If the safe level of CO in the atmosphere is less than 0.001% by volume, how man | TestHub

Passage / Comprehension

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 poisonous 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 the 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 taken from the cave measured 11.2 mL at 1atm, 273K and gave 7J on complete combustion at constant pressure. $Δ H_{comb} (CO) = - 280 kJ . mol^{- 1}$ . Neglect any use of Graham's Law.

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?

Answer:

13.00

Solution:

Given 11.2 mL air at STP gives 7 J heat:

$n (CO) = \frac{7}{280000} = 2.5 \times 1 0^{- 5} m o l$

Total moles of air:

$n = \frac{P V}{RT} = \frac{1 \times 0.0112}{0.082 \times 273} = 5 \times 1 0^{- 4} m o l$

Initial CO %:

$\frac{2.5 \times 1 0 ^{- 5}}{5 \times 1 0 ^{- 4}} \times 100 = 5%$

Each sucking reduces cave air to half, and fresh air enters, so the CO concentration becomes:

$5% (\frac{1}{2})^{n}$

Safe level:

$5 (\frac{1}{2})^{n} < 0.001$

$n > 12.3 \Rightarrow n = 13$

Chemistry - Thermochemistry Question with Solution | TestHub

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