Question

The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times.
The value of x is :

Answer 1

Graham's law says that a gas's rate of diffusion or effusion is inversely related to its molecular weight squared. If one gas has four times the molecular weight of another, it will diffuse through a porous plug or escape through a tiny puncture in a vessel at half the pace (heavier gases diffuse more slowly). Years later, the kinetic theory of gases offered a comprehensive theoretical explanation of Graham's law.

Complete answer:
The proportionality constant between the molar flow owing to molecular diffusion and the gradient in the concentration of the species is known as diffusivity, mass diffusivity, or diffusion coefficient (or the driving force for diffusion). Diffusivity appears in Fick's law and a number of other physical chemistry equations. In a multi-species system, the diffusivity is usually set pairwise for a particular pair of species. The faster two substances diffuse into each other, the higher their diffusivity (relative to one another). In most cases, a compound's diffusion coefficient in air is 10,000 times greater than in water.
$\text{Diffusion coefficient }\alpha \text{ mean free path x meanspeed}$
$D\propto \bar{l}\times \bar{c}$
But $\bar{l}=\dfrac{1}{\sqrt{2}\pi {{\sigma }^{2}}}\dfrac{RT}{P{{N}_{0}}}$
or $\bar{l}\propto \dfrac{T}{P}$ and $\bar{c}=\sqrt{\dfrac{8RT}{\pi M}}$
or $\bar{c}\propto \sqrt{\dfrac{T}{M}}$
$\therefore D\propto \dfrac{T}{P}\cdot \sqrt{\dfrac{T}{M}}$
$D\propto \dfrac{{{T}^{3/2}}}{P\sqrt{M}}$
It is usually given as
$\dfrac{{{D}_{i}}}{{{D}_{f}}}=\dfrac{{{\lambda }_{i}}\sqrt{{{v}_{i}}}}{{{\lambda }_{f}}\sqrt{{{v}_{f}}}}$
We write it as
$\dfrac{{{D}_{i}}}{{{D}_{f}}}=\dfrac{\dfrac{{{T}_{i}}}{{{P}_{i}}}\sqrt{\dfrac{{{T}_{i}}}{M}}}{\dfrac{{{T}_{f}}}{{{P}_{f}}}\sqrt{\dfrac{{{T}_{f}}}{M}}}$
Now according to the question
$\dfrac{{{D}_{i}}}{{{D}_{f}}}=\dfrac{{{P}_{f}}\cdot {{T}_{i}}\sqrt{{{T}_{i}}}}{{{P}_{i}}\cdot {{T}_{f}}\sqrt{{{T}_{f}}}}$
Also
$\dfrac{{{D}_{i}}}{{{D}_{f}}}=\dfrac{2{{P}_{i}}\cdot {{T}_{i}}\sqrt{{{T}_{i}}}}{{{P}_{i}}\cdot 4{{T}_{i}}\sqrt{4{{T}_{i}}}}$
Cancelling the common terms
$\dfrac{{{D}_{i}}}{{{D}_{f}}}=\dfrac{2}{8}$
Upon simplification
$\dfrac{{{D}_{i}}}{{{D}_{f}}}=\dfrac{1}{4}$
$\Rightarrow {{D}_{f}}=4{{D}_{i}}$
Hence x = 4.

Note:
Diffusion is the net movement of anything (for example, atoms, ions, molecules, and energy) from a higher to a lower concentration region. Diffusion is triggered by a concentration gradient. Many areas, including physics (particle diffusion), chemistry, biology, sociology, economics, and finance, employ the notion of diffusion (diffusion of people, ideas, and price values). The basic concept of diffusion, however, is the same in all of them: a material or collection that is experiencing diffusion expands out from a place or location where it has a larger concentration.