Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Advanced 2026Paper 1
Exam
JEE Advanced
Year
2026
Paper
Paper 1
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2026 · Paper 1 · Chemistry · Thermodynamics (C)

List-I contains various physical/chemical processes, and List-II...

List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy (ΔH\Delta H) and entropy (ΔS\Delta S). Match each entry in List-I to the appropriate entry in List-II, and choose the correct option.
List-IList-II
(P) Physisorption(1) ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0
(Q) Diamond ⟶\longrightarrow Graphite(2) ΔH<0\Delta H < 0 and ΔS<0\Delta S < 0
(R) Denaturation of protein(3) ΔH<0\Delta H < 0 and ΔS=0\Delta S = 0
(S) Propene ⟶\longrightarrow Cyclopropane(4) ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0
(5) ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0

Options

  1. AP →\rightarrow 2; Q →\rightarrow 3; R →\rightarrow 5; S →\rightarrow 4
  2. BP →\rightarrow 4; Q →\rightarrow 3; R →\rightarrow 5; S →\rightarrow 1
  3. CP →\rightarrow 2; Q →\rightarrow 5; R →\rightarrow 1; S →\rightarrow 4
  4. DP →\rightarrow 2; Q →\rightarrow 5; R →\rightarrow 1; S →\rightarrow 3

Correct answer

C. P →\rightarrow 2; Q →\rightarrow 5; R →\rightarrow 1; S →\rightarrow 4

Step-by-step solution

(P) Physisorption is an exothermic process because attractive forces are formed between the adsorbate and the adsorbent, so ΔH<0\Delta H < 0. The gas molecules are restricted on the solid surface, decreasing randomness, so ΔS<0\Delta S < 0. This matches (2).

(Q) Diamond ⟶\longrightarrow Graphite is an exothermic process because graphite is the thermodynamically more stable allotrope of carbon at standard conditions (ΔH<0\Delta H < 0). Graphite has a layered structure and is less dense, so it has higher entropy than the rigid 3D network of diamond (ΔS>0\Delta S > 0). This matches (5).

(R) Denaturation of protein involves the breaking of hydrogen bonds and the unfolding of its specific 3D structure into a more random coil. Breaking these interactions requires energy (ΔH>0\Delta H > 0) and the unfolding increases randomness (ΔS>0\Delta S > 0). This matches (1).

(S) Propene ⟶\longrightarrow Cyclopropane is an endothermic process because cyclopropane has significant ring strain compared to the open-chain propene, making it less stable (ΔH>0\Delta H > 0). The formation of a rigid ring restricts the degrees of freedom (such as rotation around single bonds), decreasing entropy (ΔS<0\Delta S < 0). This matches (4).

Thus, P →\rightarrow 2, Q →\rightarrow 5, R →\rightarrow 1, S →\rightarrow 4.

Answer: P →\rightarrow 2; Q →\rightarrow 5; R →\rightarrow 1; S →\rightarrow 4
Practice Thermodynamics (C) on Quantrex Academy →

More from Thermodynamics (C)

An ideal gas ( 0.5 mol), initially at 2 bar pressure, is compressed at a constant… JEE Advanced 2026Consider the reaction 2H_2S(g) + 3O_2(g) arrow 2H_2O(l) + 2SO_2(g) The magnitude of… JEE Main 2026Match List - I with List - II. Given V_1 and V_2 are initial and final volumes… JEE Main 2026Arrange the following isothermal processes in order of the magnitude of the work ( p - V… JEE Main 2026Consider the reaction X leftharpoons Y at 300 K. If Δ H^ θ and K are 28.40 kJ mol ^ -1… JEE Main 2026Consider the following data for the reaction X_2(g) + Y_2(g) leftharpoons 2XY(g) at 600 K… JEE Main 2026The correct order of molar heat capacities measured at 298 K and 1 bar is : JEE Main 2026If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat… JEE Main 2026 Full Thermodynamics (C) list All JEE Advanced PYQs