For a simple compressible system, $v, s, p$ and $T$ are specific volume, specific entropy, pressure and temperature, respectively. As per Maxwell's relations, $\bigg( \dfrac{\partial v}{\partial s} \bigg) _p$ is equal to $\bigg( \dfrac{\partial s}{\partial T} \bigg) _p \\$ ... $ - \bigg( \dfrac{\partial T}{\partial v} \bigg) _p \\$ $\bigg( \dfrac{\partial T}{\partial p} \bigg) _s$

asked
Feb 9, 2019
in Thermodynamics
Arjun
24.6k points