**Cyclical process**

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

**Reversible adiabatic process.**

In an adiabatic process, no heat is exchanged between the system and the surroundings $dq_{rev}=0$ \begin{equation} \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. \begin{equation} \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} \end{equation}

**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: \begin{equation} \Delta S=\int_{1}^{2}\frac{dq_{ rev}}{T}=\frac{1}{T}\int_{1}^{2}dq_{rev}=\frac{q_{rev}}{T} \end{equation}

**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. \begin{equation} \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: \begin{equation} \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: $dq_{rev}=C_pdT$

\begin{equation} \Delta S=\int_{T_1}^{T_1}\frac{C_p}{T}dT \end{equation}

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

\begin{equation} \Delta S=C_p\int_{T_1}^{T_2}\frac{dT}{T}=C_pln\frac{T_2}{T_1} \end{equation}