The mean free path of nitrogen molecules at a pressure of $$1$$ atm and temperature $$0$$ºC is $$0.8$$×$$10$$−$$7m$$. If the number density of molecules is $$2.7$$×$$1025$$ per $$m3$$, then the molecular diameter is $$3.2$$×N Ao. Find N.

λ$$=12$$$$π$$$$nd2$$⇒$$d=12$$πλ$$n=12$$$$π$$$$0.8$$×$$10-72.7$$×$$1025Substituting$$ the values, we get,⇒$$d=0.32$$×$$10$$−$$9m=3.2Ao$$

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The mean free path of nitrogen molecules at a pressure of 1 atm and temperature $0 º C$ is $0.8 \times 10^{- 7} m$ . If the number density of molecules is $2.7 \times 10^{25} p e r m^{3}$ , then the molecular diameter is $(3.2 \times N) \overset{o}{A}$ . Find N.

Moderate

Solution

$λ = \frac{1}{\sqrt{2} π n d^{2}}$
$\Rightarrow d = \sqrt{\frac{1}{\sqrt{2} π λ n}} = \sqrt{\frac{1}{\sqrt{2} π (0.8 \times 10^{- 7}) 2.7 \times 10^{25}}}$
Substituting the values, we get,
$\Rightarrow d = 0.32 \times 10^{- 9} m = 3.2 \overset{o}{A}$

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