Question

During one of his adventures,chacha chaudhary got trapped in an underground cave which was sealed two hundred years back.the air inside was poisonous,having some amount of carbon monoxide in addition to $O_2$ and $N_2$.sabu,being huge,could not enter the cave.so,in order to save chacha chaudhary he started sucking the poisionous air out of the cave by mouth.each time he used to fill his lungs with cave air and exhale it out in the surroundings.in the meantime,fresh air from surroundings effused into the cave till the pressure was again one atmosphere.each time sabu sucked out some air,the pressure in the cave dropped to half its initial value of one atmosphere
An initial sample of air is taken from the measured 11.2mL at 1 atm and 273 K and gave 7J on complete combustion at constant pressure
(i)if the safe level of CO in the atmosphere is less than 0.001% by volume,how many times does sabu need to suck out air in order to save chacha chaudhary?
(ii)sabu should rescue chacha chaudhary within 6 minutes else he will die,precious 80 seconds are wasted in thinking of a way to rescue him.at maximum,how much time should each cycle of inhaling-exhaling take?
$△H_{comb}\left(CO\right)=-280kJmo{l}^{-1}$.neglect any use of graham's Law.

Answer 1

$\because 280\times {10}^{3}J$ heat is given by combustion of $CO=1㏖$

$\therefore 7J$ heat is given by combustion of $CO=\frac{7}{280\times {10}^3}$

$=2.5\times {10}^{-5}㏖$

Initial moles of air$=\frac{11.2}{22400}=5\times {10}^{-4}㏖$

$\therefore$% of $CO$ initially in air$=\frac{2.5\times {10}^{-5}}{5\times {10}^{-4}}\times 100=5$

Now in one inhaling by Sabu only half of the total air is taken out as during inhaling pressure drops to half because

$P⋉n$ (v and T are constant)

$\frac{P}{2}⋉(n-a)$

where n mole of poisonous air are present in cave and a moles of poisonous air are inhaled by Sabu

$a=\frac{n}{2};$

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 $\frac{1}{2}$ in each inhaling or $50$% $CO$ is taken out in one inhaling.

Thus to reduce $CO$ from $5$% to $0.001$% $13$ times inhaling is necessary by Sabu which gives $0.00061$% $CO$ in air.

b) Sabu has only $10$ minutes time i.e., $600$ sec in which he wastes $80$ seconds 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.

Thus, Answer: (i) $13$ times, (ii) $40$ sec.