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How do you derive the Sackur-Tetrode equation?

I know that you must start off with the multiplicity of a mono-atomic ideal gas:

$$\Omega_{N}~=~\frac{1}{N!}\frac{V^{N}}{b^{3N}}\frac{\pi^{3N/2}}{(3N/2)!}\sqrt{2mU}^{3/N}.$$

Can someone please explain the process of how you derive the Sackur-Tetrode equation from this?

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I posted the following solution on this board wanting to get opinions of the validity of a solution using only the microcanonical ensemble:

Simpler derivation of Sackur-Tetrode equation

I still haven't received any comments, but if you have any, feel free to chime in.

The Sarkur-Tetrode equation is the following without the 5/2 constant term: $$ kn \ln \frac V {n\lambda^3} $$ https://en.wikipedia.org/wiki/Sackur%E2%80%93Tetrode_equation

Where $\lambda^3$ is the thermal wavelength cubed or the quantum volume for one particle.

Since each particle has a volume of $\lambda^3$,

N, the total of number of positions in the volume for a particle = $\frac V {\lambda^3}$

n = the total number of particles.

Using the binomial distribution, the definition of S from Boltzmann's equation is:

1) $$S = k\ln \Omega = k\ln \left[\frac {N!}{n!(N-n)!}\right]$$

Substituting for $N = \frac V {\lambda^3}$

2) $$S = k \ln \left\{\frac {\frac V {\lambda^3}!}{n!(\frac V {\lambda^3}-n)!}\right\}$$

Use Stirling's Approximation

3) $$S = k \left\{ \frac V {\lambda^3} \ln \left(\frac V {\lambda^3}\right) - \left(\frac V {\lambda^3} - n \right) \ln \left(\frac V {\lambda^3}-n\right) - n \ln (n)\right\}$$

Use approximation $$\ln \left(\frac V {\lambda^3}-n\right) = \ln\left(\frac V {\lambda^3}\right) $$ for $V \gg \lambda^3 $

4)$$ S = k \left\{(n ) \ln \left(\frac V {\lambda^3}-n\right) - n \ln (n)\right\}$$

Use approximation $$\frac {\frac V {\lambda^3}-n}n = \frac V {\lambda^3n}$$ for $V \gg \lambda^3$

5) $$S = kn \ln \frac V {n\lambda^3} $$

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    $\begingroup$ The Sackur-Tetrode equation only applies to indistinguishable particles, however your derivation is for distinguishable ones. $\endgroup$ – user400188 Apr 25 '19 at 5:22
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I think the $\Omega_N$ in your question is the number of states inside an energy sphere $E$;

Now we need to consider micro-canonical ensemble, which means we need to get the number of states in an energy shell between $E$ and $E+\Delta$:

$$\Omega'=\frac{1}{N!h^{3N}} V^N \frac{\pi^{3N/2}}{(3N/2)!}\{[2m(E+\Delta)]^{3N/2}-(2mE)^{3N/2}\}$$

Which is equal to:

$$\Omega'=\left[\frac{V}{h^3}(2\pi \ mE)^{3/2}\right]\frac{\Delta}{E}\frac{1}{N!(\frac{3N}{2}-1)!}$$

Then, use the Boltzmann's formula:

$S=k_B \log\Omega'$

Drop some small terms, like $\log(\frac{\Delta}{E})$; and use the Stirling's formula: $\log n!=n \log(n)-n$, after some algebra, you should be able to get the result:

$$S=Nk_B\log\left[\frac{V}{N^{5/2}h^3}\left(\frac{4\pi}{3}mE\right)^{3/2}\right]+\frac{5}{2}Nk_B$$

Hope be helpful.

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