Physics · chapter PYQs

Kinetic Theory, on the exam screen.

6 questions from NEET 2026, Re-NEET 2026, NEET 2020, 6 minutes, marked +4 / −1. See them with solutions.

Kinetic Theory · PYQs

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Question 1 · NEET 2026 · Q7 · +4 / −1
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 $\left(\frac{v_{rms}^{Ar}}{v_{rms}^{Cl}}\right)$ is: (Atomic mass of argon = 40.0 u and molecular mass of chlorine = 70.0 u)
Answer and explanation
Answer: (1)
Since $v_{rms} = \sqrt{\frac{3RT}{M}}$ , at the same temperature $v_{rms} \propto \frac{1}{\sqrt{M}}$ . Therefore $\frac{v_{rms}^{Ar}}{v_{rms}^{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}} = \sqrt{\frac{70}{40}} = \frac{\sqrt{7}}{2}$ , option (1). The 2:1 mass ratio of the gases does not affect the rms speed, which depends only on temperature and molar mass.
Question 2 · Re-NEET 2026 · Q2 · +4 / −1
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:
Answer and explanation
Answer: (4)
Mean free path is $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$ , so $\lambda \propto \frac{1}{d^2 n}$ . Taking the ratio, $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A}$ . Substituting $\lambda_A/\lambda_B = \frac{1}{2}$ and $d_A = 2d_B$ gives $\frac{1}{2} = \frac{n_B}{4n_A}$ , hence $n_A = \frac{1}{2}n_B$ . NCERT Class 11, Kinetic Theory (mean free path).
Question 3 · Re-NEET 2026 · Q45 · +4 / −1
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:
Answer and explanation
Answer: (1)
Equipartition assigns $\frac{1}{2}k_BT$ to each translational and rotational mode but a full $k_BT$ to each vibrational mode, so $C_V = \frac{R}{2}(3 + 3 + 2f) = (3+f)R$ and $C_P = C_V + R = (4+f)R$ . Setting $\frac{4+f}{3+f} = \frac{8}{7}$ gives $28 + 7f = 24 + 8f$ , so $f = 4$ . NCERT Class 11, Kinetic Theory (law of equipartition of energy).
Question 4 · NEET 2020 · Q150 · +4 / −1
A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27°C. Its density is (R = 8.3 J mol $^{-1}$ K $^{-1}$ )
Answer and explanation
Answer: (1)
The density $\rho$ of an ideal gas is given by $\rho = \frac{P \times M}{R \times T}$ , where P is the pressure, M is the molar mass, R is the gas constant, and T is the temperature in Kelvin. Substituting the values, $P = 249 \times 10^3$ Pa, $M = 2 \times 10^{-3}$ kg/mol, $R = 8.3$ J/mol K, and $T = 27 + 273 = 300$ K, we get $\rho = \frac{249 \times 10^3 \times 2 \times 10^{-3}}{8.3 \times 300} \approx 0.2$ kg/m $^3$ . Chapter: Kinetic Theory.
Question 5 · NEET 2020 · Q157 · +4 / −1
The mean free path for a gas, with molecular diameter d and number density n can be expressed as
Answer and explanation
Answer: (1)
The kinetic-theory expression for mean free path is $\lambda$ = 1/( $\sqrt{}$ 2 n $\pi$ d $^{2}$ ), where n is number density and d is molecular diameter. Chapter: Kinetic Theory.
Question 6 · NEET 2020 · Q178 · +4 / −1
The average thermal energy for a mono-atomic gas is (kB is Boltzmann constant and T, absolute temperature)
Answer and explanation
Answer: (1)
The average thermal energy for a mono-atomic gas is given by $\frac{3}{2} k_B T$ . This is derived from the equipartition theorem, which states that each degree of freedom contributes $\frac{1}{2} k_B T$ to the energy, and a mono-atomic gas has three translational degrees of freedom. Chapter: Kinetic Theory.