Question

When , $p$ and $M$ represent rate of diffusion, pressure and molecular mass, respectively, then the ratio of the rates of diffusion $(\dfrac{{{r}_{A}}}{{{r}_{B}}})$ of two gases A and B is given as:
A) ${{(\dfrac{{{p}_{A}}}{{{p}_{B}}})}^{\dfrac{1}{2}}}(\dfrac{{{M}_{A}}}{{{M}_{B}}})$
B) $(\dfrac{{{p}_{A}}}{{{p}_{B}}}){{(\dfrac{{{M}_{A}}}{{{M}_{B}}})}^{\dfrac{1}{2}}}$
C) ${{(\dfrac{{{p}_{A}}}{{{p}_{B}}})}^{\dfrac{1}{2}}}(\dfrac{{{M}_{B}}}{{{M}_{A}}})$
D) $(\dfrac{{{p}_{A}}}{{{p}_{B}}}){{(\dfrac{{{M}_{B}}}{{{M}_{A}}})}^{\dfrac{1}{2}}}$

Answer 1

Hint: We know that how diffusion works as well as how it affects molecular mass, Diffusion is a process of passive transport in which molecules move from an area of higher concentration to one of lower concentration. This results in Rate of diffusion proportional to pressure, whereas if you see rate of diffusion is indirectly proportional to Square root of Molecular Mass
Complete step-by-step answer:
We have a formula or we can say a equation for rate of Diffusion, which is givne by $r\alpha p$
Also we can also write it as $r\alpha \dfrac{p}{\sqrt{M}}$
Therefore by removing proportionality constant $\therefore r=\dfrac{1}{\sqrt{M}}$
For Gas A, the rate of diffusion can be given by
${{r}_{A}}\alpha \dfrac{{{p}_{A}}}{\sqrt{{{M}_{A}}}}$………………………… equation $(i)$
For Gas B, the rate of diffusion can be given by
${{r}_{B}}\alpha \dfrac{{{p}_{B}}}{\sqrt{{{M}_{B}}}}$………………………… equation $(ii)$
Dividing Equation $(i)$ by Equation $(ii)$ , we get
$\dfrac{{{r}_{A}}}{{{r}_{B}}}=\dfrac{{{p}_{A}}}{{{p}_{B}}}\times \sqrt{\dfrac{{{M}_{B}}}{{{M}_{A}}}}$
Or it can be written as
$\dfrac{{{r}_{A}}}{{{r}_{B}}}=\dfrac{{{p}_{A}}}{{{p}_{B}}}\times {{\left( \dfrac{{{M}_{B}}}{{{M}_{A}}} \right)}^{\dfrac{1}{2}}}$
Therefore, the ratio of the rates of diffusion $(\dfrac{{{r}_{A}}}{{{r}_{B}}})$ of two gases A and B is given as $(\dfrac{{{p}_{A}}}{{{p}_{B}}}){{(\dfrac{{{M}_{B}}}{{{M}_{A}}})}^{\dfrac{1}{2}}}$

So, Option (D) is Correct Option i.e. $(\dfrac{{{p}_{A}}}{{{p}_{B}}}){{(\dfrac{{{M}_{B}}}{{{M}_{A}}})}^{\dfrac{1}{2}}}$

Additional Information: The motion of molecules and the mass of molecules give rise to kinetic energy. The average kinetic energy of molecules gives rise to the phenomenon of temperature that means, at constant temperature, small molecules move faster than big ones to have the same kinetic energy. Which molecule will diffuse faster, one that is going fast of course it is going to be the fast one: the diffusivity of a substance is inversely proportional to molar mass.

Note: ote that the \[r\text{ }=\dfrac{1}{\sqrt{M}}\] diffusion is dependent on the inverse square root of the molar mass. Given the same temperature (equivalent average kinetic energy), heavier molecules will be moving more slowly on average. And the more slowly they move, the slower they will diffuse.