The mean free path of CO under these conditions is 0.52 m

Rationale

The mean free path, , is given by the equation :

   [ m ]

where, in this instance,

  P = (1 x 10-4 / 760) x 101325 = 1.333 x 10-2 N m-2

and

  σ = 0.42 nm2 = 4.2 x 10-19 m2

Substitution gives

  λ = ( 1.38x10-23 x 300 ) / (1.414 x 1.33x10-2 x 4.2x10-19 )

⇒     λ = 0.52 m

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