A Statistical Mechanical Derivation of the van der Waals Equation of ...

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 wellknown
partition function is
Q
IG
1 ⎛ V ⎞
⎜ ⎟
N!
⎝ Λ ⎠
=
3
N
(1)
where the thermal deBroglie wavelength is defined as
Λ =
2
h
2πmk
B
T
(2)
where h is Planck’s constant, k B 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,
Q
1 ⎛V
⎜
N!
⎝
−
Λ
=
3
Nb ⎞
⎟
⎠
N
(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,
N
E vdw
= −a
(4)
V
This energetic term modifies the partition function to become
1

Q
vdw
=
1 ⎛V
⎜
N!
⎝
N
− Nb ⎞ ⎛ NE ⎞ ⎛ − ⎞ ⎛
⎜ −
vdw 1 V Nb N
⎟ exp
⎟ = ⎜ ⎟ exp
⎜
3 3
Λ ⎠ ⎝ k
BT
⎠ N!
⎝ Λ ⎠ ⎝Vk
N
2
B
a ⎞
⎟
T ⎠
(5)
As always, we require the natural log of the partition function
2
⎛V
− Nb ⎞ N a
ln ( Qvdw
) = −ln( N!
) + N ln⎜
+
3
⎟
(6)
⎝ Λ ⎠ Vk T
We use Stirling’s approximation for the natural log of the factorial of a large number
B
2
⎛V
− Nb ⎞ N a
ln ( Qvdw
) = −N
ln( N ) + N + N ln⎜
+
3
⎟
(7)
⎝ Λ ⎠ Vk T
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.
( Q)
⎛ ∂ ln ⎞ k
BT
a k
BT
a
p = k
BT⎜
⎟ = − = −
(8)
2
2
⎝ ∂V
⎠ V
N , T
V Vm
− b V
− b ⎛ ⎞
m
N
⎜ ⎟
⎝ N ⎠
V
where we have defined a molecular volume, V m
≡ . This is the van der Waals equation of
N
state.
Now let’s derive other thermodynamic functions. We will start with the molecular
Helmholtz free energy,
A
=
⎡
⎛V
− Nb ⎞ Na ⎤
( Q) = −k
T − ln( N ) + 1+
+ ⎥ ⎦
A k
BT
= − ln
N , T B ⎢
ln⎜
⎟
(9)
N N
⎣
⎝ Λ ⎠ Vk
BT
m 3
The molecular internal energy is
B
E
m
=
E
N
=
k
BT
N
2
( Q)
⎛ ∂ ln
⎜
⎝ ∂T
⎞
⎟
⎠
N , V
=
3
2
k T −
B
Na
V
=
3
2
k T −
B
a
V
m
(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
µ = −k
⎛ ∂ ln
T⎜
⎝ ∂N
( Q)
The molecular entropy is
B
⎞
⎟
⎠
T , V
= −k
B
⎡ ⎛Vm
− b ⎞
T ⎢ln⎜
⎟ −
3
⎣ ⎝ Λ ⎠ V
m
b 2a
+
− b V k T
m
B
⎤
⎥
⎦
(11)
2

