Basically, which is the correct formula for thermal energy, and is this the same as kinetic energy? My notes are pretty conflicting on this topic, and I'm getting pretty confused.
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3$\begingroup$ Actually E is not equal to any of them unless you don't specify the degree of freedom in a system. For our convenience we ignore the rotational motion, which provides E=(3/2)kt. $\endgroup$– Harshal GajjarCommented Nov 15, 2014 at 17:09
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5$\begingroup$ This question appears to be off-topic because it shows insufficient prior research. $\endgroup$– JamalSCommented Nov 15, 2014 at 17:13
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4$\begingroup$ tom's answer correctly explains why both expressions have legitimate uses; this question was improperly closed. $\endgroup$– rob ♦Commented Nov 16, 2014 at 0:35
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1$\begingroup$ Some questions are closed here due to "insufficient prior research," as this one was. By that metric, it is irrelevant whether or not the question is understandable and has been answered. $\endgroup$– BMSCommented Nov 16, 2014 at 4:16
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3 Answers
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Prahar is correct that generally we have an energy contribution of ${1 \over 2} kT$ per degree of freedom in a system - so that atoms in a gas of atoms (e.g. Helium) will have an average energy of ${3 \over 2} kT$.
Often people talk about thermal energy being '$kT$' because of the exponential expression in
$N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}}$
where $N_i$ is the number of atoms in a state $i$ with degeneracy $g_i$ and energy $E_i$ above the ground state which has $N_0$ atoms and a degeneracy of $g_0$.
So often people compare the energy of a state $E_i$ with the 'available thermal energy' $kT$, because the term
$e^{-{E_i \over kT}}$
is dominant in the expression above and if $kT$ << $E_i$ then the ratio of population in state $i$ to the population in the ground state ($N_i$/$N_0$) will be small or perhaps close to zero.
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$\begingroup$ I still don't get it. I saw many derivations that state $PV=NkT$ (e.g. youtube.com/watch?v=JOs8UQSWmos at ~28:00) for a gas with $N$ particles in a volume $V$ of pressure $P$. This means the average energy per particle is $kT$ (this is even explicitely said in the linked video). I don't understand why and where the contradiction with $3/2 kT$ comes up. It would be great to elaborate on the $PV=NkT$ equation in this answer. $\endgroup$– divBCommented May 15, 2020 at 16:54
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1$\begingroup$ @divB that is a good question. I will have to get back to you about it and think about it $\endgroup$– tomCommented May 18, 2020 at 14:57
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$\begingroup$ @divB sorry, I have not recently had enough time to think about this.... do ping me if I have not got back to you within a couple of weeks. please $\endgroup$– tomCommented May 28, 2020 at 18:05
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$\begingroup$ @divB $PV$ is not the internal energy of the gas; the internal energy $U$ is given by $(3/2)NkT$ (for a mono-atomic gas [no internal degrees of freedom]); c.f. web.mit.edu/16.unified/www/FALL/thermodynamics/notes/… $\endgroup$– DaveCommented Oct 13, 2021 at 22:42
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The thermal energy of a system is $$ E = f \frac{1}{2} k T $$ where $f$ is the number of degrees of freedom of the theory - which is roughly speaking the number of dimensions it is allowed to move in.
For instance, if you are talking about an atom in 3 space dimensions, then the atom can move along the 3 axes and hence $f=3\implies E = \frac{3}{2} kT$. If I have a gas with $N$ atoms each of which can move in 3 dimensions, then $f = 3 N \implies E = \frac{3N}{2} kT$.
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kT is the energy of collision between two particles, since each particle carries (on average) 1/2kT energy in the direction of the collision. Thus, 1/2kT+1/2kT=kT.
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$\begingroup$ Hi, Welcome to Physics SE! Please use MathJax to write equations, it's much easier to read! $\endgroup$– user191954Commented Jun 2, 2018 at 11:08
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$\begingroup$ And if particles do not collide, their total energy is still this sum ;-) $\endgroup$ Commented Jun 2, 2018 at 11:19