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1 bar is 100,000 pascals.

I'm assuming you're working from a near-vacuum and initially at room temp.

Also, I'm using what's called the ideal-gas equation. It is an approximation that works best for gasses at low pressures and high temperatures. The equation is:

pV=nRT

p is pressure (in pascals)

V is volume (in meters cubed)

n is the amount of a substance (in moles)

R is a constant, 8.3145J/mol*K

T is temperature (in Kelvin. room temperature is kelvin is about 295)

If we try to solve for n now, we won't get it. We have another variable we need to take care of, T. So we...

Use this other equation:

(PV)/T of the initial state equals (pV)/T of the final state:

p1*V1/T1=p2*V2/T2

solve to T2:

T2=(p2*V2*T1)/(p1*V1)

Plug in values:

T2=(100,000Pa*.000032865m^3*295K)/(1pa*.000032865m^3)

which is saying that your new temp is 295000K (which is, like 600,000ºF)

And that's where I thought about quiting.

But no:

I plugged this value into the initial equation: and got that 0.00000134 moles of Nitrogen.

Nitrogen's molar mass is 14.0674 which means that:

.00001877 grams of nitrogen are required.

So, this is the long way. I didn't think to look up the density of Nitrogen at 1bar and see how much 32865 cubic millimeters of it would be.

the density of nitrogen at 101321Pa is 1.251g/L

which is .000001251 g/mm^3

multiply that by 32865 mm^3

So, .04 grams of nitrogen!

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