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Obtain the mean free path of nitrogen mo...

Obtain the mean free path of nitrogen molecule at `0^(@)C` and 1.0 atm pressure. The molecular diameter of nitrogen is 324 pm (assume that the gas is ideal).

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Given `T=0^(@)C=273K, P=1.0"atm"=1.01 xx 10^(5)Pa and d=324"pm"=324 xx 10^(-12)m`.
For ideal gas `PV=Nk_(B)T, :. (N)/(V)=(P)/(k_(B)T)`.
Using Eq. mean free path
`lambda=(1)/(sqrt(2)pid^(2)((N)/(V)))=(k_(B)T)/(sqrt(2)pid^(2)P)`
`=((1.38 xx 10^(-23)J//K)(273K))/(sqrt(2)pi(324 xx 10^(-12)m)^(2)(1.01 xx 10^(5)Pa))`
`=0.8xx10^(-7)m`
Note that this is about 247 times molecular diameter.
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