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NEET Thermodynamics MCQs with answers

276 questions on Thermodynamics for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. A gas absorbs 200 J of heat and does 100 J of work on its surroundings. The change in its internal energy is:

  1. 100 J
  2. 50 J
  3. 200 J
  4. 300 J
Show answer and explanation

Answer: (A) 100 J

First law: ΔU = Q − W = 200 − 100 = 100 J.

medium

2. A Carnot engine works between a source at 227 °C and a sink at 127 °C. Its efficiency is:

  1. 20 %
  2. 80 %
  3. 100 %
  4. 44.1 %
Show answer and explanation

Answer: (A) 20 %

Use kelvin: T₁ = 500 K, T₂ = 400 K. η = 1 − T₂/T₁ = 1 − 400/500 = 20%.

hard

3. A diatomic ideal gas (γ = 7/5) initially at 127 °C undergoes an adiabatic expansion until its pressure drops to 1/128 of its initial value. What is the final temperature of the gas?

  1. 200 K
  2. 150 K
  3. 100 K
  4. 250 K
Show answer and explanation

Answer: (C) 100 K

For an adiabatic process, T^γ · P^(1 - γ) = constant, so T₂ = T₁ · (P₂/P₁)^((γ - 1)/γ). Here, T₁ = 127 + 273 = 400 K, P₂/P₁ = 1/128 = 2^(-7), and (γ - 1)/γ = (7/5 - 1)/(7/5) = 2/7. Thus, T₂ = 400 · (2^(-7))^(2/7) = 400 · 2^(-2) = 400 / 4 = 100 K.

medium

4. In a laboratory experiment, a gas thermometer is calibrated using the triple point of water (273.16 K). The pressure of the gas at constant volume is recorded as 80.0 kPa at the triple point. If the gas pressure rises to 100.0 kPa when placed in a hot liquid bath, what is the temperature of the bath?

  1. 341.45 K
  2. 218.53 K
  3. 373.15 K
  4. 320.00 K
Show answer and explanation

Answer: (A) 341.45 K

For a constant-volume ideal gas thermometer, T = 273.16 × (P / P_tr). Substituting P = 100.0 kPa and P_tr = 80.0 kPa gives T = 273.16 × (100.0 / 80.0) = 273.16 × 1.25 = 341.45 K.

easy

5. Which of the following conditions is strictly necessary for a thermodynamic process to be completely reversible?

  1. The process must be carried out rapidly so that no heat exchange occurs with the surroundings.
  2. A finite temperature difference must exist between the system and its surroundings to drive heat flow.
  3. The system must always be in non-equilibrium states during the transformation.
  4. Dissipative effects such as friction and viscosity must be completely absent.
Show answer and explanation

Answer: (D) Dissipative effects such as friction and viscosity must be completely absent.

A process is reversible only if it is quasi-static (infinitely slow, passing through equilibrium states) and purely non-dissipative. The presence of dissipative forces like friction, turbulence, or electrical resistance makes a process intrinsically irreversible.

hard

6. A fixed mass of an ideal gas is used in a constant-volume gas thermometer. If the ratio of pressure at the boiling point of a certain liquid to that at the triple point of water (273.16 K) is 1.375, what is the boiling point of the liquid?

  1. 402.1 K
  2. 348.2 K
  3. 375.6 K
  4. 415.8 K
Show answer and explanation

Answer: (C) 375.6 K

For an ideal gas at constant volume, T = 273.16 × (P / P_tr). Substituting P / P_tr = 1.375 gives T = 273.16 × 1.375 = 375.595 K ≈ 375.6 K.

medium

7. A sample of an ideal gas undergoes a reversible process where its pressure and volume are related by P·V² = constant. What is the molar heat capacity of the gas if it is monoatomic with C_v = 1.5 R?

