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+ <h3 id="back-link"><a href="../../#5.3">$\leftarrow$Back</a></h3>
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+ <h3> Statement </h3>
+ <p>
+ $5.3.3.$ Estimate how many times per $1 \text{ s}$ in $1 \text{ cm}^3$ of air nitrogen molecules collide with each other and nitrogen molecules collide with oxygen molecules.
+
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ In <a href="../5.3.1">5.3.1</a> we obtained a formula for the mean free path of gas molecules
+$$ \lambda = \frac{1}{\sqrt{2}\pi d^2n}\quad(1) $$
+From this estimate it follows that the average number of collisions that will occur with a gas molecule after passing the path $dx$ will be
+$$ N = \sqrt{2}\pi d^2n\cdot dx\quad(2) $$
+From here, dividing by a small time interval $dt$, we obtain the number of collisions occurring with the molecule during a given time interval
+$$ Z = \sqrt{2}\pi d^2n \cdot\frac{dx}{dt}\quad(3) $$
+Let us rewrite expression $(3)$, taking into account the definition of velocity $u = \frac{dx}{dt}$:
+$$ \boxed{Z = \sqrt{2}\pi d^2n u}\quad(4) $$
+The average speed of gas molecules with a molar mass of $\mu$ can be found in approximation using the formula for the mean-square speed of molecules
+$$ u = \sqrt{ \frac{3RT}{\mu}}\quad(5.1) $$
+However, for a more accurate estimate we will use the average speed formula
+$$ u = \sqrt{ \frac{8RT}{\pi\mu}}\quad(5.2) $$
+According to Dalton's law, the total atmospheric pressure will be the sum of the pressures created by the molecules of nitrogen $N_2$ and oxygen $O_2$
+$$ p_a=p_{N_2}+p_{O_2}\quad(6) $$
+The atmosphere consists of $78\%$ nitrogen $N_2$ and $21\%$ oxygen $O_2$, with $1\%$ occupied by other gases</p><p>
+Then, from the Ideal Gas Equation, we obtain the concentration of nitrogen and oxygen molecules, respectively
+$$ n_{N_2}=\frac{0.78\cdot p_a}{RT}\quad(7) $$
+
+$$ n_{O_2}=\frac{0.2\cdot p_a}{RT}\quad(8) $$
+Hence, the number of collisions of nitrogen molecules with each other per unit of time. Considering that each oscillation occurs between two molecules simultaneously, we multiply by $\frac{1}{2}$:
+$$ Z_{NN}=\frac{N\cdot Z_N}{2}=\frac{0.78\cdot p_a}{RT}\cdot \frac{\sqrt{2}}{2}\pi d_N^2\cdot \sqrt{ \frac{8RT}{\pi\mu _N}}\quad(9) $$
+<div class="scroll-wrapper">
+$$ \boxed{Z_{NN}=1.56p_a\cdot d_N^2\cdot \sqrt{\frac{\pi}{RT\mu_N}}}\approx6\cdot10^{28}\text{ s}^{-1}\cdot\text{cm}^{-3}\quad(10) $$
+</div>
+Similarly, the number of collisions of nitrogen molecules with oxygen molecules
+$$ Z_{ON}=N\cdot Z_O=\frac{0.21\cdot p_a}{RT}\cdot \sqrt{2} \pi d_O^2 \cdot \sqrt{ \frac{8RT}{\pi\mu _O}}\quad(11) $$
+<div class="scroll-wrapper">
+$$ \boxed{Z_{ON}=0.84p_a\cdot d_O^2\cdot \sqrt{\frac{\pi}{RT\mu_O}}}\approx3\cdot10^{28}\text{ s}^{-1}\cdot\text{cm}^{-3}\quad(12) $$
+</div>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$ \nu_1\approx6\cdot10^{28}\text{ s}^{-1}\cdot\text{cm}^{-3}; ~\nu_2\approx3\cdot10^{28}\text{ s}^{-1}\cdot\text{cm}^{-3}. $$
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+ Aliaksandr Melnichenka<br>
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