To solve this problem, we need to understand the equilibrium between SrCO$_3$(s), SrO(s), and CO$_2$(g).$\\$The reaction is:$\\$SrCO$_3$(s)$\rightleftharpoons$SrO(s) + CO$_2$(g)$\\$Given:$\\$• Initial pressure of CO$_2 = 0.4$ atm$\\$• Volume of the container $V_1 = 20$ L$\\$• Temperature $T = 400$ K$\\$• Equilibrium constant $K_p = 1.6$ atm$\\$At equilibrium, the pressure of CO$_2$ is $P_{CO_2}. \\$Since SrO is in excess, it will not affect the equilibrium.$\\$The equilibrium condition is given by:$\\ P_{CO_2} = K_p = 1.6$ atm$\\$Initially, the pressure of CO$_2$ is 0.4 atm. As the volume decreases,$\\$the pressure will increase until it reaches equilibrium at 1.6 atm.$\\$Using the ideal gas law, $PV = nRT,$ we can find the initial moles of CO$_2: \\ n_{initial} = \frac{P_{initial} \cdot V_1}{RT} = \frac{0.4 \cdot 20}{R \cdot 400} \\$At equilibrium, the moles of CO$_2$ will be:$\\ n_{equilibrium} = \frac{P_{equilibrium} \cdot V_{equilibrium}}{RT} = \frac{1.6 \cdot V_{equilibrium}}{R \cdot 400} \\$Since the moles of CO$_2$ are conserved, we equate the two expressions:$\\ \frac{0.4 \cdot 20}{R \cdot 400} = \frac{1.6 \cdot V_{equilibrium}}{R \cdot 400} \\$Simplifying, we find:$\\ 0.4 \cdot 20 = 1.6 \cdot V_{equilibrium} \\ 8 = 1.6 \cdot V_{equilibrium} \\ V_{equilibrium} = \frac{8}{1.6} \\ V_{equilibrium} = 5$ L$\\$Therefore, the maximum volume of the container when the pressure of CO$_2 \\$attains its maximum value is 5 L. This corresponds to Option 4.