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 $D A$, $$\begin{array}{rlrl} & & p_A V_A & =p_D V_D \\ \therefore & & T_A & =T_D \\ \text{ or } & \Delta U & =0\end{array}$$ Further, volume of gas is decreasing. Therefore, work is done on the gas or work done by gas is negative. Therefore, $Q is negative or heat is lost. (iii) This question covers almost all the concepts of first law of thermodynamics.$