Thermodynamics — NEET UG Questions

25 NEET UG practice questions on Thermodynamics, part of Chemistry. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. Which of the following statements correctly distinguishes a reversible process from an irreversible process?

  1. In a reversible process, the system is always in equilibrium with the surroundings; in an irreversible process, the system is not in equilibrium during the process.
  2. In a reversible process, the entropy of the universe increases; in an irreversible process, the entropy of the universe decreases.
  3. In a reversible process, the work done is always greater than that in an irreversible process.
  4. In a reversible process, the system returns to its initial state without any change in the surroundings; in an irreversible process, the surroundings are unchanged.

Answer: In a reversible process, the system is always in equilibrium with the surroundings; in an irreversible process, the system is not in equilibrium during the process.

A reversible process is an idealised process that occurs through a continuous series of equilibrium states, where the system and surroundings are in mutual equilibrium at every stage. In contrast, an irreversible process involves finite driving forces (e.g., temperature or pressure gradients) and the system is not in equilibrium during the process. The other options are incorrect: entropy of the universe increases for irreversible processes and remains constant for reversible processes; work done comparison depends on the type of process; and in a reversible process the surroundings may change unless the process is cyclic.

2. The second law of thermodynamics states that for a spontaneous process:

  1. A) the total entropy of the universe remains constant
  2. B) the total entropy of the universe increases
  3. C) the entropy of the system always decreases
  4. D) the Gibbs free energy of the system decreases

Answer: B) the total entropy of the universe increases

The second law states that the entropy of an isolated system (or the universe) increases for a spontaneous process. Option A describes a reversible process; option C is not always true (system entropy may increase or decrease, but universe entropy increases); option D is a consequence for constant T,P but is not the second law itself.

3. Which of the following processes is accompanied by a positive (increase) in entropy?

  1. A) Freezing of water to ice at 0°C
  2. B) Dissolution of sugar in water
  3. C) Condensation of steam to liquid water
  4. D) Rusting of iron (Fe + O₂ → Fe₂O₃)

Answer: B) Dissolution of sugar in water

Dissolution of sugar in water increases disorder as sugar molecules disperse among water molecules, leading to higher entropy. Freezing, condensation, and rusting all involve a decrease in randomness (formation of a more ordered solid/liquid or consumption of gas), so their entropy change is negative.

4. Entropy is a thermodynamic quantity that is best described as a measure of:

  1. A) the heat content of a system
  2. B) the disorder or randomness in a system
  3. C) the internal energy of a system
  4. D) the temperature of a system

Answer: B) the disorder or randomness in a system

Entropy (S) is a measure of the degree of disorder or randomness in a system. Higher entropy indicates greater disorder. Heat content is related to enthalpy (H), internal energy (U) is a different state function, and temperature measures average kinetic energy, not disorder.

5. The standard Gibbs free energy change (ΔG°) for a reaction at 298 K is -5.7 kJ mol⁻¹. What is the equilibrium constant (K) of the reaction? (R = 8.314 J mol⁻¹ K⁻¹)

  1. 0.1
  2. 1.0
  3. 10
  4. 100

Answer: 10

Using the relation ΔG° = -RT ln K, we solve for K: ln K = -ΔG°/(RT). Convert ΔG° to J: -5.7 kJ mol⁻¹ = -5700 J mol⁻¹. R = 8.314 J mol⁻¹ K⁻¹, T = 298 K. Then ln K = 5700/(8.314 × 298) ≈ 2.303, so K = e^2.303 ≈ 10. Hence the equilibrium constant is 10.

6. One mole of an ideal gas expands isothermally and irreversibly at 27°C from an initial volume of 1 L to a final volume of 4 L against a constant external pressure of 1 atm. The work done by the gas (in J) is: (Given: 1 L·atm = 101.3 J)

  1. -303.9 J
  2. +303.9 J
  3. -3 J
  4. -3 L·atm

Answer: -303.9 J

For an irreversible isothermal expansion against constant external pressure, work done by the gas is w = –P_ext ΔV = –1 atm × (4 – 1) L = –3 L·atm = –3 × 101.3 J = –303.9 J. The negative sign indicates work is done by the system.

7. For a reaction with ΔH > 0 and ΔS > 0, the reaction is spontaneous:

  1. A) only at low temperatures
  2. B) only at high temperatures
  3. C) at all temperatures
  4. D) at no temperature

Answer: B) only at high temperatures

ΔG = ΔH – TΔS. With both ΔH and ΔS positive, at low T the TΔS term is small so ΔG > 0 (non-spontaneous). At sufficiently high T, TΔS exceeds ΔH, making ΔG < 0 (spontaneous). Hence spontaneity occurs only at high temperatures.

8. Which of the following processes is exothermic (ΔH < 0)?

  1. Melting of ice
  2. Combustion of methane
  3. Evaporation of water
  4. Dissolution of ammonium nitrate in water

Answer: Combustion of methane

Exothermic processes release heat and have a negative ΔH. Combustion of methane is a classic exothermic reaction. Melting of ice, evaporation of water, and dissolution of ammonium nitrate are endothermic (positive ΔH).

9. According to the third law of thermodynamics, the entropy of a perfect crystalline substance at absolute zero temperature is:

  1. maximum
  2. negative
  3. zero
  4. positive

Answer: zero

The third law of thermodynamics states that the entropy of a perfectly ordered crystalline substance is exactly zero at absolute zero temperature (0 K). This provides a reference point for absolute entropy values.

10. The standard Gibbs free energy change (ΔG°) is related to the standard cell potential (E°) and equilibrium constant (K) by which of the following correct relations?

  1. A) ΔG° = nFE° and ΔG° = –RT ln K
  2. B) ΔG° = –nFE° and ΔG° = –RT ln K
  3. C) ΔG° = –nFE° and ΔG° = RT ln K
  4. D) ΔG° = nFE° and ΔG° = RT ln K

Answer: B) ΔG° = –nFE° and ΔG° = –RT ln K

The correct relationships are ΔG° = –nFE° (where n is number of moles of electrons, F is Faraday constant) and ΔG° = –RT ln K. These link the thermodynamic spontaneity (negative ΔG°) with positive E° and K > 1.

11. Which one of the following is an intensive property?

  1. A. Mass
  2. B. Volume
  3. C. Density
  4. D. Enthalpy

Answer: C. Density

Intensive properties do not depend on the amount of substance. Density (mass/volume) is intensive because it remains constant regardless of sample size. Mass, volume, and enthalpy are extensive properties.

12. Which of the following is a state function?

  1. A. Work
  2. B. Heat
  3. C. Internal energy
  4. D. Path

Answer: C. Internal energy

State functions depend only on the current state of the system, not on the path taken. Internal energy is a state function, whereas work and heat are path functions. 'Path' is not a thermodynamic property.

More Chemistry topics

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