Two-D Gas: Entropy Calculation & Temperature/Chemical Potential

[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.

 

[quasar987]

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

 

[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]

 

[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.

 

[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

 

[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:
\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

 

[petmal]

Mistake in the equation

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.

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]