Scaled Particle Theory: Rigid Spheres Equation of State

In summary: Instead, they claim that the work function ##W(\lambda)## that goes like the surface area of the sphere ##\lambda^2## is a probability function, but they don't justify this. Additionally, they claim that the conditional probability of finding at least one molecular center in the ##[\lambda,\lambda+d\lambda]## shell is ##4\pi \lambda^2 \rho G(\lambda) d\lambda##, but again, they don't justify this.
  • #1
itaischles
16
2
Hi all,
I'm reading an article titled "statistical mechanics of rigid spheres" by Reiss, Frisch and Lebowitz (1959). It tries to capture the equation of state of rigid spheres.
In their article they write the probability that at a random point in the fluid the center of the nearest molecule is a distance ##\lambda ## away and in the range ##[\lambda,\lambda+d\lambda]##. They separate this probability to the product of the probability of finding a cavity of radius ##\lambda##, denoted ##p_0(\lambda)## times the conditional probability of finding at least one molecular center in the ##[\lambda,\lambda+d\lambda]## shell given that there is a cavity of radius ##\lambda## there. So far so good.
My problem is that they assign things that doesn't look like probabilities to the two product terms. First, they claim that ##p_0(\lambda)=e^{-W(\lambda)/kT}## where ##W(\lambda)## is the reversible work needed to create the cavity. Since we don't know anything about this ##W(\lambda)## why do we assume that ##p_0(\lambda)## is normalized? Then they say that the conditional probability of finding at least one molecular center in the ##[\lambda,\lambda+d\lambda]## shell is ##4\pi \lambda^2 \rho G(\lambda) d\lambda## where ##\rho## is the average fluid density and ##G(\lambda)## is the radial distribution function at the cavity's surface. This too is not normalized since if we look for example at an ideal gas (where ##G(\lambda)=1##) and integrate over ##\lambda## between 0 and ##\infty## we get ##\rho V=N## where ##V## is the volume and ##N## the number of particles in the system. Can anybody comment on that please? Thanks!
 
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  • #2
itaischles said:
My problem is that they assign things that doesn't look like probabilities to the two product terms. First, they claim that ##p_0(\lambda)=e^{-W(\lambda)/kT}## where ##W(\lambda)## is the reversible work needed to create the cavity. Since we don't know anything about this ##W(\lambda)## why do we assume that ##p_0(\lambda)## is normalized?
Can you find a possible ##W(\lambda)## where ##p_0(\lambda)## is not normalized?

Then they say that the conditional probability of finding at least one molecular center in the ##[\lambda,\lambda+d\lambda]## shell is ##4\pi \lambda^2 \rho G(\lambda) d\lambda## where ##\rho## is the average fluid density and ##G(\lambda)## is the radial distribution function at the cavity's surface. This too is not normalized since if we look for example at an ideal gas (where ##G(\lambda)=1##) and integrate over ##\lambda## between 0 and ##\infty## we get ##\rho V=N## where ##V## is the volume and ##N## the number of particles in the system.
Isn't ##\rho## the particle number density?
 
  • #3
Regarding the first reply: let us take, for example, a work function ##W(\lambda)## that goes like the surface area of the sphere ##\lambda^2##. It is reasonable since the work needed to create a cavity of size 0 is 0 and it monotonically increases as ##\lambda## get's bigger, "exploding" at infinity. Then ##p_0(\lambda)## is monotonically decreasing, starting at 1 (when ##\lambda=0##) and reaching 0 at infinity. Integrating this over the range ##0<\lambda<\infty## gives some number, not necessarily 1.

Regarding the second comment, ##\rho## is, indeed, the particle number density, as I have written. Is there anything I miss here?
 
