Calculate the mean free path of nitrogen molecule at 27°C when pressure is 1.0 atm. Given, diameter of nitrogen molecule = 1.5 A∘, kB= 1.38 x 10-23 J K-1. If the average speed of nitrogen molecule is 675 m s-1. The time taken by the molecule between two successive collisions is

Here, T=27°C=27+273=300 KP =1 atm= 1.01 × 105 Nm-2 , d=1.5 A∘=1.5 ×10-10 m , KB=1.38×10-23J K-1 , λ=?From λ=kBT2πd2p=1.38×10-23×3001.414×3.141.5×10-102×1.01×105=4.1×10-7 mTime interval between two successive collisionst=distancespeed=λv=4.1×10-7675=0.6×10-9 s

Calculate the mean free path of nitrogen molecule at 27°C when pressure is 1.0 atm. Given, diameter of nitrogen molecule = 1.5 A∘, kB= 1.38 x 10-23 J K-1. If the average speed of nitrogen molecule is 675 m s-1. The time taken by the molecule between two successive collisions is

Here, T=27°C=27+273=300 KP =1 atm= 1.01 × 105 Nm-2 , d=1.5 A∘=1.5 ×10-10 m , KB=1.38×10-23J K-1 , λ=?From λ=kBT2πd2p=1.38×10-23×3001.414×3.141.5×10-102×1.01×105=4.1×10-7 mTime interval between two successive collisionst=distancespeed=λv=4.1×10-7675=0.6×10-9 s

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Calculate the mean free path of nitrogen molecule at $27 ° C$ when pressure is $1.0 a t m$ . Given, diameter of nitrogen molecule $= 1.5 \overset{\circ}{A}$ , $k_{B} = 1.38 x 10^{- 23} J K^{- 1}$ . If the average speed of nitrogen molecule is $675 m s^{- 1}$ . The time taken by the molecule between two successive collisions is

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$0.4 n s$

$0.8 n s$

$0.3 n s$

$0.6 n s$

answer is A.

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Detailed Solution

Here, $T = 27 ° C = 27 + 273 = 300 K$

$P = 1 a t m$ $= 1.01 \times 10^{5} N m^{- 2}$ , $d = 1.5 \overset{\circ}{A} = 1.5 \times 10^{- 10} m,$ $K_{B} = 1.38 \times 10^{- 23} J K^{- 1}$ $, λ = ?$