Confusion regarding heat capacity $C_V $ for a Van der Waals gas

Question

Suppose we have the following cycle in the $P-V$ diagram for a Van der Waals gas:

The internal energy , and equation of state for a VdW gas being:

$$ U = C_V T- a\dfrac{N^2}{V} $$ $$ P + a\dfrac{N^2}{V^2} = \dfrac{Nk_B T}{V-Nb} $$

and the heat capacities being linked by the Mayer relation:

$$ C_P - C_V = TV \dfrac{\alpha^{2}}{\kappa_T} $$

so one could calculate the coefficients $\alpha , \kappa_T$ from the equation of state.

My textbook says that heat transfered from $A \rightarrow B$ is:

$$ Q_{AB} = \Delta U - \underbrace{W_{AB}}_\text{$=0$} = \int^B_A C_V dT = C_V (T_2 - T_1) $$

It can be show that $C_V$ is volume independent for a VdW gas but here we assume that $C_V$ is temperature independent , how does one justify that ?

Continuing, if we calculate the heat transfered by going from $C$ to $A$:

$$ Q_{CA} = \Delta U - W_{AC} = \int^A_C C_P dT \neq C_P \Delta T $$

but this time $C_P$ is not temperature independent , which makes sense given the various terms in the Mayer relation depend on $T$.

Summary: How does one justify $C_V$ being temperature independent for $A \rightarrow B$?

Answer 1

The internal energy , and equation of state for a VdW gas being: $$U=C_VT-a\frac{N^2}{V}$$

From this definition, we can write $dU_V$ the infinitesimal change in internal energy with fixed volume as-

$$dU_V=C_VdT+TdC_V+0$$

Now we know that at constant volume, heat absorbed by the system is given as: $dQ=C_VdT$ and $dQ=dU$

Therfore we can see that the term $TdC_V$ is zero, or that $C_V$ is temperature independent.