3.4 Van der Waals Equation of Statepersonal.rhul.ac.uk/uhap/027/PH4211/PH4211_files/slides3.pdf · 3.4 Van der Waals Equation of State ... expression for z of the form e ... We know

4211 Statistical Mechanics 1 Week 3

3.4 Van der Waals Equation of State 3.4.1 Approximating the Partition Function Rather than perform an exact calculation in powers of a small parameter (the density), we shall adopt a different approach by making an approximation to the partition function, which should be reasonably valid at all densities. The approximation is based on the single-particle partition function. Shall obtain an equation of state approximating behaviour of real gases. Originally proposed by van der Waals in his Ph. D. Thesis in 1873.


In the absence of interactions the single-particle partition function is

3

Vz =Λ .

Recall: factor V comes from integration over the position coordinates. ► How to treat inter-particle interactions – in an approximate way? Interaction U(r) comprises: i ) a strong repulsive core at short separations ii) a weak attractive tail at large separations. The key is to treat these two parts separately/differently:


i) The repulsive core effectively excludes regions of space from the integration over position coordinates. This may be accounted for by replacing V by V − Vex where Vex is the volume excluded by the hard core. ii) The attractive long tail is accounted for by including a factor in the expression for z of the form

E kTe− where 〈E〉 is some average of the attractive part of the potential. ► Thus we arrive at the approximation

ex3

E kTV Vz e−−=Λ .


We have approximated the interaction by a mean field assumed to apply to individual particles. This allows us to keep the simplifying feature of the free-particle calculation where the many-particle partition function factorizes into a product of single-particle partition functions. Accordingly, this is referred to as a mean field calculation.


3.4.2 Van der Waals Equation The equation of state is found by differentiating the free energy expression:

,

ln ln

T N T

Z zp kT NkTV V

∂ ∂= =∂ ∂ .

Now the logarithm of z is ( )exln ln 3lnz V V E kT= − − Λ−

so that

ex

dlndT

Ez NkTp NkT NV V V V

∂= = −∂ −

since we allow the average interaction energy to depend on volume (density). This equation may be rearranged as

ex

ddE NkTp NV V V

+ =−


or

( )ex

ddE

p N V V NkTV

⎛ ⎞+ − =⎜ ⎟

⎝ ⎠ .

This is similar to the ideal gas equation except that the pressure is increased and the volume decreased from the ideal gas values. These are characteristic parameters. They account, respectively, for the attractive long tail and the repulsive hard core in the interaction. Conventionally we express the parameters as aN2/V2 and Nb, so that the equation of state is

( )2

2

Np a V Nb NkTV

⎛ ⎞+ − =⎜ ⎟

⎝ ⎠

and this is known as the van der Waals equation of state.


3.4.3 Microscopic ‘estimation’ of Parameters In mean field, the repulsive and the attractive parts of the inter-particle interaction were treated separately. How are the two parameters of the van der Waals equation related to the parameters of the Lennard-Jones inter-particle interaction? i) The repulsion is strong. We accounted for this by saying that there is zero probability of two particles being closer than σ. Then that region of co-ordinate space is excluded; form of the potential in the excluded region (U(r) very large) does not enter the discussion. Interaction → boundary condition. Thus the excluded volume will be

3ex

23

V Nπσ= , the total hard core volume.


ii) The attractive part of the potential is weak. Here there is very little correlation between the positions of the particles; we therefore treat their distribution as approximately uniform. The mean interaction for a single pair of particles 〈E1〉 is then

E1 = 1V

4πr 2U r( )drσ

∞

∫

= 1V

4πr 2 4ε σr

⎛⎝⎜

⎞⎠⎟

12

− σr

⎛⎝⎜

⎞⎠⎟

6⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪dr = − 32πσ 3

9Vσ

∞

∫ ε .

Now there are N(N – 1)/2 pairs, each interacting through U(r), so neglecting the 1, the total mean energy per particle is

E = E1 N 2

= −16πσ 3

9NV

ε .


In the van der Waals equation it is the derivative of this quantity we require. Thus we find

N

d EdV

= 169πσ 3 N

V⎛⎝⎜

⎞⎠⎟

2

ε . These results give the correct assumed N and V dependence of the parameters used in the previous section. So finally we identify the van der Waals parameters a and b as

a = 16

9πσ 3ε , b = 2

3πσ 3.


3.4.4 Virial Expansion It is a straightforward matter to expand the van der Waals equation as a virial series. We express p/kT as

2

2

1 2

1 .

p N aNkT V Nb kTV

N N a NbV V kT V

−

= −−

⎛ ⎞⎛ ⎞ ⎛ ⎞= − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

and this may be expanded in powers of N/V to give 2 3 4

2 3p N N a N Nb b bkT V V kT V V

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠K .

