ) dvx :
(8)
V 0
Note that we are integrating only over positive velocities as it is only molecules with a
velocity greater than zero that make contact with the wall within our time `=vx . Since we
have an isotropic velocity distribution, we can rewrite our pressure as
Z 1
P V = mN
vx2 P (vx ) dvx = N m vx2 ;
(9)
n (vx )
1
where vx2 is the expectation value of vx2 for a single molecule. As we mentioned above
this expression places no restrictions on velocities in either the y or z directions. So equation
(9) could have also been written as
Z 1Z 1Z 1
P V = mN
vx2 P (v) dvx dvy dvz = N m vx2
(10)
1
1
1
Since v2 = vx2 + vy2 + vz2 ; it is clear that equation (10) leads to an isotropic velocity
distribution and consistent with our original assumptions vx2 = vy2 = vz2 : This allows
us to write
P V = N m vx2 = N3 m v2 = 2N
(11)
3 hKEi ;
2
where hKEi is the expectation for the kinetic energy of an individual molecule.
1.2
Pressure versus temperature in an ideal gas
To make any further progress, we need to have more detailed knowledge of the probability
distribution P (v). If our gas is in thermal contact with a large thermal bath at a constant
temperature T , then it can be shown that this distribution is proportional to the Boltzmann
factor
P / e E=kT ;
(12)
where E is the energy of each individual particle, T is the absolute temperature, and k is the
Boltzmann constant. In our case we are assuming no long range interactions between the gas
molecules, thus E is simply the kinetic energy, and we nd
2
2
2
2
P (v) / e mv =2kT = e m(vx +vy +vz )=2kT / P (v ) P (v ) P (v ) :
(13)
x
y
z
For this case, the functional forms for P and P are identical. We should note that this
probability distribution has all of our required properties as it is both spatially uniform and
isotropic. To obtain the expectation value v2 or vx2 ; we can perform this integration in
spherical coordinates for the velocities. However for simplicity and pedagogical reasons we
will perform the integration in velocity Cartesian coordinates. In these coordinates we de ne
the normalization constant A such that
Z 1
Z
2
2
mvx
=2kT
3
A
e
dvx = 1 ! A
e mv =2kT d3 v = 1:
(14)
1
Now we have
vx2
vx2
= A3
=
Z
A
1
Z
A
vx2
= A
Z
1
1
Z
1
1
vx2 e
1
Z 1
1
e
Z
1
vx2 e
2
2
m(vx
+vy
+vz2 )=2kT
1
2
mvx
=2kT
mvz2 =2kT
dvx
A
Z
1
e
dvx dvy dvz
2
mvy
=2kT
dvy
1
dvz
1
vx2 e
2
mvx
=2kT
(15)
dvx :
1
To perform the integration over vx we simply integrate by parts by de ning
2
kT mvx2 =2kT
e
;
(16)
u = vx ! du = dvx ; dv = vx e mvx =2kT dvx !v =
m
so that
Z 1
2
kT
kT
vx2 =
A
e mvx =2kT dvx =
:
(17)
m
m
1
Substituting these results back into the expression for the pressure, equation (9), leads
to
P V = N kT ;
(18)
which is the Ideal Gas Law. Next we note that the average kinetic energy/gas molecule is
simply
hKEi =
m
2
vx2 + vy2 + vz2
3
= 32 kT .
(19)