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What will be the effect on the rate of diffusion on addition of an inert gas to the gaseous mixture?

I think the rate of diffusion should increase as the addition of extra gas will increase the inside pressure. But the given answer contradicts my proposed explanation. Where am I wrong? And why is the rate of diffusion decreasing?

My question does not ask about the effect of addition of an inert gas on a reaction equilibrium.

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  • $\begingroup$ Is this from a book? If you add that source and the exact quote, other people with a similar problem will also be able to find this question. $\endgroup$ – Gaurang Tandon Mar 9 '18 at 10:14
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The inter-diffusion caused by two gasses is described by the Stefan-Maxwell equation.

If $x_1,x_2$ are the mole fractions of the two gasses, $\bar v$, the average speed and $\lambda$ the mean free path then

$$D_{1,2}= \frac{x_2}{2}\bar v_1\lambda_1+ \frac{x_1}{2}\bar v_2\lambda_2$$

where $D_{1,2}$ is the inter-diffusion coefficient. Substituting for the mean free paths does not lead to a useful result because terms that involve collision between molecules of the same kind cannot have any extra effect compared to when only one gas is present, and so these are ignored. The result is

$$D_{1,2}= \frac{1}{\pi\sigma_{1,2}^2(n_1+n_2)} \left( \frac{2k_BT}{\pi\mu} \right)^{1/2}$$

where $\sigma_{1,2}$ is the sum of the radii of the two molecule types and $n_1, n_2$ number of molecules/m$^3$ of each, the reduced mass is $\mu$ kg ($\mu=m_1m_2/(m_1+m_2)$).

This equation shows that the inter-diffusion depends on the total concentration at a given temperature, a result that is close to that observed experimentally. So your intuition was correct.

(ref chapter (II). E. Moelwyn-Hughes, 'Physical Chemistry')

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