Does a Well-Defined Entropy Exist for Non-Ideal Gases A and B?

[Adam564]

Homework Statement

Two non-ideal gases, A and B, whose internal energies only depend on temperature obey the following equations of state $$p=\alpha_A\frac{NT}{V^2}$$ and $$p=(\beta_B\frac{N}{V}T)^{1/2}$$, respectively. Here, $$\alpha_A$$ and $$\beta_B$$ are some constants. Determine for both gases individually whether a well-defined entropy exists. If not, what does that imply?

Relevant Equations

$$dU=TdS-pdV$$
$$dS=\frac{dU}{T}+\frac{p}{T}dV$$

The conclusion of my attempt I am listing below is that there do exist entropies for both but I am not sure.
$$dU=TdS-pdV$$
$$dS=\frac{dU}{T}+\frac{p}{T}dV$$
Therefore, gas A:
$$S=\frac{{\Delta}U}{T}+\alpha_A(\frac{-N}{{\Delta}V})$$
Gas B:
$$S=\frac{{\Delta}U}{T}+\frac{1}{\sqrt{T}}\sqrt{\beta_B}2\sqrt{N{\Delta}V}$$

 

[Chestermiller]

Homework Statement:: Two non-ideal gases, A and B, whose internal energies only depend on temperature obey the following equations of state $$p=\alpha_A\frac{NT}{V^2}$$ and $$p=(\beta_B\frac{N}{V}T)^{1/2}$$, respectively. Here, $$\alpha_A$$ and $$\beta_B$$ are some constants. Determine for both gases individually whether a well-defined entropy exists. If not, what does that imply?
Relevant Equations:: $$dU=TdS-pdV$$
$$dS=\frac{dU}{T}+\frac{p}{T}dV$$

The conclusion of my attempt I am listing below is that there do exist entropies for both but I am not sure.
$$dU=TdS-pdV$$
$$dS=\frac{dU}{T}+\frac{p}{T}dV$$
Therefore, gas A:
$$S=\frac{{\Delta}U}{T}+\alpha_A(\frac{-N}{{\Delta}V})$$
Gas B:
$$S=\frac{{\Delta}U}{T}+\frac{1}{\sqrt{T}}\sqrt{\beta_B}2\sqrt{N{\Delta}V}$$

Your work doesn't seem mathematically correct to me.

It seems to me that gas B would have an internal energy that depends not just on temperature.