$$\frac{{{T_1} - {T_2}}}{{\;{T_1}}}$$  is maximum in case (B).

$$\eqalign{ & \Delta Q = \Delta U + \Delta W \cr & \Rightarrow \Delta W = \Delta Q - \Delta U = 110 - 40 = 70\,J \cr} $$

According to question,
Net work done = Area enclosed in $$pV$$ curve i.e. $$\Delta ABC$$
Area of $$\Delta ABC = \frac{1}{2} \times 5 \times {10^{ - 3}} \times 4 \times {10^5}\,J$$
$$ = {10^3}J = 1000\,J$$

$$\eqalign{ & - 20 = \Delta U + 50 \Rightarrow \Delta U = - 70. \cr & {\text{From}}\,\,2 \to 1,\,\Delta U = 70\,kJ. \cr & {\text{Now}}\,\,10 = 70 + W \Rightarrow W = - 60\,kJ. \cr} $$

Work done in the cyclic process = Area bounded by the closed configuration = Area of closed configuration
$$ = \frac{1}{2} \times {V_0} \times {p_0} - \frac{1}{2} \times {V_0} \times {p_0} = 0\left( {{\text{zero}}} \right).$$

In cyclic process, change in total internal energy is zero.
$$\eqalign{ & \Delta {U_{{\text{cyclic}}}} = 0 \cr & \Delta {U_{BC}} = n{C_v}\,\Delta T \cr & = 1 \times \frac{{5\,R}}{2}\,\Delta T \cr} $$
Where, $${C_v}$$ = molar specific heat at constant volume.
For $$BC,\,\Delta T = - 200\,K$$
$$\therefore \,\,\Delta {U_{BC}} = - 500\,R$$

As we know,
$$\eqalign{ & \Delta Q = \Delta U + \Delta W\,\,\left( {{{\text{1}}^{{\text{st}}}}{\text{ law of thermodynamics}}} \right) \cr & \Rightarrow \Delta Q = \Delta U + P\Delta V \cr & {\text{or}}\,\,150 = \Delta U + 100\left( {1 - 2} \right) = \Delta U - 100 \cr & \therefore \Delta U = 150 + 100 = 250\,J \cr} $$
Thus the internal energy of the gas increases by $$250\,J$$

For an adiabatic process, there should not be any exchange of heat between the system and its surroundings. All walls of the container must be perfectly insulated. In adiabatic changes, gases obey Poisson's law, i.e., $$p{V^\gamma } = {\text{constant}}.$$    In an isochoric process, volume remains constant and for isobaric process, pressure remains constant.

$$\eqalign{ & {P^3}{V^5} = {\text{constant}} \cr & \Rightarrow \,\,P{V^{\frac{5}{3}}} = {\text{constant}} \cr & \Rightarrow \,\,\gamma = \frac{5}{3} \cr} $$
⇒ monoatomic gas
For adiabatic process

$$\eqalign{ & W = \frac{{{P_f}{V_f} - {P_i}{V_i}}}{{1 - \gamma }} \cr & = \frac{{\frac{1}{{32}} \times {{10}^5} \times 8 \times {{10}^{ - 3}} - {{10}^5} \times {{10}^{ - 3}}}}{{1 - \frac{5}{3}}} \cr & \therefore \,\,W = \frac{{25 - 100}}{{\frac{{\left( {3 - 5} \right)}}{3}}} \cr & = \frac{{75 \times 3}}{2} \cr & = 112.5\,J \cr} $$
From first law of thermodynamics $$q = \Delta U + w$$
$$\eqalign{ & \therefore \,\,\Delta U = - w \cr & \therefore \,\,\Delta U = - 112.5\,J \cr} $$
Now applying first law of thermodynamics for process 1 & 2 and adding $${q_1} + {q_2} = \Delta U + {P_i}\left( {{V_f} - {V_i}} \right)$$
$$\eqalign{ & = - 112.5 + {10^5}\left( {8 - 1} \right) \times {10^{ - 3}} \cr & = 587.55 \cr} $$

Inisothermal process temperature remains constant.
i.e., $$\Delta T = 0.$$   Hence according to
$$C = \frac{Q}{{m\Delta T}} \Rightarrow {C_{_{{\text{iso}}}}} = \infty $$