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

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Homework Help Overview

The discussion revolves around the entropy calculation for a two-dimensional gas consisting of N particles within an area A and energy U. Participants are exploring the implications of the entropy formula and its relation to temperature and chemical potential.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how to apply the Clausius entropy principle and modify the theory for a two-dimensional system, expressing confusion about the degrees of freedom. Other participants provide feedback on formatting issues and clarify the entropy expression.

Discussion Status

Some participants have offered guidance on correcting the formatting of the entropy equation, while others have suggested potential methods for calculating temperature and chemical potential based on the entropy expression. There is an ongoing exploration of the relationship between entropy and other thermodynamic variables.

Contextual Notes

Participants are navigating issues related to formatting in the forum, which has affected the clarity of the equations presented. There is also a mention of using resources from previous lectures and textbooks to aid understanding.

Sojourner01
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"The entropy of a two-dimensional gas of N particles in an area A of energy U is given by:

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

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:
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You got all the slashes inverted. make the substitution / --> \ and it should work
 
This is extremely frustrating. The code has been changed but the forum isn't updating the graphic. I'll post it again.

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

strange about non-updating graphic. it's the first time I see this happening.
 
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 [tex](\frac{\partial U}{\partial S}})_V,N[/tex]
 
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 [itex]\alpha[/itex], the inverse temperature [itex]\beta[/itex] and the free expansion coefficient [itex]\gamma[/itex]:
[tex]\frac{\partial S}{\partial N}=\alpha[/tex]

[tex]\frac{\partial S}{\partial U}=\beta[/tex]

[tex]\frac{\partial S}{\partial V}=\gamma[/tex]

You just need to know [tex]\beta=1/kT[/tex]
 
Last edited:
Mistake in the equation

quasar987 said:
[tex]S=Nk\left[ \ln \left(\frac{A}{N}\right) + \ln \left(\frac{mU}{2 \pi \hbar^2 N}\right) +2\right][/tex]

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:

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

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