Step 1: Define Boyle's Temperature (\(T_B\)). Boyle's temperature is the specific temperature at which a real gas behaves most like an ideal gas over a range of pressures. At this temperature, the effects of attractive forces and repulsive forces approximately cancel each other out.
Step 2: Analyze the behavior of a real gas below Boyle's Temperature. Below \(T_B\), the attractive forces between gas molecules are dominant at moderate pressures. An increase in pressure makes it easier for molecules to come together, causing the volume to decrease more than it would for an ideal gas. This makes the product PV decrease initially. At very high pressures, repulsive forces become dominant, causing the volume to be larger than for an ideal gas, and the PV product increases. Thus, the PV vs P curve shows a dip (minimum).
Step 3: Analyze the behavior of a real gas above Boyle's Temperature. Above \(T_B\), the kinetic energy of the molecules is high, and the repulsive forces (due to molecular size) are dominant at all pressures. An increase in pressure results in a PV value that is always greater than that of an ideal gas, and PV continuously increases with P.
Step 4: Evaluate the options. (A) Incorrect. PV first decreases, then increases below \(T_B\). (B) Incorrect. PV continuously increases above \(T_B\). (C) Correct. This accurately describes the behavior below Boyle's temperature, with an initial decrease followed by an increase. (D) Incorrect. PV is constant for an ideal gas, not a real gas above \(T_B\).
Match List - I with List - II.

1.24 g of $ {AX}_2 $ (molar mass 124 g mol$^{-1}$) is dissolved in 1 kg of water to form a solution with boiling point of 100.105$^\circ$C, while 2.54 g of $ {AY}_2 $ (molar mass 250 g mol$^{-1}$) in 2 kg of water constitutes a solution with a boiling point of 100.026$^\circ$C. $ K_{b(H_2O)} = 0.52 \, \text{K kg mol}^{-1} $. Which of the following is correct?
For the reaction:

The correct order of set of reagents for the above conversion is :