From statistical mechanical theory, a simple model for a hypothetical hard-sphere liquid (spherical molecules of finite size without attractive intermolecular forces) gives the following expression for the Helmholtz free energy $A$ with its natural variables $T$, $V$, and $n$ as the independent variables:
$$ A = -nRT \ln{\left(cT^{\frac{3}{2}}\left( \dfrac{V}{b}-b \right)\right)}-nRT+na\text{.} $$
To calculate $C_P$ from Helmholtz free Energy, I followed the following approach:
Step 1: Calculate $S$ $$ \begin{align} S &= -\left( \dfrac{\partial A}{\partial T} \right)_{V,n}\\ &= nR \ln{\left(cT^{\frac{3}{2}}\left( \dfrac{V}{b}-b \right)\right)} + \dfrac{5}{2}nR\text{.} \end{align} $$
Step 2: Calculate $U$ $$ \begin{align} U &= A + TS\\ &= \dfrac{3}{2}nRT + na\text{.} \end{align} $$
Step 3: Calculate $C_V$ $$ \begin{align} C_V &= \left( \dfrac{\partial U}{\partial T} \right)_{V,n}\\ &= \dfrac{3}{2}nR\text{.} \end{align} $$
Step 4: Calculate $C_P$ $$ \begin{align} C_P &= C_V + nR\\ &= \dfrac{5}{2}nR \text{.} \end{align} $$
Is there a direct method to obtain $C_P$ from $A$.