matter how it is oriented. Because of this perfect symmetry, there
is thus no way to tell whether it is spinning or not, and in fact we
find that it can’t rotate. The diatomic gas, on the other hand, can
rotate end over end about the x or y axis, but cannot rotate about
the z axis, which is its axis of symmetry. It has a total of five de-
grees of freedom. A polyatomic molecule with a more complicated,
asymmetric shape, l/3, can rotate about all three axis, so it has a
total of six degrees of freedom.
Because a polyatomic molecule has more degrees of freedom than
a monoatomic one, it has more possible states for a given amount of
energy. That is, its entropy is higher for the same energy. From the
definition of temperature, 1/T= dS/dE, we conclude that it has a
lower temperature for the same energy. In other words, it is more
difficult to heatnmolecules of difluoroethane than it is to heatn
atoms of helium. When the Cl ́ement-Desormes experiment is carried
out, the resultbtherefore depends on the shape of the molecule!
Who would have dreamed that such simple observations, correctly
interpreted, could give us this kind of glimpse of the microcosm?
Lets go ahead and calculate how this works. Suppose a gas is
allowed to expand without being able to exchange heat with the
rest of the universe. The loss of thermal energy from the gas equals
the work it does as it expands, and using the result of homework
problem 2 on page 347, the work done in an infinitesimal expansion
equalsPdV, so
dE+PdV = 0.
(If the gas had not been insulated, then there would have been a
third term for the heat gained or lost by heat conduction.)
From section 5.2 we haveE= (3/2)PV for a monoatomic ideal
gas. More generally, the equipartition theorem tells us that the 3
simply needs to be replaced with the number of degrees of freedomα,
so dE= (α/2)PdV+ (α/2)VdP, and the equation above becomes
0 =
α+ 2
2PdV+
α
2VdP.
Rearranging, we have
(α+ 2)
dV
V
=−α
dP
P.
Integrating both sides gives
(α+ 2) lnV =−αlnP+ constant,
and taking exponentials on both sides yields
Vα+2∝P−α.We now wish to reexpress this in terms of pressure and temper-
ature. EliminatingV∝(T/P) gives
T∝Pb,
whereb= 2/(α+ 2) is equal to 2/5, 2/7, or 1/4, respectively, for a
monoatomic, diatomic, or polyatomic gas.336 Chapter 5 Thermodynamics