S E − A E − ⎡ ⎛ − ⎞⎤
= = = = ⎢
5 + ln⎜
b
m
Am
Vm
S
m
k
B
⎟
3 ⎥
(12)
N NT T ⎣2
⎝ Λ ⎠⎦
The molecular enthalpy is
H
m
H E + PV
3
= = = Em
+ PVm
= k
BT
− 2
N N
2
a
V
m
k
BTVm
+
V − b
m
(13)
The molecular Gibbs free energy is
G H − TS
⎡ V ⎛V
− b ⎞⎤
a
= H TS k T
(14)
⎣
⎦ Vm
m
m
Gm
= =
m
−
m
= −
B ⎢1 − + ln⎜
⎟ − 2
3 ⎥
N N
Vm
− b ⎝ Λ ⎠
In addition to the basic thermodynamic properties, we can also calculate a variety of
thermodynamic properties from the partial derivatives.
⎛ ∂p
⎞
⎜ ⎟
⎝ ∂T
⎠
V m
k
B
=
V − b
m
(15)
⎛ ∂p
⎞
⎜
V
⎟
⎝ ∂
m ⎠
T
= −
k
+ 2
2
3
( V − b) m
B
T
a
V
m
(16)
⎛ ∂V
⎜
⎝ ∂T
m
⎞
⎟
⎠
p
⎛ ∂p
⎞
⎜ ⎟
⎝ ∂T
⎠
= −
⎛ ∂p
⎞
⎜
V
⎟
⎝ ∂
m ⎠
Vm
T
=
1
T a
− 2
V − b k
m
( V − b)
m
3
BVm
(17)
The molecular constant volume-heat capacity is
C
v
⎛ ∂E
⎞
m
= ⎜ ⎟ = k
B
(18)
⎝ ∂T
⎠V
2
m
3
The molecular constant-pressure heat capacity is
C
p
⎛ ∂H
⎞
⎜ ⎟
⎝ ∂T
⎠
m
m
m
= = + p
(19)
p
⎛ ∂E
⎞
⎜ ⎟
⎝ ∂T
⎠
p
⎛ ∂V
⎞
⎜ ⎟
⎝ ∂T
⎠
p
Now we need, the first term on the RHS. Going to the Tables of P.W. Bridgman, we have
3

⎛ ∂Em
⎞
⎜ ⎟
⎝ ∂T
⎠
p
= C
V
⎛ ∂p
⎞
+ T⎜
⎟
⎝ ∂T
⎠
Vm
⎛ ∂Vm
⎞
⎜ ⎟
⎝ ∂T
⎠
p
⎛ ∂Vm
⎞
− p⎜
⎟
⎝ ∂T
⎠
p
(20)
Substituting equation (20) into equation (19) yields
C
p
⎛ ∂p
⎞ ⎛ ∂Vm
⎞ 3
1
= CV
+ T⎜
⎟ ⎜ ⎟ = k
B
+ k
B
(21)
2
⎝ ∂T
⎠V
⎝ ∂T
⎠ p
2
a( Vm
− b)
m
1−
2
k TV
B
3
m
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
∆E
m
= E
V
m
− E
L
m
⎛ 1
= a⎜
L
⎝Vm
1
−
V
V
m
⎞
⎟
⎠
(22)
The enthalpy of vaporization is
∆H
m
= H
V
m
− H
L
m
= ∆E
m
+
p∆V
m
⎛ 1
= a⎜
⎝V
L
m
1
−
V
V
m
⎞
⎟
+
⎠
p∆
V L
( V −V
)
m
m
(23)
The entropy of vaporization is
V
V L
⎛V
⎞
m
− b
∆S
= − = ⎜ ⎟
m
S
m
S
m
k
B
ln L
(24)
⎝ Vm
− b ⎠
The molecular Gibbs free energy of vaporization is
∆G
m
= G
V
m
− G
L
m
= k
B
L
V
L
⎡ ⎛ V ⎞
⎤ ⎛ ⎞
m
− b Vm
Vm
1 1
T ⎢ln ⎜
⎟ + − ⎥ + 2a
⎜ −
⎟ (25)
V
V
L
L V
⎢⎣
⎝Vm
− b ⎠ Vm
− b Vm
− b⎥⎦
⎝Vm
Vm
⎠
Since the molecular Gibbs free energy of vaporization is zero at equilibrium, this gives a
condition for vapor-liquid equilibrium, namely,
L
V
L
⎡ ⎛V
⎞
⎤ ⎛ 1 1 ⎞
m
− b Vm
Vm
k ⎢ln ⎜ ⎟ + − ⎥ + 2 ⎜ ⎟
BT
a
−
= 0
(26)
V
V
L
L V
⎢⎣
⎝Vm
− b ⎠ Vm
− b Vm
− b⎥⎦
⎝Vm
Vm
⎠
The chemical potentials should also be equal at equilibrium
4

