Temperature dependent heat capacity and entropy
this problem is often presented as a simple demonstration of relationship between the heat capacity at constant volume $C_V$ and entropy $S$.
Let's say the heat capacity $C_V$ is equal to $a T + b T^3$ where $a$ and $b$ are constants. We then have $$ \mathrm{d} S = \frac{C_V (T)}{T} \mathrm{d} T \quad \implies \quad S (T) = a T + \frac{b}{3} T^3 $$
However, this got me thinking: we know, that strictly speaking, entropy is a two-variable function, so, for example, $$ \mathrm{d} S = \frac{C_V}{T} \mathrm{d} T + \frac{\alpha}{\beta_T} \mathrm{d} V $$ so even if none of the constants depend on volume and only $C_V$ depends on temperature, entropy will have a volume-dependent "constant" (with respect to temperature) term.
Alternatively if we express $\mathrm{d} S$ via $\mathrm{d} T$ and $\mathrm{d} p$ (both this and previous relations for $\mathrm{d} S$ are obtainable through crunching the <PERSON> relations and using the definitions for $\alpha$, $C_V$ and $\beta_T$), $$ \mathrm{d} S = \left( \frac{C_V}{T} + \frac{\alpha^2 V}{\beta_T} \right) \mathrm{d} T - \alpha V \mathrm{d} p $$
So this time around, the situation is much more complicated: using the equation of state we have to express $V$ as a function of pressure and temperature for this to be a reasonable differential equation. Even in the simplest case, we get $$ \mathrm{d} S = \left( \frac{C_V (T)}{T} + \frac{\alpha^2 N k_B T}{\beta_T p} \right) \mathrm{d} T - \frac{\alpha N k_B T}{p} \mathrm{d} p $$
In order for $\mathrm{d} S$ to be exact differential, the following must be true $$ \frac{\partial}{\partial p} \left( \frac{C_V (T)}{T} + \frac{\alpha^2 N k_B T}{\beta_T p} \right) = - \frac{\partial}{\partial T} \frac{\alpha N k_B T}{p} $$ $$ \frac{\alpha^2 N k_B T}{\beta_T p^2} = \frac{\alpha N k_B}{p} $$
And we already see this is not possible with constant $\alpha$ and $\beta_T$.
So my question is: what justifies the simple "only $T$ dependance" picture that is usually presented? 749