Q29 (2026)
The mean free path of a molecule of diameter $5 \times 10^{-10} \text{ m}$ at the temperature $41^{\circ}\text{C}$ and pressure $1.38 \times 10^{5} \text{ Pa}$, is given as ____ m.
$(\mathrm{Given~k_{B}} = 1.38\times 10^{-23}\text{ J / K}).$
- $2\sqrt{2} \times 10^{-10}$
- $10\sqrt{2} \times 10^{-8}$
- $2\sqrt{2} \times 10^{-8}$
- $2 \times 10^{-8}$
### Related Formula
$$\lambda = \frac{k_B T}{\sqrt{2} \pi \sigma^2 P}$$
where,
$\lambda$ = mean free path
$k_B$ = Boltzmann constant
$T$ = absolute temperature in Kelvin
$\sigma$ = diameter of the molecule
$P$ = pressure of the gas
### Step 1: Extract and Convert Given Data
$d = \sigma = 5 \times 10^{-10} \text{ m}$
$T = 41^\circ\text{C} = 273 + 41 = 314 \text{ K}$
$P = 1.38 \times 10^5 \text{ Pa}$
$k_B = 1.38 \times 10^{-23} \text{ J/K}$
### Step 2: Substitution and Calculation
$$\lambda = \frac{1.38 \times 10^{-23} \times 314}{\sqrt{2} \times \pi \times (5 \times 10^{-10})^2 \times 1.38 \times 10^5}$$
Cancel out $1.38$ from numerator and denominator:
$$\lambda = \frac{10^{-23} \times 314}{\sqrt{2} \times 3.14 \times 25 \times 10^{-20} \times 10^5}$$
$$\lambda = \frac{314 \times 10^{-23}}{\sqrt{2} \times 3.14 \times 25 \times 10^{-15}}$$
$$\lambda = \frac{100 \times 10^{-23}}{\sqrt{2} \times 25 \times 10^{-15}}$$
$$\lambda = \frac{4 \times 10^{-8}}{\sqrt{2}} = 2\sqrt{2} \times 10^{-8} \text{ m}$$
### Pattern Recognition
The temperature $314 \text{ K}$ and $\pi \approx 3.14$ are intentionally matched to cancel out and give a factor of 100. $1.38$ cancels completely. The rest is simply handling powers of 10.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory
