NEET (UG) · Physics, Class 11
Behaviour of Perfect Gas and Kinetic Theory: NEET PYQ, 2018–2026
Behaviour of Perfect Gas and Kinetic Theory (Physics, Class 11) has been asked 11 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.6 questions a paper, 3% of all Physics questions. It is the 17th most-asked of 20 Physics chapters. NEET 2026 had 2 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 1 |
| 2020 | 3 |
| 2021 | 1 |
| 2023 | 1 |
| 2024 | 1 |
| 2025 | 2 |
| 2026 | 2 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Ideal gas equation4
- Kinetic theory3
- Degrees of freedom2
- Mean free path2
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10 Behaviour of Perfect Gas and Kinetic Theory questions to practise.
Our own questions, written in the NEET pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Behaviour of Perfect Gas and Kinetic Theory
A single argon atom is free to move inside a container and does not rotate in any way relevant to its energy. How many degrees of freedom does this monatomic atom have?
Answer: (A)
Principle: the degrees of freedom of a molecule are its independent ways to store energy. For a monatomic gas atom, these are the three translational motions only, giving an average energy of (3/2)k_BT and a molar heat capacity of 3R/2 at constant volume. Rotational degrees of freedom are not counted for a point-like atom.
Q2 · Physics · Behaviour of Perfect Gas and Kinetic Theory
An open flask, in contact with the atmosphere at constant pressure, is heated slowly from 300 K to 400 K. Treating the air as an ideal gas, what fraction of the air originally in the flask escapes?
Answer: (A)
The flask is open, so pressure P and volume V stay fixed while the number of moles changes. From PV = nRT, n = PV/(RT), so n ∝ 1/T. Then n₂/n₁ = T₁/T₂ = 300/400 = 3/4. The fraction that escapes is 1 − 3/4 = 1/4.
Q3 · Physics · Behaviour of Perfect Gas and Kinetic Theory
An air bubble is released at the bottom of a lake 30 m deep, where the water temperature is 280 K. As it rises slowly, it reaches the surface, where the temperature is 300 K. Treat the air as an ideal gas and the bubble as a sphere that stays in thermal equilibrium with the water. Take atmospheric pressure = 1.0 × 10⁵ Pa, density of water = 1000 kg m⁻³ and g = 10 m s⁻². What is the ratio of the bubble's radius at the surface to its radius at the bottom?
Answer: (A)
For a fixed amount of gas, PV/T is constant. The pressure at the bottom is P₁ = P₀ + ρgh = 1 × 10⁵ + 1000 × 10 × 30 = 4 × 10⁵ Pa. At the surface it is P₂ = 1 × 10⁵ Pa. Then V₂/V₁ = (P₁/P₂)(T₂/T₁) = 4 × (300/280) = 30/7 ≈ 4.29. Since V ∝ r³, r₂/r₁ = (30/7)^(1/3) ≈ 1.62. Pressure raises the volume by 4, the temperature adds only a small further increase, and the final step is the cube root.
Q4 · Physics · Behaviour of Perfect Gas and Kinetic Theory
Why is a rigid diatomic molecule such as oxygen treated as having 5 degrees of freedom at room temperature rather than 6?
Answer: (A)
Principle: each independent quadratic energy term carries k_BT/2 on average. A rigid diatomic molecule has 3 translational degrees of freedom. Its rotations about two perpendicular axes (both perpendicular to the bond) are active. Rotation about the bond axis has negligible moment of inertia, so it contributes nothing. Total at moderate temperature: 3 + 2 = 5.
Q5 · Physics · Behaviour of Perfect Gas and Kinetic Theory
Gas A and gas B have molecules of the same effective diameter. Each molecule of B is four times as massive as a molecule of A. Both gases are at the same temperature and pressure. The ratio of the mean free path of A to that of B is:
Answer: (A)
From p = nk_BT, equal p and T mean equal number density n for both gases. The mean free path λ = 1/(√2 π d² n) depends only on n and d. Both are equal here, so λ_A/λ_B = 1. Mass affects the average speed, and therefore the collision frequency (ν = v̄/λ), but not λ itself.
