NEETAceToday's NEET Daily

Kinetic-theory assumptions and pressure: NEET MCQs with answers

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

Practise Kinetic Theory of Gases freeAll Kinetic Theory of Gases questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. The pressure exerted by a gas on the walls of its container is due to:

  1. Weight of the gas
  2. Attraction between molecules
  3. Repulsion between the walls
  4. Change of momentum of molecules colliding with the walls
Show answer and explanation

Answer: (D) Change of momentum of molecules colliding with the walls

Molecules hitting a wall reverse their normal momentum. The momentum transferred per unit time per unit area is the gas pressure.

NCERT Class 11

medium

2. A gas molecule of mass m collides elastically with a stationary vertical wall of a container at an angle of 60° to the normal of the wall with speed v. The magnitude of impulse imparted to the wall by this single collision is:

  1. m v
  2. 2 m v
  3. m v √3
  4. (1/2) m v
Show answer and explanation

Answer: (A) m v

The component of velocity perpendicular to the wall is v cos(60°) = v/2. The change in momentum of the molecule along the normal is Δp = m(v/2) - (-m(v/2)) = m v. By Newton's third law, the magnitude of impulse delivered to the wall is m v.

hard

3. Equal masses of Helium gas (molar mass 4 g/mol) and Oxygen gas (molar mass 32 g/mol) are placed in separate containers of identical volume at the same temperature. What is the ratio of the pressure exerted by Helium to that exerted by Oxygen?

  1. 2√2 : 1
  2. 1 : 8
  3. 8 : 1
  4. 4 : 1
Show answer and explanation

Answer: (C) 8 : 1

From the kinetic theory expression, P = (1/3) (M_total / V) ⟨v²⟩. Since equal masses are taken, M_total and V are identical for both gases, so P is directly proportional to ⟨v²⟩. For an ideal gas, ⟨v²⟩ = 3 R T / M_mol. Therefore, P_He / P_O2 = (M_mol,O2) / (M_mol,He) = 32 / 4 = 8 / 1. Hence, the ratio of pressures is 8 : 1.

medium

4. A cubical container of volume V contains an ideal gas at pressure P. If a flat partition dividing the container into two equal halves is introduced without altering the temperature or the total number of molecules, the pressure in each half will be:

  1. P
  2. P/2
  3. 2P
  4. 4P
Show answer and explanation

Answer: (A) P

When the partition is introduced, each half receives N/2 molecules in a volume of V/2. The number density in each compartment remains n' = (N/2)/(V/2) = N/V = n. Since temperature (and hence v_rms) is unchanged, the pressure in each compartment remains P.

easy

5. Regarding the molecular nature of an ideal gas according to the kinetic model, which statement is CORRECT?

  1. The volume occupied by the gas molecules themselves is approximately 25% of the container volume.
  2. Molecules experience strong repulsive forces at all times during their free motion.
  3. Molecules travel in curved trajectories between collisions under mutual molecular attraction.
  4. Molecules move in straight lines with constant speed between successive collisions.
Show answer and explanation

Answer: (D) Molecules move in straight lines with constant speed between successive collisions.

In kinetic theory, intermolecular forces are neglected except during collisions. Therefore, in between collisions, molecules experience zero net force and travel in straight lines with constant speed.

hard

6. An ideal gas is maintained in a container. If the root-mean-square speed of the gas molecules is increased by 20% while the volume is kept constant, by what percentage does the gas pressure increase?

  1. 20%
  2. 40%
  3. 44%
  4. 21%
Show answer and explanation

Answer: (C) 44%

Pressure is proportional to the square of v_rms when volume and number of particles are constant: P ∝ (v_rms)². If v_rms increases by 20%, the new speed is 1.2 v_rms. The new pressure becomes P' = (1.2)² P = 1.44 P. The percentage increase is (1.44 - 1) × 100% = 44%.

medium

7. A spherical container of radius R holds N molecules of a gas at pressure P. If the radius of the sphere is doubled while the temperature and the number of molecules remain constant, the new pressure will be:

  1. P/8
  2. P/4
  3. P/2
  4. P/16
Show answer and explanation

Answer: (A) P/8

The volume of a sphere is V = (4/3)πR³. When R is doubled, the new volume becomes V' = 8V. At constant temperature, the product PV is constant (or P ∝ 1/V). Therefore, the new pressure is P' = P/8.

medium

8. A container has a mixture of two gases: gas 1 with molecule mass m₁ and number density n₁, and gas 2 with molecule mass m₂ and number density n₂. If both gases are in thermal equilibrium so that their mean translational kinetic energies per molecule are equal, what is the total pressure exerted on the container walls?

