cyclical process

The entropy change in a cyclic process is zero. The initial and final states coincide. $$\Delta S=\oint\frac{dq}{T}=0$$

In an adiabatic process, no heat is exchanged between the system and the surroundings $dq_{rev}=0$ \Delta S=\int\frac{dq_{rev}}{T}=0 \end{equation }

Reversible phase shift at constant T and P.

A phase change takes place at constant temperature and pressure. The temperature is outside the integral and the heat exchanged in the phase change is equal to the enthalpy change. $$\Delta S=\int_{1}^{2}\frac{dq_{rev}}{T}=\frac{1}{T}\int_{1}^{2}dq_{rev }=\frac{q_{rev}}{T}=\frac{\Delta H}{T}$$

Isothermal reversible process.

The temperature remains constant during this process and the enthalpy change when going from state 1 to state 2 is given by the following equation: $$\Delta S=\int_{1}^{2}\frac{dq_{ rev}}{T}=\frac{1}{T}\int_{1}^{2}dq_{rev}=\frac{q_{rev}}{T}$$

Reversible change of state in an ideal gas.

In this case none of the thermodynamic variables is held constant. We will use the equation of state to relate them and the first principle to obtain the heat exchanged. \begin{eqnarray} dU & = & dq_{rev}-PdV\\ P & = & \frac{nRT}{V}\\ dq_{rev} & = & dU+nRT\frac{dV}{V}= C_vdT+nRT\frac{dV}{V} \end{eqnarray} Dividing the last equation by T and integrating gives the entropy change. \Delta S=\int_{1}^{2}C_v\frac{dT}{T}+nR\int_{1}^{2}\frac{dV}{V} \end{equation } $C_v$ in general depends on T for perfect gases. If we consider the particular case in which it is constant, we can write the above equation as follows: \Delta S=C_vln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1} \end{equation }

Constant pressure heating.

At constant pressure the heat exchanged is given by the equation: $dq_{rev}=C_pdT$

$$\Delta S=\int_{T_1}^{T_1}\frac{C_p}{T}dT$$

If $C_p$ is constant in that temperature interval, it goes out of the integral.

$$\Delta S=C_p\int_{T_1}^{T_2}\frac{dT}{T}=C_pln\frac{T_2}{T_1}$$