The system shown in the figure is in equilibrium, where `A` and `B` are isomeric liquids and form an ideal solution at `TK`. Standard vapour pressures of `A` and `B` are `P_(A)^(0)` and `P_(B)^(0)`, respectively, at `TK`. We collect the vapour of `A` and `B` in two containers of volume `V`, first container is maintained at `2 T K` and second container is maintained at `3T//2`. At the temperature greater than `T K`, both `A` and `B` exist in only gaseous form.
We assume than collected gases behave ideally at `2 T K` and there may take place an isomerisation reaction in which `A` gets converted into `B` by first-order kinetics reaction given as:
`Aoverset(k)rarrB`, where `k` is a rate constant.
In container (`II`) at the given temperature `3T//2`, `A` and `B` are ideal in nature and non reacting in nature. A small pin hole is made into container. We can determine the initial rate of effusion of both gases in vacuum by the expression

`r=K.(P)/(sqrt(M_(0)))`
where `P=` pressure differences between system and surrounding
`K=` positive constant
`M_(0)=` molecular weight of the gas
Vapours of `A` and `B` are passed into a container of volume `8.21 L`, maintained at `2T K`, where `T=50 K` and after `5 min`, moles of `B=8//3`. The pressure developed into the cotainer after two half lives is

Options

A. `3 atm`
B. `4 atm`
C. `5 atm`
D. `0.5 atm`

Explanation

`(n_(A))/(n_(B))=(2)/(1)`
`underset(8//3)(A)overset(K)rarrunderset(4//3)(B)`
`t=5 min (8)/(3)-x (4)/(3)+x=(8)/(3)impliesx=(4)/(3)`
`=4//3implies t=t_(1//2)=5 min`
`implies` At `t=10 min underset(x//3)(A)overset(K)rarrunderset(10//3)(B)`
`implies P=(4xx0.0821xx100)/(8.21)=4 atm`