Правка разделов «Statement», «Theoretical Background: Degrees of Freed», «1. Monatomic Gas (e.g., , , )»

Valter правка от 7 авг. 2026 г., 19:17

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en/5.6.1.md +22 −27

		@@ -1,47 +1,46 @@
			### Statement

		−	$5.6.1.$ [Insert the problem statement]
		−
		−	### Solution
		−
		−	### Statement
		−
			$5.6.1.$ The average energy of a single gas molecule over a wide temperature range is determined quite accurately by the formula $\bar{\varepsilon} = \frac{i}{2}kT$, where $i$ is the number of degrees of freedom of the molecule, which is equal to the number of coordinates determining its position. Using this formula, find the average energy of $H_2$, $N_2$, $H_2O$, and $CH_4$ molecules at a temperature $T$.

			### Theoretical Background: Degrees of Freedom & Molecular Models

		−	In thermodynamics, a degree of freedom ($i$) is an independent coordinate required to completely specify the position and orientation of a molecule in space. It represents the number of independent ways a molecule can move and store kinetic energy.
		+	In thermodynamics, a "degree of freedom ($i$)"" is an independent coordinate required to completely specify the position and orientation of a molecule in space. It represents the number of independent ways a molecule can move and store kinetic energy.

			At normal temperatures, we consider translational motion (the center of mass moving along coordinate axes) and rotational motion (tumbling around axes). Let's visualize all three fundamental models:

		−	1. Monatomic Gas (e.g., $He$, $Ne$, $Ar$)
		+	#### 1. Monatomic Gas (e.g., $He$, $Ne$, $Ar$)
			A single atom is treated as a point mass. It can move in 3D space, but its moment of inertia is practically zero, so rotation does not store thermodynamic energy.

		−	![Monatomic Translation](Degree-of-freedom-1.png)
		−	* Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes, as shown by the pink arrows).
		−	* Rotational ($i_{\text{rot}}$): 0
		−	* Total $i = 3$
		+	![Monatomic Translation\|750x399, 50%](../../img/5.6.1/Degree-of-freedom-1.png)

		−	2. Diatomic Gas (e.g., $H_2$, $N_2$)
		+	1. Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes, as shown by the pink arrows).
		+	2. Rotational ($i_{\text{rot}}$): 0
		+	3. Total $i = 3$
		+
		+	#### 2. Diatomic Gas (e.g., $H_2$, $N_2$)
			Two atoms connected by a rigid bond (a "dumbbell" model). Its motion is divided into two parts.

		−	![Diatomic Translation](Degree-of-freedom-3.png)
		+	![Diatomic Translation \|750x455, 50%](../../img/5.6.1/Degree-of-freedom-3.png)
		+
			First, the entire center of mass can move in 3D space.
		−	* Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes).
		+	1. Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes).

		−	![Diatomic Rotation](Degree-of-freedom-4.png)
		+	![Diatomic Rotation \|750x440, 50%](../../img/5.6.1/Degree-of-freedom-4.png)
		+
			Second, it can rotate end-over-end. As shown in the diagram above, if the molecule lies along the $Y$-axis, it can rotate independently around the $X$ and $Z$ axes (indicated by the green arrows). Rotation strictly around its own bond axis ($Y$-axis in this specific orientation) has a negligible moment of inertia and is therefore ignored.
		−	* Rotational ($i_{\text{rot}}$): 2
		−	* Total $i = 5$

		−	3. Polyatomic Non-linear Gas (e.g., $H_2O$, $CH_4$)
		+	2. Rotational ($i_{\text{rot}}$): 2
		+	3. Total $i = 5$
		+
		+	#### 3. Polyatomic Non-linear Gas (e.g., $H_2O$, $CH_4$)
			These molecules possess a complex 3D structure with substantial mass distribution around all three spatial axes.

