Timeline for The number of degrees of freedom of a monatomic gas
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2013 at 18:23 | answer | added | Paul J. Gans | timeline score: 8 | |
| Mar 10, 2013 at 16:27 | history | edited | Qmechanic♦ |
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| Mar 10, 2013 at 16:21 | comment | added | Cheeku | $N_f$ is just the number of molecules. Each of them has the energy $3kT/2$, hence the expression. Degrees of freedom is 3 | |
| Mar 10, 2013 at 16:21 | answer | added | zonksoft | timeline score: 2 | |
| Mar 10, 2013 at 16:20 | comment | added | Andrew | @Cheeku So each atom in my monatomic system has only translations (three of them). But I would like to compute, for example, the average kinetic energy of the entire system: $\langle K \rangle = 3N_f kT/2$, in which I think $N_f$ is indeed the number of degrees of freedom of the entire $N$ atoms/molecules, not just one of them. | |
| Mar 10, 2013 at 16:14 | comment | added | Cheeku | First thing that you need to understand is that $N$ is number of atoms in the molecule and not in the whole sample. | |
| Mar 10, 2013 at 16:11 | history | asked | Andrew | CC BY-SA 3.0 |