During one of his adventure Chacha Chaudhary got trapped in an underground cave which was sealed two hundred years back. The air inside was poisonous and contains `CO` in addition to `O_(2)` and `N_(2)`. Sabu, being huge, could not enter cave. In order to save Chacha Chaudhary he started sucking the poisonous air out of the cave by mouth. In each cycle he used to fill his lungs with cave air and exhale it out in the surroundings. In the mean time fresh air `(N_(2)+O_(2))` from the surrounding effused into cave till the pressure was 1 atmosphere. Each time Sabu sucked air,the pressure in the cave dropped ito `1//2 atm`. An initial sample of air taken from the cave measured `11.2mL` at STP and give `7J` on complete combusion at constant pressure.
`(a)` If the safe level of `CO` required in cave for life is less thatn `0.001%` by volume, how many times does Sabu need to suck out air in order to save Chacha Chaudhary ?
`(b)` Sabu should rescue Chacha Chaudhary within 10 minutes else he will die. Precious 80 second are wasted in thinking of a way to rescue him. At maximum how much time should each cycle of inhaling`-`exhaling take. Given, `DeltaH_(comb)CO=-280kJ mol^(-1)` ? Neglect Graham's law effect during operations.

Explanation

`:' 280xx10^(3) J`h heat is given by combustino of `CO=1 mol e`
`:. 7J` heat is given by combustion of `CO`
`=(7)/(280xx10^(3))=2.5xx10^(-5)mol e`
Initial moles of air `=(11.2)/(22400)=5xx10^(-4)mol e`
`:. % ` of `CO` initially in air `=(2.5xx10^(-5))/(5xx10^(-4))xx100=5`
Now in one inhaling by Sabu only half of the total air is taken out as during inhaling pressure drops to half because.
`P prop n (V` and `T` are constant `)`
`P prop n`
`(P)/(2) prop (n-a)`
Where `n` moles of poisonous air are present in cave and `a` moles of poisonous air are inhaled by Sabu
`:. a=(n)/(2),`
`(a)` Thus half moles of poisonous air are given out and the pure air again makes the total mole `n` by diffusing in cave as pressure becomes `1 atm`.
Thus `%` of `CO` in poisonous air is reduced by `1//2` in each inhaling or `50% CO` is taken out in one inhaling. Tus to reduce `CO` from `5%` to `0.001% 13` times inhaling is necessary by Sabu which gives `0.0061 % CO` in air .
`(b)` Sabu has only 10 minutes time. `i.e., 600 sec` in which he wastes 80 seconds in thinking thus he is left only `520 sec` in which he has to inhale and exhale 13 times or he should go for `40 sec` for one inhaling and exhaling.