ANSWER:
The arguments developed for the time-averaged force
exerted on one wall of a cubical 3D container apply equally well
for the force acting on the boundary of a 1D ``box'' - namely,
for a single particle bouncing back and forth,
,
where L is the length of the ``box''.
In this case there is only one direction of motion
(degree of freedom) so we can drop the x subscript on vx.
The EQUIPARTITION THEOREM says that the thermal
average of the kinetic energy
associated with
this translational degree of freedom is
, giving
.
If we substitute this back into the formula for
we get
. For N particles all
doing this at the same time, we just multiply by N, giving
or (ignoring the fact that the
actual force F at any instant fluctuates minutely about the average
force
)
.
This is the 1D equivalent of the IDEAL GAS LAW.
Note that the only difference between this derivation and the one for
a 3D box is that here we didn't have to introduce the notion of
pressure as the force per unit area. This is simpler!
ANSWER:
Again the 1D case is much simpler than the 3D case
because we don't have to worry about
composing v2 out of
vx2 + vy2 + vz2
(or out of kx, ky and kz).
There is just one direction, just one velocity component,
one momentum component and one wavelength to worry about
making commensurate with the length of the ``box''.
Thus the requirement that an integer n half-wavelengths
fit evenly into
gives
and so
or
. Thus the possible values of
the speed v are evenly spaced every
and the distribution of speeds varies only as the
Boltzmann factor
. This gives immediately
where A is a normalization constant that does not depend on v.
You get full credit for this result, but here's how to get A:
Let
where
is treated as a constant.
Now,
or
so
and we have
You can look up the definite integral; its value is
,
giving
.
3D:
1D:
ANSWER:
As stated in the textbook, the most probable speed for an
ideal gas molecule in 3D is
(i.e. when
).
For the 1D gas, however,
is missing that
extra factor of v2 that forces its value to zero at v=0,
so the resultant speed distribution has its maximum
at v=0:
.
ANSWER:
The qualitative picture was drawn in class; this is all about
quantitative estimates of how hot the gas has to get before
rotational and vibrational degrees of freedom are excited.
For this you need to remember that (i) the energy of either
is always given by
where
J-s; and
(ii) that the lowest nonzero angular momentum allowed
for any object (including a molecule) is
.
You already know how to find the resonant frequency of a pair
of masses connected by a spring; you need only the effective
``spring constant'' for the N2 bond, namely
N/m,
and the mass of a 14N atom:
kg.
So
s-1. The corresponding energy is
giving
.
At lower temperatures the vibrational degrees of freedom
will not be thermally excited.
Now for the rotational ones: the distance between the two
14N nuclei in an N2 molecule is about 1 Å
m.
Crudely representing the molecule as two point masses an Angstrom apart,
the moment of inertia is about
J-s2. Now,
gives
s-1 and so
J.
A state with this size energy will get excited only if the temperature
is comparable to or greater than
or
K. Hmm, looks like I was mistaken when I said in class
that the rotational degrees of freedom are only excited at high
temperature. Considering that nitrogen freezes solid at 68 K,
it is safe to say that N2 gas has its rotational
heat capacity at all temperatures, giving a total heat capacity of
from
room temperature up to about 900 K, and then
.
The binding energy of an N2 molecule is around 11 eV
(or
J) so at temperatures above about
the molecules will
dissociate into nitrogen atoms. Then each atom will contribute
only its translational degrees of freedom, giving
.
(I'm not sure I believe this myself; at any rate the atoms are
apt to ionize by that temperature.)
Sketch: 1