Tuesday, July 08, 2008

Question for today, July 8th 2008

How much space would Earth's atmosphere occupy if it was stored in a sufficiently large vessel at sea level conditions of temperature and pressure? Assume things like gravity do not have an effect, thus the container behaves like a regular balloon.

Update: and if you know this number... how does that compare to the volume of the atmosphere as is now?

2 comments:

John Dougan said...

Mass of Atmosphere = 5.1480 * 10^18 Kg
(http://en.wikipedia.org/wiki/Earth's_atmosphere)

Density of Atmosphere at STP = 1.29 * 10^-3 g/cm^3
(http://hypertextbook.com/facts/1999/MesahHarwood.shtml)

Volume of Atmosphere at STP = (5.1480 * 10^18) * 1000 / (1.29 * 10^-3)
= 3.99 * 1^24 cm^3

Andres said...

Nice :)... also we can reduce those cm^3 a bit, and since it's all metric it's so easy...

From cm to km there are 5 orders of magnitude, and therefore 3 * 5 = 15 orders of cubed magnitudes. Therefore, your answer is also

3.99 x 10^9 km^3

Andres.

PS: I had used 1.2 kg / m^3 for the density at sea level... similar results.