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How can internal energy be $\Delta{U} = nC_{v}\Delta{T}$?

Here's what my teacher did
From Law of equipartion of energy
Energy of one molecule having f Degree of freedom $= \frac{1}{2}KTf$
$K=\frac{R}{N_A}$
No of molecules $= nN_{A}$
Total energy $= \frac{nRTf}{2}$
$$U = \frac{nRTf}{2}$$ $$\Delta{U} = \frac{nfr\Delta{T}}{2}$$ Now since we didn't say anything about the process being isothermic/isobaric/isochoric this formula should be universal
BUT we also know that $C_{v} = \frac{fR}{2}$
$$\Delta{U} = nC_{v}\Delta{T}$$ $C_v$: Molar specific heat at constant volume

Now the process is under the assumption of constant volume, but didn't we say anything about volume being constant before derving this? Is it universal or only applicable for isochoric?

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  • $\begingroup$ U of an ideal gas depends only on temperature. $\endgroup$ Sep 21, 2021 at 2:16
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    $\begingroup$ I call this the cruelest equation in introductory thermodynamics. The general equation (for temperature-independent properties) is $\Delta U=C_V\Delta T+(\alpha TK-P)\Delta V$, which includes the constant-volume heat capacity, the constant-pressure thermal expansion coefficient, and the constant-temperature bulk modulus. These are just material properties that appear in the expression; they don’t imply any type of process. It’s just bad luck for the new student that $\alpha TK-P=0$ for an ideal gas, leaving only a confusing $C_V$. $\endgroup$ Sep 21, 2021 at 3:45

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Is it universal or only applicable for isochoric?

No, $\Delta U=nC_{v}\Delta T$ it is not universal (See comments by @Chemomechanics and @Chet Miller). It applies to an ideal gas, regardless of process, because an ideal gas behaves according to the following specific equation of state

$$P\Delta V=nR\Delta T$$

When you combine the equation of state for an ideal gas with the first law for a closed system

$$\Delta U=Q-W$$

and include the following relationship for an ideal gas (for derivation see: https://www.vedantu.com/chemistry/heat-capacity)

$$R=C_{p}-C_v$$

You will find that $\Delta U=nC_{v}\Delta T$ will always apply to an ideal gas for any process (not just a constant volume process).

As an example, consider a constant pressure process. The first law becomes

$$\Delta U=nC_p\Delta T – P\Delta V$$

Substituting $P\Delta V=nR\Delta T$ for an ideal gas

$$\Delta U=nC_p\Delta T – nR\Delta T$$

Substituting $R=C_{p}-C_v$

$$ \Delta U=nC_p\Delta T – n(C_p-C_v)\Delta T$$

Gives you

$$\Delta U=nC_v\Delta T$$

The same conclusion can be reached for an ideal gas undergoing other non isochoric processes.

Hope this helps.

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The energy is a state parameter. There is no process involved in the evaluation of the energy in a specific state. You don't need to and you did not assume any process when you evaluate the energy. The heat and work are process parameters so their value depends on process but not so for energy. This is true for any system, even if it is more complicated than ideal gas. The energy in a specific state does not depend on the process that brought the system in that state.

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The formula is universal. Total energy is a state function i.e. independent of path. So, we did derive it while keeping the volume constant but the formula would be applicable in any process.

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    $\begingroup$ Only for an ideal gas. $\endgroup$ Sep 21, 2021 at 2:22
  • $\begingroup$ @ChetMellier Yes, should have added that. I thought it's already implied that we are talking about ideal gases only. $\endgroup$
    – Satya
    Sep 21, 2021 at 2:37

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