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.

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    <h3 id="back-link"><a href="/#5.3">$\leftarrow$Back</a></h3>

    <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

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 $\frac{1}{2}$:
$$Z_{NN}=\frac{N\cdot Z_N}{2}=\frac{0.78\cdot p_a}{RT}\cdot \frac{\sqrt{2}}{2}\pi d_N^2\cdot \sqrt{ \frac{8RT}{\pi\mu _N}}\quad(9)$$

Similarly, the number of collisions of nitrogen molecules with oxygen molecules $$Z_{ON}=N\cdot Z_O=\frac{0.21\cdot p_a}{RT}\cdot \sqrt{2} \pi d_O^2 \cdot \sqrt{ \frac{8RT}{\pi\mu _O}}\quad(11)$$

    <h4>Answer</h4>
    <p>
        $$ \nu_1\approx6\cdot10^{28}\text{ s}^{-1}\cdot\text{cm}^{-3}; \~\nu_2\approx3\cdot10^{28}\text{ s}^{-1}\cdot\text{cm}^{-3}. $$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
	  Aliaksandr Melnichenka<br>
	</p>


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