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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 33
�e positions of the turning points depend on the van derWaals parameters as
well as the temperature.�e predictions of these equation are best explored by
inserting some typical values.
For N2, a = 1.352 atmdm6mol−2 and b = 0.0387 dm3mol−1. At 298 K the two
extrema occur at Vm = 0.237 dm3mol−1 and Vm = 0.0211 dm3mol−1; using
these values in eqn 1.5 gives compression factors of 0.962 and −3.82.�e latter
value is physically unreasonable, but the former is acceptable and for this value
the pressure is calculated as 99.3 atm.
For Ne a = 0.205 atmdm6mol−2 and b = 0.0167 dm3mol−1. At 298 K one
of the extrema occurs at a negative value of Vm and the other occurs at 6.92 ×
10−3 dm3mol−1. However, this latter value gives a negative value of Z, which
is not acceptable.�erefore, there are no physically plausible conditions under
which Z is a minimum for Ne at 298 K.
Answers to integrated activities
I1.2 According to the equipartition theorem (�e chemist’s toolkit 7 in Topic 2A),
each quadratic contribution to the energy of a molecule contributes 12 kT to
the average energy per molecule. Translational kinetic energy is a quadratic
term, and because translation is possible in three dimensions, the quoted en-
ergy density of 0.15 J cm−3 is the result of three such contributions.
�e rotation of a molecule about an axis is also a quadratic contribution to the
energy, and in general three such contributions are expected corresponding to
rotation about three mutually perpendicular axes. However, linear molecules,
such as diatomics, do not show rotation about their long axes, so there are
only two contributions.�e contribution of rotation to the energy density will
therefore be 23 of that due to translation
total energy denisty =
trans.
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
(0.15 J cm−3)+
rot.
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
2
3 × (0.15 J cm−3) = 0.25 J cm−3