# What is the volume of a parcel of air of a giving weight at average sea level pressure

I know this is probably some basic math once you have the figures, but my Google-fu is failing me.

Say I have 5 pounds of air. Air meaning, the typical composition of the atmosphere of Earth. And by 5 pounds I mean the weight of the mass of the air, not it's pressure or anything like that.

I want to know approximately how much volume that amount of air would take up at a given pressure -- specifically, sea level pressure. I know that can vary -- lets just say its a typical warm summer day, or just an average. I'm not looking for exact numbers.

Actually, I'd rather a subjective interpretation of the exact figure if you can offer one. Something like 'about the volume of the air in a 12x12 room', or what have you. I have no idea how big 5 pounds of air is.

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Check in Wolfram Alpha: wolframalpha.com/input/?i=5+pounds+of+air –  mmc Jul 13 '11 at 19:53
@mmc: hah, for me it assumed 5 GBP (the currency) of air, and it doesn't know the current market price. Anyway, eventually W|A tells me that it is roughly the size of a 4' by 4' by 4' cube. –  Willie Wong Jul 13 '11 at 20:00
It says the the density of the air is 0.001275 grams/cm^3 at 0C. Thats pretty cold.. it would be slightly larger at room temp. Any idea how much? :) –  InfinitiesLoop Jul 13 '11 at 20:12
Yes, by the ratio of the absolute temperatures. ~ 25 L/mol at 30 °C (summer is hot here). –  Edgar Bonet Jul 13 '11 at 20:25

The easy way to answer this is to just look up the density of air $\rho$ under whatever conditions you need it and then calculate $V=m/\rho$.

If you don't have a ready source for the density you need, though, you can use the ideal gas law,

$$PV=nRT$$

Here $n$ is the number of moles of the gas, which you can get by dividing the mass by the average molar mass of the gas:

$$n=\frac{m}{\mu}$$

The average molar mass of air would be a weighted average of the molar masses of its constituent gases. This is something you look up for the composition of air under your specified conditions, but it'll be around 29 g/mol. This at least does not depend directly on temperature.

If you need more accuracy, you could use something like the Van der Waals equation.

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Ah, makes sense. The Wolfram Alpha answer gave me the result I needed, but I was really after how to calculate it. Thanks. –  InfinitiesLoop Jul 13 '11 at 21:52

Air has a density of 1.29 grams/liter at $0^oC$ $(273^oK)$ at sea level. Using the formula which David so kindly provided, we readily see that the same mass of air grows in volume in direct proprtion to absolute ($^oK$) temperature. A warm sunny day is about $77^oF$ which is about $25^C$, or $298^oK$.

So, the volume of 1.29 grams of air at sea level on a 25C day would be (1 liter)(298K/273K) = 1.0916 liters. This makes the density of air at that warmer condition = 1.29 grams/ 1.0916 liters = 1.182 grams/liter.

(1 liter air)/(1.182grams) x (454.5 grams/lb.) x (5 lbs.) = (1923 liters air) / (5 lbs. air). That's 1.923 cubic meters of air: about 106% of the air you could hold in a 4ft x 4ft x 4ft box.

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