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In the following derivation, I just calculated the absolute internal energy of an ideal gas as a function of its temperature. But my teacher said that we can never calculate the absolute internal energy in thermodynamics. Then how is the below proof correct?

Proof to calculate absolute internal energy:

We know that from the law of equipartition of energy, average energy ($E_\mathrm{avg}$) can be written as, \begin{align} E_\mathrm{avg} &= \frac{f}{2}k_\mathrm bT \\ E_\mathrm{avg}(n\, \mathrm{mol}) &= nN_A\frac{f}{2}k_\mathrm bRT \\ &=\frac{f}{2}nRT \tag{1} \end{align}

Therefore, average energy for n moles of gas molecules is $\frac{f}{2}nRT$.

In case of an ideal gas, $$E_\mathrm{total} = K.E$$ Since there are no interactive forces, $P.E = 0$ and other energies are negligible. Therefore internal energy is equal to the kinetic energy.

From (1), we get: $$U_\mathrm{ideal} = K.E_\mathrm{ideal} = \frac{f}{2}nRT$$

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    $\begingroup$ "other energies are negligible. Therefore internal energy is equal to the kinetic energy." - that's not true, by any means - it's the kinetic energy atom that is negligible in comparison with energies inherent to atom itself. $\endgroup$
    – Mithoron
    Apr 18, 2021 at 13:08
  • $\begingroup$ It is true using classical thermodynamics that $\Delta U = q+w$ so changes only. However, thermodynamics predates our understanding of what molecules are and using Statistical Mechanics we can find the internal energy as you have done without using classical thermodynamics. See the link above for some more info in my comment there. $\endgroup$
    – porphyrin
    Apr 18, 2021 at 13:10
  • $\begingroup$ So, is it actually possible to calculate absolute internal energy or did I go wrong somewhere? Please answer @porphyrin. Yes or No? $\endgroup$
    – user281837
    Apr 18, 2021 at 14:00
  • $\begingroup$ I am talking about an ideal gas. $\endgroup$
    – user281837
    Apr 18, 2021 at 14:00
  • $\begingroup$ So therefore there is only kinetic energy to consider and for an ideal gas the internal energy is proportional to the temperature as you have found (from kinetic theory of gasses) so 'yes', but not from classical thermodynamics so 'no' from this view point. So your teacher is correct. $\endgroup$
    – porphyrin
    Apr 18, 2021 at 15:50

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