∆µ = µ
V
− µ
L
= k
B
L
⎡ ⎛V
⎞
⎤ ⎛ ⎞
m
− b b b 1 1
T ⎢ln ⎜
⎟ + − ⎥ − 2a
⎜ −
⎟ (27)
V
V
L
V L
⎢⎣
⎝Vm
− b ⎠ Vm
− b Vm
− b⎥⎦
⎝Vm
Vm
⎠
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
k
BT
a k
BT
a
∆p = pV
− pL
= − − + = 0
(28)
V
2 L
2
V −
V
−
L
m
b V Vm
b V
m
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.
m
⎡
k
BT
⎢
⎢⎣
V
+ k
B
L
m
⎡ ⎛V
⎢ln
⎜
⎢⎣
⎝V
1 ∂V
− b ∂T
L
m
V
m
L
m
sat
− b ⎞
⎟ +
− b ⎠ V
−
V
V
m
V
m
V
1 ∂Vm
− b ∂T
b
−
− b V
L
m
sat
−
b ⎤ ⎛ 1
⎥ + 2a⎜
− b⎥
⎜ V
⎦ ⎝Vm
V 2
( ) ∂
L
V − b T
sat ( V − b)
m
b
2
∂V
∂V
V
m
∂T
V
m
sat
+
1
−
V
L 2
m
m
b
∂V
L
m
∂T
2
sat
∂V
L
m
∂T
⎞
⎟ = 0
⎟
⎠
sat
⎤
⎥
⎥⎦
(29)
Next we differentiate equation (28)
−
V
V
L
L
∂Vm
k
B
2 a ∂Vm
k
BT
∂Vm
k
B a ∂Vm
+ +
+
− − 2
= 0
2 2
L
∂T
V −
L 3
m
b ∂T
m
m
sat
Vm
sat
(30)
V
V
3
( ) ∂ −
V
−
∂
L
V b T V
sat m
b V T
sat ( V − b)
m
k
B
T
Now we solve these two equations for the two partial derivatives. Take equation (30) and
combine like terms.
⎛
⎜
⎜
⎝
k
T
a ⎞ V
L
∂V
⎛ k T a ⎞
m
B
∂Vm
k
B
k
B
+ 2 ⎟ + ⎜ − ⎟ = −
(31)
m ⎠ ⎝ m
m ⎠
− B
2
2 3
m
V
V 3
( V b) V ⎟ T ⎜ L 2 L
L
V
−
∂ ( V b) V ⎟ ∂T
Vm
− b Vm
− b
sat −
sat
5

6
( )
( ) ⎟ ⎟ ⎠
⎞
⎜
⎜
⎝
⎛
−
−
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
−
−
=
∂
∂
3
2
3
2
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
b
V
T
k
T
V
V
a
b
V
T
k
b
V
k
b
V
k
T
V
(32)
Next we combine terms in equation (29)
( ) ( )
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
= −
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
−
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
b
V
b
b
V
b
b
V
b
V
k
T
V
V
a
b
V
Tb
k
b
V
T
k
T
V
V
a
b
V
Tb
k
b
V
T
k
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
2
2
2
2
2
2
(33)
( )
( ) ⎟ ⎟ ⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
=
∂
∂
2
2
2
2
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
b
V
Tb
k
b
V
T
k
T
V
V
a
b
V
Tb
k
b
V
T
k
b
V
b
b
V
b
b
V
b
V
k
T
V
(34)
Now we equate (32) and (34), and solve to obtain
( ) ( )
( )
( )
( )
( ) ⎟ ⎟ ⎠
⎞
⎜
⎜
⎝
⎛
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
−
−
=
∂
∂
3
2
3
2
2
2
2
2
2
2
3
2
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
b
V
T
k
V
a
b
V
T
k
V
a
b
V
Tb
k
b
V
T
k
V
a
b
V
Tb
k
b
V
T
k
V
a
b
V
Tb
k
b
V
T
k
b
V
b
b
V
b
b
V
b
V
k
V
a
b
V
T
k
b
V
k
b
V
k
T
V
(35)
We can obtain the corresponding partial derivative for the liquid phase by substituting equation
(35) into equation (32).