Question

: During one of his adventures, Chacha Chaudhary got trapped in an underground cave which was sealed two hundred year back. The air inside the cave was poisonous, having some amount of $CO(5\% \,by\,volume)$ in addition to ${O_{2\,}}$and ${N_2}$. The safe limit of $CO$ in the atmosphere is less than $0.001\% $ by volume. Sabu, being huge, could not enter into the cave, so in order to save Chacha Choudhary started sucking the poisonous air out of the cave by mouth. Each time, he filled his lunge with cave air and exhaled it out in the surroundings. In the meantime fresh air from the surrounding effused into the cave till the pressure was again one atmosphere. If Sabu sucked out half the air present in the cave, how many times does Sabu need to suck the air out of the cave in order to save Chacha Chaudhary?

Answer 1

Hint: $Pressure \propto \,number\,of\,moles/concentration$, so as the pressure is dropping the value of number of moles also drops which is what is required to happen. So we can say that the value of new concentration every time Sabu sucks the air from the cave will be equal to initial concentration divided by $\dfrac{1}{{{2^n}}}$, where n is the number of time Sabu sucks the air. So we need to find out how much pressure needs to drop , so that the concentration becomes less than $0.001\% $

Formula used:
$Concentration\,of\,gas\,after\,nth\,inhalation = \dfrac{{initial\,concentration}}{{{2^n}}}$

Complete step by step answer:
Let us consider the gases to be ideal and hence we can apply the Gas law.
The formula for Gas law is:

$PV = nRT$

Where, P = pressure,R = gas constant ,T = temperature,n = number of moles,V = Volume
We consider that the volume and pressure is kept constant as the physical parameters of the cave do not change.
hence, we can say that;

$Pressure \propto \,number\,of\,moles/concentration$

In the question it is given to us that the value of pressure drops by half each time Sabu sucks out the air from the cave.
This Implies that the number of moles or the concentration of the gas in the case also reduces by half each time Sabu sucks out the air.
This could be represented by frames as,
The concentration of gas in the cave after Sabu Sucking the air out of the cave for the $nth$ time will be

$\dfrac{1}{{{2^n}}}$ of the initial concentration.

Where, nis the number of times Sabu has sucked the air out of the cave.

Hence, we can use the above statement and frame it into a mathematical equation which would help us determine the value of $n$ at which $CO$ levels become safe.

$Concentration\,of\,gas\,after\,nth\,inhalation = \dfrac{{initial\,concentration}}{{{2^n}}}$

Now, we need the value of concentration to become, $0.001\% $ or less than that.
The value of Initial concentration has been given to us as $5\% \,$

Substituting these values in the above equation , we get

$0.001\% = \dfrac{{5\% }}{{{2^n}}}$

Solving the equation for n we get n = 12.28

Hence, the$13th$ time Sabu Suck’s out the air from the Cave, Chacha Chaudhary will be safe.

Note: This solution is only viable if two assumptions are made
1. The gas is considered to be ideal. We assume that there is no interaction between the gas molecules and each gas molecule behaves independently of each other.
2. We have to keep temperature and pressure as constant, this was possible as we considered a cave which has its physical dimensions set and no change in temperature could take place.