Kinetic Theory NEET PYQs (2005–2026) with Solutions

10 solved Kinetic Theory PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumMean free path of gas molecules

   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 number densities of the gases $A$ and $B$ are $n_A$ and $n_B$, respectively, the correct option is:

   1. AOption A: $n_A = n_B$
   2. BOption B: $n_A = 2n_B$
   3. COption C: $n_A = \frac{1}{4}n_B$
   4. DOption D: $n_A = \frac{1}{2}n_B$

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   Correct answer: (D) $n_A = \frac{1}{2}n_B$

   Explanation

   Mean free path: $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$, so $\lambda \propto \frac{1}{d^2 n}$. $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A}$. With $\lambda_A = \frac{1}{2}\lambda_B$ and $d_A = 2d_B$: $\frac{1}{2} = \frac{n_B}{4n_A}$. So $4n_A = 2n_B$, i.e. $n_A = \frac{1}{2}n_B$, option (D).

2. Question 2 (NEET 2026)

   NEET 2026EasyKinetic Theory of Gases

   A flask contains argon and chlorine in the ratio of $2:1$ by mass. The temperature of the mixture is $27^\circ C$. The ratio of root mean square speed of the molecules of the two gases $\left(\dfrac{v_{\mathrm{rms}}^{\mathrm{Ar}}}{v_{\mathrm{rms}}^{\mathrm{Cl}}}\right)$ is: (Atomic mass of argon $=40.0\,u$ and molecular mass of chlorine $=70.0\,u$)

   1. AOption A: $\dfrac{\sqrt{7}}{2}$
   2. BOption B: $\dfrac{7}{4}$
   3. COption C: $\dfrac{7}{2}$
   4. DOption D: $\dfrac{2}{\sqrt{7}}$

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   Correct answer: (A) $\dfrac{\sqrt{7}}{2}$

   Explanation

   For gases at the same temperature: Thus, $v_{rms} \propto \dfrac{1}{\sqrt{M}}$ Therefore, Substituting: Hence, $\dfrac{v_{rms}^{Ar}}{v_{rms}^{Cl}}=\dfrac{\sqrt7}{2}$ Therefore, option (A) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumLaw of equipartition of energy

   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 $C_P/C_V$ of the gas is $8/7$, then the value of $f$ is:

   1. AOption A: $4$
   2. BOption B: $3$
   3. COption C: $2$
   4. DOption D: $1$

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   Correct answer: (A) $4$

   Explanation

   By equipartition, each translational and each rotational degree of freedom contributes $\frac{1}{2}k_BT$, and each vibrational mode contributes $k_BT$ (kinetic + potential). Energy per mole: $U = \left(\frac{3}{2} + \frac{3}{2} + f\right)RT = (3 + f)RT$, so $C_V = (3 + f)R$ and $C_P = C_V + R = (4 + f)R$. $\frac{C_P}{C_V} = \frac{4 + f}{3 + f} = \frac{8}{7} \Rightarrow 28 + 7f = 24 + 8f \Rightarrow f = 4$. Option (A).

4. Question 4 (NEET 2025)

   NEET 2025HardEquation of State

   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 oxygen withdrawn from the cylinder is nearly equal to : [Given, R = 100/12 J mol⁻¹ K⁻¹ and molecular mass of O₂ = 32, 1 atm pressure = 1.01 × 10⁵ N/m]

   1. AOption A: 0.125 kg
   2. BOption B: 0.144 kg
   3. COption C: 0.116 kg
   4. DOption D: 0.156 kg

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   Correct answer: (C) 0.116 kg

   Explanation

   Using PV = nRT for final condition: P = 11 × 1.01 × 10⁵ Pa, V = 30 × 10⁻³ m³, T = 300 K. Final moles ≈ 14.57 mol. Initial moles = 18.20 mol. Moles withdrawn ≈ 3.63 mol. Mass withdrawn = 3.63 × 32 g ≈ 116 g = 0.116 kg.

5. Question 5 (NEET 2023)

   NEET 2023MediumEquation of State

   Which amongst the following options is correct graphical representation of Boyle’s Law?

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (B) Option (2)

   Explanation

   According to Boyle’s law, at constant temperature: P ∝ 1/V Hence, a graph between pressure (P) and reciprocal of volume (1/V) is a straight line passing through the origin. For higher temperatures, the slope increases. Therefore, option (B) is the correct graphical representation.

