Kinetic Theory: NEET previous year questions

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.

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).

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).

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.

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.

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.