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In my understanding $C_V$ which is defined when the volume is constant is written as $$C_V=\frac{q_V}{\Delta T}$$

Thus, $$q_V={nC_V\,\Delta T}$$ since for an isochoric process work done is zero

$$\Delta U=q+w$$ (which is zero) $$\Delta U=nC_V\,\Delta T$$

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Because $\Delta U$ is a state function! It does not depends on how you progress.

For $$V=\text{const}\Rightarrow w=0$$ $$U=q+w=q=nC_V\,\Delta T$$ For $$p=\text{const}\Rightarrow w=-p\,\Delta V=-nR\,\Delta T$$ Since $nR=\text{const}$, only $T$ changes which makes $V$ changes $$\Delta U=q+w\\ \Leftrightarrow nC_p\,\Delta T-p\,\Delta V\\ \Leftrightarrow nC_p\,\Delta T-nR\,\Delta T\\ \Leftrightarrow n(C_p-R)\,\Delta T\\ \Leftrightarrow nC_V\,\Delta T$$ So $\Delta U=nC_V\,\Delta T$ when $p=\text{const}$. Only for $T=\text{const}$ $$\Delta U=0$$ For $p$ or $V=\text{const}$ $$\Delta U=nC_V\,\Delta T$$

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The below equation is definition of the molar heat capacity.

$$C_V=\frac 1n \left(\frac{\partial U}{\partial T}\right)_V$$

The condition $\mathrm{d}V=0$ is not required for ideal gases, as their internal energy depends only on temperature. It should be therefore obvious why the defining equation is used so often.

Note that the increase of internal energy and temperature during adiabatic processes is achieved even at $q=0$, so $$C_V=\frac 1n \left(\frac{\delta q}{\text{d}T}\right)$$ cannot be used for definition of $C_V$ of real gases.

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