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Second law; reversible and irreversible processes: NEET MCQs with answers

56 questions on this topic of Thermodynamics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The Second Law of Thermodynamics fundamentally provides information about which of the following aspects of a physical process?

  1. The direction in which a thermodynamic process can take place spontaneously
  2. The total conservation of energy involved during the process
  3. The exact rate at which heat is transferred between bodies
  4. The microscopic kinetic energy distribution of the system molecules
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Answer: (A) The direction in which a thermodynamic process can take place spontaneously

While the First Law states energy conservation, it does not explain why heat flows spontaneously only from hot to cold bodies. The Second Law of Thermodynamics dictates the feasible direction of spontaneous physical processes.

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2. A Carnot engine works between a source at 527 °C and a sink at 77 °C. Its efficiency is:

  1. 28.1 %
  2. 56.3 %
  3. 85.4 %
  4. 43.8 %
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Answer: (B) 56.3 %

Use kelvin: T₁ = 800 K, T₂ = 350 K. η = 1 − T₂/T₁ = 1 − 350/800 = 56.3%.

hard

3. A reversible heat engine converts 1/6th of heat input into work. When the temperature of the sink is reduced by 62 °C (or 62 K), its efficiency becomes doubled. The original temperatures of the source and the sink are respectively:

  1. 372 K and 310 K
  2. 360 K and 300 K
  3. 400 K and 338 K
  4. 420 K and 350 K
Show answer and explanation

Answer: (A) 372 K and 310 K

Initially, η = 1 - T₂/T₁ = 1/6 ⇒ T₂/T₁ = 5/6. When sink temperature is reduced by 62 K, η' = 2 × (1/6) = 1/3. So 1 - (T₂ - 62)/T₁ = 1/3 ⇒ (T₂ - 62)/T₁ = 2/3. Subtracting the two equations gives 62/T₁ = 5/6 - 2/3 = 1/6, which gives T₁ = 372 K. Then T₂ = (5/6) × 372 = 310 K.

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4. A Carnot engine works between a source at 127 °C and a sink at 27 °C. Its efficiency is:

  1. 78.7 %
  2. 75 %
  3. 100 %
  4. 25 %
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Answer: (D) 25 %

Use kelvin: T₁ = 400 K, T₂ = 300 K. η = 1 − T₂/T₁ = 1 − 300/400 = 25%.

easy

5. Which of the following fundamental principles is an immediate consequence of the Kelvin-Planck statement of the second law of thermodynamics?

  1. Work cannot be completely converted into heat in a cyclical process.
  2. Heat cannot be transferred from a body at a higher temperature to a body at a lower temperature.
  3. The total internal energy of an isolated system undergoing any cyclic process remains constant.
  4. A heat engine operating in a complete cycle cannot produce a net amount of work by exchanging heat with a single thermal reservoir.
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Answer: (D) A heat engine operating in a complete cycle cannot produce a net amount of work by exchanging heat with a single thermal reservoir.

The Kelvin-Planck statement asserts that no process is possible whose sole result is the absorption of heat from a single reservoir and its complete conversion into work. Thus, a heat engine cannot deliver net work while exchanging heat with only one reservoir.

hard

6. A Carnot refrigerator operates between 0 °C and 50 °C. The amount of mechanical work required to freeze 1.5 kg of water already at 0 °C into ice at 0 °C is approximately (Latent heat of fusion of ice L = 3.34 × 10⁵ J/kg):

  1. 2.73 × 10⁵ J
  2. 5.01 × 10⁵ J
  3. 9.17 × 10⁴ J
  4. 1.83 × 10⁵ J
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Answer: (C) 9.17 × 10⁴ J

The heat extracted is Q₂ = m L = 1.5 × 3.34 × 10⁵ = 5.01 × 10⁵ J. The coefficient of performance is α = T₂ / (T₁ - T₂) = 273 / (323 - 273) = 273 / 50 = 5.46. Work required W = Q₂ / α = (5.01 × 10⁵) / 5.46 ≈ 9.17 × 10⁴ J.

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7. A Carnot engine works between a source at 527 °C and a sink at 27 °C. Its efficiency is:

  1. 31.3 %
  2. 94.9 %
  3. 37.5 %
  4. 62.5 %
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Answer: (D) 62.5 %

Use kelvin: T₁ = 800 K, T₂ = 300 K. η = 1 − T₂/T₁ = 1 − 300/800 = 62.5%.

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8. A Carnot engine works between a source at 527 °C and a sink at 127 °C. Its efficiency is:

  1. 75.9 %
  2. 100 %
  3. 25 %
  4. 50 %
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Answer: (D) 50 %

Use kelvin: T₁ = 800 K, T₂ = 400 K. η = 1 − T₂/T₁ = 1 − 400/800 = 50%.

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9. A Carnot engine works between a source at 327 °C and a sink at 127 °C. Its efficiency is:

  1. 61.2 %
  2. 66.7 %
  3. 33.3 %
  4. 16.7 %
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Answer: (C) 33.3 %

Use kelvin: T₁ = 600 K, T₂ = 400 K. η = 1 − T₂/T₁ = 1 − 400/600 = 33.3%.

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10. Consider the following statements regarding heat engines and refrigerators: Statement I: The coefficient of performance of a refrigerator can never be greater than 1. Statement II: The efficiency of any reversible heat engine operating between two given temperatures is always strictly less than 1. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (C) Statement I is incorrect but Statement II is correct

The coefficient of performance COP = T₂/(T₁ - T₂) can be greater than 1 if T₁ - T₂ < T₂, which is very common. Statement II is correct because efficiency η = 1 - (T₂/T₁), and since T₂ cannot be 0 K in practice, η is always strictly less than 1 (or 100%).

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11. The coefficient of performance (α) of a refrigerator working between a cold reservoir at temperature T₂ and a hot reservoir at temperature T₁ is related to the efficiency (η) of a Carnot heat engine working between the same two temperatures by:

  1. α = η / (1 - η)
  2. α = (1 - η) / η
  3. α = 1 / η
  4. α = 1 - η
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Answer: (B) α = (1 - η) / η

For a Carnot refrigerator, α = T₂ / (T₁ - T₂). For a Carnot engine, η = (T₁ - T₂) / T₁. Dividing these gives α = (T₂ / T₁) / [(T₁ - T₂) / T₁] = (1 - η) / η.

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12. Two Carnot engines 1 and 2 operate between the same source temperature T₁ and sink temperature T₂. Engine 1 uses an ideal monoatomic gas (γ = 5/3) and engine 2 uses an ideal diatomic gas (γ = 7/5) as the working substance. The relation between their thermal efficiencies η₁ and η₂ is:

  1. η₁ = η₂
  2. η₁ < η₂
  3. η₁ > η₂
  4. η₁ = (5/7) η₂
Show answer and explanation

Answer: (A) η₁ = η₂

According to Carnot's theorem, the efficiency of a reversible heat engine operating between two given temperatures depends solely on the temperatures of the source and the sink, and is completely independent of the nature of the working substance.

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