Explain reversible and irreversible processes. Describe the working of carnot engine. Obtain an expression for the efficiency?

Reversible process : A process that can be retraced back in the opposite direction in such a way that the system passes through the same states as in the direct process and finally the system and surroundings return to their original states is called a reversible process.
Examples :
1) Peltier effect and seebeck effect
2) Fusion of ice and vaporization of water
Irreversible process : A process that cannot be retraced back in the opposite direction is called as irreversible process.
Examples : 1) Work done against friction
2) Diffusion of gases
3) Joule heating
Carnot engine: A reversible heat engine operating between two temperatures is called a carnot engine. The cycle operating it is known as carnot cycle. Carnot cycle contains the following four steps

1) In isothermal expansion from A to B
$\text{ Work done }{\mathrm{W}}_1={\mathrm{Q}}_1={\mathrm{RT}}_1\log \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$
2) In adiabatic expansion from B to C
$\text{ Work done }{\mathrm{W}}_2=\frac{\mathrm{R}\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)}{\mathrm{\gamma }-1} ...\left(2\right)$
3) In isothermal compression from C to D
Workdone W₃
$={\mathrm{Q}}_2={\mathrm{RT}}_2\log \left(\frac{{\mathrm{V}}_4}{{\mathrm{V}}_3}\right)=-{\mathrm{RT}}_2\log \left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_4}\right). ...\left(3\right)$
4) In adiabatic compression from D to A
${\mathrm{W}}_4=\frac{\mathrm{R}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)}{\mathrm{\gamma }-1}=\frac{-\mathrm{R}\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)}{\mathrm{\gamma }-1} ...\left(4\right)$
* In adiabatic expansion from B to C
${\mathrm{T}}_1{\mathrm{V}}_2^{\mathrm{\gamma }-1}={\mathrm{T}}_2{\left({\mathrm{V}}_3\right)}^{\mathrm{\gamma }-1} ...\left(5\right)$
* In adiabatic compression from D to A
$\begin{array}{l}{\mathrm{T}}_1{\left({\mathrm{V}}_1\right)}^{\mathrm{\gamma }-1}={\mathrm{T}}_2{\left({\mathrm{V}}_4\right)}^{\mathrm{\gamma }-1 } ...\left(6\right)\\ \frac{5}{6}\Rightarrow {\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)}^{\mathrm{\gamma }-1}={\left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_4}\right)}^{\mathrm{\gamma }-1}\\ \Rightarrow \frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{{\mathrm{V}}_3}{{\mathrm{V}}_4} ...\left(7\right)\end{array}$
From (3) & (7)
${\mathrm{W}}_3=-{\mathrm{RT}}_2\log \left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_4}\right)=-{\mathrm{RT}}_2\log \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$
Total workdone $\left(\mathrm{W}\right)={\mathrm{W}}_1+{\mathrm{W}}_2+{\mathrm{W}}_3+{\mathrm{W}}_4$
$\mathrm{W}=\mathrm{R}\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)\log \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$
$\mathrm{Efficiency}\left(\mathrm{\eta }\right)=\frac{\mathrm{W}}{{\mathrm{Q}}_1}=\frac{\mathrm{R}\left({\mathrm{T}}_1-{\mathrm{T}}_2\right){\log }_{\mathrm{e}}\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)}{{\mathrm{RT}}_1{\log }_{\mathrm{e}}\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)}$
$\begin{array}{l}\mathrm{\eta }=\frac{{\mathrm{T}}_1-{\mathrm{T}}_2}{{\mathrm{T}}_1}\\ \mathrm{\eta }=1-\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\end{array}$