# Density of air using ideal gas law

I need to find the density of air using (I think) the ideal gas law. I have calculated $V$ of an ideal gas at s.t.p. to be $.0224m^3$. I am asked to find the density of air knowing it's temperature ($20^\circ$) and atmospheric pressure. I have looked up online however I feel the methods I find are skipping steps from where I am currently. Where do I go from here? $$pV = nRT$$

• number density, or mass density? Apr 22, 2016 at 12:06
• Solving for n after you input p V and T seems like a good next step. Apr 22, 2016 at 13:06

Simplify the equation to make it more convenient to you by using the definition of a mole. A mole , generally denoted by 'n' is the mass of the substance taken divided my its molecular weight. On solving the only unknown in this equation , you get the mass of air contained in the volume you obtained. Now , you can just plug in this value into the definition of density,since density is the mass per unit volume. It would be wise to make sure the units of M and m are in the units you want the answer in or usually , we take the SI units. As pentane already mentioned, evaluating n and then solving for m is also a valid option. Note that as Ilja said the density of a substance doesn't depend on the mass taken or the volume independently, it is the mass of the gas contained per volume of that gas (in this case , air). The reason you are calculating mass here IS for the very same reason.So should you take a different mass or volume of air , the density doesn't change as the mass changes with volume to keep their ratio constant. I hope you get the general idea.

The ideal gas law says you something about $p$, $V$ and $T$ in terms of the number of particles. That's nice, because it holds approximately for all gases. For a given $p$ and $T$ you know the number per volume - it is independent of the type of gas!

Then why is the density of gases different? - because the mass per particle is different.
So now you have to rewrite your equation: the $n$ has to go, you want to have $m$ instead, and the molar mass. And then you can bring $m/V$ to one side, and there you are. Do you see the way now?

• I have the molar mass of air, $M_{air}$ = 28.8x$10^{-3}$, and I divided that by $N_A$ to get $m$ = 4.78x$10^{-26}$. Do I then use the ideal gas law to get a volume for this or what V do I use? Apr 22, 2016 at 11:59
• the ideal gas law is still valid; you have just to replace the $n$ and bring the desired $m/V$ on one side.
– Ilja
Apr 22, 2016 at 12:13
• as to your calculation: no, M is not 28g, it 28 g/mole... and you need just this to replace the $n$ (which is the number of moles). You calculated the mass of one particle; which is interesting to have a feeling for the magnitude, but you don't need it.
– Ilja
Apr 22, 2016 at 12:14