<p>Your assumption about constant pressure can't actually exist while everything is closed. I am not using this assumption here because it is more correct and a bit easier:</p>
<p><span class="math-container">$$\pu{10 feet} = \pu{3.048 m}$$</span>
<span class="math-container">$$(\pu{10 feet})^3 = \pu{28.3 m^3} = \pu{28300 L}$$</span></p>
<p>We can calculate the amount of substance using ideal gas equation:
<span class="math-container">$$pV = nRT$$</span>
<span class="math-container">$$n = \frac{pV}{RT} = \frac{101.325\cdot 28300}{8.31\cdot 293.15}\text{ mol} = \pu{1177 mol}$$</span></p>
<p>There is 78 percent nitrogen, 21 percent oxygen and 1 percent of other gases:
<span class="math-container">$$n(\ce{N2}) = 0.78\cdot n = \pu{918.06 mol}$$</span>
<span class="math-container">$$n(\ce{O2}) = 0.21\cdot n = \pu{247.17 mol}$$</span>
<span class="math-container">$$n(\text{oth.}) = 0.01\cdot n = \pu{11.77 mol}$$</span></p>
<p>The amount of oxygen is constant, but we want to reach 5%. Thus, the total amount of substance after the insertion of nitrogen is:
<span class="math-container">$$n'(\text{together}) = n'(\ce{N2}) + n'(\ce{O2}) + n'(\text{oth.}) = \frac{n'(\ce{O2})}{0.05} = \frac{n(\ce{O2})}{0.05} = \frac{\pu{247.17 mol}}{0.05} = \pu{4943.4 mol}$$</span></p>
<p>Also, <span class="math-container">$n'(\text{oth.}) = n(\text{oth.})$</span>, thus:
<span class="math-container">$$n'(\ce{N2}) = n'(\text{together}) - n'(\ce{O2}) - n'(\text{oth.}) = \pu{4943.4 mol} - \pu{247.17 mol} - \pu{11.77 mol} = \pu{4684.46  mol}$$</span></p>
<p>So the solution is <span class="math-container">$\pu{4684 mol}$</span>, but we can't imagine that easily. Mass of <span class="math-container">$\pu{4684 mol}$</span> of nitrogen is:
<span class="math-container">$$m'(\ce{N2}) = M(\ce{N2}) \cdot n'(\ce{N2}) = 2 \cdot 14.007 \cdot \pu{4684.46 g} = \pu{131.23 kg}$$</span></p>
<p>Also, the volume of such nitrogen at normal pressure is equal to:
<span class="math-container">$$V'(\ce{N2}) = \frac{n'(\ce{N2})\cdot RT}{p} = \frac{4684.46\cdot 8.31\cdot 293.15}{101.325}\text{ L} = \pu{112600 L} = \pu{112.6 m^3}$$</span></p>
<p>Thus, we need to add <span class="math-container">$\pu{112.6 m^3}$</span> of nitrogen to reach 5% concentration of oxygen. The answer seems reasonable because we need to add approximately 4 times of the original volume in order to lower the concentration 4 times.</p>

CC BY-SA 4.0
2021-06-02T08:54:21.213Z
Can "112.6 L" possibly be correct? (10 feet/3.281)³ is 27.83 m³ or 27,830 liters, and 21-5=15% of that is still 4,452 liters. Am I missing something?
uhoh
CC BY-SA 4.0
2021-06-02T11:07:52.660Z
@uhoh Thank you, corrected in the text (I replaced liters with cubic meters).
User123
CC BY-SA 4.0
2021-06-05T14:25:27.503Z
Wow thats an awesome and very detailed answer, thanks so much for taking the time!
Tim Menser
CC BY-SA 4.0
2021-06-05T14:30:48.670Z
Wouldn't the Oxygen in a real world scenario simply be pressed "out" of them room by the Nitrogen? If the room only holds 28.3m3 of total volume and you release 28.3 m3 of Nitrogen you should start to force out the normal room air with your new gas and lower the oxygen level. So while your answer is correct for a closed system one with a pressure exhaust would need a much lower amount right?
Tim Menser
CC BY-SA 4.0
2021-06-05T17:21:32.250Z
@TimMenser Imagine a bottle filled with nitrogen and oxygen. If you add the nitrogen, the mass of oxygen stays the same. I imagined that scenario because closed room with possible gas escape doesn't exist. (Closed room = no interchange, oxygen escape = interchange -> contradiction). But yes, with a pressure exhaust we would need lower amount.
User123
CC BY-SA 4.0
2021-06-01T21:29:05.360Z
2021-06-02T16:07:15.453Z
2021-06-02T16:07:15.453Z
User123
<p>Imagine a room of 10x10 feet in size with a 10 feet high ceiling. The air in the room consists of roughly 78 percent nitrogen and 21 percent oxygen. The windows are shut closed and the gap under the door is sealed. How much nitrogen would you need to add, in order to reduce the oxygen level to less than 5%? For the sake of simplicity lets assume that the pressure in the room should stay the same as before, and that the surplus air just escapes.</p>

CC BY-SA 4.0
2021-06-02T11:08:01.200Z
This isn't Earth science, this is a homework question. Boo.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-06-05T14:23:22.723Z
Do you know a better place to ask that? It is not a homework question, its just an interesting thought.
Tim Menser
CC BY-SA 4.0
2021-06-05T17:34:51.987Z
I am voting to reopen this question because I believe this isn't a homework question. He says: "For the sake of simplicity lets assume that the pressure in the room should stay the same as before." If this were a homework question, the teacher clearly wouldn't give that (the problem gets harder, not simpler : )
User123
<oxygen><nitrogen>
How much nitrogen would you need to replace all oxygen in a room of 1,000 cubic feet?
Tim Menser
# How much nitrogen would you need to replace all oxygen in a room of 1,000 cubic feet?

