diff --git a/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md b/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md
index c365e38..894bc36 100755
--- a/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md
+++ b/uni/mmme/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md
@@ -40,7 +40,7 @@ The system is not in equilibrium if parts of the system are at different conditi
 
 #### Adiabatic
 
-A process in which does not cross the system boundary
+A process in which heat does not cross the system boundary
 
 ## Perfect (Ideal) Gasses
 
@@ -204,6 +204,41 @@ Entropy is defined as the following, given that the process s reversible:
 
 $$S_2 - S_1 = \int\! \frac{\mathrm{d}Q}{T}$$
 
+#### Change of Entropy of a Perfect Gas
+
+Consider the 1st corollary of the 1st law:
+
+$$\mathrm dq + \mathrm dw = \mathrm du$$
+
+and that the process is reversible:
+
+\begin{align*}
+\mathrm ds &= \frac{\mathrm dq} T \bigg|_{rev} \\
+\mathrm dq = \mathrm ds \times T \\
+\mathrm dw &= -p\mathrm dv \\
+\end{align*}
+
+The application of the 1st corollary leads to:
+
+$$T\mathrm ds - p\mathrm dv = \mathrm du$$
+
+Derive the change of entropy
+
+\begin{align*}
+\mathrm ds &= \frac{\mathrm du}{T} + \frac{p \mathrm dv}{T} \\
+\\
+\mathrm du &= c_v \mathrm{d}T \\
+\frac p T &= \frac R v \\
+\\
+\mathrm ds &= \frac{c_v\mathrm{d}T}{T} \frac{R\mathrm dv}{v} \\
+s_2 - s_1 &= c_v\ln\left(\frac{T_2}{T_1}\right) + R\ln\left(\frac{v_2}{v_1}\right)
+\end{align*}
+
+There are two other forms of the equation that can be derived:
+
+$$s_2 - s_1 = c_v\ln\left(\frac{p_2}{p_1}\right) + c_p\ln\left(\frac{v_2}{v_1}\right)$$
+$$s_2 - s_1 = c_p\ln\left(\frac{T_2}{T_1}\right) - R\ln\left(\frac{p_2}{p_1}\right)$$
+
 ### Heat Capacity and Specific Heat Capacity
 
 Heat capacity is quantity of heat required to raise the temperature of a system by a unit
@@ -375,3 +410,125 @@ The 1st Law of Thermodynamics can be thought of as:
 
