Since a reversible cycles could be constructed such that $\sigma_{ab}=0,\>\sigma_{cb}=0$ then
$$ \int_1^2(\frac{\delta Q}{T})_a=\int_1^2(\frac{\delta Q}{T})_c $$
and so we can define an extensive property, entropy, whose change is given by
$$ S_2-S_1=\int_1^2(\frac{\delta Q}{T})_{rev} $$
Further, this renders a differential equation which will produce two $Tds$ equations, which serve as the basis for many of our derivations.
$$ \delta Q_{rev}=TdS $$
$$ \delta Q_{rev}=dU+\delta W_{rev}=dU+pdV $$
$$ Tds=du+pdv $$
$$ dH=dU+pdV+Vdp $$
$$ Tds=dh-vdp $$
Considering a change in phase from saturated liquid to saturated vapor
$$ p=c, T=c\Rarr ds=\frac{dh}{T} $$
$$ s_g-s_f=\frac{h_g-h_f}{T} $$
Considering an incompressible liquid
$$ du=c_TdT,\>v=c\Rarr ds=\frac{c_TdT}{T} $$
$$ s_2-s_1=\int_{T_1}^{T_2}\frac{c_T}{T}dT $$
$$ c_T=const $$
$$ s_2-s_1=c_T\ln\frac{T_2}{T_1} $$
The change in entropy of an ideal gas may be calculated using specific heats and pressures or volumes, however, these calculations are already completed, and may be used by referencing $s^o(T)$ tables. Alternatively, we may assume constant specific heats to simplify the calculations
$$ Tds=du+pdv,\>du=c_vdT,\>p=\frac{RT}{v} $$
$$ \Delta s=\int_1^2\frac{c_v(T)}{T}dT+R\ln\frac{v_2}{v_1} $$
$$ Tds=dh-vdp,\>dh=c_pdT,\>v=\frac{RT}{p} $$
$$ \Delta s=\int_1^2\frac{c_p(T)}{T}dT+R\ln\frac{p_2}{p_1} $$