Timeline for Simpler derivation of Sackur-Tetrode equation
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 27, 2016 at 8:13 | comment | added | Ted Yu | I trust you are right, but I still can't get over the fact that U is independent of N. If U = $\frac 3 2 N k_B T$, it would cancel all the N dependence inside the log in the case where it is constant P. But I guess in this case, it is constant E (or U), so I get it. Still stumped on the other question though. | |
| Jan 25, 2016 at 21:00 | comment | added | CR Drost | @TedYu: You need that term because the expression you're deriving is $k N \ln(\alpha U^{3/2} V N^{-5/2})$ and the derivative of $N \log N$ is $\log N + 1$, so you really get $k \ln (\alpha U^{3/2} V) + \frac{-5}2 k(\ln N + 1),$ the extra $-5/2 k$ needs to be balanced by a $+5/2 k$ to get the result you seek. | |
| Jan 25, 2016 at 18:14 | comment | added | Ted Yu | I posed the question here if you are interested: physics.stackexchange.com/questions/231080/… | |
| Jan 25, 2016 at 18:09 | comment | added | Ted Yu | 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. | |
| Jan 25, 2016 at 17:15 | comment | added | CR Drost | @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. | |
| Jan 25, 2016 at 8:19 | comment | added | Ted Yu | 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? | |
| Aug 20, 2015 at 17:29 | vote | accept | Ted Yu | ||
| Aug 18, 2015 at 18:53 | comment | added | Ted Yu | 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? | |
| Aug 17, 2015 at 18:18 | history | answered | CR Drost | CC BY-SA 3.0 |