# Partial pressure of oxygen at a certain altitude

songoku
Homework Statement:
The concentration of oxygen O2 at the atmospheric pressure is 20.9% v/v. The partial pressure of this concentration is 21.2 kPa. At an altitude of 6962 m above sea level (approximately 7000 m), what is the partial pressure of oxygen?
a. 44 kPa
b. 9.33 kPa
c. 0.21 kPa
d. 0.44 kPa
e. 21 kPa
Relevant Equations:
Not sure (maybe ideal gas or P = ρgh)
The pressure of oxygen at sea level = ##\frac{20.9}{100} ~\text{x} ~(21.2 ~\text{x} ~ 10^3) = 4430.8~ \text{Pa}##

Then I do not know how to calculate the pressure at altitude 7000 m. I tried using P = ρgh (taking ρ as density of air = 1.3 kg/m3) then subtract the result from 4430.8 Pa but got negative result so I guess the formula I used is wrong but I do not know what other formula to use.

Thanks

Homework Helper
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The pressure of oxygen at sea level = ##\frac{20.9}{100} ~\text{x} ~(21.2 ~\text{x} ~ 10^3) ##
Umm... why? Isn't the ##21.2\times10^3## the given partial pressure of oxygen at sea level?

Homework Helper
Then I do not know how to calculate the pressure at altitude 7000 m. I tried using P = ρgh (taking ρ as density of air = 1.3 kg/m3)
By the time you are seven kilometers up, the air will be remarkably thin. Its mass density will no longer be 1.3 kg/m3. A simple linear pressure reduction formula will already have broken down.

The reduction in pressure going up another 1 meters (for instance) will be proportional to the pressure at the current altitude -- the percentage decrease for each incremental meter is fixed. Ignoring temperature changes with altitude, the pressure decline will be exponential.

It is possible to Google this and get a formula for pressure decline with altitude. Or you could attack the problem from first principles. Or you could incrementally calculate the pressure decrease at 1000 meter increments. Maybe use a spreadsheet to do even smaller increments.

songoku
Umm... why? Isn't the ##21.2\times10^3## the given partial pressure of oxygen at sea level?
I thought it is the total pressure of air dan oxygen is only 20.9 % of this value

By the time you are seven kilometers up, the air will be remarkably thin. Its mass density will no longer be 1.3 kg/m3. A simple linear pressure reduction formula will already have broken down.

The reduction in pressure going up another 1 meters (for instance) will be proportional to the pressure at the current altitude -- the percentage decrease for each incremental meter is fixed. Ignoring temperature changes with altitude, the pressure decline will be exponential.

It is possible to Google this and get a formula for pressure decline with altitude. Or you could attack the problem from first principles. Or you could incrementally calculate the pressure decrease at 1000 meter increments. Maybe use a spreadsheet to do even smaller increments.
I found out from one of thread here in PF and through google, there is formula called barometric formula:

##P = P_o ~ e^{- \frac{Mgh}{RT}}## where ##M## = molecular mass of oxygen (in this case) and ##T## = temperature of air at height ##h##

Is this the formula I should use? If yes, then:
M = 32 g/mol = 32 x 10-3 kg/mol
g = 9.8 m/s2
h = 7000 m
R = 8.31
T = I do not know

Do I need another formula to determine the temperature at altitude of 7000 m?

Thanks

Homework Helper
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I thought it is the total pressure of air dan oxygen is only 20.9 % of this value
The question states:
The concentration of oxygen O2 at the atmospheric pressure is 20.9% v/v.
The partial pressure of this concentration is 21.2 kPa.

This is saying that of the 1atm of pressure at sea level, 21.2 kPa is from the oxygen molecules there. Note that this is roughly 20.9% of 1atm.

For the air pressure at 7000m, I would use https://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html.

songoku
The question states:
The concentration of oxygen O2 at the atmospheric pressure is 20.9% v/v.
The partial pressure of this concentration is 21.2 kPa.

This is saying that of the 1atm of pressure at sea level, 21.2 kPa is from the oxygen molecules there. Note that this is roughly 20.9% of 1atm.

For the air pressure at 7000m, I would use https://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html.

The formula in the link is ##P=P_o~(1-2.25577 ~ \text{x} 10^{-5}~h)^{5.25588}##

Using h = 7000 m, I get P = 8.59 kPa (not in the option given by the question, the closest is 9.33 kPa)

And what is the name of the formula? I want to see the derivation so I understand where all the numbers come from

So I can not use ##P = P_o ~ e^{- \frac{Mgh}{RT}}##?

Thanks

Homework Helper
Gold Member
2022 Award
The formula in the link is ##P=P_o~(1-2.25577 ~ \text{x} 10^{-5}~h)^{5.25588}##

Using h = 7000 m, I get P = 8.59 kPa (not in the option given by the question, the closest is 9.33 kPa)

And what is the name of the formula? I want to see the derivation so I understand where all the numbers come from

So I can not use ##P = P_o ~ e^{- \frac{Mgh}{RT}}##?

Thanks
I found another reference, https://www.mide.com/air-pressure-at-altitude-calculator. It does not say what formula it uses, but it gives the same result. (You have to specify a temperature, but it seems to be the sea level temperature.)

I note that for more modest altitudes you could approximate the formula using ##(1-x)^n \approx e^{-nx}##. Using that with the constants in the formula gives about 9.24kPa.

Yes, you could use the other formula, involving the temperature at altitude, but you would have to look up what that would be. It should not give a different result.

songoku
Thank you very much for all the help and explanation jbriggs444 and haruspex