How long could one survive in a perfectly airtight room? Given Inhaled air contains 21% oxygen while exhaled breath contains approximately 16% oxygen and 5% carbon dioxide The minimum oxygen level required for survival is 19 %. Dimensions of room in meters: 5 X 5 X 3

1. How long could one survive in a perfectly airtight room?
Given
 Inhaled air contains 21% oxygen while exhaled breath
contains approximately 16% oxygen and 5% carbon
dioxide
 The minimum oxygen level required for survival is 19 %.
 Dimensions of room in meters: 5 𝑋 5 𝑋 3
Answer:
1) Volume air in room.
5 𝑚 𝑋 5 𝑚 𝑋 3 𝑚 = 75 m3
1 m3
= 1000 L
Therefore,
753
= 75000 L
2) Volume of oxygen in the room.
We know that,
100% Air = 78% Nitrogen + 21% Oxygen + other gases
Therefore,
The volume of oxygen in the room =
21 %
100 %
× 75000 𝐿 = 15750 𝐿
3) Total volume of air consumed per minute.
We can say about ten breaths are taken per minute.
~10 breaths of 0.5 L per min= 5.0 L/min
The total volume of air consumed per minute is 5.0 min
4) Amount of oxygen consumed per one min.
As exhaled air contains about 16% oxygen, the percentage of
oxygen reduced from the air is 5% of the volume of oxygen
consumed.

2. The volume of oxygen consumed per minute
=
(𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛
× 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑏𝑟𝑒𝑎𝑡ℎ)
100%
=
(5% × 5 𝐿)
100%
= 0.25 L per min
Therefore, we can say that a healthy adult consumes 0.25 L of
oxygen per minute
5) Amount of oxygen available for breathing.
The minimum oxygen level required by humans in the air is 19%.
The percentage of oxygen available in the room is = 21%-19%
=2%
Therefore, 2% of 75000 L oxygen =
2% × 75000 𝐿
100%
= 1500 𝐿
So, about 1500 L of oxygen is available for breathing.
6) the time required to consume 1500 L of oxygen is
=
2% 𝑜𝑓 75000 𝐿 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛
𝑜𝑥𝑦𝑔𝑒𝑛 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑝𝑒𝑟 𝑡𝑖𝑚𝑒
=
1500 𝐿
0.25 𝐿/𝑚𝑖𝑛
= 6000 min
That is,
= 100 hours
=~4 days
How about carbon dioxide generated per breath? As high
carbon dioxide levels could be dangerous too.