Partition Function Homework: Solve for N Particles in Spherical Container

In summary: Well, if V(r) \rightarrow \infty, then e^{-\beta V(r)} \rightarrow 0. So that means that the integral over d^3 r only needs to cover the region where V(r) < 0, which is the region 0 < r < R0. So you only need to integrate over that region.
  • #1
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Homework Statement



An ideal monatomic gas at the temperature T is confined in a spherical container of radius R. There are N molecules of mass m in the gas. Molecules move in a spherically symmetric potential V(r) where r is the distance from the center of the container. The potential V(r) is given by a piecewise function:
V(r) = -Uo*r/R for r<Ro
= 0 for Ro<r<R
= ∞ for r>R

U0 is a positive constant (i.e., “attractive well”). Thus, there is a potential well (a trap) of radius R0 inside the container. Notice also that the interaction between the molecule and the cavity wall is hard-core like.

a) Calculate the partition function Z.

b) What is the probability distribution function (probability per unit volume), P(r), that a particle is located at distance r from the center of spherical container?

c) What is the probability, Ptrap(r), that a particle is located inside the potential well (i.e. at a distance r smaller than R0 from the center of the spherical container).

d) What is the average number of molecules inside the trap, Ntrap?

e) Calculate the internal energy E=<H(r,p)> and the heat capacity Cv of this gas.

f) Calculate the pressure p exerted by the molecules on the wall of the spherical container.

Homework Equations



H = p^2/(2*m) + V(r)

The Attempt at a Solution



ℤ =1/(N!*h^(3N)) ∫ exp(-β*p^2/(2*m)∫exp(-β*V(r))

I get stuck with the integration of the potential term. I don't know if you would need to integrate each part of the piecewise separately?If someone could help me out with this, I would be extremely grateful![/B]
 
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  • #2
muskie25 said:

Homework Statement



An ideal monatomic gas at the temperature T is confined in a spherical container of radius R. There are N molecules of mass m in the gas. Molecules move in a spherically symmetric potential V(r) where r is the distance from the center of the container. The potential V(r) is given by a piecewise function:
V(r) = -Uo*r/R for r<Ro
= 0 for Ro<r<R
= ∞ for r>R

U0 is a positive constant (i.e., “attractive well”). Thus, there is a potential well (a trap) of radius R0 inside the container. Notice also that the interaction between the molecule and the cavity wall is hard-core like.

a) Calculate the partition function Z.

b) What is the probability distribution function (probability per unit volume), P(r), that a particle is located at distance r from the center of spherical container?

c) What is the probability, Ptrap(r), that a particle is located inside the potential well (i.e. at a distance r smaller than R0 from the center of the spherical container).

d) What is the average number of molecules inside the trap, Ntrap?

e) Calculate the internal energy E=<H(r,p)> and the heat capacity Cv of this gas.

f) Calculate the pressure p exerted by the molecules on the wall of the spherical container.

Homework Equations



H = p^2/(2*m) + V(r)

The Attempt at a Solution



ℤ =1/(N!*h^(3N)) ∫ exp(-β*p^2/(2*m)∫exp(-β*V(r))

I get stuck with the integration of the potential term. I don't know if you would need to integrate each part of the piecewise separately?If someone could help me out with this, I would be extremely grateful![/B]

Let's simplify to the case [itex]N=1[/itex]. Then [itex]Z = \frac{1}{h^3} \int d^3 p \int d^3 r e^{-\beta (\frac{p^2}{2m} + V(r))}[/itex]. The integral can be factored as:

[itex]Z = \frac{1}{h^3} (\int d^3 p e^{-\beta \frac{p^2}{2m}}) (\int d^3 r e^{-\beta V(r)})[/itex].
 
  • #3
stevendaryl said:
Let's simplify to the case [itex]N=1[/itex]. Then [itex]Z = \frac{1}{h^3} \int d^3 p \int d^3 r e^{-\beta (\frac{p^2}{2m} + V(r))}[/itex]. The integral can be factored as:

[itex]Z = \frac{1}{h^3} (\int d^3 p e^{-\beta \frac{p^2}{2m}}) (\int d^3 r e^{-\beta V(r)})[/itex].

I don't understand the ∞ in the piecewise. I feel like that would just cause the whole thing to go to ∞.

I don't know how to generate equations in this forum, so I will just try to explain my thought process.

The container is spherical, thus need to use spherical coordinates for V(r).

The first integral involving the momentum is well known, so that doesn't need to be "solved" by me. The second integral (the piecewise) needs to be separated into the three pieces of the function. The first going 0<r<Ro, which is the region closest to the center of the sphere (the trap), the second pieces has Ro<r<R, which goes from the potential trap to the edge of the sphere, and then R<r<∞. But, putting an exp(∞) into the integral causes it to go to infinity, and adding infinity to anything is causing the whole thing to go to ∞... What am I missing here?
 
