Van der Waals equation of state Adrian Down November 30, 2005
1 1.1
Van der der Waa Waals ls equa equati tion on of of stat state e Review
Last lecture, we introduced the two corrections in the Van der Waals approximation. 1.1. 1.1.1 1
Volum olume e
The effective volume available to the molecules is reduced by the effective volume of each molecule, denoted by b.
V = V − N b N n = V − N b
1.1. 1.1.2 2
Pres Pressu sure re
We also found that there is a correction to the Helmholtz free energy, N 2 a ∆F = − V
1.2 1.2
Van der der Waa Waals ls equ equat atio ion n
For the ideal gas, the free energy is given by F =
N τ log
−
1
nQ n
+1
Introducing the two corrections discussed above, F VDW =
log −Nτ
nQ (V − N b) N
+1
−
N 2 a V
The pressure is found from the derivative, p =
−
∂F VDW ∂V
= N τ · τ,N
=
N
n (V − N b) Q 2
Nτ V − N b
−
n Q − · N
N 2 a V 2
N a V 2
Some algebraic manipulation gives the Van der Waals equation for a gas.
N 2 a p + 2 V
1.3 •
(V − N b) = N τ
Notes The VDW equation is the same form as the ideal gas equation, pV = Nτ
•
The correction to the pressure increases the effective pressure. The attraction in the Van der Waals gas reduces the effective pressure on any container containing a gas, since the molecules are drawn towards the center of the gas. This leads to an alternative derivation of the pressure correction which is more qualitative. – The mean separation r between the molecules is proportional to 1 1
n3
.
– The Van der Waals force is proportional to
1 r6
∝
n 2 . 2
– Hence the additional pressure is proportional to an2 = a N V
2
•
•
•
The corrections are carried through the derivation and turn out to be independent of each other in the final result. a and b are usually found by experiment.
This result allows us to predict the location of critical points and phase transitions in a gas, as demonstrated in the next section. 2
2 2.1
Critical behavior PV diagram
Recall the PV diagram for a substance. The coexistence curve slopes downwards, plateaus, and then decreases further for increasing V . The height of this plateau increases with temperature. The width of this plateau corresponds to the intersection of the coexistence curve with a concave-downward parabola. There is a particular value of τ for which the coexistence curve intercepts the parabola at exactly one point. The point of intersection along this isotherm is the critical point. Define the values of the thermodynamic parameters at this point to be pc , V c , and τ c . Since the critical point is a point of inflection on the PV diagram, at this point, the following conditions are satisfied.
∂p ∂V
=0
τ
2
∂ p ∂V 2
=0
τ
These conditions, along with the VDW equation, allow to find conditions for the critical point.
2.2 2.2.1
VDW equation Take derivatives
Our goal is to express all quantities in the VDW equation of state in terms of N and a and b. Taking derivatives of the VDM equation, N τ N 2 a p = − V − N b V 2 2 N 2 a ∂p Nτ =− + =0 (V − N b)2 ∂V τ V 3
This gives the relation
2Na V 3
=
τ (V − N b)2
3
(1)
The second derivative must also be equal to 0.
∂ 2 p ∂V 2
= τ
2Nτ (V − N b)3
−
6bN 2 a V 4
=0
This yields a second relation, 3Na V 4 2.2.2
=
τ (V − N b)3
(2)
Solve for V c
The most obvious course of action is to divide (1) by (2), which yields 2V c = V c − N b 3 Hence measuring V c gives b, V c = 3Nb 2.2.3
(3)
Solve for τ c
Now, substitute (3) into (1). Rearranging to solve for τ c , 2N a 2 · 2Nb 3 3 27N b Simplifying, the critical temperature is given by
τ c =
τ c = 2.2.4
8 a 27 b
(4)
Solve for pc
Insert (3) and (4) into the original expression for p. 8 N ab 27
N 2 a − pc = 2Nb 9N 2 b2 4 a 3 a = − 27 b2 27 b2
The critical pressure is then pc =
a 27b2
4
(5)
2.2.5
Dimensionless equation of state
We can now insert the three critical parameters into the VDW equation to recast the equation in a surprising way. Our goal is to eliminate a, b, and N from the Van der Waals equation of state. There is no direct motivation for the following algebraic manipulations, other than the final result.
p +
2
V c 3b
27b2 pc V 2
(3)
V c
V −
=
3
(3)
(5)
V c τ 3b
(6)
(3)
To eliminate b, consider (5) divided by (4), which gives pc a = 27b2 τ c
·
27b 1 = 8a 8b
(7)
With this substitution, (6) =
8 V c τ pc 3 τ c
(7)
Divide (6) by pc vc , which gives
p 3 V c 2 + 2 pc V
8 τ V 1 − = 3 3 τ c V c
This unexpected formulation of the VDW equation involves only the reduced parameters, pˆ =
p pc
ˆ = V
V V c
τˆ =
τ τ c
With these substitutions,
3 pˆ + ˆ 2 V
Note.
•
ˆ− V
1 8 = τˆ 3 3
This is a universal result that applies to any gas to the extent that it obeys the VDW equation of state. In essence, this result says that all gasses behave the same way. •
ˆ , or τˆ , the gasses If two (or more) gasses share the same values of p ˆ , V are said to be in corresponding states. 5
2.3
Plateaus on the PV diagram
The location of the plateaus on the PV diagram can also be determined from ˆ 2 to obtain the above result. Multiply the equation by V
ˆ 2 + 3 pˆV
ˆ− V
1 8 2 = τˆ ˆ V 3 3
ˆ . In general, these For a given value of τˆ , this is a cubic equation in V will give “s-shaped” curves with two turning points. Connecting the two turning points for a given value of ˆτ gives a straight lines which correspond to the plateau of the isotherm on the PV diagram. These s-shaped curves only occur for τ < τ c . The parts of these curves in the region of the plateau give rise to metastable states.
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