The van der Waals gas
Assuming that from the notes you can write the partition function as
Z=
1 −3N
N (N − 1) N −2
λ
V
VN +
N!
2
Z
d3 x1 d3 x2 f12
where f12 = f (|x1 − x2 |) depends only on the separation between x 1 and
x2 . This suggests a change of variable to center of mass coordinates.
Write x = (x1 − x2 )/2 and r = x1 − x2 . Then,
Z
d3 x1 d3 x2 f12 −→ V
Z
d3 r e−βV (r) − 1
where e−βV (r) − 1 = f (r).
Writing,
1 −3N
λ
N!
Z =
=
(
V
N
1 −3N N
λ
V
N!
(
N 2 V N −1
+
2
Z
3
d rf (|r|)
N2
1+
κ(β, a, L)
2V
)
)
where κ(β, a, L) = 4π r 2 drf (|r|) = 4π r 2 dr(e−βV (|r|) − 1). This integral
can be evaluated to give κ(β, a, L) = −A + Bβ with A = 4πa 3 /3, B =
4π
3
3
3 ((a + L) − a )V0 . Then writing,
R
F
R
=
−1
ln Z
β
=
N2
−1
ln Z 0 1 +
κ(β)
β
2V
(
!)
where Z 0 = λ−3N V N /N ! is the partition sum for the perfect gas. Assuming
that V0 and a are small so that N 2 κ/2V ≤ 1. Then,
F ∼
1 N 2κ
1
1 N2
−1
ln Z 0 − ·
= − ln Z 0 −
(−A + Bβ).
β
β 2V
β
β 2V
The equation of state for a gas with interactions of this type is
δF
P =−
δV
=
T
N kT
N2
−
(−A + Bβ)
V
2βV 2
1
which, substituting the expression for A and B, gives,
N kT
+
P =
V
or
N 2B
P+
2V 2
N 2A
2V 2
!
!
· kT −
NA
V −
2
N 2B
2V 2
= N kT.
This is the van der Waals equation of state. It describes a realistic gas a
2
low density with a shallow attractive potential ( N2Vκ ≤ 1).
2