For the process AB,

$$\begin{aligned} &\begin{aligned} & d V=0 \Rightarrow d W=0 \quad (\because \text { volume is constant) }\\ & d Q=d U+d W=d U \\ \Rightarrow \quad & d Q=d U=\text { Change in internal energy. } \end{aligned}\\ \end{aligned}$$

Hence, in this process heat supplied is utilised to increase, internal energy of the system.

Since, $p=\left(\frac{n R}{V}\right) T$, in isochoric process, $T \propto p$. So temperature increases with increases of pressure in process $A B$ which inturn increases internal energy of the system i.e., $d U>0$. This imply that $d Q>0$. So heat is supplied to the system in process $A B$.

(b) For the process $C D$, volume is constant but pressure decreases.

Hence, temperature also decreases so heat is given to surroundings.

(c) To calculate work done by the engine in one cycle, we calculate work done in each part separately.

$$\begin{aligned} W_{A B} & =\int_A^B p d V=0, W_{C D}=\int_{V_C}^{V_D} p d V=0 \quad (\because dV=0)\\ W_{B C} & =\int_{V_B}^{V_C} p d V=k \int_{V_B}^{V_C} \frac{d V}{V^\gamma}=\frac{k}{1-\gamma}\left[V^{1-\gamma}\right]_{V_B}^{V_C} \\ & =\frac{1}{1-\gamma}[p V]_{V_B}^{V_C}=\frac{\left(p_C V_C-p_B V_B\right)}{1-\gamma} \\ \text{Similarly,}\quad W_{D A} & =\frac{p_A V_A-p_D V_D}{1-\gamma} \quad [\because \text{BC is adiabatic process}] \end{aligned}$$

$$\because$$ B and C lies on adiabatic curve BC.

$$\begin{aligned} \therefore \quad p_B V_B^\gamma & =p_C V_C^\gamma \\ p_C & =p_B\left(\frac{V_B}{V_C}\right)^\gamma=p_B\left(\frac{1}{2}\right)^\gamma=2^{-\gamma} p_B \end{aligned}$$

Similarly, $$p_D=2^{-\gamma} p_A$$

$$\begin{aligned} &\text { Total work done by the engine in one cycle } A B C D A \text {. }\\ &\begin{aligned} W & =W_{A B}+W_{B C}+W_{C D}+W_{D A}=W_{B C}+W_{D A} \\ & =\frac{\left(p_C V_C-p_B V_B\right)}{1-\gamma}+\frac{\left(p_A V_A-p_D V_D\right)}{1-\gamma} \\ W & =\frac{1}{1-\gamma}\left[2^{-\gamma} p_B\left(2 V_B\right)-p_B V_B+p_A V_A-2^{-\gamma} p_B\left(2 V_B\right)\right] \\ & =\frac{1}{1-\gamma}\left[p_B V_B\left(2^{-\gamma+1}-1\right)-p_A V_A\left(2^{-\gamma+1}-1\right)\right. \\ & =\frac{1}{1-\gamma}\left(2^{1-\gamma}-1\right)\left(p_B-p_A\right) V_A \\ & =\frac{3}{2}\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]\left(p_B-p_A\right) V_A \end{aligned} \end{aligned}$$

(a) For process $A B$,

Volume is constant, hence work done $d W=0$

Now, by first law of thermodynamics,

$$\begin{aligned} & d Q=d U+d W=d U+0=d U \\ & =n C V d T=n C V\left(T_B-T_A\right) \\ & =\frac{3}{2} R\left(T_B-T_A\right) \quad (\because n=1)\\ & =\frac{3}{2}\left(R T_B-R T_A\right)=\frac{3}{2}\left(p_B V_B-p_A V_A\right) \\ & \text { Heat exchanged }=\frac{3}{2}\left(p_B V_B-p_A V_A\right) \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { (b) For process BC,}\quad p & =\text { constant } \\ d Q & =d U+d W=\frac{3}{2} R\left(T_C-T_B\right)+p_B\left(V_C-V_B\right) \\ & =\frac{3}{2}\left(p_C V_C-p_B V_B\right)+p_B\left(V_C-V_B\right)=\frac{5}{2} p_B\left(V_C-V_B\right) \end{aligned} \end{aligned}$$

Heat exchanged $=\frac{5}{2} p_B\left(V_C-V_A\right) \quad \left(\because p_B=p_C\right.$ and $\left.p_B=V_A\right)$

(c) For process $C D$, Because $C D$ is adiabatic, $d Q=$ Heat exchanged $=0$

(d) $D A$ involves compression of gas from $V_D$ to $V_A$ at constant pressure $p_A$.

