**Dependence of S on T**

Dividing the ratio $dS=dq/T$ by dT and keeping the pressure constant gives the desired ratio.

\begin{equation} \left(\frac{\partial S}{\partial P}\right)_T=\frac{C_P}{T} \end{equation}

**Dependence of S with respect to P**

Using the Gibbs equation $dG=-SdT-VdP$ and one of Maxwell's relations, we obtain:

\begin{equation} \left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P=-\alpha V \end{equation}