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Question : 10 of 14
Marks: +1, -0
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 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2=28.0 uN_2 = 28.0 \, \mathrm{u}).
Solution:  
Here, Pressure of nitrogen = 2 atm
=2×1.013×105 N m−2=2 \times 1.013 \times 10^{5} \, \mathrm{N} \, \mathrm{m}^{-2} =2.026×105 N m−2=2.026 \times 10^{5} \, \mathrm{N} \, \mathrm{m}^{-2}
Temperature of nitrogen =17∘ C=17+273=290 K= 17^{\circ} \, \mathrm{C} = 17 + 273 = 290 \, \mathrm{K}
Molecular diameter d =2×1=2 A˚=2×10−10 m= 2 \times 1 = 2 \, \text{\AA} = 2 \times 10^{-10} \, \mathrm{m}
Molecular mass of nitrogen =28 g=28×10−3 kg=28 \, \mathrm{g} = 28 \times 10^{-3} \, \mathrm{kg}
Mean free path = ?
Collision frequency f=?f = ?
Mean free path λ=kT2πd2P\lambda = \frac{k T}{\sqrt{2} \pi d^{2} P}
=1.38×10−23×2901.414×3.14(2×10−10)2×2.026×105= \frac{1.38 \times 10^{-23} \times 290}{1.414 \times 3.14 (2 \times 10^{-10})^{2} \times 2.026 \times 10^{5}}
=1.38×29×10−71.414×3.14×4×2.026= \frac{1.38 \times 29 \times 10^{-7}}{1.414 \times 3.14 \times 4 \times 2.026} =1.11×10−7 m=1.11 \times 10^{-7} \, \mathrm{m}
rms velocity of nitrogen vrms=3RTMv_{\mathrm{rms}} = \sqrt{\frac{3 R T}{M}} =3×8.31×29028×10−3=508.24 m/s= \sqrt{\frac{3 \times 8.31 \times 290}{28 \times 10^{-3}}} = 508.24 \, \mathrm{m/s}
Collision frequency =vrmsλ=508.241.11×10−7= \frac{v_{\mathrm{rms}}}{\lambda} = \frac{508.24}{1.11 \times 10^{-7}} =4.58×109=4.58 \times 10^{9}
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