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It makes sense to me conceptually that $dU=C_VdT$ but is that always the case because I have seen that result derived thus: $$dU=U(V,T)=(\frac{\partial U}{\partial T})_VdT+(\frac{\partial U}{\partial V})_TdV$$

(Not the typical expression for internal energy $dU=TdS-pdV$

But for a perfect gas $(\frac{\partial U}{\partial V})_T=0$ so for a perfect gas internal energy is only a function of $T$: $U=U(T)$. Thus, it follows that;

$$dU=(\frac{\partial U}{\partial T})_VdT=C_VdT$$

However, is this only true for a perfect gas? I ask this because, the knowledge of this equality is required to show that entropy is a state function and I was wondering whether the proof that uses the above argument is only valid for a perfect gas or whether it's valid more generally?

I will show you the proof now:

$$dU=dq-pdV=C_VdT$$ $$dq=C_VdT+pdV$$ $$\frac{dq}{T}=\frac{C_V}{T}dT+\frac{nR}VdV$$ (using pV=nRT)

Then if you differentiate appropriately you can show it's an exact differential. However, is this only valid for a perfect gas?

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The statement in the title question "internal energy [can] always be expressed as the product of the constant volume heat capacity and the change in temperature" would only apply to a calorically perfect gas.

If you instead said "an infinitesimal change in internal energy can be expressed as the product of the constant volume heat capacity and an infinitesimal change in temperature", that is only true for a thermally perfect gas.

For real gas, the heat capacities depend upon both temperature and pressure.

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For a real gas (and for single phase liquids and solids of constant composition), the effect of temperature and molar volume on molar internal energy is given by: $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_T\right]dV$$where, in general, Cv is a function of both temperature and pressure. Note that, for an ideal gas, the term in brackets is zero and Cv becomes dependent only on temperature.

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