Question

For an ideal monoatomic gas, match the following graphs for constant mass in different process.

	List-I		List-II
(P)		1.
(Q)		2.
(R)		3.
(S)		4.

• A. P-1; Q-2; R-3; S-4
• B. P-2; Q-3; R-4; S-1
• C. P-2; Q-4; R-3; S-1
• D. P-4; Q-2; R-3; S-1

Answer 1

Answer: (A)

P–1; Q–2; R–3; S–4 (option 1)

Answer 2

We can “translate” the process‐curves from one pair of variables to another using the ideal‐gas law

$$PV=nRT\quad \Longrightarrow \quad T=\frac{PV}{nR}\,,$$

or, equivalently,

$$P=\frac{nRT}{V}\quad \mbox{and}\quad \rho=\frac{m}{V}\,.$$

Thus a process given as a pressure–volume curve will appear different when the state–variable pair is changed. In the problem one is given a closed (triangular) cycle represented in four different “pictures” (with different variables) so that if the process is “translated” using the state–equation the “shape” of a given segment may become “curved” or “straight” in the new variables.

A short outline of the reasoning is as follows:

• Among the four curves in List–I the one labeled (Q) is already plotted as “$P$ versus $T$”. Its description (“two straight‐line segments and one curved”) exactly matches that of graph “2” in List–II. Hence

  $$Q\to2.$$

• The curve labeled (S) in List–I is drawn as “density ($\rho$) versus $T$”. In List–II only the graph labeled “4” is of the same variables. Hence

  $$S\to4.$$

• The curve labeled (R) is “$V$ versus $T$” with the straight‐line portions and a dashed line from the origin. When one “translates” (R) to the $P$–$T$ plane using $P=nRT/V$ the two segments which were originally straight become “curved” (because $V$ is a linear function of $T$) and only one segment remains a straight line; this matches the description of graph “3” in List–II. Thus

  $$R\to3.$$

• Finally the (P) curve in List–I is “$P$ versus $V$” where one segment is curved and the other two are straight. On converting via $T=PV/(nR)$ one obtains a $P$–$T$ representation that comes out as in graph “1” of List–II (with a dashed line from the origin as a “marker” of the transformation). Thus

  $$P\to1.$$

Thus the overall matching is

P–1; Q–2; R–3; S–4.

Among the given choices the correct answer is:

• P–1; Q–2; R–3; S–4

Explanation (minimal):

1. The process (Q) is already $P$–$T$ with two straight segments and one curved – it exactly matches graph 2.
2. The (S) curve is a $\rho$–$T$ graph; the only graph with those variables is 4.
3. Transforming the (R) $V$–$T$ cycle (with two straight parts) to $P$–$T$ by $P=nRT/V$ gives one straight segment (A–B) and two curved parts, which is graph 3.
4. Finally, (P) is the $P$–$V$ cycle; its conversion using $T=PV/(nR)$ leads to the appearance of graph 1.