# Two-dimensional gas?

1. Oct 18, 2006

### Sojourner01

"The entropy of a two-dimensional gas of N particles in an area A of energy U is given by:

$$S=Nk[ln \frac{A}{N} + ln \frac{mU]{2 \pi \hbar^2 N} +2]$$

Calculate the temperature of tge gas and the chemical potential."

I have absolutely no idea how to even begin. There was some bit of some previous lectures that used the clausius entropy principle to derive some partial differentials expressing different parameters in terms of entropy, but other than that, I'm lost. The two-dimensional bit is even more confusing. I gather that this affects the number of degrees of freedom, but how one goes about modifying the theory to aco**** for this, I have no clue.

edit: gah. I have no idea how the tex formatting screwed up. I've done it, as far as I can see, exactly how the help topic says.

Last edited: Oct 18, 2006
2. Oct 18, 2006

### quasar987

You got all the slashes inverted. make the substitution / --> \ and it should work

3. Oct 18, 2006

### Sojourner01

This is extremely frustrating. The code has been changed but the forum isn't updating the graphic. I'll post it again.

$$S=Nk[ln \frac{A}{N} + ln \frac{mU]{2 \pi \hbar^2 N} +2]$$

4. Oct 18, 2006

### quasar987

$$S=Nk\left[ \ln \left(\frac{A}{N}\right) + \ln \left(\frac{mU}{2 \pi \hbar^2 N}\right) +2\right]$$

strange about non-updating graphic. it's the first time I see this happening.

5. Oct 18, 2006

### Sojourner01

Thanks.

After all that, I believe I've worked out the problem using something I dug out of Carrington's Basic Thermodynamics.

Supposedly, temperature in an isolated 'fluid' is simply equal to $$(\frac{\partial U}{\partial S}})_V,N 6. Oct 19, 2006 ### Galileo The expression for the entropy (as function of the number of particles N, energy U, and volume V) will allow you to calculate the affinity $\alpha$, the inverse temperature $\beta$ and the free expansion coefficient $\gamma$: [tex]\frac{\partial S}{\partial N}=\alpha$$

$$\frac{\partial S}{\partial U}=\beta$$

$$\frac{\partial S}{\partial V}=\gamma$$

You just need to know $$\beta=1/kT$$

Last edited: Oct 19, 2006
7. Mar 1, 2007

### petmal

Mistake in the equation

It is probably too late, but I am quite sure that the equation should look like this:

]$$S=Nk\left[ \ln \left(\frac{A}{N}\right) + \ln \left(\frac{2\pi mU}{\hbar^2 N}\right) +2\right]$$