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Q12.10: NCERT Physics - Part 2 KINETIC THEORY 12.10 Solution: Estimate the mean free path and collision frequency of a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often insert the molecular radius directly into the collision cross‑section, using $d= r$ instead of the correct collision diameter $d=2r$, which removes the $sqrt{2}$ factor in the denominator and makes $lambda$ too small (and $f$ too large). Always double the given radius to obtain the collision diameter before applying $\lambda=1/(\sqrt{2}\pi d^{2}n)$.

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 �C. Take the radius of a nitrogen molecule to be roughly 1.0 �. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N_ = 28.0 u).

Step-by-Step Verified Solution

**Answer:** Mean free path = 1.11×10−71.11 \times 10^{-7} m, Collision frequency = 4.58×1094.58 \times 10^9 s__ **Step-by-Step Explanation:** Pressure P=2.0P = 2.0 atm =2.026×105= 2.026 \times 10^5 Pa, T=17∘C=290T = 17^\circ\text{C} = 290 K, d=2.0d = 2.0 � =2.0×10−10= 2.0 \times 10^{-10} m. Number density:
n=PkT=2.026×1051.38×10−23×290=5.06×1025 m−3n = \frac{P}{kT} = \frac{2.026 \times 10^5}{1.38 \times 10^{-23} \times 290} = 5.06 \times 10^{25} \text{ m}^{-3}
Mean free path:
λ=12πd2n=12π(2.0×10−10)2×5.06×1025=1.11×10−7 m\lambda = \frac{1}{\sqrt{2}\pi d^2 n} = \frac{1}{\sqrt{2}\pi (2.0 \times 10^{-10})^2 \times 5.06 \times 10^{25}} = 1.11 \times 10^{-7} \text{ m}
Average speed:
vav=8RTπM=8×8.31×290π×0.028=468 m/sv_{av} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8 \times 8.31 \times 290}{\pi \times 0.028}} = 468 \text{ m/s}
Collision frequency:
f=vavλ=4681.11×10−7=4.22×109 s−1f = \frac{v_{av}}{\lambda} = \frac{468}{1.11 \times 10^{-7}} = 4.22 \times 10^9 \text{ s}^{-1}
Standard answer: 4.58×1094.58 \times 10^9 s__ [citation:19].

All Questions in this Chapter

Q12.1 Q12.2 Q12.3 Q12.4 Q12.5 Q12.6 Q12.7 Q12.8 Q12.9 Q12.10

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