Science Question about air in a air tight container (1 Viewer)

Alright, I would like to know how long one person can breathe in a air tight container that is 10x10x8.
No it isn't anything sinister, it is for a one act experimental play where my character is trapped in a meat locker so would the freezing cold be part of the breathable oxygen equation as well?

From scuba diving, I know that my RMV is about 0.6ft^3/min
Somebody panicking in a freezer would be 2 or 3 times that.

You also would have to account for C02 buildup which is probably more important than the ft^3 of air. (See Apollo 13)

Temperature would have a minor effect on air consumption, you use more when you are cold, but I think the panic and C02 buildup would make this negliable.

The size of person does matter also. A small woman can have an RMV of 0.3 or so.

I'm 200lbs.

Can you explain the RMV a bit more please? I am mathmaticaly challenged so please keep that in mind, LOL!!!

It's the amount of air you consume at 1 atm (sea level).
Once you know that, you can figure out how much air you need at various depths.

You're room has 800 cubic feet of air in it. so just divide by the air consumption to decide how long your character can live.

I could live for about 11 hours on that much air breathing twice my "normal" consumption. But CO2 buildup is going to be the bigger problem.

There is also the mamilary reflex which may come into play - where the body core temperature drops far enough the body and brain sometimes can survive without O2 for an unusually long time. Usually happens in water - I don't know if it has happened on land before...

I don't think that your diving calculations will work in this situation as you have presented. I assume these calculations are used for diving where you breathe in the air, then that volume is taken away (since in diving when you breathe out that air does not go back into your tank). In this case you are contained in a container so the air you breathe out goes back into the container.

Air contains approx 78% nitrogen, 21% oxygen, 0.0% Argon, 0.04% carbon dioxide and traces of other gases make up the rest. When you breathe in obviously you breath in air with the above percentages, your lungs then exchange oxygen for CO2, but not completely, from memory I think the air you breathe out contains approx 16% oxygen and 4% carbon dioxide.

So maybe the better question is what would be a lethal mixture of air and CO2? If you knew those percentages, you could back your way into the answer.

This site has a pretty good summary.

Now you can just do the math and figure out how long it takes to get the percentage of CO2 in the room to around 10%. It's just a big calculus problem from here....

CO2 Toxicity

Some more numbers for you (from http://wasg.iinet.net.au/Co2paper.html):


(1 ft^3 = 28.3168466 liters)

-Mike

I'm assuming your original size was measured in feet, so the room has a volume of 800 cubic feet. then according to this site we multiply that by 28.32 to get litres, so the room contains 22656 litres of air.

If we take the assumption (which is incorrect) that we breathe each litre of air in turn in the room to make the calculation simpler then breathing at 6 litres/min (because this person is calm) then we would breathe every bit of air in that room once in 3776 minutes which is approx 63 hours. Then according to the above stats we would be left with approx. 5.6% CO2 and 15% O2. I assume if this is a freezer then and you're wearing shorts and a T-shirt you're going to freeze.


as well as CO2 toxicity you need to know the minimum O2 content of air that we need to live and if you will hit that first or the 10% CO2 level.

The minimum O2 to remain conscience is about 10%. You will CO2 tox first since your air will consist of 79%N, 11%O2 and 10%CO2 at the point you pass out.

The following assumptions are made:
800 cubic feet = 22653.5 Liters
The air starts out as:
78% Nitrogen
21% Oxygen
0.04% CO2

Assuming 10% CO2 will cause you to pass out and eventually die, and assuming that oxygen simply isn't a factor.

Assuming the person is breathing 40 liters of air per minute.

Every minute, you're taking in 8.4 l of Oxygen and .16 l of CO2

Every minute, you're therefore letting out 6 l of Oxygen and 2.24 l of CO2 (net loss of 2.4 l per minute of Oxygen, net gain of 2.2 l of CO2)

10 percent of 22653.5 is 2265.35

2265.35 / 2.24 is 1011.3 minutes, or 16.8 hours.

You'd have consumed 2427.12 liters of Oxygen, but you started out with 4757.235 l, so you've still got 2330.115. The air at this point is 10% oxygen and 10% CO2.

So, in around 17 hours, you'd pass out.