Shouldn't 1 mole of any ideal gas at open space occupy more than 22.4 L volume at S.T.P. due to diffusion?
Here is the calculation for the volume occupied by one mole ($n$) of an ideal gas at STP (temperature $T = 273.15\ \mathrm K$, pressure $P = 1\ \mathrm{bar}$):  

gas constant: $R = 0.08314\ \mathrm{L\ bar\ mol^{-1}\ K^{-1}}$
ideal gas law: $PV = nRT$  

$$\begin{align}
V &amp;= \frac{nRT}{P}\\[6pt]  
&amp;= \mathrm{\frac{1\ mol\ \times\ 0.08314\ L\ bar\ mol^{-1}\ K^{-1} \times\ 273.15\ K}{1\ bar}}\\[6pt] 
&amp;= 22.71\ \mathrm L
\end{align}$$  

Another way of looking at this of course is that any 22.71 liter chunk of particulate-free air, of arbitrary shape, will contain about 1 mole of air. Molecules of the different gases composing the air will diffuse in to and out of this region, but it will always contain about 1 mol of air at STP.