Question

Given,$20\,d{m^3}$ of $S{O_2}$ diffuses through a porous partition in $60$ seconds? What volume be ${O_2}$ will diffuse under similar conditions in $30$ second?

Answer 1

Hint: We have to know that Diffusion is a process where one material moves from a region of high concentration to a region of lower concentration. This implies that molecules go through a particular medium. The rate of gases and molar masses could be explained using Graham’s law.

Complete answer:
As per Graham's Law, at constant temperature and pressure, particles or atoms with lower molecular mass would radiate quicker when compared to the higher molecular mass particles or molecules. Thomas even discovered the rate at which they escape by diffusion. As such, it expresses that the pace of radiation of gas is indirectly related to the square root of its atomic mass. This equation is by and large utilized while looking at the rates of two unique gases at equivalent temperature and pressure. We can express the formula for Graham’s law of diffusion as,
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]
Here, the molar mass of gas 1 is ${M_1}$
The molar mass of gas 2 is ${M_2}$.
The rate of diffusion of gas 1 is $Rat{e_1}$.
The rate of diffusion of gas 2 is $Rat{e_2}$.
Let us now calculate the volume of oxygen using Graham's law of diffusion.
We know the molecular mass of oxygen is $32g/mol$.
We know the molecule mass of sulfur dioxide is $64g/mol$.
We are provided with volume and time for diffusion of sulfur dioxide and time for the diffusion of oxygen. We can modify the equation of Graham’s law of diffusion as,
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \dfrac{{\dfrac{{{V_1}}}{{{t_1}}}}}{{\dfrac{{{V_2}}}{{{t_2}}}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]
Using the above formula, we can calculate the volume of oxygen as,
\[\dfrac{{\dfrac{{{V_1}}}{{{t_1}}}}}{{\dfrac{{{V_2}}}{{{t_2}}}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \]
\[\dfrac{{\dfrac{{20}}{{60}}}}{{\dfrac{{{V_2}}}{{30}}}} = \sqrt {\dfrac{{32}}{{64}}} \]
\[{V_2} = 14.1\,d{m^3}\]
The volume be ${O_2}$ will diffuse under similar conditions in $30$ second is $14.1\,d{m^3}$.

Note:
 We have to remember Graham’s law of effusion and Graham’s law of diffusion are the same. This law describes that effusion rate (or) diffusion rate is inversely related to the molecular mass of the gas. We also have to remember that lighter gas molecules would travel at a faster rate when compared to heavier gas molecules.