Q6 · Physics · Behaviour of Perfect Gas and Kinetic Theory
A gas mixture of helium (molar mass 4.00 g mol⁻¹) and oxygen (molar mass 32.0 g mol⁻¹) behaves ideally. At a total pressure of 1.00 atm and a temperature of 300 K, its density is 0.325 g L⁻¹. Taking R = 0.0821 L atm mol⁻¹ K⁻¹, what is the partial pressure of oxygen in the mixture?
Answer: (A)
Write PV = nRT as PM = ρRT, where M is the average molar mass. Then M = ρRT/P = 0.325 × 0.0821 × 300 / 1.00 ≈ 8.0 g mol⁻¹. For mole fraction x of oxygen, M = 32x + 4(1 − x) = 4 + 28x, so 28x = 4 and x = 1/7. By Dalton's law, the partial pressure of oxygen is x × P_total = (1/7)(1.00 atm) ≈ 0.14 atm.
Q7 · Physics · Behaviour of Perfect Gas and Kinetic Theory
A gas is confined to move in a plane, so each molecule can move only in two independent directions and cannot rotate. In thermal equilibrium at absolute temperature T, what is the average energy per molecule of this gas? (k_B is the Boltzmann constant.)
Answer: (B)
The law of equipartition states that each quadratic degree of freedom has an average energy of ½k_BT in thermal equilibrium. Here the molecules have only two translational degrees of freedom in the plane and no rotational ones. The total average energy per molecule is therefore 2 × ½k_BT = k_BT.
Q8 · Physics · Behaviour of Perfect Gas and Kinetic Theory
Two gases have the same number of molecules per unit volume, but the molecules of the second gas have twice the diameter of those of the first. The ratio of the mean free path in the first gas to that in the second is:
Answer: (A)
The mean free path is l = 1/(√2 n π d²). With the same n, l is inversely proportional to d². Doubling the diameter makes the collision cross-section four times as large, so the second gas has a quarter of the mean free path: the ratio is 4 : 1.
Q9 · Physics · Behaviour of Perfect Gas and Kinetic Theory
A fixed mass of an ideal gas undergoes a quasi-static process in which its pressure is directly proportional to its volume (p ∝ V). During the process its volume increases to four times its initial value. Assuming the molecular diameter stays constant, what is the ratio (final/initial) of the number of collisions a molecule makes per second?
Answer: (C)
The collision frequency is z = v_avg/λ, with v_avg ∝ √T and λ = 1/(√2 π d² n) ∝ V for a fixed mass of gas. Since p ∝ V, the ideal gas law gives T ∝ pV ∝ V². When V becomes 4V, T becomes 16T, so v_avg ∝ √T becomes 4 times larger. Meanwhile n falls to one quarter, so λ becomes 4 times larger. Hence z_final/z_initial = 4/4 = 1. Equivalently, z ∝ n·v ∝ (1/V)(V) = constant.
Q10 · Physics · Behaviour of Perfect Gas and Kinetic Theory
A gas of rigid non-linear polyatomic molecules, such as ammonia, is in thermal equilibrium at absolute temperature T. What is the average total kinetic energy of one molecule? (k_B is the Boltzmann constant.)
Answer: (B)
Equipartition assigns ½k_BT to each quadratic degree of freedom. A rigid non-linear molecule has 3 translational and 3 rotational degrees of freedom, so it has 6 in total. The average energy is 6 × ½k_BT = 3k_BT. Note that this is the energy of one molecule, not the molar energy.
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About Behaviour of Perfect Gas and Kinetic Theory in NEET.
How many questions come from Behaviour of Perfect Gas and Kinetic Theory in NEET?
Behaviour of Perfect Gas and Kinetic Theory (Physics, Class 11) has been asked 11 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.6 questions a paper, 3% of all Physics questions. It is the 17th most-asked of 20 Physics chapters. NEET 2026 had 2 from it.
Which topics of Behaviour of Perfect Gas and Kinetic Theory are asked most in NEET?
Ideal gas equation (4), Kinetic theory (3), Degrees of freedom (2), Mean free path (2), counted over the papers from 2018–2026.
Is Behaviour of Perfect Gas and Kinetic Theory important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 17 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
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