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Class XI · Physics

Kinetic Theory of Gases: NEET MCQ practice

59 practice questions on Kinetic Theory of Gases, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

59

Practice questions

41

Easy

17

Medium

1

Hard

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What Kinetic Theory of Gases covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Equation of state of a perfect gas, work done in compressing a gas
  • Kinetic theory of gases- assumptions, concept of pressure
  • Kinetic interpretation of temperature, RMS speed of gas molecules
  • Degrees of freedom, law of equipartition of energy, specific heat capacities of gases
  • Mean free path, Avogadro's number

Sample questions on Kinetic Theory of Gases

Try each one, then open the answer and explanation.

Question 1 · Practice question

A constant volume gas thermometer shows pressure reading of 50cm and 90cm of mercury at 0°C and 100° C respectively. When the pressure reading is 60 cm of mercury, the temperature is

  1. A.25°C
  2. B.40°C
  3. C.15°C
  4. D.12.5°C
Show answer

Answer: A. 25°C

For a constant volume gas thermometer, pressure is directly proportional to absolute temperature, which means it increases linearly with Celsius temperature between fixed points. Using the given pressures at 0°C (50 cm Hg) and 100°C (90 cm Hg), we set up the linear relation: (T - 0)/(100 - 0) = (60 - 50)/(90 - 50). Solving gives T = 25°C. The common mistake is to mistakenly assume a direct proportionality without using the two reference points, possibly leading to 40°C if misapplying ratios.

Question 2 · Practice question

A constant pressure air thermometer gave a reading of 47.5 units of volume when immersed in ice cold water, and 67 units in a boiling liquids. The boiling point of the liquid will be

  1. A.135°C
  2. B.125°C
  3. C.112° C
  4. D.100°C
Show answer

Answer: C. 112° C

A constant pressure gas thermometer follows Charles's law: volume is directly proportional to absolute temperature. Using V = V₀(1 + t/273), where V₀ is volume at 0°C and t is temperature in °C, substituting V = 67 and V₀ = 47.5 gives t ≈ 112°C. The tempting 100°C option assumes the liquid is water, but here the boiling liquid is unknown and calculation shows it's higher than 100°C.

Question 3 · Practice question

A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

  1. A.4 R T
  2. B.5 R T
  3. C.15 R T
  4. D.11 R T
Show answer

Answer: D. 11 R T

The total internal energy equals the sum of energies of each component, calculated using U = n × (f/2)RT, where f is degrees of freedom. For O₂ (diatomic, no vibration) f = 5, giving 2 × (5/2)RT = 5RT. For Ar (monatomic) f = 3, giving 4 × (3/2)RT = 6RT. Total = 11RT. A common mistake is using f = 5 for Ar or f = 3 for O₂, which would yield totals like 4RT or 15RT.

Question 4 · Practice question

The ratio of translational and rotational kinetic energies at 100 K temperature is 3 : 2. Then the internal energy of one mole gas at that temperature is [R = 8.3 J/mol-K]

  1. A.1175J
  2. B.1037.5 J
  3. C.2075 J
  4. D.4150J
Show answer

Answer: C. 2075 J

For a diatomic gas at moderate temperature, translational kinetic energy is (3/2)RT and rotational is RT, giving ratio 3:2. The internal energy is sum of these, i.e., (5/2)RT. Substituting R=8.3 and T=100 gives 2075J. The wrong options come from forgetting rotational contribution or miscalculating the factor.

Question 5 · Practice question

12 gm He and 4 gm H 2 is filled in a container of volume 20 litre maintained at temperature 300 K. The pressure of the mixture is nearly :

  1. A.3 atm
  2. B.5 atm
  3. C.6.25 atm
  4. D.12.5 atm
Show answer

Answer: C. 6.25 atm

The pressure is found using the ideal gas law P = nRT/V. First, find total moles: moles of He = 12/4 = 3, moles of H₂ = 4/2 = 2, total n = 5. With R = 0.0821 L·atm·mol⁻¹·K⁻¹, T = 300 K, V = 20 L, we get P ≈ 6.16 atm, nearest to 6.25 atm. A common mistake is forgetting to convert both gases to moles properly, which could yield 3 atm (using only He) or 5 atm (mixing up mass and moles).

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