A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle.${C_v} = \frac{3}{2}R$

$(a)$ $A$ to $B$ : constant volume

$(b)$ $B$ to $C$ : constant pressure

$(c)$ $C$ to $D$ : adiabatic

$(d)$ $D$ to $A$ : constant pressure

$(a)$ For process $A$ to $B$ volume is constant hence work done $d W=0$ $\Rightarrow$ From first law of thermodynamics,

$d \mathrm{Q}=d \mathrm{U}+d \mathrm{~W}$ $\therefore d \mathrm{Q}=d \mathrm{U}+0$ $\therefore d \mathrm{Q}=d \mathrm{U}$ $=n \mathrm{C}_{\mathrm{V}} d \mathrm{~T}$$=n \mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$ $=(1)\left(\frac{3}{2} \mathrm{R}\right)\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$ $=\frac{3}{2}\left(\mathrm{RT}_{\mathrm{B}}-\mathrm{RT}_{\mathrm{A}}\right)$ $=\frac{3}{2}\left(\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}-\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}\right)$ $\text { because in equation of gas } \mathrm{PV}=\mathrm{RT} \text { at point } \mathrm{A}, \mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}=\mathrm{RT}_{\mathrm{A}} \text { and at point } \mathrm{B}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{RT}_{\mathrm{B}}$ $\therefore \text { Heat exchanged }=\frac{3}{2}\left(\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}-\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}\right)$

$\therefore$ Heat exchanged $=\frac{3}{2}\left(\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}-\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}\right)$

$(b)$ For process $B$ to $C$ pressure is constant hence work done,

$d \mathrm{~W}=\mathrm{P} \Delta \mathrm{V}$ $\therefore d \mathrm{~W}=\mathrm{P}_{\mathrm{B}}\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)$ $\Rightarrow \mathrm{From} \text { first law of thermodynamics, }$ $d \mathrm{Q}=d \mathrm{U}+d \mathrm{~W}$ $=\frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{C}}-\mathrm{T}_{\mathrm{B}}\right)+\mathrm{P}_{\mathrm{B}}\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)$ $\therefore d \mathrm{Q} =\frac{3}{2}\left(\mathrm{RT}_{\mathrm{C}}-\mathrm{RT}_{\mathrm{B}}\right)+\mathrm{P}_{\mathrm{B}}\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)$

$\quad=\frac{3}{2}\left(\mathrm{P}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}-\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}\right)+\mathrm{P}_{\mathrm{B}}\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)$

$\quad=\frac{5}{2} \mathrm{P}_{\mathrm{B}}\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right) \quad\left[\because \mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{C}} \text { and } \mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{A}}\right]$

$\therefore \text { Heat exchanged }=\frac{5}{2} \mathrm{P}_{\mathrm{B}}\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}\right)$

$(c)$ For process $\mathrm{C}$ to $\mathrm{D}$, it is adiabatic process so heat exchanged $d \mathrm{Q}=0$

$(d)$ For process $D$ to $A$ at constant pressure $P_{A}$ the gas compressed from volume $V_{D}$ to $V_{A}$. The heat exchanged will be as shown in (b) $P_{A}$ instead of $P_{B}, V_{A}$ instead of $V_{C}$ and $V_{A}$ instead of $V_{D}$ have to be taken.

$\therefore$ Heat exchanged $d \mathrm{Q}=\frac{5}{2} \mathrm{P}_{\mathrm{A}}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}\right)$