Answer

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Hint: Look at the values given in the question, this gives you a clear idea of what the question is asking. Since we have been given volume, pressure, temperature and Kp, this is a question for chemical equilibrium.

Complete step by step answer:

- Let us first write down the values given in the question.

Volume = 20 L

Temperature = 400 K

Pressure (initial) of \[C{{O}_{2}}\]= 0.4 atm

> As we know, volume and pressure are inversely related. So, when volume decreases, the pressure will increase.

From the reaction

\[SrC{{O}_{3}}(s)\rightleftharpoons SrO(s)+C{{O}_{2}}(g)\]

Given, Kp = 1.6 atm.

Since \[SrC{{O}_{3}}\] and SrO are solids, they don’t contribute to pressure.

Therefore, Kp depends only on carbon dioxide.

So, Kp = pressure of carbon dioxide at equilibrium

Hence, we can say that the maximum pressure of carbon dioxide after compression is 1.6 atm.

So, Pressure (final) = 1.6 atm

According to Boyle’s law,

\[\begin{align}& {{\text{P}}_{\text{i}}}{{\text{V}}_{\text{i}}}\text{=}{{\text{P}}_{\text{f}}}{{\text{V}}_{\text{f}}} \\

& \text{Vf = }\dfrac{{{\text{P}}_{\text{i}}}{{\text{V}}_{\text{i}}}}{{{\text{P}}_{\text{f}}}}\text{=}\dfrac{\text{0}\text{.4 x 20}}{\text{1}\text{.6}}\text{ = 5L} \\

\end{align}\]

Therefore, the answer is – option (a) – The maximum volume of the container, when the pressure of carbon dioxide attains its maximum value, will be 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: \[{{K}_{p}}\] is a gas equilibrium constant. It is calculated from the partial pressures of a reaction equation. \[{{K}_{p}}\]is used to express the relationship between product and reactant pressures.

Complete step by step answer:

- Let us first write down the values given in the question.

Volume = 20 L

Temperature = 400 K

Pressure (initial) of \[C{{O}_{2}}\]= 0.4 atm

> As we know, volume and pressure are inversely related. So, when volume decreases, the pressure will increase.

From the reaction

\[SrC{{O}_{3}}(s)\rightleftharpoons SrO(s)+C{{O}_{2}}(g)\]

Given, Kp = 1.6 atm.

Since \[SrC{{O}_{3}}\] and SrO are solids, they don’t contribute to pressure.

Therefore, Kp depends only on carbon dioxide.

So, Kp = pressure of carbon dioxide at equilibrium

Hence, we can say that the maximum pressure of carbon dioxide after compression is 1.6 atm.

So, Pressure (final) = 1.6 atm

According to Boyle’s law,

\[\begin{align}& {{\text{P}}_{\text{i}}}{{\text{V}}_{\text{i}}}\text{=}{{\text{P}}_{\text{f}}}{{\text{V}}_{\text{f}}} \\

& \text{Vf = }\dfrac{{{\text{P}}_{\text{i}}}{{\text{V}}_{\text{i}}}}{{{\text{P}}_{\text{f}}}}\text{=}\dfrac{\text{0}\text{.4 x 20}}{\text{1}\text{.6}}\text{ = 5L} \\

\end{align}\]

Therefore, the answer is – option (a) – The maximum volume of the container, when the pressure of carbon dioxide attains its maximum value, will be 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: \[{{K}_{p}}\] is a gas equilibrium constant. It is calculated from the partial pressures of a reaction equation. \[{{K}_{p}}\]is used to express the relationship between product and reactant pressures.

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