# Gas with potential wall

1. Jun 14, 2013

### aaaa202

1. The problem statement, all variables and given/known data
See the attached picture

2. Relevant equations
The equations given should be the relevant ones.

3. The attempt at a solution
I feel I got a bad intuition and understanding of this problem. I could eliminate T in favor of S and then express x and T by U and S. Is this how its supposed to be done? I don't think so, because I wonna use the information that the wall slides without friction at some point. Anyone have an idea what to do?
Also I don't agree with the expression for the entropy given. If you do a phase space integral to find the partition function you have to integrate over the walls position too, so you end up with an expression where T is not just raised to the power of 3N/2 but more (after all the 3N/2 is just an ideal gas in a container with fixed walls). Don't YOU agree on this?

#### Attached Files:

• ###### Gas.png
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2. Jun 14, 2013

### Mute

The information about the wall's position is contained in the volume in the equation for the entropy. That is, the volume of the gas will be the surface area of the wall times a linear length which contains the variable x. Your attachment doesn't specify what x is measured relative to, so I can't say for certain what that length scale is.

So, your equation for entropy seems like it should be fine, or at least approximately fine. Now you just need to use it to determine the equilibrium condition.

To do that, note that your problem states that system is isolated and the energy, U, is known. So, it sounds like the energy isn't changing, is that correct? If so, your equation for U then relates the temperature to the wall's position.

3. Jun 14, 2013

### aaaa202

The problem is exactly that the problem only specifies that the mean energy is known. But I suppose that doesn't make a difference. The mean of the energy will in any case be the same at all times if the system is isolated, so indeed my equation for U relates the position of the wall to the temperature. Problem is then just that I have to specify the temperature too. Do I get that from the equation for the entropy? I still don't see how the expression for the entropy accomodates for the energy from the wall. If you calculate the partition function you would get an integral of exp(-x^2/T) which would make the partition function and thus the entropy different from the one of an ideal gas. I don't know - maybe it's gibberish.
Also x is measured from the bottom of the container.

4. Jun 14, 2013

### aaaa202

The problem is exactly that the problem only specifies that the mean energy is known. But I suppose that doesn't make a difference. The mean of the energy will in any case be the same at all times if the system is isolated, so indeed my equation for U relates the position of the wall to the temperature. Problem is then just that I have to specify the temperature too. Do I get that from the equation for the entropy? I still don't see how the expression for the entropy accomodates for the energy from the wall. If you calculate the partition function you would get an integral of exp(-x^2/T) which would make the partition function and thus the entropy different from the one of an ideal gas. I don't know - maybe it's gibberish.
Also x is measured from the bottom of the container.

5. Jun 14, 2013

### Mute

If $x$ is the distance from the leftmost edge of the box to the movable wall, then the volume of the gas should just be $V = Ax$, where A is the surface area of the movable wall. Then, the formula you are given would be $S \sim N \ln(x T^{3/2})$. If that's the equation the problem gives you, you can just work with it by assumption to derive the final answer.

I don't have time to try and work out the partition function myself, so I can't really answer whether or not the factor $\exp(-x^2/T)$ that you calculate is correct or if gets cancelled out by something, etc.

If U is not fixed, then I would guess that you need to use both the second law and the fact that the system will likely try to minimize its energy. By minimizing the energy, you could find a relation between x and T that you could insert into $S$, which you could then maximize to find the final temperature and position.

You should post your calculation of the partition function. Perhaps someone else here will have time to go through it with you and see if they agree that the entropy expression given by the problem does not sufficiently describe the system.

6. Jun 14, 2013

### aaaa202

Here is my calculation of the partition function. The crucial point is that I sum over all positions of the wall (we might aswell extend the integral to infinity) and multiply it by all states of each atom. I should think that is combinatorically correct, but its probably wrong.

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