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Kinetic interpretation of temperature and RMS speed: NEET MCQs with answers

89 questions on this topic of Kinetic Theory of Gases (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. An ideal gas is compressed isothermally such that its pressure is increased by a factor of 4. The root mean square speed of the gas molecules will:

  1. Remain unchanged
  2. Be halved
  3. Be doubled
  4. Increase by a factor of 4
Show answer and explanation

Answer: (A) Remain unchanged

The root mean square speed depends solely on the absolute temperature and molar mass through v_rms = √(3RT/M). Since the process is isothermal, the temperature remains constant, and therefore v_rms remains unchanged.

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2. The rms speed of the molecules of a gas at 300 K is 300 m/s. Its rms speed at 4800 K is:

  1. 1200 m/s
  2. 4800 m/s
  3. 75 m/s
  4. 76800 m/s
Show answer and explanation

Answer: (A) 1200 m/s

v_rms ∝ √T, so v = 300 × √(4800/300) = 1200 m/s.

hard

3. A fixed mass of an ideal gas undergoes a thermodynamic process obeying the relation P · V² = constant. If the gas expands to twice its initial volume, its root mean square speed becomes:

  1. 1/√2 times the initial value
  2. 1/2 times the initial value
  3. √2 times the initial value
  4. 2 times the initial value
Show answer and explanation

Answer: (A) 1/√2 times the initial value

For an ideal gas, P = nRT/V. Substituting into P · V² = constant gives (nRT/V) · V² = nRT · V = constant, so T · V = constant. Thus, T ∝ 1/V. When volume doubles (V' = 2V), temperature becomes T' = T/2. Since v_rms ∝ √T, the new rms speed is v_rms' = v_rms / √2.

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4. The rms speed of the molecules of a gas is 500 m/s at 200 K. Its rms speed at 800 K is:

  1. 500 m/s
  2. 1000 m/s
  3. 2000 m/s
  4. 250 m/s
Show answer and explanation

Answer: (B) 1000 m/s

v_rms = √(3RT/M) ∝ √T. So v = 500 × √(800/200) = 1000 m/s.

Quick tip: v_rms ∝ √T, with T in kelvin.

NCERT Class 11, Chapter 12

easy

5. At a given temperature T, the mean square speed (⟨v²⟩) of the molecules of an ideal gas of molar mass M is directly proportional to:

  1. T / M
  2. √(T / M)
  3. T² / M
  4. T / √M
Show answer and explanation

Answer: (A) T / M

By definition, the mean square speed is ⟨v²⟩ = (v_rms)² = 3RT/M. Hence, the mean square speed is directly proportional to T/M.

hard

6. The escape speed from the surface of the Moon is approximately 2.4 km/s. If the surface temperature of the Moon reaches 400 K during the lunar day, what is the root-mean-square speed of hydrogen molecules (H₂, molar mass = 2 × 10⁻³ kg/mol) at this temperature? (Take R = 8.31 J/(mol·K))

  1. 2.23 km/s
  2. 1.58 km/s
  3. 3.16 km/s
  4. 0.71 km/s
Show answer and explanation

Answer: (A) 2.23 km/s

Using v_rms = √(3RT/M): v_rms = √[(3 × 8.31 × 400) / (2 × 10⁻³)] = √[9972 / (2 × 10⁻³)] = √(4.986 × 10⁶) ≈ 2.23 × 10³ m/s = 2.23 km/s. (Since this is comparable to the escape velocity of 2.4 km/s, hydrogen easily escapes the lunar atmosphere).

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7. Gas molecules have an rms speed of 500 m/s at 400 K. If the temperature becomes 1200 K, the rms speed becomes:

  1. 500 m/s
  2. 866 m/s
  3. 1500 m/s
  4. 289 m/s
Show answer and explanation

Answer: (B) 866 m/s

v_rms = √(3RT/M) ∝ √T. So v = 500 × √(1200/400) = 866 m/s.

Quick tip: v_rms ∝ √T, with T in kelvin.

NCERT Class 11, Chapter 12

medium

8. The rms speed of the molecules of a gas at 300 K is 300 m/s. Its rms speed at 75 K is:

  1. 18.8 m/s
  2. 600 m/s
  3. 150 m/s
  4. 75 m/s
Show answer and explanation

Answer: (C) 150 m/s

v_rms ∝ √T, so v = 300 × √(75/300) = 150 m/s.

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9. The rms speed of the molecules of a gas is 300 m/s at 200 K. Its rms speed at 1200 K is:

  1. 122 m/s
  2. 735 m/s
  3. 300 m/s
  4. 1800 m/s
Show answer and explanation

Answer: (B) 735 m/s

v_rms = √(3RT/M) ∝ √T. So v = 300 × √(1200/200) = 735 m/s.

Quick tip: v_rms ∝ √T, with T in kelvin.

NCERT Class 11, Chapter 12

medium

10. If the absolute temperature of an ideal gas kept in a closed rigid vessel is increased by 44%, the percentage increase in the root-mean-square speed of its molecules is:

  1. 44%
  2. 20%
  3. 22%
  4. 12%
Show answer and explanation

Answer: (B) 20%

The rms speed is proportional to √T. If temperature T' = 1.44 T, then v'_rms = √(1.44 T) = 1.2 √(T) = 1.2 v_rms. The percentage increase is [(1.2 - 1.0)/1.0] × 100% = 20%.

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11. Consider the following two statements regarding an ideal gas: Statement I: At absolute zero (0 K), the translational kinetic energy of ideal gas molecules reduces to zero. Statement II: The root-mean-square speed of gas molecules depends on pressure when the temperature of the gas is kept constant. In the light of the above statements, 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: (B) Statement I is correct but Statement II is incorrect.

Statement I is correct because E_tr = (3/2) k_B T vanishes as T → 0 K in classical kinetic theory. Statement II is incorrect because v_rms = √(3RT/M), which depends solely on absolute temperature T and molar mass M, and does not change with pressure at constant temperature.

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12. The rms speed of the molecules of a gas is 600 m/s at 400 K. Its rms speed at 800 K is:

  1. 424 m/s
  2. 849 m/s
  3. 600 m/s
  4. 1200 m/s
Show answer and explanation

Answer: (B) 849 m/s

v_rms = √(3RT/M) ∝ √T. So v = 600 × √(800/400) = 849 m/s.

Quick tip: v_rms ∝ √T, with T in kelvin.

NCERT Class 11, Chapter 12

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