van der Waals equation 177at the critical point. The first and second derivatives of eqn. (4.7.35) with respect to
Vyield two equations in the two unknownsVandT:
−
NkT
(V−Nb)^2+
2 aN^2
V^3= 0
2 NkT
(V−Nb)^3−
6 aN^2
V^4= 0. (4.7.37)
Solving these equations leads to the critical volumeVcand critical temperatureTc:
Vc= 3Nb, kTc=8 a
27 b. (4.7.38)
Substitution of the critical volume and temperature into the van der Waals equation
gives the critical pressurePc:
Pc=
a
27 b^2. (4.7.39)
Let us now consider the behavior of a particular thermodynamic quantity as the
critical point is approached. Because we are interested in the relationship between
pressure and volume as the critical point is approached, it is usefulto study the
isothermal compressibility, defined to be
κT=−1
V
(
∂V
∂P
)
T=−
1
V(∂P/∂V)
. (4.7.40)
AtV=Vc, the pressure derivative gives
∂P
∂V∣
∣
∣
∣
V=Vc=−
NkT
2 N^2 b^2+
2 aN^2
27 N^3 b^3=
1
4 Nb^2(
8 a
27 b−kT)
∼(Tc−T), (4.7.41)so that
κT∼(T−Tc)−^1. (4.7.42)
This shows that atV=Vc, asTapproachesTcfrom above, the isothermal compress-
ibility diverges according to a power law. ThatκTdiverges is also confirmed experi-
mentally. The experimentally observed power-law divergence ofκTcan be expressed
generally in the form
κT∼|T−Tc|−γ, (4.7.43)
whereγis an example of what is termed acritical exponent. The van der Waals theory
clearly predicts that the value ofγ= 1.
Briefly, critical exponents describe the behavior of systems neartheircritical points.
A critical point is a point in the phase diagram where a coexistence curve terminates.