# Can a change in internal energy always be expressed as the product of the constant volume heat capacity and the change in temperature?

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?

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.