  1. (2/3) (n₁ + n₂) E_avg, where E_avg is the average kinetic energy per molecule
  2. (1/3) (n₁ + n₂) (m₁ + m₂) (v_rms)²
  3. (1/3) (n₁ m₁ + n₂ m₂) (v_rms)₁²
  4. (2/3) [ (n₁ + n₂) / (m₁ + m₂) ] E_avg
Show answer and explanation

Answer: (A) (2/3) (n₁ + n₂) E_avg, where E_avg is the average kinetic energy per molecule

By Dalton's law of partial pressures and kinetic theory, P_total = P₁ + P₂ = (2/3) n₁ E_avg + (2/3) n₂ E_avg = (2/3)(n₁ + n₂)E_avg, since the mean translational kinetic energy per molecule E_avg = (3/2) k_B T is identical for both species at thermal equilibrium.

medium

9. A vessel contains an ideal gas at pressure P. If a partition is suddenly removed so that the volume occupied by the gas doubles while its temperature is kept constant, the rate of molecular collisions per unit area on the container walls changes by a factor of:

  1. 2
  2. 1/√2
  3. 1/2
  4. 1/4
Show answer and explanation

Answer: (C) 1/2

The collision rate per unit area on a container wall is proportional to n × v_avg, where n is the number density (N/V) and v_avg is the average speed. Since the temperature is kept constant, v_avg remains unchanged. When the volume doubles, the number density n is halved, meaning the rate of collisions per unit area becomes half of its initial value.

medium

10. Given below are two statements: Statement I: The pressure of an ideal gas inside a closed container is directly proportional to the number density of the molecules at a given temperature. Statement II: Increasing the number density of molecules increases the rate of momentum transfer per unit area to the container walls. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true, and Statement II is the correct explanation of Statement I.
  2. Both Statement I and Statement II are true, but Statement II is NOT the correct explanation of Statement I.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are true, and Statement II is the correct explanation of Statement I.

The pressure P = (1/3) n m (v_rms)² is directly proportional to number density n = N/V. Pressure physically arises from the rate of momentum transferred per unit area by colliding molecules, which scales directly with number density. Thus, both statements are true and Statement II explains Statement I.

medium

11. A gas sample has a density of 1.25 kg/m³ at a pressure of 1.0 × 10⁵ N/m². The root-mean-square speed of its molecules is:

  1. 240 m/s
  2. 600 m/s
  3. 350 m/s
  4. 490 m/s
Show answer and explanation

Answer: (D) 490 m/s

Using the kinetic pressure equation P = (1/3) ρ (v_rms)², we have v_rms = √(3P/ρ). Substituting P = 1.0 × 10⁵ N/m² and ρ = 1.25 kg/m³ yields v_rms = √(3 × 10⁵ / 1.25) = √(2.4 × 10⁵) ≈ 489.9 m/s, or approximately 490 m/s.

medium

12. Which of the following correctly describes how intermolecular forces are treated in the kinetic theory of an ideal gas to deduce the expression for pressure?

  1. Attractive forces dominate at moderate distances, while repulsive forces act during collision
  2. Molecules exert no forces on one another except during elastic collisions of negligible duration
  3. Molecules exert repulsive forces inversely proportional to the cube of their separation at all times
  4. Forces between molecules are long-range and maintain an equilibrium spacing throughout the volume
Show answer and explanation

Answer: (B) Molecules exert no forces on one another except during elastic collisions of negligible duration

One of the foundational assumptions of kinetic theory is that intermolecular forces (both attractive and repulsive) are zero at all distances, except during physical contact during a collision, where repulsive forces act over an infinitesimally short time to produce an elastic bounce.

Keep going with Kinetic Theory of Gases. All 289 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Kinetic Theory of Gases