van der Waals equation 175σσFig. 4.7Two hard spheres of diameterσat closest contact. The distance between their cen-
ters is alsoσ. A sphere of radiusσjust containing the two particles is shown in cross-section.
which shows two spheres at their minimum separation, where the distance between
their centers isσ. If we now consider a larger sphere that encloses the two particles
when they are at closest contact (shown as a dashed line), then the radius of this sphere
is exactlyσ, and the its volume is 4πσ^3 /3. This is the total excluded volume for two
particles. Hence, the excluded volumeper particleis just half of this, orb= 2πσ^3 /3,
and the unperturbed configurational partition function is given approximately by
Z(0)(N,V,T) =
(
V−
2 Nπσ^3
3)N
= (V−Nb)N. (4.7.32)Therefore, the free energy, to first order, becomes
A(N,V,T)≈−
1
βln[
(V−Nb)N
N!λ^3 N]
−
aN^2
V. (4.7.33)
We now use this free energy to compute the pressure from
P=−
(
∂A
∂V
)
, (4.7.34)
which gives
P=
NkT
V−Nb−
aN^2
V^2P
kT=
ρ
1 −ρb−
aρ^2
kT. (4.7.35)
Eqn. (4.7.35) is known as thevan der Waals equation of state. Specifically, it is an
equation of state for a system described by the pair potentialu(r) =u 0 (r) +u 1 (r) to
first order in perturbation theory in the low density limit. Given the many approxi-
mations made in the derivation of eqn. (4.7.35) and the crudeness ofthe underlying