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Exercises · 12.10

Q.Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm2.0\ \text{atm} and temperature 17 ∘C17\ ^\circ\text{C}. Take the radius of a nitrogen molecule to be roughly 1.0 A˚1.0\ \text{\AA}. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2=28.0 u\text{N}_2 = 28.0\ \text{u}).

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At 2.0 atm2.0\ \text{atm} and 290 K290\ \text{K} the mean free path is λ≈1.1×10−7 m\lambda \approx 1.1\times10^{-7}\ \text{m} and the collision frequency is ν≈4.6×109 s−1\nu \approx 4.6\times10^{9}\ \text{s}^{-1}. A collision lasts about 500500 times less than the free-flight time between collisions.

Set-up

We need the number density nn, then the mean free path λ\lambda, the molecular speed, the collision frequency ν=v/λ\nu = v/\lambda, and finally the ratio of collision time to free-flight time.

λ=12 πd2n,ν=vrmsλ\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}, \qquad \nu = \frac{v_{\text{rms}}}{\lambda}

Data (SI)

  • P=2.0 atm=2.026×105 PaP = 2.0\ \text{atm} = 2.026\times10^{5}\ \text{Pa}
  • T=17 ∘C=290 KT = 17\ ^\circ\text{C} = 290\ \text{K}
  • r=1.0 A=1.0×10−10 m ⇒ d=2.0×10−10 mr = 1.0\ \text{A} = 1.0\times10^{-10}\ \text{m}\ \Rightarrow\ d = 2.0\times10^{-10}\ \text{m}
  • m=28.0×1.66×10−27=4.65×10−26 kgm = 28.0\times1.66\times10^{-27} = 4.65\times10^{-26}\ \text{kg}

Step 1 - Number density

n=PkT=2.026×105(1.38×10−23)(290)≈5.06×1025 m−3n = \frac{P}{kT} = \frac{2.026\times10^{5}}{(1.38\times10^{-23})(290)} \approx 5.06\times10^{25}\ \text{m}^{-3}

Step 2 - Mean free path

λ=12 π(2.0×10−10)2(5.06×1025)≈1.11×10−7 m\lambda = \frac{1}{\sqrt{2}\,\pi (2.0\times10^{-10})^2 (5.06\times10^{25})} \approx 1.11\times10^{-7}\ \text{m}

Step 3 - Molecular speed

vrms=3kTm=3(1.38×10−23)(290)4.65×10−26≈5.1×102 m s−1v_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3(1.38\times10^{-23})(290)}{4.65\times10^{-26}}} \approx 5.1\times10^{2}\ \text{m s}^{-1}

Step 4 - Collision frequency

ν=vrmsλ=5081.11×10−7≈4.6×109 s−1\nu = \frac{v_{\text{rms}}}{\lambda} = \frac{508}{1.11\times10^{-7}} \approx 4.6\times10^{9}\ \text{s}^{-1}

Step 5 - Collision time vs free-flight time

A collision lasts roughly the time to cross one molecular diameter: …

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