$\therefore$ Heat exchanged can be calculated by similar way as $B C_1$,

Hence, $$d Q=\frac{5}{2} p_A\left(V_A-V_D\right)$$

According to question, slope of the curve $=f(V)$, where $V$ is volume .

$$\begin{array}{rr} \therefore \quad & \text { Slope of } p=f(V) \text { curve at }\left(V_0, p_0\right)=f\left(V_0\right) \\ & \text { Slope of adiabatic at }\left(V_0, p_0\right)=k(-\gamma) V_0^{-1-\gamma}=-\gamma p_0 / V_0 \end{array}$$

Now heat absorbed in the process $p=f(V)$

$$d Q=d U+d W=n C_V d T+p d V\quad \text{... (i)}$$

$$\because \quad p V=n R T \Rightarrow T=\left(\frac{1}{n R}\right) p V$$

$$\begin{array}{ll} \Rightarrow & T=\left(\frac{1}{n R}\right) V f(V) \\ \Rightarrow & d T=\left(\frac{1}{n R}\right)\left[f(V)+V f^{\prime}(V)\right] d V\quad \text{... (ii)} \end{array}$$

$$\begin{aligned} &\text { Now from Eq. (i) }\\ &\begin{aligned} \frac{d Q}{d V} & =n C_V \frac{d T}{d V}+p \frac{d V}{d V}=n C_V \frac{d T}{d V}+p \\ & =\frac{n C_V}{n R} \times\left[f(V)+V f^{\prime}(V)\right]+p \quad \text{[from Eq. (ii)]}\\ & =\frac{C_V}{R}\left[f(V)+V f^{\prime}(V)\right]+f(V) \quad [\because p=f(V)] \end{aligned} \end{aligned}$$

$$\left.\Rightarrow \quad\left[\frac{d Q}{d V}\right]_{V=V_0}=\frac{C_V}{R}\left[f\left(V_0\right)+V_0 f^{\prime}\right]\left(V_0\right)\right]+f\left(V_0\right)$$

$$=f\left(V_0\right)\left[\frac{C_V}{R}+1\right]+V_0 f^{\prime}\left(V_0\right) \frac{C_V}{R}$$

$$\begin{aligned} \because \quad C_V & =\frac{R}{\gamma-1} \Rightarrow \frac{C_V}{R}=\frac{1}{\gamma-1} \\ \Rightarrow \quad\left[\frac{d Q}{d v}\right]_{V=V_0} & =\left[\frac{1}{\gamma-1}+1\right] f\left(V_0\right)+\frac{V_0 f^{\prime}\left(V_0\right)}{\gamma-1} \\ & =\frac{\gamma}{\gamma-1} p_0+\frac{V_0}{\gamma-1} f^{\prime}\left(V_0\right) \end{aligned}$$

Heat is absorbed where $\frac{d Q}{d V}>0$, when gas expands

Hence, $\gamma p_0+V_0 f^{\prime}\left(V_0\right)>0 \quad$ or $f^{\prime}\left(V_0\right)>\left(-\gamma \frac{p_0}{V_0}\right)$

$$\text { (a) Initially the piston is in equilibrium hence, } p_i=p_a$$

(b) On supplying heat, the gas expands from $V_0$ to $V_1$

$\therefore$ Increase in volume of the gas $=V_1-V_0$

As the piston is of unit cross-sectional area hence, extension in the spring

$$x=\frac{V_1-V_0}{\text { Area }}=V_1-V_0 \quad \text{[Area=1]}$$

$\therefore$ Force exerted by the spring on the piston

$$=F=k x=k\left(V_1-V_0\right)$$

Hence,

$$\begin{aligned} \text { final pressure } & =p_f=p_a+k x \\ & =p_a+k \times\left(V_1-V_0\right) \end{aligned}$$

(c) From first law of thermodynamics $d Q=d U+d W$

If $T$ is final temperature of the gas, then increase in internal energy

$$\begin{aligned} d U & =C_V\left(T,-T_0\right)=C_V\left(T,-T_0\right) \\ \text{We can write,} \quad T & =\frac{p_t V_1}{R}=\left[\frac{p_a+k\left(V_1-V_0\right)}{R}\right] \frac{V_1}{R} \end{aligned}$$

Work done by the gas $=p d V+$ increase in PE of the spring

$$=p_a\left(V_1-V_0\right)+\frac{1}{2} k x^2$$

Now, we can write $d Q=d U+d W$

$$\begin{aligned} & =C_V\left(T-T_0\right)+p_a\left(V-V_0\right)+\frac{1}{2} k x^2 \\ & =C_V\left(T-T_0\right)+p_a\left(V-V_0\right)+\frac{1}{2}\left(V_1-V_0\right)^2 \end{aligned}$$

This is the required relation.