NEET Kinetic Theory PYQs & Practice Questions

10 free NEET Physics questions on Kinetic Theory, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Ideal Gas Law · Mixing of Gases / Internal Energy · Ideal Gas Equation · Pressure of an Ideal Gas · Kinetic Energy and Temperature · Law of Equipartition of Energy · RMS Speed · Mean Free Path · Speed Distribution · Degrees of Freedom · Ideal Gas Laws · Real Gases and Avogadro's Law

1. Q1NEET 2026Degrees of Freedom and Specific Heat Ratio

   An ideal gas is made of polyatomic molecules. Each of the molecules has three translational, three rotational and f number of vibrational modes. If the ratio of heat capacities Cp/Cv of the gas is 8/7, then the value of f is:

   • A. 3
   • B. 2
   • C. 1
   • D. 4
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   Answer: C — 1

   Total degrees of freedom = 3 (translational) + 3 (rotational) + 2f (each vibrational mode contributes 2, for kinetic and potential energy). gamma = Cp/Cv = 1 + 2/(degrees of freedom) = 8/7, so 2/(dof) = 1/7, giving dof = 14. So 3+3+2f = 14, 2f = 8, f = 4.

2. Q2NEET 2026Mean Free Path

   The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If the number densities of gases A and B are n_A and n_B respectively, then the correct option is:

   • A. n_A = 2n_B
   • B. n_A = (1/4)n_B
   • C. n_A = (1/2)n_B
   • D. n_A = n_B
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   Answer: B — n_A = (1/4)n_B

   Mean free path λ = 1/(√2πd²n). Given λ_A = λ_B/2 and d_A = 2d_B: λ_A/λ_B = (d_B²n_B)/(d_A²n_A) = 1/2. Substituting d_A² = 4d_B²: (d_B²n_B)/(4d_B²n_A) = 1/2, so n_B/(4n_A) = 1/2, giving n_B = 2n_A, i.e., n_A = (1/2)n_B.

3. Q3NEET 2026RMS speed ratio of gas mixture

   A flask contains argon and chlorine in the ratio of 2:1 by mass. The temperature of the mixture is 27°C. The ratio of root mean square speed of the molecules of the two gases (vAr(rms)/vCl(rms)) is: (atomic mass of argon=40.0u, molecular mass of chlorine=70.0u)

   • A. 7/4
   • B. 2/√7
   • C. √(7/2)
   • D. 7/√2
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   Answer: B — 2/√7

   vrms∝1/√M, ratio=√(MCl/MAr)=√(70/40)=√(7/4)=√7/2.

4. Q4NEET 2025Ideal Gas Law

   An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given, R = 100/12 J mol⁻¹K⁻¹, and molecular mass of O2 = 32, 1 atm pressure = 1.01×10⁵ N/m]

   • A. 0.125 kg
   • B. 0.144 kg
   • C. 0.116 kg
   • D. 0.156 kg
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   Answer: B — 0.144 kg

   P2 = Pgauge + Patm = 11+1 = 12 atm. Initial pressure P1 = n1RT/V = (18.20×(100/12)×300)/(30×10⁻³) = 1.5×10⁶ N/m². Since P∝n at fixed V,T: P1/P2 = n1/n2 ⟹ n2 = n1×P2/P1 = 18.20×(12×10⁵)/(1.5×10⁶) = 14.56 mol. Δn = 18.20−14.56 = 3.64 mol. Mass withdrawn = Δn×32 = 116.48 g ≈ 0.116 kg.

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5. Q5NEET 2025Mixing of Gases / Internal Energy

   A container has two chambers of volumes V1 = 2 litres and V2 = 3 litres separated by a partition made of a thermal insulator. The chambers contain n1 = 5 and n2 = 4 moles of ideal gas at pressures P1 = 1 atm and P2 = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of

   • A. 1.3 atm
   • B. 1.6 atm
   • C. 1.4 atm
   • D. 1.8 atm
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   Answer: A — 1.3 atm

   Internal energy conservation: (f/2)P1V1+(f/2)P2V2 = (f/2)P(V1+V2). So P = (P1V1+P2V2)/(V1+V2) = (1×2+2×3)/(2+3) = 8/5 = 1.6 atm.

