Second Law of Thermodynamics – The second law of thermodynamics: NEET MCQs with Answers

9 free questions on Second Law of Thermodynamics – The second law of thermodynamics for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: The second law of thermodynamics, Entropy

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Question 1

Previous-year question
In which case is the change in entropy negative? [NEET 2019]
  1. Evaporation of water
  2. Expansion of a gas at constant temperature
  3. Sublimation of solid to gas
  4. 2H(g) → H₂(g)
Show answer
Answer: (D) 2H(g) → H₂(g)
Explanation: Combining two free hydrogen atoms into one molecule halves the number of independent gaseous particles and locks them together, so the disorder decreases and ΔS is negative. The other three processes all increase disorder.

Question 2

Previous-year question
The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is: [AIPMT Pre. 2012]
  1. 5.260 cal/(mol K)
  2. 0.526 cal/(mol K)
  3. 10.52 cal/(mol K)
  4. 21.04 cal/(mol K)
Show answer
Answer: (A) 5.260 cal/(mol K)
Explanation: ΔS = ΔH_fus/T = 1435/273 = 5.26 cal/(mol K).

Question 3

Previous-year question
If the enthalpy change for the transition of liquid water to steam is 30 kJ mol⁻¹ at 27°C, the entropy change for the process would be: [AIPMT Pre. 2011]
  1. 10 J mol⁻¹ K⁻¹
  2. 1.0 J mol⁻¹ K⁻¹
  3. 0.1 J mol⁻¹ K⁻¹
  4. 100 J mol⁻¹ K⁻¹
Show answer
Answer: (D) 100 J mol⁻¹ K⁻¹
Explanation: ΔS = ΔH/T = 30000/300 = 100 J mol⁻¹ K⁻¹.

Question 4

Previous-year question
For the gas phase reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), which of the following conditions are correct? [AIPMT 2008]
  1. ΔH < 0 and ΔS < 0
  2. ΔH > 0 and ΔS < 0
  3. ΔH = 0 and ΔS < 0
  4. ΔH > 0 and ΔS > 0
Show answer
Answer: (D) ΔH > 0 and ΔS > 0
Explanation: Dissociation of PCl₅ requires energy (endothermic, ΔH > 0) and increases gaseous moles (ΔS > 0).

Question 5

1200 J of heat flows directly from a very large block at 600 K to a very large block at 300 K, and the temperatures of the blocks do not change noticeably. What is the total entropy change of the two blocks, and what does it show?
  1. +2 J K⁻¹, because the same heat gives a larger entropy change at the lower temperature
  2. −2 J K⁻¹, because the hot block loses more entropy than the cold block gains
  3. +6 J K⁻¹, because the two entropy changes add up numerically
  4. 0, because the heat lost equals the heat gained
Show answer
Answer: (A) +2 J K⁻¹, because the same heat gives a larger entropy change at the lower temperature
Explanation: The hot block changes by −1200/600 = −2 J K⁻¹ and the cold block by +1200/300 = +4 J K⁻¹, so the total is +2 J K⁻¹. Heat added at low temperature causes a greater relative increase in disorder than the same heat at high temperature. The positive total agrees with the second law, since this flow is spontaneous.

Question 6

Two different gases are separated by a partition inside a perfectly isolated container. When the partition is removed, the gases diffuse into each other. Which statement is correct for this change?
  1. ΔU is positive, because the mixed gases occupy a larger volume
  2. ΔS is positive, and this increase of entropy is what makes the change spontaneous
  3. ΔS is zero, because no heat enters or leaves the system
  4. ΔU is negative, because a spontaneous change must lower the energy of the system
Show answer
Answer: (B) ΔS is positive, and this increase of entropy is what makes the change spontaneous
Explanation: An isolated system exchanges neither heat nor work with its surroundings, so q = 0, w = 0 and ΔU = 0. Energy therefore cannot be what drives the mixing. The disorder of the gases rises, so ΔS is positive, and in an isolated system a spontaneous change is one in which entropy increases.

Question 7

An ideal gas at 300 K, kept in thermal contact with its surroundings, expands isothermally from 2.0 L to 5.0 L against a constant external pressure of 2.0 atm. What is the entropy change of the surroundings? (1 L atm = 101.3 J)
  1. −2.03 J K⁻¹
  2. −6.08 J K⁻¹
  3. +2.03 J K⁻¹
  4. Zero
Show answer
Answer: (A) −2.03 J K⁻¹
Explanation: For an isothermal process of an ideal gas ΔU = 0, so the heat absorbed by the gas equals the work it does: q = PΔV = 2.0 × 3.0 = 6.0 L atm = 607.8 J. The surroundings lose this heat at 300 K, so ΔS(surroundings) = −607.8/300 = −2.03 J K⁻¹. The process is irreversible, so the entropy gain of the gas is larger in size than this loss.
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Question 8

Two different gases are kept in the two halves of a rigid, perfectly insulated container separated by a partition. The partition is removed and the gases mix spontaneously. Treating the container as an isolated system, which is correct?
  1. ΔU > 0 and ΔS > 0
  2. ΔU < 0 and ΔS = 0
  3. ΔU = 0 and ΔS > 0
  4. ΔU = 0 and ΔS = 0
Show answer
Answer: (C) ΔU = 0 and ΔS > 0
Explanation: An isolated system exchanges neither heat nor work with the surroundings, so q = 0, w = 0 and ΔU = 0. Energy therefore cannot drive the change. The mixing happens because disorder increases, and entropy rises in any spontaneous process in an isolated system.

Question 9

Supercooled water at −5 °C freezes spontaneously in a beaker kept in surroundings at −5 °C. Which set of entropy changes is correct?
  1. ΔS(system) < 0, ΔS(surroundings) > 0, ΔS(total) > 0
  2. ΔS(system) > 0, ΔS(surroundings) < 0, ΔS(total) > 0
  3. ΔS(system) < 0, ΔS(surroundings) < 0, ΔS(total) < 0
  4. ΔS(system) < 0, ΔS(surroundings) > 0, ΔS(total) = 0
Show answer
Answer: (A) ΔS(system) < 0, ΔS(surroundings) > 0, ΔS(total) > 0
Explanation: Freezing makes the water more ordered, so the entropy of the system falls. The heat released passes to the surroundings and raises their entropy. Because the process is spontaneous, the second law requires the total entropy change to be positive. ΔS(total) would be zero only at the true freezing point, where the phases are in equilibrium.

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