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 slid `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 pressure of `CO_(2)` attains its maximum value, will be
[Given that `: SrCO_(3)(s) hArr SrO(s) +CO_(2)(g), Kp=1.6 atm]`

Options

A. 5 litre
B. 10 litre
C. 4 litre
D. 2 litre

Answer

A

Explanation

Maximum pressure of `CO_(2)` = Pressure of `CO_(2)` at equilibrium
For the reaction
`SrCO_(3) (s) hArr 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=nRT/P" "` ...(i)
Intially, no . of moles of the gas (n)
` (PV)/(RT)= (0.4 xx20)/(RT)= 8/(RT)`
As no. of moles of the gas will remain constant , therefore, at equilibrium also,
`n=8/(RT)`
Putting this value in eqn. (i),
`V= 8/(RT)*(RT)/P = 8/P= 8/1.6=5L`