Правка разделов «Solution», «Answer»

jzmicer правка от 9 июн. 2026 г., 12:35

правка #18487 предыдущая #18486 ← раньше

en/5.4.19.md +3 −3

			### Statement

			$5.4.19.$ 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.
			$$W_{1}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{m}}S$$

			$$W_{2}=\frac{3}{2}\lambda kn\frac{\Delta T}{\delta}\sqrt{\frac{3kT}{m}}S$$

		@@ -15,13 +15,13 @@Solution
			$$\frac{W_{1}}{W_{2}}=\frac{\delta}{\lambda}$$

			$$\frac{W_{1}}{W_{2}}=4\delta \pi r^{2}n\Rightarrow$$
		−	$$r = \frac{1}{2} (\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$
		+	$$r = \frac{1}{2} \left(\frac{W_{1}}{W_{2}\pi n\delta}\right)^\frac{1}{2}$$


			However, for the evaluation, the coefficient $\frac{1}{2}$ must be discarded. Hence the final answer:

		−	$$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$
		+	$$r \approx\left(\frac{W_{1}}{W_{2}\pi n\delta}\right)^\frac{1}{2}$$

			#### Answer

		−	$$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$
		+	$$r \approx\left(\frac{W_{1}}{W_{2}\pi n\delta}\right)^\frac{1}{2}$$

	### Statement		### Statement

	$5.4.19.$ Due to a small temperature difference between two parallel plates in a rarefied gas containing $n$ particles		$5.4.19.$ 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		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		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 $δ$		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.
	$$W_{1}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{m}}S$$		$$W_{1}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{m}}S$$

	$$W_{2}=\frac{3}{2}\lambda kn\frac{\Delta T}{\delta}\sqrt{\frac{3kT}{m}}S$$		$$W_{2}=\frac{3}{2}\lambda kn\frac{\Delta T}{\delta}\sqrt{\frac{3kT}{m}}S$$

	@@ -15,13 +15,13 @@Solution
	$$\frac{W_{1}}{W_{2}}=\frac{\delta}{\lambda}$$		$$\frac{W_{1}}{W_{2}}=\frac{\delta}{\lambda}$$

	$$\frac{W_{1}}{W_{2}}=4\delta \pi r^{2}n\Rightarrow$$		$$\frac{W_{1}}{W_{2}}=4\delta \pi r^{2}n\Rightarrow$$
	$$r = \frac{1}{2} (\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$		$$r = \frac{1}{2} \left(\frac{W_{1}}{W_{2}\pi n\delta}\right)^\frac{1}{2}$$


	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:

	$$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$		$$r \approx\left(\frac{W_{1}}{W_{2}\pi n\delta}\right)^\frac{1}{2}$$

	#### Answer		#### Answer

	$$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$		$$r \approx\left(\frac{W_{1}}{W_{2}\pi n\delta}\right)^\frac{1}{2}$$