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A 20 litre container at 400 K contains $CO_2$(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of $CO_2$ attains its maximum value, will be (Given that: $SrCO_3(s) \rightleftharpoons SrO(s) + CO_2(g)$, $K_p$ = 1.6 atm)
Explanation
$P_{CO_2}$ can rise to $K_p$ = 1.6 atm; Boyle's law with fixed n gives V = 5 L.
Detailed Solution
Maximum pressure of $CO_2$ = pressure of $CO_2$ at equilibrium.
For $SrCO_3(s) \rightleftharpoons SrO(s) + CO_2(g)$: $K_p = P_{CO_2} = 1.6$ atm = maximum pressure of $CO_2$
Volume of container at this stage: $V = \frac{nRT}{P}$ ...(i)
Since the container is sealed and the reaction was not earlier at equilibrium, n = constant.
$n = \frac{PV}{RT} = \frac{0.4\times20}{RT}$ ...(ii)
Put (ii) in (i): $V = \left[\frac{0.4\times20}{RT}\right]\frac{RT}{1.6} = 5$ L