One mole of a monatomic ideal gas is taken through a cycle \(A B C D A\) as shown in the \(p\)-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.

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.