(Asked by: username_0 on 2021-06-01)

Imagine a room of 10x10 feet in size with a 10 feet high ceiling. The air in the room consists of roughly 78 percent nitrogen and 21 percent oxygen. The windows are shut closed and the gap under the door is sealed. How much nitrogen would you need to add, in order to reduce the oxygen level to less than 5%? For the sake of simplicity lets assume that the pressure in the room should stay the same as before, and that the surplus air just escapes.

**Tags:** oxygen, nitrogen

> **Comments:**
> **username_1**: This isn't Earth science, this is a homework question. Boo.  
> **username_0**: Do you know a better place to ask that? It is not a homework question, its just an interesting thought.  
> **username_2**: I am voting to reopen this question because I believe this isn't a homework question. He says: "For the sake of simplicity lets assume that the pressure in the room should stay the same as before." If this were a homework question, the teacher clearly wouldn't give that (the problem gets harder, not simpler : )  

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## 1 answers

### Answer by username_2 on 2021-06-01. Score: 3 | Accepted answer

Your assumption about constant pressure can't actually exist while everything is closed. I am not using this assumption here because it is more correct and a bit easier:

$$\pu{10 feet} = \pu{3.048 m}$$
$$(\pu{10 feet})^3 = \pu{28.3 m^3} = \pu{28300 L}$$

We can calculate the amount of substance using ideal gas equation:
$$pV = nRT$$
$$n = \frac{pV}{RT} = \frac{101.325\cdot 28300}{8.31\cdot 293.15}\text{ mol} = \pu{1177 mol}$$

There is 78 percent nitrogen, 21 percent oxygen and 1 percent of other gases:
$$n(\ce{N2}) = 0.78\cdot n = \pu{918.06 mol}$$
$$n(\ce{O2}) = 0.21\cdot n = \pu{247.17 mol}$$
$$n(\text{oth.}) = 0.01\cdot n = \pu{11.77 mol}$$

The amount of oxygen is constant, but we want to reach 5%. Thus, the total amount of substance after the insertion of nitrogen is:
$$n'(\text{together}) = n'(\ce{N2}) + n'(\ce{O2}) + n'(\text{oth.}) = \frac{n'(\ce{O2})}{0.05} = \frac{n(\ce{O2})}{0.05} = \frac{\pu{247.17 mol}}{0.05} = \pu{4943.4 mol}$$

Also, $n'(\text{oth.}) = n(\text{oth.})$, thus:
$$n'(\ce{N2}) = n'(\text{together}) - n'(\ce{O2}) - n'(\text{oth.}) = \pu{4943.4 mol} - \pu{247.17 mol} - \pu{11.77 mol} = \pu{4684.46 mol}$$

So the solution is $\pu{4684 mol}$, but we can't imagine that easily. Mass of $\pu{4684 mol}$ of nitrogen is:
$$m'(\ce{N2}) = M(\ce{N2}) \cdot n'(\ce{N2}) = 2 \cdot 14.007 \cdot \pu{4684.46 g} = \pu{131.23 kg}$$

Also, the volume of such nitrogen at normal pressure is equal to:
$$V'(\ce{N2}) = \frac{n'(\ce{N2})\cdot RT}{p} = \frac{4684.46\cdot 8.31\cdot 293.15}{101.325}\text{ L} = \pu{112600 L} = \pu{112.6 m^3}$$

Thus, we need to add $\pu{112.6 m^3}$ of nitrogen to reach 5% concentration of oxygen. The answer seems reasonable because we need to add approximately 4 times of the original volume in order to lower the concentration 4 times.

> **Comments:**
> **username_3**: Can "112.6 L" possibly be correct? (10 feet/3.281)³ is 27.83 m³ or 27,830 liters, and 21-5=15% of that is still 4,452 liters. Am I missing something?  
> **username_2**: @username_3 Thank you, corrected in the text (I replaced liters with cubic meters).  
> **username_0**: Wow thats an awesome and very detailed answer, thanks so much for taking the time!  
> **username_0**: Wouldn't the Oxygen in a real world scenario simply be pressed "out" of them room by the Nitrogen? If the room only holds 28.3m3 of total volume and you release 28.3 m3 of Nitrogen you should start to force out the normal room air with your new gas and lower the oxygen level. So while your answer is correct for a closed system one with a pressure exhaust would need a much lower amount right?  
> **username_2**: @username_0 Imagine a bottle filled with nitrogen and oxygen. If you add the nitrogen, the mass of oxygen stays the same. I imagined that scenario because closed room with possible gas escape doesn't exist. (Closed room = no interchange, oxygen escape = interchange -> contradiction). But yes, with a pressure exhaust we would need lower amount.  

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