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$\begingroup$ I am not too sure about your statement. Consider an ideal gas; the total internal energy is defined as $U = \text{PE} + \text{KE}$, where $\text{PE}$ and $\text{KE}$ are the potential and kinetic energies, respectively. Since in an ideal gas, there are no attractive or repulsive forces, $\text{PE} = 0 $. Are you saying that $U = \text{PE} + \text{KE} = 0 + 3/2kt$ for all gases? Surely this can't be true.... $\endgroup$
– Cyclopropane
Commented Jan 27, 2020 at 14:10
$\begingroup$ @DrPepper No, I do not say that. KE >= translational KE =3/2kT. ( Al as mean energy) $\endgroup$
– Poutnik
Commented Jan 27, 2020 at 14:24

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