  • #4
itaischles said:
Regarding the first reply: let us take, for example, a work function W(λ)W(\lambda) that goes like the surface area of the sphere λ2\lambda^2. It is reasonable since the work needed to create a cavity of size 0 is 0 and it monotonically increases as λ\lambda get's bigger, "exploding" at infinity. Then p0(λ)p_0(\lambda) is monotonically decreasing, starting at 1 (when λ=0\lambda=0) and reaching 0 at infinity. Integrating this over the range 0<λ<∞0
OK - so what does the paper say about W? (You may have to read between the lines.)
Is there an online copy of the paper I can see?

itaischles said:
Regarding the second comment, ρ is, indeed, the particle number density, as I have written. Is there anything I miss here?
Well you wrote:
itaischles said:
This too is not normalized since if we look for example at an ideal gas (where G(λ)=1G(\lambda)=1) and integrate over λ\lambda between 0 and ##\infty## we get ρV=N where V is the volume and N the number of particles in the system.
... so what is wrong with that?

It may be that you are expected to realize that the function should be chosen so that the result is normalized - they are building a model.
Did you read the rest of the paper to see what they came up with?
 
  • #5
Simon Bridge said:
OK - so what does the paper say about W? (You may have to read between the lines.)
Is there an online copy of the paper I can see?

There it is: ftp://213.176.96.142/aip1c48329b-957e-20141114083409.pdf
or: http://scitation.aip.org/content/aip/journal/jcp/31/2/10.1063/1.1730361

Simon Bridge said:
... so what is wrong with that?

It may be that you are expected to realize that the function should be chosen so that the result is normalized - they are building a model.
Did you read the rest of the paper to see what they came up with?

Well, for [itex]p_0(\lambda)[/itex] to be a probability function, it has to be normalized so when you integrate over all possible values of [itex]\lambda[/itex] you get 1. Here we get some number which depends on your choice of [itex]W(\lambda)[/itex]. In their article they don't say anything special about [itex]W(\lambda)[/itex] except for the fact that it is the reversible work needed to create the cavity. It could equally be volume work, surface work... whatever.
I did read the rest of the paper but still, no sign of any constraints on [itex]W(\lambda)[/itex].
 
  • #6
It looks like it comes ut in the wash though - other papers running the same thing end up deriving functions for W and G.
Do the authors not do anything like that here?
 
  • #7
The theory isn't about deriving the W function as far as I understand it. It uses a given W function to calculate all other things so it can't assume anything about it, besides the fact that it should go to [itex]\infty[/itex] when [itex]\lambda\rightarrow\infty[/itex] and to 0 when [itex]\lambda\rightarrow0[/itex].
As I understand it, other papers do the same thing only with explicit W functions as opposed to this article.
 

1. What is the Scaled Particle Theory?

The Scaled Particle Theory (SPT) is a theoretical model used to describe the thermodynamic behavior of fluids composed of rigid particles. It is based on the assumption that the particles are perfectly hard, meaning they cannot overlap or penetrate each other.

2. What is the Rigid Spheres Equation of State?

The Rigid Spheres Equation of State is a mathematical expression that relates the thermodynamic properties of a fluid composed of rigid particles, such as pressure, volume, and temperature. It is derived from the Scaled Particle Theory and is used to predict the behavior of such fluids under various conditions.

3. How is the Scaled Particle Theory applied in real-world situations?

The Scaled Particle Theory is commonly used in the study of colloidal systems, where the particles are larger than molecules but still small enough to exhibit Brownian motion. It is also applied in the study of gases, liquids, and solids, as well as in the development of new materials and industrial processes.

4. What are the limitations of the Scaled Particle Theory?

The Scaled Particle Theory has several limitations, as it is a simplified model that cannot accurately describe all fluid systems. It assumes that the particles are perfectly hard and do not interact with each other, which is not always the case in real systems. It also does not consider the effects of non-rigid particles, intermolecular forces, and thermal fluctuations.

5. How does the Rigid Spheres Equation of State compare to other equations of state?

The Rigid Spheres Equation of State is a more accurate model for fluids composed of rigid particles compared to other equations of state, such as the ideal gas law. However, it still has limitations and may not accurately predict the behavior of real fluids. Other equations of state, such as the van der Waals equation, take into account intermolecular forces and non-rigid particles, making them more suitable for certain systems.

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