Thus we immediately identify the second virial coefficient as

( )VW2

aB T bkT

= − .


This has the form as sketched for the square well potential. For this model we can find the Boyle temperature and the inversion temperature:

B

i

,

2 .

aTbkaTbk

=

=

So we conclude that for the van der Waals gas the inversion temperature is double the Boyle temperature. Incidentally, we observe that the third and all higher virial coefficients, within the van der Waals model, are constants independent of temperature.


3.5 Other Phenomenological Equations of State 3.5.1 The Dieterici equation The Dieterici equation of state is one of a number of purely phenomenological equations crafted to give reasonable agreement with the behaviour of real gases. The Dieterici equation may be written as

( )NakTVp V Nb NkTe

−− = .

As with the van der Waals equation, this equation has two parameters, a and b, that parameterise the deviation from ideal gas behaviour. The interest in the Dieterici equation is that it gives a reasonable description of fluids in the vicinity of the critical point.


3.5.2 Virial expansion – for Dieterici equation In order to obtain the virial expansion we express the Dieterici equation as

NakTVp N e

kT V Nb−

=− .

And from this we may expand to give the series in N/V 2 3 2

22 22

p N N a N a abb bkT V V kT V k T kT

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + − − +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠K

This gives the second virial coefficient to be D2

aB bkT

= − . This is the same as that for the van der Waals gas, and the parameters a and b may thus be identified with those of the van der Waals model. As a consequence, we conclude that both the van der Waals gas and the


Dieterici gas have the same values for the Boyle temperature and the inversion temperature. The third virial coefficient is given by

( )2

D 23 2 22

a abB T bk T kT

= − − ; we see that this depends on temperature, unlike that for the van der Waals equation, which is temperature-independent.


3.5.3 The Berthelot equation As with the Dieterici equation, the Berthelot equation is another of phenomenological origin. The equation is given by

( )2

2

Np V Nb NkTkTVα⎛ ⎞

+ − =⎜ ⎟⎝ ⎠ .

The parameters of the Berthelot equation are given by α and b. We observe this equation is very similar to the van der Waals equation; there is a slight difference in the pressure-correction term that accounts for the long distance attraction of the intermolecular potential. Since the Berthelot and van der waals equation are related by a kTα= it follows that the Berthelot second virial coefficient is given

by ( )B2 2B b

kTα= −

.


3.5.4 The Redlich-Kwong equation Equation proposed by Redlich and Kwong. [Chem. Rev. 44, 234 (1949)] This is another two-parameter equation.

( ) ( )( )2

1 2NkT N apV Nb V V Nb kT

= −− + .

It is instructive to compare this with the van der Waals equation, written in a similar form

( )2

2

NkT N apV Nb V

= −−

and the Berthelot equation 2

2

NkT NpV Nb V kT

α= −−

- so the R-K equation has a more complicated ‘pressure correction’ term, that includes some temperature dependence.


Expansion of the pressure in powers of the density gives the virial expansion:

( ) ( )

2 32

3 2 5 2p N a N ab Nb bkT V V VkT kT

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + − + + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠K

and so we identify the second virial coefficient to be

( )RK2 3 2

aB bkT

= −.


**3.6 Hard Sphere Gas (not covered this year) Chaikin and Lubensky: “Although this seems like an immense trivialisation of the problem, there is a good deal of unusual and unexpected physics to be found in hard-sphere models.” P. M. Chaikin and T. C. Lubensky, Principles of Condensed Matter Physics, Cambridge University Press, (1995) U r( )

0 r

σ

( )0

U r rr

σσ

=∞ <= >


There is no analytic solution to the problem. So……. Three possible approaches:

1 Mean field 2 Virial expansion 3 Molecular dynamics

Task is to solve for the equation of state and, more generally, all thermodynamic properties.


3.6.1 Mean field treatment → Clausius equation of state: p(V – Vex) = NkT . Correct in the 1-d case


3.6.2 Hard Sphere Equation of State The equation of state of a hard-sphere fluid has a very special form. Start from Helmholtz free energy F:

lnF kT Z= − . Recall that for an interacting gas Z may be written as

id NZ Z Q= where Zid is the partition function for an ideal (non-interacting) gas

id 3

1!

NVZN

⎛ ⎞= ⎜ ⎟Λ⎝ ⎠ and QN is the configuration integral

31 di jU kT

NN NQ e qV

<

−∑= ∫ .


Equation of state – find the pressure, by differentiating the free energy

,

,

id

, ,

ln

ln ln .

