Collisions of molecules of a gaseous substance

Exercise 1

Here are the diameters of some molecules:

Calculate the frequency of collisions of each molecule in its gas at $27^o C$ and $10^5 Pa$.

$z=$ $\sqrt 2 \pi d^2 \frac{P}{k\cdot T}(\frac{8RT}{\pi M} )^\frac{1}{2} \,=$ $\sqrt 2 \cdot 3,14 (2,89 \cdot 10^{-10})^2 \frac{10^5}{1,38 \cdot 10^{-23}\cdot 300}(\frac{8 \cdot 8,3 \cdot 300}{3,14 \cdot 2 \cdot 10^{-3}} )^\frac{1}{2} \,=$ $z=$ $\sqrt 2 \pi d^2 \frac{P}{k\cdot T}(\frac{8RT}{\pi M} )^\frac{1}{2} \,=$ $\sqrt 2 \cdot 3,14 (2,6 \cdot 10^{-10})^2 \frac{10^5}{1,38 \cdot 10^{-23}\cdot 300}(\frac{3 \cdot 8,3 \cdot 300}{4 \cdot 10^{-3}} )^\frac{1}{2} \,=$ $z=$ $\sqrt 2 \pi d^2 \frac{P}{k\cdot T}(\frac{8RT}{\pi M} )^\frac{1}{2} \,=$ $\sqrt 2 \cdot 3,14 (3,46 \cdot 10^{-10})^2 \frac{10^5}{1,38 \cdot 10^{-23}\cdot 300}(\frac{3 \cdot 8,3 \cdot 300}{32 \cdot 10^{-3}} )^\frac{1}{2} \,=$ $z=$ $\sqrt 2 \pi d^2 \frac{P}{k\cdot T}(\frac{8RT}{\pi M} )^\frac{1}{2} \,=$ $\sqrt 2 \cdot 3,14 (3,64 \cdot 10^{-10})^2 \frac{10^5}{1,38 \cdot 10^{-23}\cdot 300}(\frac{3 \cdot 8,3 \cdot 300}{28 \cdot 10^{-3}} )^\frac{1}{2} \,=$ $z=$ $\sqrt 2 \pi d^2 \frac{P}{k\cdot T}(\frac{8RT}{\pi M} )^\frac{1}{2} \,=$ $\sqrt 2 \cdot 3,14 (3,3 \cdot 10^{-10})^2 \frac{10^5}{1,38 \cdot 10^{-23}\cdot 300}(\frac{3 \cdot 8,3 \cdot 300}{44 \cdot 10^{-3}} )^\frac{1}{2} \,=$ $z=$ $\sqrt 2 \pi d^2 \frac{P}{k\cdot T}(\frac{8RT}{\pi M} )^\frac{1}{2} \,=$ $\sqrt 2 \cdot 3,14 (3,8 \cdot 10^{-10})^2 \frac{10^5}{1,38 \cdot 10^{-23}\cdot 300}(\frac{3 \cdot 8,3 \cdot 300}{16 \cdot 10^{-3}} )^\frac{1}{2} \,=$