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Calculate the mean free path of nitogen ...

Calculate the mean free path of nitogen at `27^(@)C` when pressure is 1.0 atm. Given, diameter of nitogen molecule = `1.5 Å, k = 1.38 xx 10^(-23) JK^(-1)`. If the average speed of nitrogen molecules is `675 ms^(-1)`, find the time taken by the molecule between two successive collsions and the frequency of collisions.

A

0.6 ns

B

0.4 ns

C

0.8 ns

D

0.3 ns

Text Solution

Verified by Experts

The correct Answer is:
A
Here, `T=27^(@)C=27+273=300K.`
`P=1 atm=1.01xx10^(5)Nm^(-2),d=1.5Ã…=1.5xx10^(-10)m,`
`k_(B)=1.38xx10^(-23)JK^(-1),lamda=?`
From `lamda=(k_(B)T)/(sqrt2pid^(2)P)=(1.38xx10^(-23)xx300)/(1.414xx3.14(1.5xx10^(-10))^(2)xx1.01xx10^(5))`
`=4.1xx10^(-7)m`
Time interval between two successive collisions
`t=("distance")/("speed")=(lamda)/(v)=(4.1xx10^(-7))/(675)~~0.6xx10^(-9)s`
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