# Volume of composition of air

My teacher stated:

At standard conditions, 1 liter of air at 21% oxygen possesses $\pu{0.21 L}$ of oxygen. Since at STP, 1 mole of gas occupies $\pu{22.4 L}$, simply divide $0.21/22.4$, to arrive at 0.0094 moles of oxygen.

I have a question regarding this. What if I trap $\pu{100 ml}$ of air in a vessel and maintain $\pu{273 K}$ and pressure $\pu{1 atm}$? So, in that case, what does it mean that oxygen has volume of $\pu{21 ml}$? Won’t the volume of oxygen and nitrogen be same inside vessel? Or am I wrong somehow?

• Note that, apparently, your exercise still uses the old definition of STP $(T=273.15\ \mathrm K$, $p = 1\ \mathrm{atm})$, which was used until 1982. The molar volume of an ideal gas at the current definition of STP $(T=273.15\ \mathrm K$, $p = 1\ \mathrm{bar})$ actually is $V_\mathrm m=22.710947(13)\ \mathrm{l\ mol^{-1}}$. – user7951 Jun 7 '18 at 13:08
• @Loong interesting comment would you happen to have a link to any primary sources? – A.K. Sep 29 '18 at 18:45

If you took the 100 mL sample of air, and took out everything but the oxygen, then the amount of oxygen trapped would fill 21 mL at STP.

One way to think of it is, even though the oxygen would expand to fill the container, its pressure would need to reduce according to Boyle's law: P1V1 = P2V2. This is where the concept of Partial Pressure comes from. But if you then compressed that oxygen so that P = 1 atm (since you specified initial pressure to be 1 atm), the oxygen would take up 21 mL of volume.

• So I calculated some values. If I want to maintain a volume of air as 100 ml at pressure 1 atm and temperature 273 K , I should have 0.00353 mole of nitrogen molecules and 0.00094 mole of oxygen molecules . And If I keep that same mole of Oxygen molecules in another vessel (whose volume can be changed) at 1 atm pressure and 273 K , the volume came out 21 ml . So correct . Thanks again. – Sandeep Jun 8 '18 at 3:07
• @ShawnDypx That's because this definition of volume composition is just a fancy way of saying mole fraction. Sigh. – Anurag B. Jun 8 '18 at 3:22
• @AnuragBaundwal yes. Indeed. – Sandeep Jun 8 '18 at 3:24
• @ShawnDypx And this is not even the "volume composition" — this definition is like saying that light can't escape a neutron star, if the neutron star is somehow compressed to the density of back hole — it's wrong because it's not a neutron star anymore and light can in fact escape an actual neutron star. They've all bent and twisted the definition, making it invalid. This "definition" of "volume composition" is only more confusing (and incorrect) compared to more fraction, if anything. Sigh. – Anurag B. Jun 8 '18 at 3:31
• Yeah , they should really say mole fraction rather than volume composition. It can be confusing . – Sandeep Jun 8 '18 at 3:33

Air is 21% oxygen by moles, not by volume. Remember that gases occupy the whole of the container in which they are kept and not just a fraction of its volume. This is because the gas molecules are constantly moving about. Even if you tried to confine oxygen (along with air) in 21% volume of a container, the oxygen molecules would move about randomly and eventually occupy the whole container. The other gases can't stop this.

Clarification: Air is actually composed of approx. 21% dioxygen by moles, and not by mass or volume. A simple calculation of the molar mass of air assuming this composition and comparing it with the actual data will verify this. Sorry for the confusion.

• Air has an oxygen content of about 21% by volume. This is wrong, hence the down-vote. – Martin - マーチン Jun 7 '18 at 15:17
• @Martin-マーチン isn't by moles and by volume the same for gases? – ypercubeᵀᴹ Jun 8 '18 at 9:56
• The correct interpretation of the ideal gas laws here is that at a constant pressure the volume occupied by a gas is proportional to the number of moles of a gas. So if we take partial pressures in the atmosphere the volume % is the same as the molar %. Removing everything but oxygen from a vessel is the wrong comparison, the right comparison is removing everything else but keeping the pressure the same in which case the volume % will equal the molar %. – matt_black Jun 8 '18 at 10:06
• Thnx @matt_black, that helped me. (I wasn't claiming that the answer is good.) – ypercubeᵀᴹ Jun 8 '18 at 10:09