  • #4
muskie25 said:
I don't understand the ∞ in the piecewise. I feel like that would just cause the whole thing to go to ∞.

I don't know how to generate equations in this forum, so I will just try to explain my thought process.

The container is spherical, thus need to use spherical coordinates for V(r).

The first integral involving the momentum is well known, so that doesn't need to be "solved" by me. The second integral (the piecewise) needs to be separated into the three pieces of the function. The first going 0<r<Ro, which is the region closest to the center of the sphere (the trap), the second pieces has Ro<r<R, which goes from the potential trap to the edge of the sphere, and then R<r<∞. But, putting an exp(∞) into the integral causes it to go to infinity, and adding infinity to anything is causing the whole thing to go to ∞... What am I missing here?

Well, if [itex]V(r) \rightarrow \infty[/itex], then [itex]e^{-\beta V(r)} \rightarrow 0[/itex]. So that means that the integral over [itex]d^3 r[/itex] only needs to cover the region where [itex]V(r) < \infty[/itex]
 
  • #5
stevendaryl said:
Well, if [itex]V(r) \rightarrow \infty[/itex], then [itex]e^{-\beta V(r)} \rightarrow 0[/itex]. So that means that the integral over [itex]d^3 r[/itex] only needs to cover the region where [itex]V(r) < \infty[/itex]

Ahh, I was forgetting the (-) in front of the ∞. That would explain a lot! This is still a huge ℤ, so I thought that I was doing it wrong but it sounds like I was on the right track. Thanks!
 
  • #6
stevendaryl said:
Well, if [itex]V(r) \rightarrow \infty[/itex], then [itex]e^{-\beta V(r)} \rightarrow 0[/itex]. So that means that the integral over [itex]d^3 r[/itex] only needs to cover the region where [itex]V(r) < \infty[/itex]

One more quick question: Would there be a V^N term in front of everything? I know that in the classical monatomic gas solution, there is. And , considering that the pressure is a partial wrt V, I assume it needs to be there again?
 
  • #7
muskie25 said:
One more quick question: Would there be a V^N term in front of everything? I know that in the classical monatomic gas solution, there is. And , considering that the pressure is a partial wrt V, I assume it needs to be there again?

No. The [itex]V^N[/itex] term is from integrating [itex]d^3 r[/itex] (there is one such integral for each particle). If you integrate [itex]d^3 r[/itex] over a region, you get the volume of that region. But in this case, you're not integrating [itex]d^3 r[/itex], you're integrating [itex]e^{-\beta V(r)} d^3 r[/itex]. That's not going to give you the volume. (We have two different [itex]V[/itex]s here: the volume and the potential).
 
  • #8
stevendaryl said:
No. The [itex]V^N[/itex] term is from integrating [itex]d^3 r[/itex] (there is one such integral for each particle). If you integrate [itex]d^3 r[/itex] over a region, you get the volume of that region. But in this case, you're not integrating [itex]d^3 r[/itex], you're integrating [itex]e^{-\beta V(r)} d^3 r[/itex]. That's not going to give you the volume. (We have two different [itex]V[/itex]s here: the volume and the potential).

Hmm. Then I don't get how to determine the pressure without having a volume term?
 
  • #9
muskie25 said:
Hmm. Then I don't get how to determine the pressure without having a volume term?

The integral [itex]\int e^{-\beta V(r)} d^3 r[/itex] will involve [itex]R[/itex]. [itex]R[/itex] is related to the volume [itex]V[/itex] through [itex]V = \frac{4}{3} \pi R^3[/itex], or [itex]R = (\frac{3}{4\pi} V)^{\frac{1}{3}}[/itex]
 

1. What is the partition function in the context of this homework?

The partition function in this context is a mathematical function that represents the total number of ways that N particles can be distributed within a spherical container. It takes into account the energy levels of each particle and the constraints of the container.

2. How do I solve for N particles in a spherical container using the partition function?

To solve for the number of particles in a spherical container, you will need to use the partition function formula, which takes into account the energy levels and constraints of the container. You will also need to incorporate any additional information given in the homework, such as the temperature and volume of the container.

3. What are the units of the partition function?

The partition function is a dimensionless quantity, meaning it has no units. This is because it represents a ratio of the number of microstates to the total number of particles in the system.

4. Can the partition function be used for any type of container, or just a spherical one?

The partition function can be used for any type of container, as long as the container has a defined volume and the particles have a defined energy. However, the formula may differ depending on the shape and constraints of the container.

5. Is there a specific method for solving partition function homework, or can I use any approach?

There are various methods for solving partition function homework, such as using mathematical equations, statistical mechanics, or computer simulations. The approach you use may depend on the complexity of the problem and your personal preference.

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