Density of states in the ideal gas

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SUMMARY

The discussion centers on deriving the density of states (DOS) for an ideal gas, specifically focusing on the expression MB_PDF(E, T) = D * e^(-E/(kB*T)). The participants emphasize that the DOS cannot be expressed solely in terms of energy (E), pi, Boltzmann's constant (kB), and temperature (T) without including Planck's constant (h). The conversation also touches on the relationship between DOS and extensive entropy, highlighting that while Boltzmann's counting of states initially appeared inconsistent with Clausius' entropy, Gibbs resolved this by accounting for indistinguishable particles.

PREREQUISITES
  • Understanding of Maxwell-Boltzmann (MB) distribution
  • Familiarity with density of states (DOS) concepts
  • Knowledge of statistical mechanics and partition functions
  • Basic grasp of thermodynamic entropy and its formulations
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  • Study the derivation of the density of states for ideal gases, focusing on the role of Planck's constant.
  • Explore the implications of Gibbs' paradox in statistical mechanics.
  • Learn about the relationship between partition functions and density of states in thermodynamic systems.
  • Investigate the connection between entropy formulations by Boltzmann and Clausius.
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Physicists, chemists, and students of statistical mechanics who are interested in the theoretical foundations of thermodynamics and the behavior of ideal gases.

rabbed
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The MB energy distribution is: MB_PDF(E, T) = 2*sqrt(E/pi) * 1/(kB*T)^(3/2) * e^(-E/(kB*T))
How do I arrive at the density of states which hides inside the expression 2*sqrt(E/pi) * 1/(kB*T)^(3/2) ? I've only seen DOS that have "h" in them.. I want it to contain only E, pi, kB and T.. This is how far I've gotten (using a momentum vector):

V = 4*pi*p^3/3
dV = 4*pi*p^2*dp dV = 4*pi*(2*m*E)*sqrt(m/(2*E))*dE (since p = sqrt(2*m*E) and dp = sqrt(m/(2*E))*dE) dV = 2*pi*(2*m)^(3/2)*sqrt(E)*dE How do I get rid of the m and how do I get in kB and T?

On the same theme:

Is there a DOS-expression D for the ideal gas which will both fit into MB_PDF(E, T) = D * e^(-E/(kB*T)) / Z (where Z normalizes the distribution) as well as giving an extensive entropy S = kB*ln(D) ? It should be the same quantity, right?
 
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rabbed said:
How do I arrive at the density of states which hides inside the expression
You can't. The best you can do is arrive at the density of states divided by the partition function.

rabbed said:
I've only seen DOS that have "h" in them.. I want it to contain only E, pi, kB and T..
That doesn't make sense. The density of states depends on Planck's constant, so there is no way to write the DOS without it. Even Gibbs, way before Planck, figured out that there was a parameter ##h## that needed to be included in the partition function. It cancels out if you take the DOS divided by Z, see my comment above.
 
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Thanks, looks like an interesting discussion that I will read.

So the way I've understood this is: Boltzmann's counting of states didn't give the same (extensive) entropy as the entropy of Clausius. Then Gibbs fixed this (Gibbs paradox) by considering indistinguishable particles?

This document seem to say that Boltzmann counting does give extensive entropy after all: https://www.hindawi.com/journals/jther/2016/9137926/

Does Gibbs give a DOS-expression D for the ideal gas which will both fit into MB_PDF(E, T) = D * e^(-E/(kB*T)) / Z (where Z normalizes the distribution) as well as giving an extensive entropy S = kB*ln(D) ?
Should it be the same quantity in both these places and in that case, does any h-parameters cancel out in the entropy expression also?

Just trying to sort things out.. :)
 

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