# Simpler derivation of Sackur-Tetrode equation

Is there a reason the following derivation for the Sackur-Tetrode equation is not common? I am teaching a lower undergraduate level class and would like to derive it with simpler terms of only using relative volume.

The equation is: $$kn \ln \frac V {n\lambda^3},$$ where $\lambda^3$ is the thermal wavelength cubed, or the quantum volume for one particle.

Since each particle has a volume of $\lambda^3$, the total of number of positions in the volume for a particle is $N = \frac V {\lambda^3}$, and $n$ is the total number of particles.

1. Using the binomial distribution, the definition of $S$ from Boltzmann's equation is: $$S = k\ln \Omega = k\ln \biggl[\frac {N!}{n!(N-n)!}\biggr]$$
2. Substituting for $N = \frac V {\lambda^3}$, $$S = k \ln\left[\frac {\bigl(\frac V {\lambda^3}\bigr)!}{n!\bigl(\frac V {\lambda^3}-n\bigr)!}\right]$$
3. Using Stirling's approximation: $$S = k \biggl[ \frac V {\lambda^3} \ln \biggl(\frac V {\lambda^3}\biggr) - \biggl(\frac V {\lambda^3} - n \biggr) \ln \biggl(\frac V {\lambda^3}-n\biggr) - n \ln (n)\biggr]$$
4. Using the approximation $\ln \bigl(\frac V {\lambda^3}-n\bigr) = \ln\bigl(\frac V {\lambda^3}\bigr)$ for $\frac V {\lambda^3} \gg n$ $$S = k \biggl[ n \ln \biggl(\frac V {\lambda^3}\biggr) - n \ln (n)\biggr]$$
5. Manipulate algebra. $$S = kn \ln \frac V {n\lambda^3}$$

I'd like to use this in my class because it is simpler and cleaner and develops a sort of chemical intuition based on binomial distribution. However, I want to know if this is correct. I have seen this equation without the $5/2$ term before.

Edit: The physical assumption that I think might be invalid is the use of N as $\frac V {\lambda^3}$. The binomial distribution is valid when you have n particles that fit into N DISCREET positions. That would work fine for a box of volume V with N slots separated by partitions. But in this case, the volume V does not have any partitions and is continuous. In this case, a volume $\lambda^3$ can fit into positions that are not discreetly defined. This creates a technically infinite number of N's. So, would this still be a valid $\Omega$ to be used in the Boltzmann's equation?

• Hi Ted, Mathjax is used on this site for expressions such as the ones above and, as it makes any possible answers more likely, imo, it might be worth your while performing a quick edit. Best of luck with your question. – user81619 Aug 11 '15 at 18:38
• Thanks for the tip. Just learned Mathjax which was pretty easy. Hopefully, some people will respond. – Ted Yu Aug 12 '15 at 18:19
• Also, I think that people answering your questions may be able to copy and paste your hard work, increasing the chances of an answer. – user81619 Aug 12 '15 at 18:34
• People are voting to put your question on hold, which may be because it's not really clear why you suspect this derivation may not be correct. If you could expand on that, I think it would improve the question considerably. We tend to be hard on questions that are just asking us to check something, partly as a measure to discourage people looking for homework help. – David Z Aug 12 '15 at 18:41
• P.S. I fixed up the math formatting for you. – David Z Aug 12 '15 at 18:48

Issues with that derivation:

• You're missing the extra term $\frac 52 k N,$ which may matter if you have to do any work with chemical potentials.
• Your students will not necessarily know why to parcel the space into volumes of size $\lambda^3$. Starting from the definition of entropy and deriving that the thermal volume $\lambda^3$ is important seems preferable.
• Your students may benefit from knowing that half-factorials exist via $n! = \int_0^\infty dx~x^n~e^{-x},$ that they start from $(-1/2)! = \sqrt{\pi},$ and that the volume of an $n$-ball of radius $r$ is $\pi^{n/2} ~ r^n / (n/2)!.$ It is only about fifteen minutes or less on a chalkboard of mathematical trickery, but it can help to facilitate some later calculations, plus an "oh, that's how that works" explaining why Gaussians have these mysterious factors of $\sqrt{\pi}$ in them.
• The fact that quantum mechanics makes particles even-in-principle indistinguishable is huge, and it resolves the Gibbs paradox, leading to the Sackur-Tetrode equation. Robbing a student of this fact, and the amount of brain-warp that it creates, is not a bad thing necessarily, but it's also not necessarily a good thing. This is an opportunity to plant a seed of cognitive dissonance that only really will get resolved when the student learns quantum field theory. It is maybe even the central problem with particle-centric understandings of the world.
• If you're teaching statistical mechanics, your students just barely are beginning to understand what temperature really is. I'm not sure I personally would make it a "load-bearing beam" of their understanding of statistical mechanics. For learning thermodynamics prior to a statistical mechanics course, sure, just treat temperature as a phenomenological given, we can measure it with a thermometer, who cares what it is? But now that you have an ability to talk about how $\beta = (k_B T)^{-1}$ deals much better with "negative temperatures" than $T$ does, everything that the student knows from earlier thermodynamics courses is painfully incomplete and you may not want to rest too much material on a good understanding of temperature and the thermal wavelength.

With all of this said: your approach is certainly simple and clean, and might make a good heuristic introduction to the topic if you do not want to devote a whole lecture to the Sackur-Tetrode equation.

• Thanks for your comments, and it is great to get your perspective. My students are chemical engineering majors, and the biggest challenge is to help them understand the 2nd law and why heat is related to entropy (or randomness). I will use the concept of the quantum volume, $\lambda^3$ to relate this. The one issue I see from your feedback is that you need entropy to derive quantum volume. The thermal wavelength can be derived by $\Lambda = \frac h p$, $E_k = \frac {p^2} {2m}$, $E_k = \pi k_B T$. Do you mean one of these equations require S to derive? – Ted Yu Aug 18 '15 at 18:53
• CR Drost: I've been thinking about this comment you made: "You're missing the extra term (5/2)kN, which may matter if you have to do any work with chemical potentials." Are you referring to the fact that I do not derive the correct chemical potential definition if I leave out the 5/2 term for entropy? – Ted Yu Jan 25 '16 at 8:19
• @TedYu I believe so. Certainly the expression $\mu = T {\left({\partial S\over\partial N}\right)}_{E,V}$ would seem to depend crucially on it. – CR Drost Jan 25 '16 at 17:15
• The chemical potential of an ideal monoatomic gas should be: $\mu = \tau ln \frac {n} {n_Q}$web.mit.edu/ndhillon/www/Teaching/Physics/bookse5.html Curiously, I get this result if I leave out the 5/2 term with your above definition of chemical potential. Which is why I am confused about your comment. I don't understand why you need the 5/2 term to get the correct chemical potential. – Ted Yu Jan 25 '16 at 18:09
• I posed the question here if you are interested: physics.stackexchange.com/questions/231080/… – Ted Yu Jan 25 '16 at 18:14