Added 5.3.1-5.3.2
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| + | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
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| + | <p class="author"> | ||
| + | Solutions of Savchenko Problems in Physics <br> | ||
| + | <i><b>knowledge must be free</b></i> | ||
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| + | <h3 id="back-link"><a href="../../#5.3">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $5.3.1.$ At atmospheric pressure and a temperature of $0 \,^{\circ}C$, the mean free path of a hydrogen molecule is $0.1 ~\mu m$. Estimate the diameter of this molecule. | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | To find the mean free path $\lambda$, it is necessary to find the volume that a particle must travel to meet another. <p></p> | ||
| + | Each particle has a volume | ||
| + | $$ V=\frac{1}{n}\quad(1) $$ | ||
| + | In this case, a particle with an effective cross-section $S=\pi(2r)^2=\pi d^2$? cross-section at which it touches at least the edge of another particle, will pass through a volume | ||
| + | $$ V = \lambda \pi d^2\quad(2) $$ | ||
| + | If we assume that all other particles are at rest, then | ||
| + | $$ \lambda = \frac{1}{\pi d^2n}\quad(3) $$ | ||
| + | If the beam particle is part of an established equilibrium system with identical particles, then the square of the relative velocity is equal to: | ||
| + | $$ \overline{{v}_{\mathrm{relative}}^2}=\overline{({v}_1-{v}_2)^2}=\overline{{v}_1^2+{v}_2^2-2{v}_1\cdot{v}_2}\quad(4) $$ | ||
| + | At equilibrium, the values of the velocities | ||
| + | ${\displaystyle {\overline { {v} _{1}\cdot {v} _{2}}}=0}$, and the relative velocity is equal | ||
| + | at equilibrium, the values of the velocities | ||
| + | $$ {\displaystyle v_{\rm {rel}}={\sqrt {\overline {\ {v} _{\rm {relative}}^{2}}}}={\sqrt {\overline {\ {v} _{1}^{2}+\ {v} _{2}^{2}}}}={\sqrt {2}}v.}\quad(5) $$ | ||
| + | The mean free path of a particle in this case is then defined as | ||
| + | $$ \boxed{\lambda = \frac{1}{\sqrt{2}\pi d^2n}}\quad(6) $$ | ||
| + | From the formula for the mean free path we obtain that | ||
| + | $$ d=\frac{1}{\sqrt{\sqrt{2}\pi \lambda n}}\quad(7) $$ | ||
| + | The concentration of particles can be expressed through the Ideal Gas Law | ||
| + | $$ p=nkT_0\Rightarrow \boxed{n=\frac{p}{kT_0}}\quad(8) $$ | ||
| + | Substituting $(8)$ into $(7)$, we get | ||
| + | $$ \boxed{d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}}\quad(9) $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}$$ | ||
| + | </p> | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
| + | Aliaksandr Melnichenka<br> | ||
| + | </p> | ||
| + | |||
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| <meta name="author" content="Aliaksandr Melnichenka"> | |||
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| <header style="text-align:center;"> | |||
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| <div id="logo"> | |||
| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | |||
| </div> | |||
| </a> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#5.3">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $5.3.1.$ At atmospheric pressure and a temperature of $0 \,^{\circ}C$, the mean free path of a hydrogen molecule is $0.1 ~\mu m$. Estimate the diameter of this molecule. | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| To find the mean free path $\lambda$, it is necessary to find the volume that a particle must travel to meet another. <p></p> | |||
| Each particle has a volume | |||
| $$ V=\frac{1}{n}\quad(1) $$ | |||
| In this case, a particle with an effective cross-section $S=\pi(2r)^2=\pi d^2$? cross-section at which it touches at least the edge of another particle, will pass through a volume | |||
| $$ V = \lambda \pi d^2\quad(2) $$ | |||
| If we assume that all other particles are at rest, then | |||
| $$ \lambda = \frac{1}{\pi d^2n}\quad(3) $$ | |||
| If the beam particle is part of an established equilibrium system with identical particles, then the square of the relative velocity is equal to: | |||
| $$ \overline{{v}_{\mathrm{relative}}^2}=\overline{({v}_1-{v}_2)^2}=\overline{{v}_1^2+{v}_2^2-2{v}_1\cdot{v}_2}\quad(4) $$ | |||
| At equilibrium, the values of the velocities | |||
| ${\displaystyle {\overline { {v} _{1}\cdot {v} _{2}}}=0}$, and the relative velocity is equal | |||
| at equilibrium, the values of the velocities | |||
| $$ {\displaystyle v_{\rm {rel}}={\sqrt {\overline {\ {v} _{\rm {relative}}^{2}}}}={\sqrt {\overline {\ {v} _{1}^{2}+\ {v} _{2}^{2}}}}={\sqrt {2}}v.}\quad(5) $$ | |||
| The mean free path of a particle in this case is then defined as | |||
| $$ \boxed{\lambda = \frac{1}{\sqrt{2}\pi d^2n}}\quad(6) $$ | |||
| From the formula for the mean free path we obtain that | |||
| $$ d=\frac{1}{\sqrt{\sqrt{2}\pi \lambda n}}\quad(7) $$ | |||
| The concentration of particles can be expressed through the Ideal Gas Law | |||
| $$ p=nkT_0\Rightarrow \boxed{n=\frac{p}{kT_0}}\quad(8) $$ | |||
| Substituting $(8)$ into $(7)$, we get | |||
| $$ \boxed{d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}}\quad(9) $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}$$ | |||
| </p> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| Aliaksandr Melnichenka<br> | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
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