Question

20 L of $ {\text{S}}{{\text{O}}_2} $ diffuses through a porous partition in 60 seconds. Volume of $ {{\text{O}}_2} $ diffuse under similar conditions in 30 seconds will be:

Answer 1

Hint :Diffusion is defined as the migration of a chemical from a high-concentration area to a low-concentration area. Diffusion occurs in liquids and gases when particles migrate randomly from one location to another. Diffusion is the mechanism through which chemicals flow into and out of cells in living organisms.

Complete Step By Step Answer:
Graham's Law, sometimes known as Graham's Law of Effusion, was developed by Thomas Graham in the year 1848. Thomas Graham studied the effusion process and identified an essential characteristic: lighter gas molecules move quicker than heavier gas molecules.
Graham's Law states that at constant pressure and temperature, smaller molecular mass molecules or atoms will effuse quicker than greater molecular mass molecules or atoms. Thomas even discovered the pace at which they disperse via diffusion. In other words, the rate of gas effusion is inversely proportional to the square root of the molecular mass of the gas.
It can be written as
 $ \dfrac{{{\text{ Rate}}{{\text{ }}_1}}}{{{\text{ Rate}}{{\text{ }}_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
 $ {M_1} $ represents the molar mass of gas 1
 $ {M_2} $ represents the molar mass of gas 2
 $ Rat{e_1} $ represents the rate of effusion of the first gas
 $ Rat{e_2} $ represents the rate of effusion for the second gas
The rate of diffusion is defined as the volume diffused per unit time and is inversely related to molar mass squared. The rate of diffusion, the volume diffused, the time required for diffusion, and the molar mass of $ S{O_2} $ are represented by r, V, t, and M, whereas the rate of diffusion, the volume diffused, the time required for diffusion, and the molar mass of $ {O_2} $ are represented by r', V', t', and M', respectively.
Substituting all the values we get
 $ \Rightarrow \dfrac{{{r_{S{O_2}}}}}{{{r_{{O_2}}}}} = \sqrt {\dfrac{{{M_{{O_2}}}}}{{{M_{S{O_2}}}}}} $
Hence
 $ \dfrac{{{V_{S{O_2}}}/60}}{{{V_{{O_2}}}/30}} = \sqrt {\dfrac{{32}}{{64}}} $
Also
 $ \Rightarrow \dfrac{{20 \times 30}}{{60 \times {V_{{{\text{O}}_2}}}}} = \sqrt {\dfrac{1}{2}} $
Rearranging the values to get
 $ \Rightarrow {{\text{V}}_{{{\text{O}}_2}}} = 10\sqrt 2 $
 $ {V_{{O_2}}} = \sqrt {200} $
Hence
 $ \Rightarrow {V_{{O_2}}} = 14.14{\text{d}}{{\text{m}}^3} $
14.14 L is the correct answer.

Note :
Effusion is the process of air escaping or leaking through a hole with a diameter less than the mean free path of molecules. Because collisions between molecules in these regions are insignificant at these conditions, all molecules that reach the hole will pass through.