Download Degrees of freedom

Degrees of freedom
In the section headed the Kinetic theory of matter we were only considering monatomic
gases. We must now extend the ideas to cover gases of higher atomicity, that is, molecules
with more than one atom per molecule.
If we go back to the kinetic theory formula, PV = ⅓ [mNc2] = RT, you can see that
RT = 2/3 [average kinetic energy of the molecules] since kinetic energy = ½ mNc2.
Therefore
Average kinetic energy of the gas molecules = 3/2 RT
We have considered the motion of these molecules to be in three directions, in other words
we say that the molecule has three degrees of freedom. It is therefore sensible to suppose
that one-third of the total energy is associated with each degree of freedom, and this is
known as Boltzmann’s law of equipartition of energy. Thus each degree of freedom has an
amount of energy ½ RT associated with it.
If a gas has its temperature raised at constant volume the energy input is the increase in
kinetic energy of the gas molecules. So for a unit mass and for a rise in temperature dT we
have:
CVdT = kinetic energy increase = 3/2 RdT
Where CV is the specific heat capacity of the gas at constant volume.
But CP - CV = R, so for a monatomic gas:
CV = 3/2 R
and CP = 5/2 R
(monatomic gas)
Now consider a diatomic molecule. In addition to three translational degrees of freedom it
can also rotate about three axes X, Y and Z (Figure 1).
The energy associated with axis X is very small however,
and so we say that the molecule has five degrees of
freedom.
If we assume that the energy associated with each
rotational degree of freedom is the same as that for each
translational degree of freedom then the total energy of the
molecule will be 5 x ½ RT = 5/2 RT. (The vibrational
energy of the molecule is insignificant except at very high
temperatures.)
Z
X
Y
Y
X
Z
Figure 1
Using the same argument as for the monatomic gas, we
have for the two principal molar specific heats of a diatomic gas:
CV = 5/2 R
and
CP = 7/2 R
(diatomic molecule)
at temperatures around room temperature.
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We therefore have for the ratio CP/CV. ():
For a monatomic gas,  = 5/3 = 1.66
For a diatomic gas,  = 7/5 = 1.40
For more complex molecules we have that:
CP/CV = 1+ 2/n
where n is the total number of degrees of freedom.
The table below gives the value of  for a number of common gases.
Gas
Air
Ammonia
Argon
Carbon dioxide
Carbon monoxide
Helium
Hydrogen
Oxygen
Value of 
1.410
1.31
1.66
1.30
1.40
1.66
1.41
1.40
These results show very good agreement with theory.
The preceding section has dealt with one mole of a gas. If we now consider a sample of gas
containing n moles we have:
2/3 N (½ mc2) = nRT
and this gives ½ mc2 = 3/2[R/L]T
where L is the Avogadro constant.
The quantity R/L is known as the Boltzmann constant (k) and its value can be shown to be:
1.38x10-23 J K-1
The average translational energy of a single molecule is therefore: 3/2 kT
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