O2 mass related to partial pressure

[N_Molson]

Hello, :tiphat:

I'm trying to calculate the mass of O2 needed to fill a cabin.

The atmospheric parameters are, to keep things simple for now, a total pressure of 300 mmHg and a partial pressure for O2 of 300 mmHg (100% O2). The cabin volume is 11,44 cubic meters. The temperature is 20°C.

What's the right formula to use to calculate this ? It is probably quite obvious, but I'm a little confused.

Sidenote : at 0 °C and 101.325 kPa (760 mmHg), O2 density is 1.429 g/L

:hmm:

 

[Jarvitä]

Around 6 kg.

http://www.wolframalpha.com/input/?i=(oxygen+density+at+300mmHg)+*+11.44+m**3

 

[N_Molson]

Thank you very much. That link is going to be VERY useful to me. :thumbup:

 

[RisingFury]

Use ideal gas law:

p*V = m/M * R * T
p - pressure in Pascal
V - volume
m - mass
M - molar mass
R - gas constant
T - temperature in Kelvin

 

[N_Molson]

39996.717*11.44 = m/15.9994g * 8.306 J/mol•K * 293.15 ? :hmm:

 

[Grover]

interesting, this is on my physics topic we just started

and PV=NRT is your best bet, stick with ideal gas laws for oxygen

---------- Post added at 09:07 PM ---------- Previous post was at 09:00 PM ----------

i make it 60Kg:

((101000 Pa * 11.44M^3)/(293K*1.806)) [number of oxygen moles] *0.028 [mass of oxygen mole]. it gives an air density of 5Kg/M^3... does it sound right?

 

[Jarvitä]

Or, you could use a computer simulation to obtain a precise result instead of relying on highschool-level simplifications to do it by hand.

 

[N_Molson]

Since the answer is 6.009 kgs using the calculator pointed by Jarvitä (adding the 293K parameter), I'd say you just miss it by one decimal

Also 60 kgs of atmosphere seems very heavy for a small spacecraft cabin, while 6 kgs seems right.

I managed to code it, now the O2 tank loses 6.009 kgs when you repressurize the cabin. Which means you'll be able to do it a very limited number of times, that was the main purpose :thumbup:

 

[RisingFury]

Or, you could use a computer simulation to obtain a precise result instead of relying on highschool-level simplifications to do it by hand.

No faith in the ideal gas law, huh?

The ideal gas law applies pretty well where in cases where there are few interactions between individual particles and the distance between particles is much greater then the individual particle volume...

In air at Earth's surface pressure that's a pretty fair bet.

i make it 60Kg:

((101000 Pa * 11.44M^3)/(293K*1.806)) [number of oxygen moles] *0.028 [mass of oxygen mole]. it gives an air density of 5Kg/M^3... does it sound right?

Uh uh! 300 mmHg = 40 kPa

 

[N_Molson]

Well, if I replace 101000 by 40000 in the above equation I find 24.213562 as a result...

...but 24.213562 what ? :hmm:

 

[RisingFury]

Well, if I replace 101000 by 40000 in the above equation I find 24.213562 as a result...

...but 24.213562 what ? :hmm:

My only guess is that you can't enter the numbers correctly into your calculator. Using the ideal gas law, I get 6 kg.

---------- Post added at 17:06 ---------- Previous post was at 17:01 ----------

interesting, this is on my physics topic we just started

and PV=NRT is your best bet, stick with ideal gas laws for oxygen

---------- Post added at 09:07 PM ---------- Previous post was at 09:00 PM ----------

i make it 60Kg:

((101000 Pa * 11.44M^3)/(293K*1.806)) [number of oxygen moles] *0.028 [mass of oxygen mole]. it gives an air density of 5Kg/M^3... does it sound right?

N = m/M. Usually marked n, not N...

m = p * V * M / (R * T)
p = 40 000
V = 11.4
M = 32
R = 8.3
T = 20°C = 293 K

 

[N_Molson]

My only guess is that you can't enter the numbers correctly into your calculator.

That's true. :lol: :facepalm:

So the online calculator is correct. Perfect.