For a simple compressible system v, s, p and T are specific volume, specific entropy, pressure and temperature, respectively. As per Maxwell’s relations, (∂v /∂s)_p is equal to

To find the thermodynamic relation equal to ${\left(\frac{\partial v}{\partial s}\right)}_p$, we can derive it from the fundamental thermodynamic relations and Maxwell's relations.

Let's start with the definition of enthalpy, $h$:
$h=u+pv$
where $u$ is internal energy, $p$ is pressure, and $v$ is specific volume.

The differential form of enthalpy is given by:
$dh=Tds+vdp$
where $T$ is temperature and $s$ is specific entropy.

Since enthalpy $h$ is a state function, its differential $dh$ is an exact differential of the form:
$dh={\left(\frac{\partial h}{\partial s}\right)}_{p}ds+{\left(\frac{\partial h}{\partial p}\right)}_{s}dp$

Comparing the coefficients of $ds$ and $dp$ in the two equations for $dh$, we get:
${\left(\frac{\partial h}{\partial s}\right)}_p=T$
and
${\left(\frac{\partial h}{\partial p}\right)}_s=v$

According to Euler's reciprocity relation for an exact differential of the form $dz=Mdx+Ndy$, the mixed second-order partial derivatives are equal:
${\left(\frac{\partial M}{\partial y}\right)}_x={\left(\frac{\partial N}{\partial x}\right)}_y$

Applying this mathematical theorem to $dh$, we obtain:
${\left(\frac{\partial T}{\partial p}\right)}_s={\left(\frac{\partial v}{\partial s}\right)}_p$

Therefore, as per Maxwell's relations, the term ${\left(\frac{\partial v}{\partial s}\right)}_p$ is equal to ${\left(\frac{\partial T}{\partial p}\right)}_s$.