A $20$ litre container at $400\text{K}$ contains ${\text{CO}}_2(g)$ at pressure $0.4\text{atm}$ and an excess of $\text{SrO}$ (neglect the volume of solid $\text{SrO}$). The volume of the container is now decreased by moving a piston. The maximum volume of the container, when pressure of ${\text{CO}}_2$ attains its maximum value, will be [Given: ${\text{SrCO}}_3(s)\rightleftharpoons \text{SrO}(s)+{\text{CO}}_2(g)$, $K_p=1.6\text{atm}$ ]:

1. For the reaction ${\text{SrCO}}_3(s)\rightleftharpoons \text{SrO}(s)+{\text{CO}}_2(g)$, $K_p=P_{{\text{CO}}_2}=1.6\text{atm}$.2. This is the equilibrium pressure. If the volume is decreased, the pressure increases. To maintain $K_p$, the reaction must shift to the left (${\text{CO}}_2$ consumed).3. The pressure remains constant at $1.6\text{atm}$ as long as the equilibrium is maintained by moving the piston.4. Using ideal gas law for constant amount and temperature before the reaction starts shifting: $P_1{V}_1=P_2{V}_2$.$0.4\text{atm}\times 20\text{L}=1.6\text{atm}\times V_{max}$$8=1.6V_{max}⟹V_{max}=5\text{L}$.