Question

Find the order of the mole of gases ${\text{N}}{{\text{H}}_{\text{3}}}$, ${\text{C}}{{\text{O}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ adsorbed by 1g charcoal at 300K and 1atm pressure:
(A) ${\text{C}}{{\text{O}}_{\text{2}}} > {{\text{H}}_{\text{2}}} > {\text{N}}{{\text{H}}_{\text{3}}}$
(B) ${\text{N}}{{\text{H}}_{\text{3}}}> {\text{C}}{{\text{O}}_{\text{2}}}> {{\text{H}}_{\text{2}}}$
(C) ${{\text{H}}_{\text{2}}}> {\text{C}}{{\text{O}}_{\text{2}}}> {\text{N}}{{\text{H}}_{\text{3}}}$
(D) ${\text{C}}{{\text{O}}_{\text{2}}}> {\text{N}}{{\text{H}}_{\text{3}}}> {{\text{H}}_{\text{2}}}$

Answer 1

Hint: (1) The phenomenon of adsorption refers to the attraction and retention of the molecules of a substance on the surface of a solid or liquid.
(2) The adsorption of different gases by solids depend on a number of factors, one of them is the nature of the gas being adsorbed.

Complete step by step answer: For a gas, the critical temperature refers to the maximum temperature at which it can be liquefied by pressure. This critical temperature is different for different gases. Now, what is this liquefaction of gases? When a gas is compressed at a temperature below its critical temperature, liquid is suddenly formed at a particular pressure. If the gas is compressed further, the amount of liquid will increase and the amount of gas will decrease. The pressure will remain constant until all the gas is converted into liquid form. This phenomenon in which a substance in the gaseous state is converted into the liquid state is known as liquefaction of gas.
When the critical temperature of a gas increases, its ease of liquefaction also increases. In other words, higher the critical temperature of a gas, more will be the amount of that particular gas adsorbed. Or it can also be said that a gas that is more easily liquefiable or a gas that is more soluble in water is more easily adsorbed. This is because the ease of liquefaction of a gas is directly proportional to the Van der Waals forces of attraction. Van der Waals forces increase with increase in the size of the molecule and the number of electrons present in it as large molecules have more surface area of their electron clouds.
Now, the given gas molecules are ${\text{N}}{{\text{H}}_{\text{3}}}$, ${\text{C}}{{\text{O}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ all adsorbed by 1g charcoal at 300K and 1atm pressure. Different gas molecules will be adsorbed by the same adsorbent at the same temperature to different extents and so all the given gases will be adsorbed by 1g charcoal at 300K to different extents.
We can see that the size of the molecules increases from ${{\text{H}}_{\text{2}}}$ to ${\text{C}}{{\text{O}}_{\text{2}}}$ to ${\text{N}}{{\text{H}}_{\text{3}}}$. Thus, the Van der Waals forces increase in the same order. So, the ease of liquefaction and hence the critical temperature of these gases will also increase from ${{\text{H}}_{\text{2}}}$ to ${\text{C}}{{\text{O}}_{\text{2}}}$ to ${\text{N}}{{\text{H}}_{\text{3}}}$. As a result, the amount of gas adsorbed will also increase in the same order: ${{\text{H}}_{\text{2}}}< {\text{C}}{{\text{O}}_{\text{2}}}< {\text{N}}{{\text{H}}_{\text{3}}}$ or ${\text{N}}{{\text{H}}_{\text{3}}}< {\text{C}}{{\text{O}}_{\text{2}}}< {{\text{H}}_{\text{2}}}$.

So, the option (B) is correct.

Note: The volume of different gases adsorbed by the same adsorbent at the same temperature increase with increase in their critical temperature. Since the number of moles is directly proportional to volume, so the number of moles also increases under the same conditions.