Consider the diagram p-V, where variation is shown for each process.

Process 1 is isobaric and process 2 is isothermal. Since, work done $=$ area under the $p-V$ curve. Here, area under the $p-V$ curve 1 is more. So, work done is more when the gas expands in isobaric process.

Let $p V^{1 / 2}=$ Constant $=K, p=\frac{K}{\sqrt{V}}$

(a) Work done for the process 1 to 2 ,

$$\begin{aligned} W & =\int_{V_1}^{V_2} p d V=K \int_{V_1}^{V_2} \frac{d V}{\sqrt{V}}=K\left[\frac{\sqrt{V}}{1 / 2}\right]_{V_1}^{V_2}=2 K\left(\sqrt{V_2}-\sqrt{V_1}\right) \\ & =2 p_1 V_1^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right)=2 p_2 V_2^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right) \end{aligned}$$

(b) From ideal gas equation,

$$\begin{aligned} &\begin{aligned} p V & =n R T \quad \Rightarrow \quad T=\frac{p V}{n R}=\frac{p \sqrt{V} \sqrt{V}}{n R} \\ T & =\frac{K \sqrt{V}}{n R} \quad (\mathrm{As}, p \sqrt{V}=K) \end{aligned}\\ \end{aligned}$$

$$\text{Hence,}\quad T_1=\frac{K \sqrt{V_1}}{n R} \Rightarrow T_2=\frac{K \sqrt{V_2}}{n R}$$

$$\Rightarrow \quad \frac{T_1}{T_2}=\frac{\frac{K \sqrt{V_1}}{n R}}{\frac{K \sqrt{V_2}}{n R}}=\sqrt{\frac{V_1}{V_2}}=\sqrt{\frac{V_1}{2 V_1}}=\frac{1}{\sqrt{2}} \quad\left(\because V_2=2 V_1\right)$$

$$\begin{aligned} &\text { (c) Given, internal energy of the gas }=U=\left(\frac{3}{2}\right) R T\\ &\begin{aligned} \Delta U & =U_2-U_1=\frac{3}{2} R\left(T_2-T_1\right) \\ & =\frac{3}{2} R T_1(\sqrt{V}-1) \end{aligned}\\ &\left[\because T_2=\sqrt{2} T_1 \text { from }(\mathrm{b})\right] \end{aligned}$$

$$ \begin{aligned} \Delta W & =2 p_1 V_1^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right) \\ & =2 p_1 V_1^{1 / 2}\left(\sqrt{2} \times \sqrt{V_1}-\sqrt{V_1}\right) \\ & =2 p_1 V_1(\sqrt{2}-1)=2 R T_1(\sqrt{2}-1) \\ \therefore & \quad \Delta Q=\Delta U+\Delta W \\ & =\frac{3}{2} R T_1(\sqrt{2}-1)+2 R T_1(\sqrt{2}-1) \\ & =(\sqrt{2}-1) R T_1(2+3 / 2) \\ & =\left(\frac{7}{2}\right) R T_1(\sqrt{2}-1) \end{aligned}$$

This is the amount of heat supplied.

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 $f^{\prime}\left(V_0\right)>\left(-\gamma \frac{p_0}{V_0}\right)$