Calculate CO concentration at high altitude

[jwingeart]

Question:
You are cooking in a tent at 17,000ft with outside air temp at -15F. The concentration of CO in the tent is 40ppm. Calculate the equivalent CO concentration in mg/m^3.

First, I converted 17000ft to atm and got 501.5atm, which seems very wrong.

Next, I converted -15F to 247.039K.

I applied the Ideal Gas Law, PV=nRT

I tried to solve for V.

p=501.5 atm
n(molecular weight CO)=28
R (should be a constant)=.082056
T=247.039

I come up with a ridiculous answer. It seems simple, but I can't figure out where I'm going wrong.

 

[jwingeart]

Forgot to mention also, (but it's obvious) that the conditions are not at STP. (not 25C, not 1.0 atm)

 

[Chestermiller]

Show us how you calculated the pressure at that altitude?

Chet

 

[SteamKing]

Question:
You are cooking in a tent at 17,000ft with outside air temp at -15F. The concentration of CO in the tent is 40ppm. Calculate the equivalent CO concentration in mg/m^3.

First, I converted 17000ft to atm and got 501.5atm, which seems very wrong.

Why do you think altitude is equivalent to atmospheres? Your calculation assumes that there is a water column 17000 feet tall somewhere.

Next, I converted -15F to 247.039K.

FWIW, this calculation is OK.

I applied the Ideal Gas Law, PV=nRT

I tried to solve for V.

p=501.5 atm
n(molecular weight CO)=28
R (should be a constant)=.082056
T=247.039

I come up with a ridiculous answer. It seems simple, but I can't figure out where I'm going wrong.

Well, for one thing thinking that the atmospheric pressure at 17000 feet altitude is 500 atmospheres. (Hint: the atmospheric pressure decreases with altitude, although not in a linear relationship.)

 

[SCP]

First, I converted 17000ft to atm and got 501.5atm, which seems very wrong.

Check your units. At 17,000 ft, atmospheric pressure should be ~500 kPa, or 0.5 atm. If the answer you're getting is off by 3 orders of magnitude, this might be why.

 

[D H]

First, I converted 17000ft to atm and got 501.5atm, which seems very wrong.

So why didn't you stop right there?

It obviously is very wrong. When you get an answer that seems very wrong, you should stop and double check. Carrying a bad calculation forward is generally a very bad idea because that bad calculation oftentimes poisons all subsequent results.

Check your units. At 17,000 ft, atmospheric pressure should be ~500 kPa, or 0.5 atm.

That's wrong, too. One atmosphere is on the order of 100 kPa, so 500 kPa is about 5 atmospheres.

 

[SteamKing]

So why didn't you stop right there?

It obviously is very wrong. When you get an answer that seems very wrong, you should stop and double check. Carrying a bad calculation forward is generally a very bad idea because that bad calculation oftentimes poisons all subsequent results.



That's wrong, too. One atmosphere is on the order of 100 kPa, so 500 kPa is about 5 atmospheres.

It's that damn metric system confusin' everyone again.

 

[SCP]

Check your units. At 17,000 ft, atmospheric pressure should be ~500 kPa, or 0.5 atm. If the answer you're getting is off by 3 orders of magnitude, this might be why.

That's wrong, too. One atmosphere is on the order of 100 kPa, so 500 kPa is about 5 atmospheres.

Ah yes. I was thinking in millibars. Silly. In any case, at 17,000 ft you should get pressure at ~0.5 atm (or ~50 kPa).