		−	![Polyatomic Rotation](Degree-of-freedom-2.png)
		−	* Translational ($i_{\text{trans}}$): 3 (motion along $X$, $Y$, and $Z$ axes, always present).
		−	* Rotational ($i_{\text{rot}}$): 3 (as shown by the green arrows, the molecule can tumble around the $X$, $Y$, and $Z$ axes independently).
		−	* Total $i = 6$
		+	![Polyatomic Rotation \|750x399, 50%](../../img/5.6.1/Degree-of-freedom-2.png)

		+	1. Translational ($i_{\text{trans}}$): 3 (motion along $X$, $Y$, and $Z$ axes, always present).
		+	2. Rotational ($i_{\text{rot}}$): 3 (as shown by the green arrows, the molecule can tumble around the $X$, $Y$, and $Z$ axes independently).
		+	3. Total $i = 6$
		+
			### Solution

			To find the average energy of a single gas molecule using the formula:
			$$\bar{\varepsilon} = \frac{i}{2}kT$$
			we determine the number of degrees of freedom $i$ based on the geometric structures analyzed above.

			For the diatomic gases $H_2$ and $N_2$, the number of degrees of freedom is $i = 5$. Therefore, the average energy of a single molecule of these gases is:
			$$\bar{\varepsilon}_{H_2} = \bar{\varepsilon}_{N_2} = \frac{5}{2}kT$$

			For the non-linear polyatomic molecules $H_2O$ and $CH_4$, the number of degrees of freedom is $i = 6$. Therefore, the average energy of a single molecule is:
			$$\bar{\varepsilon}_{H_2O} = \bar{\varepsilon}_{CH_4} = \frac{6}{2}kT = 3kT$$

		@@ -57,7 +56,3 @@Solution
			#### Answer
			$$\bar{\varepsilon}_{H_2} = \bar{\varepsilon}_{N_2} = \frac{5}{2}kT$$
			$$\bar{\varepsilon}_{H_2O} = \bar{\varepsilon}_{CH_4} = 3kT$$
		−
		−	#### Answer
		−
		−	[Insert a concise answer or boxed result]

	@@ -1,47 +1,46 @@
	### Statement		### Statement

	$5.6.1.$ [Insert the problem statement]

	### Solution

	### Statement

	$5.6.1.$ The average energy of a single gas molecule over a wide temperature range is determined quite accurately by the formula $\bar{\varepsilon} = \frac{i}{2}kT$, where $i$ is the number of degrees of freedom of the molecule, which is equal to the number of coordinates determining its position. Using this formula, find the average energy of $H_2$, $N_2$, $H_2O$, and $CH_4$ molecules at a temperature $T$.		$5.6.1.$ The average energy of a single gas molecule over a wide temperature range is determined quite accurately by the formula $\bar{\varepsilon} = \frac{i}{2}kT$, where $i$ is the number of degrees of freedom of the molecule, which is equal to the number of coordinates determining its position. Using this formula, find the average energy of $H_2$, $N_2$, $H_2O$, and $CH_4$ molecules at a temperature $T$.

	### Theoretical Background: Degrees of Freedom & Molecular Models		### Theoretical Background: Degrees of Freedom & Molecular Models

	In thermodynamics, a degree of freedom ($i$) is an independent coordinate required to completely specify the position and orientation of a molecule in space. It represents the number of independent ways a molecule can move and store kinetic energy.		In thermodynamics, a "degree of freedom ($i$)"" is an independent coordinate required to completely specify the position and orientation of a molecule in space. It represents the number of independent ways a molecule can move and store kinetic energy.

	At normal temperatures, we consider translational motion (the center of mass moving along coordinate axes) and rotational motion (tumbling around axes). Let's visualize all three fundamental models:		At normal temperatures, we consider translational motion (the center of mass moving along coordinate axes) and rotational motion (tumbling around axes). Let's visualize all three fundamental models:

	1. Monatomic Gas (e.g., $He$, $Ne$, $Ar$)		#### 1. Monatomic Gas (e.g., $He$, $Ne$, $Ar$)
	A single atom is treated as a point mass. It can move in 3D space, but its moment of inertia is practically zero, so rotation does not store thermodynamic energy.		A single atom is treated as a point mass. It can move in 3D space, but its moment of inertia is practically zero, so rotation does not store thermodynamic energy.

