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Caluculate the collision density between $\ce{O2}$ and $\ce{N2}$ in air at $T = \pu{10 ^\circ C}$. Assume that the air is an ideal gas and that the molecules are spheres with diameters $\pu{0.36 nm}$ for $\ce{O2}$ and $\pu{0.37 nm}$ for $\ce{N2}$.

This is what I have done so far, the collision density equation is:

$$ Z_{AB} = \frac{N_A}{V} \frac{N_B}{V} π d^2_{AB} c_A$$

And since we assume that air is an ideal gas, we can use:

$$\frac{N_A}{V} = \frac{N_B}{V} = \frac {p}{kT}$$

$$\frac {p}{kT} = \frac {1.01325 \cdot 10^5}{(1.38 \cdot 10^{-23}) \cdot 283} =2.6 \cdot 10^{25}$$

The collision cross-section can also be calculated:

$$ \sigma = \pi d^2_{AB} = \pi \left((3.6 \cdot 10^{-10})+(3.7 \cdot 10^{-10})\right)^2 = 1.67 \cdot 10^{-18} m^2 $$

But to calculate the average speed I need the mass of the molecules, which I don't have and I am not sure if I can determine it from the information given either. I thought about calculating the volume of each molecule and then looking up the density of oxygen and nitrogen to get the mass, however I am not sure if this is the correct way that the problem should be solved. Is there a specific assumption I need to take in order to determine the masses?

Mathew Mahindaratne
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  • It is a common calculation see this for example https://www.tec-science.com/thermodynamics/kinetic-theory-of-gases/mean-free-path-collision-frequency/ – MaxW Apr 03 '20 at 15:33
  • The "in air" means you need more facts than the problem gives: You need to know the fractions of O2 and N2 in air. Given that you need to add that to the problem, it's probably OK to also use the (well known) masses of O2 and N2. – Bob Jacobsen Apr 03 '20 at 16:09
  • The mass of $1$ molecule $\ce{O2}$ is $\pu{32 g/6.022 10^{23} = 5.314 \times 10^{-23} g}$ – Maurice Mar 26 '22 at 17:51
  • In case that previous comment isn't obvious enough: atomic-mass-per-mole / Avogadro's number = mass / particle, and in this case each particle has twice the atomic mass of an oxygen atom. – Ray Butterworth Mar 27 '22 at 00:39

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