Kinetic Theory: NEET and JEE questions with solutions

Answer: (D) $n_A = \dfrac{1}{2}n_B$

Mean free path: $\lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n}$, so $\lambda \propto \dfrac{1}{d^2 n}$.

$$\frac{\lambda_A}{\lambda_B} = \frac{d_B^2\, n_B}{d_A^2\, n_A} = \frac{d_B^2\, n_B}{4d_B^2\, n_A} = \frac{n_B}{4n_A}$$

Given $\dfrac{\lambda_A}{\lambda_B} = \dfrac{1}{2}$:
$$\frac{n_B}{4n_A} = \frac{1}{2} \;\Rightarrow\; n_A = \frac{1}{2}n_B$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4$

Each vibrational mode contributes 2 degrees of freedom (kinetic and potential energy). The total number of degrees of freedom is
$$F = 3 + 3 + 2f$$

$$\gamma = 1 + \frac{2}{F} = \frac{8}{7} \;\Rightarrow\; \frac{2}{F} = \frac{1}{7} \;\Rightarrow\; F = 14$$

$$6 + 2f = 14 \;\Rightarrow\; f = 4$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.116$ kg

Gauge pressure $11$ atm means absolute pressure $P = 11 + 1 = 12$ atm $= 12 \times 1.01 \times 10^5 = 12.12 \times 10^5$ Pa.

Moles left, with $V = 30 \times 10^{-3}\ \text{m}^3$ and $T = 300$ K:
$$n_2 = \frac{PV}{RT} = \frac{12.12 \times 10^5 \times 30 \times 10^{-3}}{\frac{100}{12} \times 300} = \frac{36360}{2500} = 14.54\ \text{mol}$$

Moles withdrawn: $18.20 - 14.54 = 3.66$ mol.

Mass withdrawn: $3.66 \times 32 = 117$ g $\approx 0.116$ kg.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.6$ atm

The container is isolated, so no heat enters and no work is done: the total internal energy is conserved.

For an ideal gas, $U = \dfrac{f}{2}nRT = \dfrac{f}{2}pV$. With the same gas on both sides:
$$p(V_1 + V_2) = p_1V_1 + p_2V_2$$
$$p = \frac{1 \times 2 + 2 \times 3}{2 + 3} = \frac{8}{5} = 1.6\ \text{atm}$$

(The mole numbers are not needed.)

Solution by Sreeraj P, M.Sc Physics