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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 31
P1C.18 In Section 1C.2(b) on page 24 it is explained that critical behaviour is associated
with oscillations in the isotherms predicted by a particular equation of state. In
particular, at the critical point a minimum and a maximum in the isotherm
coincide at the same pressure and volume, thus creating a point of in�exion.
�is is the critical isotherm, and is at the critical temperature.
A gas which shows critical behaviour can be lique�ed by the application of
pressure alone, provided the temperature is below the critical temperature.
An equation of state should predict that, as the volume decreases eventually
the pressure increases without limit. For gas A, in the limit b/Vm ≫ 1 the
equation of state becomes pV 2m = bRT , which has the correct behaviour. As the
volume increases, the expectation is that the perfect gas law will be regained,
and for gas A the limit b/Vm ≪ 1 gives pVm = RT , as expected. For gas B in
the limit Vm ≫ b the perfect gas law results, but in the opposite limit, Vm ≪ b,
the equation of state becomes −pb = RT , implying that the pressure reaches a
constant value of −RT/b which is problematic for positive b.
For there to be (at least) two extrema in the plots ofVm against p, the derivative
dVm/dT must have two roots – that is, values at which it is = 0. �is implies
that the expression for the derivative must be a quadratic in Vm, and therefore
that the expression for Vm must be a cubic in Vm. It is evident that neither of
the proposed equations of state are cubic in Vm, and so neither is expected to
show critical behaviour.
P1C.20 In Brief Illustration 1C.5 the reduced pressure is 0.48 and the reduced temper-
ature is 1.33; the critical constants for NH3 are given in the Resource section as
pc = 111.3 atm and Tc = 405.5 K. �e pressure and temperature for the same
state as argon are therefore
p = 0.48 × (111.3 atm) = 53.42 atm T = 1.22 × (405.5 K) = 494.7 K
P1C.22 �e equation of state for a perfect gas, [1A.4–8], is pV = nRT , and the van
der Waals equation of state in terms of the volume is given by [1C.5a–23], p =
nRT/(V−b)−an2/V 2. From theResource section the van derWaals parameters
for CO2 are a = 3.610 atmdm6mol−2 and b = 4.29 × 10−2 dm3 mol−1.
Plots of p against V , Fig. 1.3, show only minor deviations between a van der
Waals gas and a perfect gas under these conditions. Plotting p against 1/V
gives a straight line for the perfect gas, and as in seen in Fig. 1.4 the deviations
for the van der Waals gas are minor.
P1C.24 �e van der Waals equation in terms of the molar volume is given by [1C.5b–
24], p = RT/(Vm − b)− a/V 2m, and the compression factor Z is given by [1C.2–
20], Z = pVm/RT , hence
Z = Vm
RT
( RT
Vm − b
− a
V 2m
) = Vm
Vm − b
− a
RTVm
(1.5)