6. Question 6 (NEET 2022)

   NEET 2022HardEquation of State

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

   1. AOption A: 5.6 m³
   2. BOption B: 5.6 × 10⁶ m³
   3. COption C: 5.6 × 10³ m³
   4. DOption D: 5.6 × 10⁻³ m³

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   Correct answer: (A) 5.6 m³

   Explanation

   Mass of water: m = 4.5 kg = 4500 g Number of moles: n = 4500/18 = 250 mol At STP, 1 mol ideal gas occupies 22.4 × 10⁻³ m³. Thus: V = 250 × 22.4 × 10⁻³ = 5.6 m³ Hence option (A) is correct.

7. Question 7 (NEET 2021)

   NEET 2021MediumKinetic Theory of Gases

   Match Column-I and Column-II and choose the correct match from the given choices.

   1. AOption A: (A)-(R), (B)-(Q), (C)-(P), (D)-(S)
   2. BOption B: (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
   3. COption C: (A)-(Q), (B)-(R), (C)-(S), (D)-(P)
   4. DOption D: (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

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   Correct answer: (D) (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

   Explanation

   Root mean square speed: vrms = √(3RT/M) → (Q) Pressure of ideal gas: P = (1/3)nmv² → (P) Average kinetic energy per molecule: K = (3/2)kBT → (S) For 1 mole of diatomic gas: U = (f/2)RT For diatomic gas, f = 5 U = (5/2)RT → (R) Hence correct matching is: (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

8. Question 8 (NEET 2020)

   NEET 2020MediumMean Free Path & Specific Heat

   The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as:

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (B) Option (2)

   Explanation

   The expression for mean free path is: $$\lambda = \frac{1}{\sqrt{2}\,n\pi d^2}$$ Hence, option (B) is correct.

9. Question 9 (NEET 2019)

   NEET 2019MediumEquipartition of Energy

   The value of $\gamma\left(=\frac{C_p}{C_v}\right)$ for hydrogen, helium and another ideal diatomic gas X (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to,

   1. AOption A: $\frac{7}{5},\;\frac{5}{3},\;\frac{7}{5}$
   2. BOption B: $\frac{7}{5},\;\frac{5}{3},\;\frac{9}{7}$
   3. COption C: $\frac{5}{3},\;\frac{7}{5},\;\frac{9}{7}$
   4. DOption D: $\frac{5}{3},\;\frac{7}{5},\;\frac{7}{5}$

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   Correct answer: (B) $\frac{7}{5},\;\frac{5}{3},\;\frac{9}{7}$

   Explanation

   For an ideal gas, $$\gamma = 1 + \frac{2}{f}$$ where $$f$$ is the degree of freedom. For hydrogen (diatomic gas): A diatomic gas without vibrational mode has: $$f = 5$$ Therefore, $$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$$ For helium (monoatomic gas): $$f = 3$$ Therefore, $$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$$ For gas X (diatomic gas with one vibrational mode active): $$f = 7$$ Therefore, $$\gamma = 1 + \frac{2}{7} = \frac{9}{7}$$ Hence, the correct answer is option (B).

10. Question 10 (NEET 2018)

    NEET 2018MediumRMS speed and escape velocity

    At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given: Mass of oxygen molecule $m = 2.76 \times 10^{-26}$ kg, Boltzmann constant $k_B = 1.38 \times 10^{-23}$ JK$^{-1}$)

    1. AOption A: $2.508 \times 10^4$ K
    2. BOption B: $8.360 \times 10^4$ K
    3. COption C: $1.254 \times 10^4$ K
    4. DOption D: $5.016 \times 10^4$ K

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    Correct answer: (B) $8.360 \times 10^4$ K

    Explanation

    For oxygen molecules to escape Earth's atmosphere, their rms speed must equal the escape velocity. Escape velocity of Earth: $$v_e = 11200\text{ m/s}$$ RMS speed is given by: $$v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}$$ Equating: $$11200 = \sqrt{\frac{3k_BT}{m}}$$ Squaring both sides: $$11200^2 = \frac{3k_BT}{m}$$ Therefore: $$T = \frac{m(11200)^2}{3k_B}$$ Substituting values: $$T = \frac{2.76 \times 10^{-26} \times (11200)^2}{3 \times 1.38 \times 10^{-23}}$$ $$T \approx 8.360 \times 10^4\text{ K}$$ Hence, option (B) is correct.