T N

T N

N

T N T N

FpV

ZkTV

Z QkTV V

∂= −∂

∂=∂

⎛ ⎞∂ ∂= +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

Note: QN is independent of temperature. This must be so, since there is no energy scale for the problem; the interaction energy is either zero or it is infinite. Thus the ratio E/kT will be temperature-independent.


The pressure of the hard-sphere gas is then given by:

( )Np kT g N VV

⎛ ⎞= +⎜ ⎟⎝ ⎠ . The function g(N/V) is found by differentiating lnQ with respect to V. We know it is a function of N and V and in the thermodynamic limit the argument must be intensive. Thus the functional form and we have the low-density ideal gas limiting value g(0) = 0. The important conclusion we draw from these arguments, and in particular from the equation above is that for a hard sphere gas the combination p/kT is a function of the density N/V. This function must depend also on the only parameter of the interaction: the hard core diameter σ.


3.6.3 Virial Expansion The virial expansion is written as

2 3

2 3p N N NB BkT V V V

⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠K ,

where the virial coefficients Bm are, in the general case, functions of temperature. However, as argued above, for the hard sphere gas the virial coefficients are temperature-independent. The virial expansion may be regarded as a low-density approximation to the equation of state – true when only a finite number of coefficients is available. If, however, all the coefficients were known, then provided the series were convergent, the sum would give p/kT for all values of the density N/V: the complete equation of state.


Now although we are likely to know the values for but a finite number of the virial coefficients, there may be ways of guessing / inferring / estimating the higher-order coefficients. We shall examine two ways of doing this.


3.6.4 Virial Coefficients The second virial coefficient for the hard sphere gas has been calculated; we found

3223

B πσ= where σ is the hard core diameter. General term of the virial expansion is Bm(N/V)m, which must have the dimensions of N/V. Thus Bm will have the dimensions of (volume)m–1. Now the only variable that the hard sphere virial coefficients depend on is σ. Thus it is clear that

3( 1)const mmB σ −= ×

where the constants are dimensionless numbers – which must be determined.


It is increasingly difficult to calculate the higher-order virial coefficients; those up to sixth order were evaluated by Rhee and Hoover in 1964, and terms up to tenth order were found by Clisby and McCoy in 2006. These are listed in the table below, in terms of the single parameter b:

3223

b B πσ= = . B2/b = 1 B3/b2 = 0.625 B4/b3 = 0.2869495 B5/b4 = 0.110252 B6/b5 = 0.03888198 B7/b6 = 0.01302354 B8/b7 = 0.0041832 B9/b8 = 0.0013094 B10/b9 = 0.0004035

Rhee & Hoover 1964

Clisby & McCoy 2006


3.6.5 Carnahan and Starling procedure Remarkable procedure: They inferred a general (approximate) expression for the nth virial coefficient, enabling them to sum the virial expansion and thus deduce an (approximate) equation of state. The virial expansion is written as

2 3

2 3 41pV N N NB B BNkT V V V

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠K

This is written out for the known values of the virial coefficients: 2 3

4 5 6

7 8 9

1 0.625 0.2869495

0.110252 0.03888198 0.01302354

0.0041832 0.0013094 0.0004035 .

pV N N Nb b bNkT V V V

N N Nb b bV V V

N N Nb b bV V V

⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠K


In terms of the dimensionless density variable y = Nb/4V this becomes

2 3 4

6

5

7 8 9

1 4 10 18.364768 28.224512 39.815147

53.34441984 68.5375488 85.8128384 105. 1

5

7 5 0

2

7 4

pV y y y y yNk

y y yT

y+

= + + + + +

+

+

+ + +K

Only terms up to y5 were known to Carnahan and Starling. Round coefficients to whole numbers: 4, 10, 18, 28, 40. Then the coefficient of yn is n(n + 3) – assume it works in general. Check this with the newly-known virial coefficients; Carnahan and Starling’s formula gives 54, 70, 88, 108; actual rounded integers are 53, 69, 86, 106. Agreement is quite good.


From this assumption, the general expression for the nth virial coefficient is

( )( ) 11

1 24

nn n

n nB b −

−

− += .

These are tabulated below, together with the true values. C+S value B2/b = 1 1 B3/b2 = 0.625 0.625 B4/b3 = 0.2869495 0.28125 B5/b4 = 0.110252 0.109375 B6/b5 = 0.03888198 0.0390625 B7/b6 = 0.01302354 0.0131836 B8/b7 = 0.0041832 0.00427246 B9/b8 = 0.0013094 0.00134277 B10/b9 = 0.0004035 0.000411987


We can now sum the infinite virial series, with the general Bn; we obtain

( )2

34 211

pV y yNkT y

+= +− .

