For the case of nitrogen under standard conditions find : (a) the mean number of collisions experienced by each molecule per second , (b) the total number of collisions occurring between the molecules within `1 cm^3` of nitrogen per second.

For the case of nitrogen under standard conditions find : (a) the mean

1.	For the case of nitrogen under standard conditions find : (a) the mean number of collisions experienced by each molecule per second , (b) the total number of collisions occurring between the molecules within `1 cm^3` of nitrogen per second.
Answer» (a) `v = (1)/(tau) = (1)/(lamda//lt v gt) = (lt v gt)/(lamda)` =`sqrt(2) pi d^2 n lt v gt = .74 xx 10^10 s^-1 (sec 2.223)` (b) Total number of collisions is `(1)/(2) nv ~~ 1.0 xx 10^29 s cm^-3` Note, the factor `(1)/(2)`. When two molecules collide we must not count it twice.

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For the case of nitrogen under standard conditions find : (a) the mean number of collisions experienced by each molecule per second , (b) the total number of collisions occurring between the molecules within `1 cm^3` of nitrogen per second.

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