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Answer 1 · 2017-09-02T10:27:59

From kinetic theory, the expression for diffusion coefficient is,

$D = \frac{1}{3} ¯ v λ$ $D = 1/3barvlamda$

But, from Maxwell's expression for mean free path,

$λ = \frac{1}{\sqrt{2} n σ^{2} π}$ $lamda = 1/(sqrt(2)nsigma^2pi)$
Where $σ$ $sigma$ is molecular diameter.

Thus, the diffusion coefficient,

$D = \frac{¯ v}{3 \sqrt{2} π σ^{2} n}$ $D = barv/(3sqrt(2)pisigma^2n)$

Since, $ρ = m n$ $rho = mn$ is the density, then,

$D = \frac{¯ v m}{3 \sqrt{2} ρ π σ^{2}}$ $D = (barvm)/(3sqrt(2)rhopisigma^2)$

Therefore, the diffusion coefficient varies inversely with density!