6. Q6NEET 2023RMS Speed and Temperature Relation

   The temperature of a gas is -50°C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

   • A. 3295 °C
   • B. 3097 K
   • C. 223 K
   • D. 699 °C
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   Answer: —

   V ∝ √T. Initial T1 = -50°C = 223 K. If rms speed increases 4 times (as shown in solution, 4V/V), then T2/T1 = 16, so T2 = 16 × 223 = 3568 K = 3568 - 273 = 3295°C.

7. Q7NEET 2022Molar Volume Calculation

   The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is:

   • A. 5.6 x 10^6 m^3
   • B. 5.6 x 10^3 m^3
   • C. 5.6 x 10^-3 m^3
   • D. 5.6 m^3
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   Answer: C — 5.6 x 10^-3 m^3

   Moles of water = 4500 g / 18 g/mol = 250 mol. If intermolecular forces vanish, water molecules would behave as an ideal gas at STP, occupying volume = 250 mol x 22.4 L/mol = 5600 L = 5.6 m^3.

8. Q8NEET 2022Molecular Volume

   The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is

   • A. 5.6 × 10⁶ m³
   • B. 5.6 × 10³ m³
   • C. 5.6 × 10⁻³ m³
   • D. 5.6 m³
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   Answer: C — 5.6 × 10⁻³ m³

   4.5 kg water = 4500 g = 250 moles. At STP as ideal gas, V = 250×22.4 L = 5600 L = 5.6 m³.

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9. Q9NEET 2021RMS Speed, Pressure, Kinetic Energy, Internal Energy

   Match Column-I and Column-II and choose the correct match from the given choices. Column-I: (A) Root mean square speed of gas molecules (B) Pressure exerted by ideal gas (C) Average kinetic energy of a molecule (D) Total internal energy of 1 mole of a diatomic gas. Column-II: (P) (1/3)nmv̄² (Q) √(3RT/M) (R) (5/2)RT (S) (3/2)kBT

   • A. (A)-(R), (B)-(Q), (C)-(P), (D)-(S)
   • B. (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
   • C. (A)-(Q), (B)-(R), (C)-(S), (D)-(P)
   • D. (A)-(Q), (B)-(P), (C)-(S), (D)-(R)
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   Answer: C — (A)-(Q), (B)-(R), (C)-(S), (D)-(P)

   Vrms = √(3RT/M) → A-Q. P = (1/3)nmVm² → B-P. Average KE = (3/2)kT → C-S. Total internal energy of diatomic gas = (5/2)RT → D-R.

10. Q10NEET 2021Matching gas kinetic theory relations

    Match Column-I and Column-II and choose the correct match. (A) Root mean square speed of gas molecules, (B) Pressure exerted by ideal gas, (C) Average kinetic energy of a molecule, (D) Total internal energy of 1 mole of a diatomic gas, with (P) (1/3)nmv̄², (Q) √(3RT/M), (R) (5/2)RT, (S) (3/2)kBT.

    • A. (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
    • B. (A)-(Q), (B)-(R), (C)-(S), (D)-(P)
    • C. (A)-(Q), (B)-(P), (C)-(S), (D)-(R)
    • D. (A)-(R), (B)-(Q), (C)-(P), (D)-(S)
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    Answer: B — (A)-(Q), (B)-(R), (C)-(S), (D)-(P)

    vrms=√(3RT/M) (Q); Pressure=(1/3)nmv̄² (P); average KE of a molecule=(3/2)kBT (S); total internal energy of 1 mol diatomic gas=(5/2)RT (R). (A)-(Q),(B)-(P),(C)-(S),(D)-(R).