The statistical mechanical derivation of the van der waals equation of state

1. 1
The Statistical Mechanical Derivation of the van der Waals Equation of State
and its associated thermodynamic properties
David Keffer
Department of Chemical Engineering
The University of Tennessee, Knoxville
dkeffer@utk.edu
started: February 23, 2005
last updated: February 23, 2005
We will perform our derivation in the canonical ensemble, where we specify the number
of molecules, N, the volume, V, and the temperature T. For a monatomic ideal gas, the well-
known partition function is
N
IG
V
N
Q ⎟
⎠
⎞
⎜
⎝
⎛
Λ
= 3
!
1
(1)
where the thermal deBroglie wavelength is defined as
Tmk
h
Bπ
=Λ
2
2
(2)
where h is Planck’s constant, kB is Boltzmann’s constant, and m is the mass of the molecule. The
van der Waals fluid differs from an ideal gas in two ways. First, the molecules have a finite
minimum molar volume, b. In other words, they can not be compressed to infinite density. This
volume occupied by the molecules is not part of the accessible volume of the system. Therefore,
the partition function becomes,
N
NbV
N
Q ⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
= 3
!
1
(3)
where we have multiplied b by N to account for the total volume occupied by molecules. The
second way in that a van der Waals fluid differs from an ideal gas is that the van der Waals
molecule experiences a net attractive force to all other molecules. This attraction is based upon a
mean field approximation. The energy associated with each molecule is linearly proportional to
the molar density of the system,
V
N
aEvdw −= (4)
This energetic term modifies the partition function to become

2. 2
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
=
TVk
aNNbV
NTk
NENbV
N
Q
B
N
B
vdw
N
vdw
2
33
exp
!
1
exp
!
1
(5)
As always, we require the natural log of the partition function
( ) ( )
TVk
aNNbV
NNQ
B
vdw
2
3
ln!lnln +⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
+−= (6)
We use Stirling’s approximation for the natural log of the factorial of a large number
( ) ( )
TVk
aNNbV
NNNNQ
B
vdw
2
3
lnlnln +⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
++−= (7)
The first thermodynamic property that we will calculate is the pressure because that is what
everyone associates with the van der Waals equation of state.
( )
22
,
ln
mm
BB
TN
B
V
a
bV
Tk
N
V
a
b
N
V
Tk
V
Q
Tkp −
−
=
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
= (8)
where we have defined a molecular volume,
N
V
Vm ≡ . This is the van der Waals equation of
state.
Now let’s derive other thermodynamic functions. We will start with the molecular
Helmholtz free energy,
( ) ( ) ⎥
⎦
⎤
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
++−−=−==
TVk
NaNbV
NTkQ
N
Tk
N
A
A
B
BTN
B
m 3, ln1lnln (9)
The molecular internal energy is
( )
m
BB
VN
B
m
V
a
Tk
V
Na
Tk
T
Q
N
Tk
N
E
E −=−=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
==
2
3
2
3ln
,
2
(10)
Unlike the ideal gas, the internal energy of a vdW fluid is not only a function of temperature but
also a function of molar volume. The chemical potential is
( )
⎥
⎦
⎤
⎢
⎣
⎡
+
−
−⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−=µ
TkV
a
bV
bbV
Tk
N
Q
Tk
Bmm
m
B
VT
B
2
ln
ln
3
,
(11)
The molecular entropy is

3. 3
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
+=
−
=
−
== 3
ln
2
5 bV
k
T
AE
NT
AE
N
S
S m
B
mm
m (12)
The molecular enthalpy is
bV
TVk
V
a
TkPVE
N
PVE
N
H
H
m
mB
m
Bmmm
−
+−=+=
+
== 2
2
3
(13)
The molecular Gibbs free energy is
m
m
m
m
Bmmm
V
abV
bV
V
TkTSH
N
TSH
N
G
G 2ln1 3
−⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
Λ
−
+
−
−−=−=
−
== (14)
In addition to the basic thermodynamic properties, we can also calculate a variety of
thermodynamic properties from the partial derivatives.
bV
k
T
p
m
B
Vm
−
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
(15)
( ) 32
2
mm
B
Tm V
a
bV
Tk
V
p
+
−
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
(16)
( )
3
2
1
mB
m
m
Tm
V
p
m
Vk
bVa
bV
T
V
p
T
p
T
V m
−
−
−
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
(17)
The molecular constant volume-heat capacity is
B
V
m
v k
T
E
C
m
2
3
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
= (18)
The molecular constant-pressure heat capacity is
p
m
p
m
p
m
p
T
V
p
T
E
T
H
C ⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
= (19)
Now we need, the first term on the RHS. Going to the Tables of P.W. Bridgman, we have

