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}

We use cookies

We use cookies on our website. Some of them are essential for the operation of the site, while others help us to improve this site and the user experience (tracking cookies). You can decide for yourself whether you want to allow cookies or not. Please note that if you reject them, you may not be able to use all the functionalities of the site.