  1. R
  2. 0.5 R
  3. 2.5 R
  4. 3.5 R
Show answer and explanation

Answer: (B) 0.5 R

For a polytropic process P·Vˣ = constant, the molar heat capacity is C = C_v + R / (1 - x). Here, x = 2 and C_v = 1.5 R. Therefore, C = 1.5 R + R / (1 - 2) = 1.5 R - R = 0.5 R.

easy

8. A gas absorbs 250 J of heat and does 100 J of work on its surroundings. The change in its internal energy is:

  1. 250 J
  2. 100 J
  3. 150 J
  4. 350 J
Show answer and explanation

Answer: (C) 150 J

First law: ΔU = Q − W = 250 − 100 = 150 J.

medium

9. Consider the following two statements regarding thermal equilibrium: Statement I: Two systems in thermal equilibrium with each other must have identical pressures and volumes. Statement II: Thermal equilibrium between two systems implies that there is no net exchange of heat energy when they are separated by a diathermic wall. 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.
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct.

Statement I is incorrect because thermal equilibrium requires only equality of temperature; pressure and volume can be completely different in the two systems. Statement II is correct because a diathermic wall permits heat flow, and at thermal equilibrium, the net heat exchange between them is zero.

medium

10. The resistance of a platinum wire in a platinum resistance thermometer is found to be 5.00 Ω at the ice point (0 °C) and 5.93 Ω at the steam point (100 °C). When placed in a hot oil bath, its resistance increases to 7.79 Ω. What is the temperature of the oil bath measured on this resistance scale?

  1. 280 °C
  2. 300 °C
  3. 320 °C
  4. 350 °C
Show answer and explanation

Answer: (B) 300 °C

The temperature on the resistance scale is given by t = [(R_t - R₀) / (R₁00 - R₀)] × 100 °C. Here, t = [(7.79 - 5.00) / (5.93 - 5.00)] × 100 = (2.79 / 0.93) × 100 = 3 × 100 = 300 °C.

medium

11. The zeroth law of thermodynamics was formulated (named) by:

  1. Joule, in 1845
  2. Clausius, in 1850
  3. R.H. Fowler, in 1931
  4. Carnot, in 1824
Show answer and explanation

Answer: (C) R.H. Fowler, in 1931

R.H. Fowler formulated the zeroth law in 1931, long after the first and second laws, which is why it is called the zeroth law.

NCERT Class 11

easy

12. A gas expands at a constant pressure of 250 kPa from V to V + 6 L. The work done by the gas is:

  1. 1.5 × 10⁶ J
  2. 41.7 J
  3. 1.5 J
  4. 1500 J
Show answer and explanation

Answer: (D) 1500 J

W = PΔV = 250 × 10³ Pa × 6 × 10⁻³ m³ = 1500 J.

easy

13. 300 J of heat is supplied to a gas, which expands and does 75 J of work. The increase in its internal energy is:

  1. 300 J
  2. 75 J
  3. 375 J
  4. 225 J
Show answer and explanation

Answer: (D) 225 J

First law: ΔU = Q − W = 300 − 75 = 225 J (W is the work done by the gas).

Quick tip: ΔU = Q − W when W is the work done BY the system.

NCERT Class 11, Chapter 11

medium

14. Two systems 1 and 2, consisting of different ideal gases with equations of state P₁V₁ = μ₁RT₁ and P₂V₂ = μ₂RT₂, are brought into thermal contact through a fixed diathermic wall. The entire composite system is thermally insulated from the surroundings. At thermal equilibrium, which condition is necessarily true?

  1. (P₁V₁) / μ₁ = (P₂V₂) / μ₂
  2. P₁ = P₂
  3. V₁ / μ₁ = V₂ / μ₂
  4. P₁V₁ = P₂V₂
Show answer and explanation

Answer: (A) (P₁V₁) / μ₁ = (P₂V₂) / μ₂

When in thermal contact through a diathermic wall, the two systems achieve thermal equilibrium, meaning T₁ = T₂. From the ideal gas law, T = PV / (μR). Equating the temperatures yields (P₁V₁) / μ₁ = (P₂V₂) / μ₂.

easy

15. A gas expands at a constant pressure of 100 kPa from 4 L to 13 L. The work done by the gas is:

  1. 1300 J
  2. 0.9 J
  3. 900 J
  4. 1700 J
Show answer and explanation

Answer: (C) 900 J

W = PΔV = (100 × 10³ Pa) × (9 × 10⁻³ m³) = 900 J.

Quick tip: kPa × L = J, because 10³ × 10⁻³ = 1.

NCERT Class 11, Chapter 11

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