Question

A $10{\text{L}}$ container at $300{\text{K}}$ contains $C{O_2}$ gas at pressure of $0.2{\text{atm}}$ and an excess solid ${\text{CaO}}$ (neglect the volume of solid ${\text{CaO}}$ ). The volume of the container is now decreased by moving the movable piston fitted in the container. What will be the maximum volume of the container when pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$ attains its maximum value.
given that ${\text{CaC}}{{\text{O}}_{\text{3}}} \rightleftarrows {\text{CaCO}}\left( {\text{s}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$ . ${{\text{K}}_{\text{p}}}{\text{ = 0}}{\text{.800atm}}$
A.${\text{5L}}$
B.${\text{2}}{\text{.5L}}$
C.${\text{1L}}$
D.The information is insufficient

Answer 1

To solve this question, it is required to have knowledge about Boyle’s law. Since we have been given volume, pressure, temperature and ${{\text{k}}_{\text{p}}}$ we can calculate the final pressure of the gas and then the volume it attains using Boyle’s law.
Formula used:
Boyle's Law: ${{\text{P}}_{\text{i}}}{{\text{V}}_{\text{i}}}{\text{ = }}{{\text{P}}_{\text{f}}}{{\text{V}}_{\text{f}}}$
where ${{\text{P}}_{\text{i}}}{\text{ and }}{{\text{P}}_{\text{f}}}$ is the initial and final pressure and ${{\text{V}}_{\text{i}}}{\text{ and }}{{\text{V}}_{\text{f}}}$ is the initial and final volume.

Complete step by step answer:
From the reaction ${\text{CaC}}{{\text{O}}_{\text{3}}} \rightleftarrows {\text{CaCO}}\left( {\text{s}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
Volume = ${\text{10L}}$
Temperature = ${\text{300K}}$
Pressure (initial) of ${\text{C}}{{\text{O}}_{\text{2}}}$= $0.2{\text{atm}}$
Given ${{\text{K}}_{\text{p}}}{\text{ = 0}}{\text{.800atm}}$
Since ${\text{CaC}}{{\text{O}}_{\text{3}}}$and ${\text{CaCO}}\left( {\text{s}} \right)$are solids, they don't contribute to pressure. Therefore ${{\text{k}}_{\text{p}}}$ depends only on carbon dioxide. So, ${{\text{k}}_{\text{p}}}$= pressure of carbon dioxide at equilibrium
Hence, we can say that the maximum pressure of carbon dioxide after compression is $0.80$.
So, pressure (final) = $0.80{\text{atm}}$
According to Boyle's Law
${{\text{P}}_{\text{i}}}{{\text{V}}_{\text{i}}}{\text{ = }}{{\text{P}}_{\text{f}}}{{\text{V}}_{\text{f}}}$
${{\text{V}}_{\text{f}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{i}}}{{\text{V}}_{\text{i}}}}}{{{{\text{P}}_{\text{f}}}}}$ = $\dfrac{{0.2 \times 10}}{{0.80}} = 2.5{\text{L}}$

Therefore, the answer is - option (B) - The maximum volume of the container, when the pressure of carbon dioxide attains its maximum value, will be ${\text{2}}{\text{.5L}}$.

Additional Information:
According to Boyle's law "the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature".

Note:
${{\text{k}}_{\text{p}}}$ is a gas equilibrium constant. It is calculated from the partial pressures of a reaction equation. ${{\text{k}}_{\text{p}}}$ is used to express the relationship between product and reactant pressures. The equilibrium constants ${{\text{k}}_{\text{p}}}$ , ${{\text{k}}_{\text{c}}}$ and ${{\text{k}}_{\text{x}}}$ are related as: ${{\text{k}}_{\text{p}}}{\text{ = }}{{\text{k}}_{\text{x}}}{\left( {\text{P}} \right)^{{\text{Delta n}}}}{\text{ = }}{{\text{k}}_{\text{c}}}{\left( {{\text{RT}}} \right)^{{\text{Delta n}}}}$