How long does it take a person at rest to breathe one mole of air if the person breathes 85.0 mL/s of air that is measured at 25 degrees Celsius and 755mmHg?


Air is not a pure compound of course, so you cannot breath a mol of air. It is approx 20% O2 and 80% N2.  Because the mol ratio is so far off you would get one mole of one long before the other. It is my assumption that oxygen is what the subject is here since we are dealing with respiration and the nitrogen just serves to dilute it, so let's go with that for our target.  Because I don't have any recent hard-copy works that give the composition of air I found a composition at The Engineering Toolbox (1) which listed air as 20.95% oxygen by volume in dry air.  The ideal gas formula is what will come into play here yet again.

PV = nRT

n = PV/(RT)

Pressure is given as 755 mmHg, volume is given as 85 ml, temperature is given as 25 C.  We need an R constant that encompasses all of these units or we need to convert any one of these units to match.  We can use R = 62.3637 (L x mmHg)/(K x mol), other R values will be available in your own text book.  In this case we need to do two conversions to use this.  Converting 85 ml to L is trivial, it is 0.085 L.  Converting 25C is as simple as most calculators, but it's pretty common knowledge, 298.5 K.  Now just plug and chug into the above equation.

n = (755 mmHg x 0.085L) / [(62.3637 (L x mmHg) / (K x mol)) x  298.5 K]

n = 64.175 / 18615.56445

n = 0.003447384 mol O2

Now, the above value is only useful to show you how many mols of oxygen there would be in 0.085 full liters of oxygen.  You're only dealing with 20.95% of that so multiply by the appropriate factor (0.2095) then divide one by your answer to get how many of these breaths it would take to get one mole of oxygen (since this is the amount of oxygen per breath).

0.003447384 mol O2 x 0.2095 = 7.2222 x 10-4 mol / breath

1 / (7.2222 x 10-4 mol / breath) = 1384.6 breaths

Once breaths is measured then it's easy to go to time, one breath per second per your problem so just multiply by one, to get the number of seconds, divide by 60 to get minutes.

1384.6 seconds / 60 sec/min = 23.07 min

Note that if you round every step as I have your answer would be way off.  I kept using the number in my calculator and only rounded for display purposes here.  An easier way to tackle this problem is to just keep in mind that using the ideal gas law, one mol of ideal gas at STP should take up a volume of exactly 24.414 L of volume, STP being 0C and 1 atm (760 mm Hg) it's not exact but you can get a rough approximation by working backwards from here since you are close to STP.  Take your volume per breath in liters, multiply by 0.2095 to get the amount of oxygen per breath, then divide the volume of one mole of ideal gas by this quantity.  This will again tell you how many breaths you need to take.  Divide by 60 to find out how many minutes this would take.

0.085 L x 0.2095 = 0.0178075

24.414 / 0.0178075 = 1370.995 breaths

1371 / 60 = 22.85 min

So you can see that even without using the explicit numbers in the problem you can get a pretty close answer without even resorting to the ideal gas equation.


After discussion with the questioner I have a different take on this problem.  One mole of air in this case is the volume of air occupied by the ideal gas volume of one mol of any gas (24.414 L) at STP.  So let's first convert that to the pressure and temperature specified.

P1V1/(RT1) = P2V2/(RT2)
Initial pressure and volume and temp specified by STP, final conditions specified in problem.

[(760 mm Hg) x (24.414 L)]/(273.5 K) = [(755 mm Hg) x V2]/(298.5 K)

20250.676 (mmHg x L)  = (755 mm Hg)V2

26.822 L = V2

Now we know how much volume 'one mol' of air will take up at the conditions specified.  Now just divide that quantity by the respiration rate (0.085 L/s), liters will cancel, seconds will swap to the top and you will get the time in seconds that it takes to breath in this much air, divide by 60 to get a more manageable unit in minutes.

(26.822 L) / (0.085 L/s) = 315.55 sec

(315.55 sec) / (60 sec/min) = 5.26 min

Hopefully this answer is more in line with the difficulty of the problem and it has the advantage of not relying on an outside source for any information (such as the percent oxygen in air).



(1) "Air Composition."  The Engineering Toolbox. 29 Sept. 2008  <>

Chang, Raymond. Chemistry. 6th ed. USA, McGraw-Hill. 1998



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