 > The internal energy of a closed system remains unchanged if it
 > [thermally isolated](#thermally-insulated-and-isolated-systems) from its surroundings
+
+# Polytropic Processes
+
+A polytropic process is one which obeys the polytropic law:
+
+$$pv^n = k \text{ or } p_1v_1^n = p_2v_2^n$$
+
+where $n$ is a constant called the polytropic index, and $k$ is a constant too.
+
+A typical polytropic index is between 1 and 1.7.
+
+<details>
+<summary>
+
+#### Example 1
+
+Derive
+
+$$\frac{p_2}{p_1} = \left(\frac{T_2}{T_1}\right)^{\frac{n}{n-1}}$$
+
+</summary>
+
+\begin{align*}
+p_1v_1^n &= p_2v_2^n \\
+pv &= RT \rightarrow v = R \frac{T}{p} \\
+\frac{p_2}{p_1} &= \left( \frac{v_1}{v_2} \right)^n \\
+                &= \left(\frac{p_2T_1}{T_2p_1}\right)^n \\
+                &= \left(\frac{p_2}{p_1}\right)^n \left(\frac{T_1}{T_2}\right)^n \\
+\left(\frac{p_2}{p_1}\right)^{1-n} &= \left(\frac{T_1}{T_2}\right)^n \\
+\frac{p_2}{p_1} &= \left(\frac{T_1}{T_2}\right)^{\frac{n}{1-n}} \\
+                &= \left(\frac{T_2}{T_1}\right)^{\frac{n}{n-1}} \\
+\end{align*}
+
+<details>
+<summary>
+
+How did you do that last step?
+
+</summary>
+
+For any values of $x$ and $y$
+
+\begin{align*}
+\frac x y &= \left(\frac y x \right) ^{-1} \\
+\left(\frac x y \right)^n &= \left(\frac y x \right)^{-n} \\
+\left(\frac x y \right)^{\frac{n}{1-n}} &= \left(\frac y x \right)^{\frac{-n}{1-n}} \\
+ &= \left(\frac y x \right)^{\frac{n}{n-1}} \\
+\end{align*}
+
+</details>
+</details>
+
+# Isentropic
+
+*Isentropic* means constant entropy:
+
+$$\Delta S = 0 \text{ or } s_1 = s_2 \text{ for a precess 1-2}$$
+
+A process will be isentropic when:
+
+$$pv^\gamma = \text{constant}$$
+
+This is basically the [polytropic law](#polytropic-processes) where the polytropic index, $n$, is
+always equal to $\gamma$.
+
+<details>
+<summary>
+
+Derivation
+
+</summary>
+
+\begin{align*}
+0 &= s_2 - s_1 = c_v \ln{\left(\frac{p_2}{p_1}\right)} + c_p \ln{\left( \frac{v_2}{v_1} \right)} \\
+0 &= \ln{\left(\frac{p_2}{p_1}\right)} + \frac{c_p}{c_v} \ln{\left( \frac{v_2}{v_1} \right)} \\
+  &= \ln{\left(\frac{p_2}{p_1}\right)} + \gamma \ln{\left( \frac{v_2}{v_1} \right)} \\
+  &= \ln{\left(\frac{p_2}{p_1}\right)} + \ln{\left( \frac{v_2}{v_1} \right)^\gamma} \\
+  &= \ln{\left[\left(\frac{p_2}{p_1}\right)\left( \frac{v_2}{v_1} \right)^\gamma\right]} \\
+ e^0  = 1 &= \left(\frac{p_2}{p_1}\right)\left( \frac{v_2}{v_1} \right)^\gamma \\
+&\therefore p_2v_2^\gamma = p_1v_1^\gamma \\
+\\
+pv^\gamma = \text{constant}
+\end{align*}
+
+</details>
+
+During isentropic processes, it is assumed that no heat is transferred into or out of the cylinder.
+It is also assumed that friction is 0 between the piston and cylinder and that there are no energy
+losses of any kind.
+
+This results in a reversible process in which the entropy of the system remains constant.
+
+An isentropic process is an idealization of an actual process, and serves as the limiting case for
+real life processes.
+They are often desired and often the processes on which device efficiencies are calculated.
+
+## Heat Transfer During Isentropic Processes
+
+Assume that the compression 1-2 follows a polytropic process with a polytropic index $n$.
+The work transfer is:
+
+$$W  = - \frac{1}{1-n} (p_2V_2 - p_1V_1) = \frac{mR}{n-1} (T_2-T_1)$$
+
+Considering the 1st corollary of the 1st law, $Q + W = \Delta U$, and assuming the gas is an ideal
+gas [we know that](#obey-the-perfect-gas-equation) $\Delta U = mc_v(T_2-T_1)$ we can deduce:
+
+\begin{align*}
+Q &= \Delta U - W = mc_v(T_2-T_1) - \frac{mR}{n-1} (T_2-T_1) \\
+  &= m \left(c_v - \frac R {n-1}\right)(T_2-T-1)
+\end{align*}
+
+Now, if the process was *isentropic* and not *polytropic*, we can simply substitute $n$ for
+$\gamma$ so now:
+
+$$Q = m \left(c_v - \frac R {\gamma-1}\right)(T_2-T-1)$$
+
+But since [we know](#relationship-between-specific-gas-constant-and-specific-heats) $c_v = \frac R {\gamma - 1}$:
+
+$$Q = m (c_v-c_v)(T_2-T_1) = 0 $$
+
+This proves that the isentropic version of the process adiabatic (no heat is transferred across the
+boundary).