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 container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of CO 2 attains its maximum value, will be: Given   that : SrCO 3 s ⇌ SrO s + CO 2 g ,   K p = 1

Step-by-step solution

Max. pressure of CO 2 = Pressure of CO 2 at equilibrium. For reaction, SrCO 3 s   ⇌ SrO s + CO 2 K p = P CO 2 = 1 .6 atm = maximum pressure of CO 2 . The volume of the container at this stage. V = nRT P ......(i) Since the container is sealed and the reaction was not earlier at equilibrium ∴    n = constant n = PV RT = 0 .4 × 20 RT ......(ii) Put equation (ii) in equation (i) V = 0 .4 × 20 RT   RT 1 .6 = 5   L