In terms of the number density N/V this gives the equation of state as

( )( )

2 28

34

11

N b NV V

b NV

bp NkT V

⎧ ⎫+⎪ ⎪= +⎨ ⎬−⎪ ⎪⎩ ⎭ .

Incidentally, the universal function g(n) in the equation of state is then given by

( )2

3( ) 8( )

1 4bn bng n n

bn+=− .


Conventionally the equation of state is expressed and plotted in terms of the ‘packing fraction’ V0/V, where V0 is the volume occupied by the hard spheres V0 = Nπσ3/4 so that

30 3

4 8V N b NV V V

πσ= = . Then we obtain

( )( )0 0

0

28 83 9

323

11

V VV V

VV

p NkT V

⎧ ⎫+⎪ ⎪= +⎨ ⎬−⎪ ⎪⎩ ⎭

.

This is plotted in the figure below. For comparison we have also shown some data points obtained by molecular dynamics simulations.


0.1 0.2 0.3 0.4 0.5 0.6 0.7

5

10

00

15pVNkT

V V/ 0

1ideal gas

Carnhan + Starling

- molecular dynamics

At higher densities the data points from molecular dynamics simulations fall consistently below the curve of Carnahan and Starling. This indicates the shortcoming of their method. Inspired as the C+S procedure is, it is not quite correct. And the discrepancy is expected to become greater at the higher densities. Perhaps the problem is to be expected – after all, even the lower order virial coefficients have been approximated, when the coefficients of the powers of y were truncated to integers.


A more systematic way at arriving at an equation of state is the Padé method. 3.6.6 Padé approximants The equation of state of the hard sphere gas takes the form

( )0pV f V VNkT

= where f is a universal function of its argument. So if the function is determined then the hard sphere equation of state is known. The virial series gives f as a power series in its argument. And in reality one can only know a finite number of these terms. The Carnahan and Starling procedure takes the known terms, ‘guesses’ the (infinite number of) higher-order terms and then sums the series. The figure above indicates that the result is good, but it could be better.


For the Carnahan and Starling equation of state the function f may be written as

2 320 823 9 27

2 3843 27

1( )1 2

x x xf xx x x

+ + −=− + − .

In this form we observe that f (x) is the quotient of two polynomials. And this leads us naturally to the Padé method. One knows f (x) to a finite number of terms. In the Padé method the function f (x) is approximated by the quotient of two polynomials

,( )( ) ( )( )

nn m

m

P xf x F xQ x

≈ = . Here Pn(x) and Qm(x) are polynomials of degrees n and m respectively.


20 1 2

20 1 2

( ) ,

( ) .

nn n

mm m

P x p p x p x p xQ x q q x q x q x

= + + + +

= + + + +

KK

Without loss of generality we may (indeed it is convenient to) restrict q0 = 1. The terms of Pn(x) and Qm(x) may be determined so long as f (x) is known to at least n + m terms. In other words if f (x) is known to n + m terms, then Fn,m(x) agrees with the known terms of the series for f(x); moreover the quotient generates a series of higher order terms as well. The hope is that this series will be a good approximation to the true (but unknown) f(x). Power series of f (x) Qm(x) – Pn(x) begins with the term xm + n + 1. Different m, n subject to m + n = N. – See Reichl for (some) details.


Thus Ree and Hoover (1964) – (i.e. before Clisby and McCoy’s extra virial coeffs) constructed

( ) ( )( )

0 0 0

0 0

2 3

2

2.66667 0.451605 0.3286091

1 1.49731 0.578226

V V VV V V

V VV V

pVNkT

+ += +

− +

0.1 0.2 0.3 0.4 0.5 0.6 0.7

5

10

00

15pVNkT

V V/ 0

3, 2 Padé

which is seriously better than the Carnhan + Starling expression.


3.6.7 Phase transition

0.1 0.2 0.3 0.4 0.5 0.6 0.7

5

10

00

15pVNkT

V V/ 0 We note that a random assembly of spheres will pack to V0/V = 0.638 so this is the greatest density possible for the fluid. However a close-packed lattice (fcc or hcp) will pack more densely, to 0.7405. Thus there should be a phase transition to a solid phase.…..


3d hard spheres 2d hard discs

The lack of an attractive part of the inter-particle interaction means that there will be no gas-liquid transition; attraction is needed to have a self-bound fluid phase coexisting with a dispersed gas phase.

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3.4 Van der Waals Equation of Statepersonal.rhul.ac.uk/uhap/027/PH4211/PH4211_files/slides3.pdf · 3.4 Van der Waals Equation of State ... expression for z of the form e ... We know