Edits to “Statement”, “Answer”
en/5.4.19.md
+5 −2
| @@ -1,6 +1,9 @@ | |||
| ### Statement | |||
| − | $5.4.19.$ | ||
| + | $5.4.19.$ Due to a small temperature difference between two parallel plates in a rarefied gas containing $n$ particles | ||
| + | per unit volume, a heat flux $W_1$ has appeared. When the gas pressure increases, the heat flow first | ||
| + | increases, and then, when it reaches the value $W_2$, it ceases to depend on the gas pressure. Explain this | ||
| + | effect. Estimate the radius of the gas molecules. The distance between the plates $δ$ | ||
| ### Solution | |||
| This effect can be explained by the fact that as the pressure increases, the mean free path of the molecules, determined by their effective cross-section and concentration, decreases, and when it becomes less than the distance between the plates, the heat flux begins to be determined by this length and ceases to depend on the pressure, since $\lambda\sim\frac{1}{n}\Rightarrow\lambda\sim\frac{1}{p}$ , and the heat flux in the general case is $W\sim n\Rightarrow W\sim p$. We will use the formula from the solution to problem 5.4.16. | |||
| $$W_{1}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{m}}S$$ | |||
| $$W_{2}=\frac{3}{2}\lambda kn\frac{\Delta T}{\delta}\sqrt{\frac{3kT}{m}}S$$ | |||
| $$\frac{W_{1}}{W_{2}}=\frac{\delta}{\lambda}$$ | |||
| $$\frac{W_{1}}{W_{2}}=4\delta \pi r^{2}n\Rightarrow$$ | |||
| $$r = \frac{1}{2} (\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | |||
| However, for the evaluation, the coefficient $\frac{1}{2}$ must be discarded. Hence the final answer: | |||
| $$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | |||
| @@ -21,4 +24,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | ||
| @@ -1,6 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $5.4.19.$ |
$5.4.19.$ Due to a small temperature difference between two parallel plates in a rarefied gas containing $n$ particles | ||
| per unit volume, a heat flux $W_1$ has appeared. When the gas pressure increases, the heat flow first | |||
| increases, and then, when it reaches the value $W_2$, it ceases to depend on the gas pressure. Explain this | |||
| effect. Estimate the radius of the gas molecules. The distance between the plates $δ$ | |||
| ### Solution | ### Solution | ||
| This effect can be explained by the fact that as the pressure increases, the mean free path of the molecules, determined by their effective cross-section and concentration, decreases, and when it becomes less than the distance between the plates, the heat flux begins to be determined by this length and ceases to depend on the pressure, since $\lambda\sim\frac{1}{n}\Rightarrow\lambda\sim\frac{1}{p}$ , and the heat flux in the general case is $W\sim n\Rightarrow W\sim p$. We will use the formula from the solution to problem 5.4.16. | This effect can be explained by the fact that as the pressure increases, the mean free path of the molecules, determined by their effective cross-section and concentration, decreases, and when it becomes less than the distance between the plates, the heat flux begins to be determined by this length and ceases to depend on the pressure, since $\lambda\sim\frac{1}{n}\Rightarrow\lambda\sim\frac{1}{p}$ , and the heat flux in the general case is $W\sim n\Rightarrow W\sim p$. We will use the formula from the solution to problem 5.4.16. | ||
| $$W_{1}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{m}}S$$ | $$W_{1}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{m}}S$$ | ||
| $$W_{2}=\frac{3}{2}\lambda kn\frac{\Delta T}{\delta}\sqrt{\frac{3kT}{m}}S$$ | $$W_{2}=\frac{3}{2}\lambda kn\frac{\Delta T}{\delta}\sqrt{\frac{3kT}{m}}S$$ | ||
| $$\frac{W_{1}}{W_{2}}=\frac{\delta}{\lambda}$$ | $$\frac{W_{1}}{W_{2}}=\frac{\delta}{\lambda}$$ | ||
| $$\frac{W_{1}}{W_{2}}=4\delta \pi r^{2}n\Rightarrow$$ | $$\frac{W_{1}}{W_{2}}=4\delta \pi r^{2}n\Rightarrow$$ | ||
| $$r = \frac{1}{2} (\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | $$r = \frac{1}{2} (\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | ||
| However, for the evaluation, the coefficient $\frac{1}{2}$ must be discarded. Hence the final answer: | However, for the evaluation, the coefficient $\frac{1}{2}$ must be discarded. Hence the final answer: | ||
| $$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | $$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | ||
| @@ -21,4 +24,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$ | ||