4. 4
p
m
p
m
V
V
p
m
T
V
p
T
V
T
p
TC
T
E
m
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
(20)
Substituting equation (20) into equation (19) yields
( )
3
2
21
1
2
3
mB
m
BB
p
m
V
Vp
TVk
bVa
kk
T
V
T
p
TCC
m
−
−
+=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+= (21)
The constant-pressure heat capacity is a function of temperature and molar volume.
Now let’s look at vapor-liquid equilibrium. The internal energy of vaporization is
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−=−=∆ V
m
L
m
L
m
V
mm
VV
aEEE
11
(22)
The enthalpy of vaporization is
( )L
m
V
mV
m
L
m
mm
L
m
V
mm VVp
VV
aVpEHHH −∆+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−=∆+∆=−=∆
11
(23)
The entropy of vaporization is
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
=−=∆
bV
bV
kSSS L
m
V
m
B
L
m
V
mm ln (24)
The molecular Gibbs free energy of vaporization is
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
=−=∆ V
m
L
m
L
m
L
m
V
m
V
m
V
m
L
m
B
L
m
V
mm
VV
a
bV
V
bV
V
bV
bV
TkGGG
11
2ln (25)
Since the molecular Gibbs free energy of vaporization is zero at equilibrium, this gives a
condition for vapor-liquid equilibrium, namely,
0
11
2ln =⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
V
m
L
m
L
m
L
m
V
m
V
m
V
m
L
m
B
VV
a
bV
V
bV
V
bV
bV
Tk (26)
The chemical potentials should also be equal at equilibrium

5. 5
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
=µ−µ=µ∆ L
m
V
m
L
m
V
m
V
m
L
m
BLV
VV
a
bV
b
bV
b
bV
bV
Tk
11
2ln (27)
This expression doesn’t look the same as equation (26), but if you get a common denominator in
equations (26) and (27) then it turns out that they are identical equations.
So at VLE, given the temperature, we have three unknowns: the vapor molar volume, the
liquid molar volume, and the vapor pressure. The two molar volumes are found by
simultaneously solving equation (27) and the condition of mechanical equilibrium, namely
022
=+
−
−−
−
=−=∆
L
m
L
m
B
V
m
V
m
B
LV
V
a
bV
Tk
V
a
bV
Tk
ppp (28)
Once the molar volumes are known, the vapor pressure can be obtained by evaluating equation
(8) at either molar volume.
For a certain application, we may want to know the change in the molar volume as a
function of temperature along the two-phase boundary of the phase diagram, also known as the
saturation line. In order to obtain an analytical expression for this derivative, we must
differentiate both equation (27) and (28) with respect to temperature. We will then obtain two
derivatives, one for the liquid and one for the vapor. We will combine the two equations to solve
for the two derivatives individually. We begin by differentiating equation (27) with respect to
temperature along the saturation line.
( ) ( )
0
11
2ln
11
22
22
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
∂
∂
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
∂
∂
−
+
∂
∂
−
−
∂
∂
−
−
∂
∂
−
sat
L
m
L
msat
V
m
V
m
L
m
V
m
V
m
L
m
B
sat
L
m
L
msat
V
m
V
msat
V
m
V
msat
L
m
L
m
B
T
V
VT
V
V
a
bV
b
bV
b
bV
bV
k
T
V
bV
b
T
V
bV
b
T
V
bVT
V
bV
Tk
(29)
Next we differentiate equation (28)
( ) ( )
022 3232
=
∂
∂
−
−
−
∂
∂
−
+
∂
∂
+
−
+
∂
∂
−
−
sat
L
m
L
m
L
m
B
sat
L
m
L
m
B
sat
V
m
V
m
V
m
B
sat
V
m
V
m
B
T
V
V
a
bV
k
T
V
bV
Tk
T
V
V
a
bV
k
T
V
bV
Tk
(30)
Now we solve these two equations for the two partial derivatives. Take equation (30) and
combine like terms.
( ) ( ) bV
k
bV
k
T
V
V
a
bV
Tk
T
V
V
a
bV
Tk
V
m
B
L
m
B
sat
L
m
L
m
L
m
B
sat
V
m
V
m
V
m
B
−
−
−
=
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
− 3232
22 (31)

6. 6
( )
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
−
−
=
∂
∂
32
32
2
2
L
m
L
m
B
sat
V
m
V
m
V
m
B
V
m
B
L
m
B
sat
L
m
V
a
bV
Tk
T
V
V
a
bV
Tk
bV
k
bV
k
T
V
(32)
Next we combine terms in equation (29)
( ) ( )
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−=
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
−
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
bV
b
bV
b
bV
bV
k
T
V
V
a
bV
Tbk
bV
Tk
T
V
V
a
bV
Tbk
bV
Tk
L
m
V
m
V
m
L
m
B
sat
V
m
V
m
V
m
B
V
m
B
sat
L
m
L
m
L
m
B
L
m
B
ln
22
2222
(33)
( )
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
=
∂
∂
22
22
2
2
ln
L
m
L
m
B
L
m
B
sat
V
m
V
m
V
m
B
V
m
B
L
m
V
m
V
m
L
m
B
sat
L
m
V
a
bV
Tbk
bV
Tk
T
V
V
a
bV
Tbk
bV
Tk
bV
b
bV
b
bV
bV
k
T
V
(34)
Now we equate (32) and (34), and solve to obtain
( ) ( )
( )
( )
( )
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
−
−
=
∂
∂
32
32
22
22
2232
2
2
2
2
2
ln
2
L
m
L
m
B
V
m
V
m
B
L
m
L
m
B
L
m
B
V
m
V
m
B
V
m
B
L
m
L
m
B
L
m
B
L
m
V
m
V
m
L
m
B
L
m
L
m
B
V
m
B
L
m
B
sat
V
m
V
a
bV
Tk
V
a
bV
Tk
V
a
bV
Tbk
bV
Tk
V
a
bV
Tbk
bV
Tk
V
a
bV
Tbk
bV
Tk
bV
b
bV
b
bV
bV
k
V
a
bV
Tk
bV
k
bV
k
T
V
(35)
We can obtain the corresponding partial derivative for the liquid phase by substituting equation
(35) into equation (32).