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| ### Statement |
| ### Statement |
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| $5.3.13.$ [Insert the problem statement] |
| $5.3.13.$ The thermal conductivity of gases $A_1$ and $A_2$ is equal to $k_1$ and $k_2$, respectively. Determine the thermal |
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| conductivity of a mixture in which $A_1$ gas molecules are α times larger than $A_2$ gas molecules. The |
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| temperature of the gases is the same, the gases are monatomic. The molar mass of gases is $µ_1$ and $µ_2$, |
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| respectively. |
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| ### Solution |
| ### Solution |
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| Определим коэффициенты теплопроводности газов в смеси обозначим их $\kappa'_{1}$ и $\kappa'_{2}$.Так как газы одноатомные и температуры одинаковые можно считать, что: |
| Let's determine the thermal conductivity coefficients of gases in a mixture, denote them $\kappa'_{1}$ and $\kappa'_{2}$. Since the gases are monatomic and the temperatures are the same, we can assume that: |
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| $$\kappa\sim \frac{1}{\sigma\cdot \sqrt{\mu}}$$ |
| $$\kappa\sim \frac{1}{\sigma\cdot \sqrt{\mu}}$$ |
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| Значит после всех преобразований можно получить: |
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| This means that after all the transformations we can obtain: |
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| $$\frac{\kappa_{1}}{\kappa_{2}}=\sqrt{\frac{\mu_{2}}{\mu_{1}}}\cdot \frac{r_{2}^2}{r_{1}^2}$$ |
| $$\frac{\kappa_{1}}{\kappa_{2}}=\sqrt{\frac{\mu_{2}}{\mu_{1}}}\cdot \frac{r_{2}^2}{r_{1}^2}$$ |
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| Где $r_{1}$ и $r_{2}$ радиусы атомов газа. |
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| Where $r_{1}$ and $r_{2}$ are the radii of the gas atoms. |
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| $$r_{2}= r_{1}\cdot (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^{\frac{1}{2}}$$ |
| $$r_{2}= r_{1}\cdot (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^{\frac{1}{2}}$$ |
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| Концентрации газов в смеси относятся как $$\frac{n_{1}}{n_{2}}=\alpha$$ |
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| Число столкновений молекул газа 1 и газа 2 в смеси: |
| The concentrations of gases in a mixture are related as $$\frac{n_{1}}{n_{2}}=\alpha$$ |
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| $$\nu_{1}=4\cdot \pi\cdot r_{1}^{2}\cdot n_{1}\cdot <\upsilon_{кв_{1}}> + \frac{n_{1}}{\alpha}\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{кв_{1}}>$$ |
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| $$\nu_{2}=4\cdot \pi\cdot r_{2}^{2}\cdot n_{2}\cdot <\upsilon_{кв_{2}}> + n_{2}\cdot \alpha\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{кв_{2}}>$$ |
| The number of collisions of molecules of gas 1 and gas 2 in the mixture: |
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| Запишем отношения теплопроводностей "до" к "после", после преобразований получим: |
| $$\nu_{1}=4\cdot \pi\cdot r_{1}^{2}\cdot n_{1}\cdot <\upsilon_{sq_{1}}> + \frac{n_{1}}{\alpha}\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{sq_{1}}>$$ |
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| $$\nu_{2}=4\cdot \pi\cdot r_{2}^{2}\cdot n_{2}\cdot <\upsilon_{sq_{2}}> + n_{2}\cdot \alpha\cdot \pi\cdot (r_{1} + r_{2})^{2}\cdot <\upsilon_{sq_{2}}>$$ |
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| Let's write down the ratio of thermal conductivities "before" to "after", after transformations we get: |
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| $$\frac{\kappa_{1}}{\kappa'_{1}}=\frac{4\cdot \pi\cdot r_{1}^{2}\cdot n_{1} + \frac{n_{1}}{\alpha}\cdot \pi\cdot (r_{1} + r_{2})^{2}}{4\cdot \pi\cdot r_{1}^{2}\cdot n_{1}}$$ |
| $$\frac{\kappa_{1}}{\kappa'_{1}}=\frac{4\cdot \pi\cdot r_{1}^{2}\cdot n_{1} + \frac{n_{1}}{\alpha}\cdot \pi\cdot (r_{1} + r_{2})^{2}}{4\cdot \pi\cdot r_{1}^{2}\cdot n_{1}}$$ |
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| Подставим выражение для $r_{2}$, закончим преобразования и выразим новую теплопроводность газа 1: |
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| We substitute the expression for $r_{2}$, complete the transformations and express the new thermal conductivity of gas 1: |
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| $$\kappa'_{1} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}}$$ |
| $$\kappa'_{1} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}}$$ |
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| Аналогично для газа 2: |
| Similarly for gas 2: |
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| $$\kappa'_{2} = \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$ |
| $$\kappa'_{2} = \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$ |
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| Пусть создан градиент температур $\frac{\partial T}{\partial x}$ внутри смеси. Поток тепла через смесь равен: |
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| Let a temperature gradient $\frac{\partial T}{\partial x}$ be created within the mixture. The heat flux through the mixture is: |
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| $$q_{12} = \kappa_{12}\cdot \frac{\partial T}{\partial x}\cdot S = q_{1} + q_{2} = \kappa'_{1}\cdot \frac{\partial T}{\partial x}\cdot S + \kappa'_{2}\cdot \frac{\partial T}{\partial x}\cdot S$$ |
| $$q_{12} = \kappa_{12}\cdot \frac{\partial T}{\partial x}\cdot S = q_{1} + q_{2} = \kappa'_{1}\cdot \frac{\partial T}{\partial x}\cdot S + \kappa'_{2}\cdot \frac{\partial T}{\partial x}\cdot S$$ |
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| Отсюда: |
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| From here: |
| $$\kappa_{12} = \kappa'_{1} + \kappa'_{2}$$ |
| $$\kappa_{12} = \kappa'_{1} + \kappa'_{2}$$ |
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| $$\kappa_{12} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}} + \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$ |
| $$\kappa_{12} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}} + \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$ |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $$\kappa_{12} = \frac{\kappa_{1}}{1 + \frac{1}{4\cdot \alpha}\cdot [1 + (\frac{\kappa_{1}}{\kappa_{2}}\cdot \sqrt{\frac{\mu_{1}}{\mu_{2}}})^\frac{1}{2}]^{2}} + \frac{\kappa_{2}}{1 + \frac{\alpha}{4}\cdot [1 + (\frac{\kappa_{2}}{\kappa_{1}}\cdot \sqrt{\frac{\mu_{2}}{\mu_{1}}})^\frac{1}{2}]^{2}}$$ |
