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A 20 litre container at 400K contains \[C{{O}_{2}}(g)\] at pressure 0.4 atm and an excess of SrO. The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of carbon dioxide attains its maximum value, will be:
(A) 5L
(B) 10L
(C) 4L
(D) 2L
Given that: \[SrC{{O}_{3}}(s)\rightleftharpoons SrO(s)+C{{O}_{2}}(g)\]; \[{{K}_{p}}\]= 1.6atm

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Last updated date: 29th May 2024
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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.