What is the dependency of rate of diffusion ( Graham's law) on mean free path, molecular size , average velocity, collision frequency and pressure? I have searched everywhere on internet but just found that rate of diffusion is inversely proportional to square root of density or molecular mass.

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    $\begingroup$ A proper textbook on physical chemistry should explain this in detail. You might have better luck searching in one than on the internet $\endgroup$ – orthocresol Aug 18 '18 at 14:31

My reference is Concepts in Thermal Physics, Blundell and Blundell, Ch. 7, 2nd ed. Everything here is from there.

Consider a gas trapped in a rigid box. We begin by evaluating the molecular flux incident on one part of one wall of the box. Let there be $n$ particles per unit volume, so $n$ is the particle density of the gas in the box. Let $A$ be the area of a small patch of the wall, and let $\theta$ be the angle between the particle's direction of approach and the normal to the wall. Let $f(v)$ denote the speed distribution function of the particles, which we assume to be a Maxwell-Boltzmann distribution.

Consider all the molecules with speed between $v$ and $v+\text{d}v$ approaching the wall at an angle between $\theta$ and $\theta+\text{d}\theta$, which we write in shorthand as ($v,\theta$). There are $n$ particles per unit volume, and a fraction $f(v)\,\text{d}v$ of them are moving within the desired speed range. A further fraction $\sin\theta\,\text{d}\theta/2$ of them are moving in the desired direction, so the particle density for ($v$, $\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2.$$ (To derive the fraction $\sin\theta\,\text{d}\theta/2$, we consider a unit sphere with surface area $4\pi$. The surface area of the differential annulus is $2\pi\sin\theta$ (circumference) $\times\;\text{d}\theta$ (differential width), so the fraction of surface area occupied by the annulus is as stated.)

Now consider how many particles traveling at ($v,\theta$) hit the patch of wall in a time $\text{d}t$. These particles must be contained within a volume $$A\,\times\,v\,\text{d}t\cos\theta$$ (distance particle travels normal to the wall in a time $\text{d}t$), for all particles traveling at ($v,\theta$) must impinge on the wall within $\text{d}t$. Actually this area is slightly enlarged by a width $v\sin\theta\,\text{d}t$ uniformly, but this contribution to the area is of order $\text{d}t^2$ and vanishes in the small $\text{d}t$ limit.

All in all, then, the number of molecular collisions in $\text{d}t$ on a patch of wall of area $A$ at ($v,\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2\,\times\,Av\,\text{d}t\cos\theta.$$

The flux $\Phi$ is the rate of molecular collisions per unit area, so we divide by $A$ and $\text{d}t$, and integrate over all $v$ and $\theta$. This yields $$\Phi = \int_0^\infty\text{d}v\int_0^{\pi/2}\text{d}\theta\,nvf(v)\sin\theta/2\cos\theta = \frac{1}{4}n\langle v\rangle = \frac{p}{\sqrt{2\pi mk_BT}},$$ assuming an ideal gas and given the average speed of gases within the Maxwell-Boltzmann distribution.

By definition of the molecular flux, the rate of effusion is simply $\Phi A'$ for $A'$ the cross-sectional area of the hole out of which gas effuses, and the resulting dependencies can be read off from the equation for $\Phi$.

Interesting notes.

  • Due to the extra factor of $v$ in the integral for the molecular flux, the distribution of speeds of the effusing particles is $v^3e^{-\beta\epsilon}$ , rather than the standard $v^2e^{-\beta\epsilon}$. This extra factor of $v$ comes about because speedier molecules hit the walls of the box faster than do slower ones.
  • The hole through which particles effuse must be small, and the length scale that governs smallness is the mean free path. Make the hole too large, and the box eventually reaches equilibrium with the outside world. Consider a system of two boxes placed side by side, and a small hole between the neighboring walls of the boxes. For holes small enough that effusive behavior is observed, at equilibrium the molecular fluxes must balance and hence $$\frac{p_1}{\sqrt{T_1}} = \frac{p_2}{\sqrt{T_2}}.$$ On the other hand, if the holes are too large, then we simply have regular thermodynamic equilibrium, with $$\frac{p_1}{T_1} = \frac{p_2}{T_2}.$$