One mole of a monatomic ideal gas is taken through a cycle A B C D...

Internal energy $\propto T \propto p V$ This is because U=$\frac{n f}{2}$R T=$\frac{f}{2}$p V Here, $n=$ number of moles $f=$ degree of freedom If the product $p V$ increases, then internal energy will increase and if product decreases, the internal energy will decrease. Further, work is done on the gas, if volume of gas decreases. For heat exchange. Q=W+$\Delta U Work done is area under$p-V$graph. If volume increases work done by gas is positive and if volume decreases work done by gas is negative. Further$\Delta U$is positive if product of$p V$is increasing and$N U$is negative, if product of$p V$is decreasing. If heat is taken by the gas$Q$is positive and if heat is lost by the gas$Q$is negative. Keeping the above points in mind the answer to this question is as under. (A)$\rightarrow($\mathrm{p}$, $\mathrm{r}$, $\mathrm{t}$)$(B)$\rightarrow($\mathrm{p}$, $\mathrm{r}$)$(C)$\rightarrow ($q, s) (D)$\rightarrow($\mathrm{r}$, $\mathrm{t}$)$Analysis of Question (i) Calculation wise, question is slightly lengthy. Otherwise question is theory based and simple. (ii) In process$D A$,$$\begin{array}{rlrl} & & p_A V_A & =p_D V_D \\ \therefore & & T_A & =T_D \\ \text{\text{ or } } & \Delta U & =0\end{array}$$Further, volume of gas is decreasing. Therefore, work is done on the gas or work done by gas is negative. Therefore, Q is negative or heat is lost. (iii) This question covers almost all the concepts of first law of thermodynamics.