	![Monatomic Translation](Degree-of-freedom-1.png)		![Monatomic Translation\|750x399, 50%](../../img/5.6.1/Degree-of-freedom-1.png)
	* Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes, as shown by the pink arrows).
	* Rotational ($i_{\text{rot}}$): 0
	* Total $i = 3$

	2. Diatomic Gas (e.g., $H_2$, $N_2$)		1. Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes, as shown by the pink arrows).
			2. Rotational ($i_{\text{rot}}$): 0
			3. Total $i = 3$

			#### 2. Diatomic Gas (e.g., $H_2$, $N_2$)
	Two atoms connected by a rigid bond (a "dumbbell" model). Its motion is divided into two parts.		Two atoms connected by a rigid bond (a "dumbbell" model). Its motion is divided into two parts.

	![Diatomic Translation](Degree-of-freedom-3.png)		![Diatomic Translation \|750x455, 50%](../../img/5.6.1/Degree-of-freedom-3.png)

	First, the entire center of mass can move in 3D space.		First, the entire center of mass can move in 3D space.
	* Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes).		1. Translational ($i_{\text{trans}}$): 3 (motion along the $X$, $Y$, and $Z$ axes).

	![Diatomic Rotation](Degree-of-freedom-4.png)		![Diatomic Rotation \|750x440, 50%](../../img/5.6.1/Degree-of-freedom-4.png)

	Second, it can rotate end-over-end. As shown in the diagram above, if the molecule lies along the $Y$-axis, it can rotate independently around the $X$ and $Z$ axes (indicated by the green arrows). Rotation strictly around its own bond axis ($Y$-axis in this specific orientation) has a negligible moment of inertia and is therefore ignored.		Second, it can rotate end-over-end. As shown in the diagram above, if the molecule lies along the $Y$-axis, it can rotate independently around the $X$ and $Z$ axes (indicated by the green arrows). Rotation strictly around its own bond axis ($Y$-axis in this specific orientation) has a negligible moment of inertia and is therefore ignored.
	* Rotational ($i_{\text{rot}}$): 2
	* Total $i = 5$

	3. Polyatomic Non-linear Gas (e.g., $H_2O$, $CH_4$)		2. Rotational ($i_{\text{rot}}$): 2
			3. Total $i = 5$

			#### 3. Polyatomic Non-linear Gas (e.g., $H_2O$, $CH_4$)
	These molecules possess a complex 3D structure with substantial mass distribution around all three spatial axes.		These molecules possess a complex 3D structure with substantial mass distribution around all three spatial axes.

	![Polyatomic Rotation](Degree-of-freedom-2.png)		![Polyatomic Rotation \|750x399, 50%](../../img/5.6.1/Degree-of-freedom-2.png)
	* Translational ($i_{\text{trans}}$): 3 (motion along $X$, $Y$, and $Z$ axes, always present).
	* Rotational ($i_{\text{rot}}$): 3 (as shown by the green arrows, the molecule can tumble around the $X$, $Y$, and $Z$ axes independently).
	* Total $i = 6$

			1. Translational ($i_{\text{trans}}$): 3 (motion along $X$, $Y$, and $Z$ axes, always present).
			2. Rotational ($i_{\text{rot}}$): 3 (as shown by the green arrows, the molecule can tumble around the $X$, $Y$, and $Z$ axes independently).
			3. Total $i = 6$

	### Solution		### Solution

	To find the average energy of a single gas molecule using the formula:		To find the average energy of a single gas molecule using the formula:
	$$\bar{\varepsilon} = \frac{i}{2}kT$$		$$\bar{\varepsilon} = \frac{i}{2}kT$$
	we determine the number of degrees of freedom $i$ based on the geometric structures analyzed above.		we determine the number of degrees of freedom $i$ based on the geometric structures analyzed above.

	For the diatomic gases $H_2$ and $N_2$, the number of degrees of freedom is $i = 5$. Therefore, the average energy of a single molecule of these gases is:		For the diatomic gases $H_2$ and $N_2$, the number of degrees of freedom is $i = 5$. Therefore, the average energy of a single molecule of these gases is:
	$$\bar{\varepsilon}_{H_2} = \bar{\varepsilon}_{N_2} = \frac{5}{2}kT$$		$$\bar{\varepsilon}_{H_2} = \bar{\varepsilon}_{N_2} = \frac{5}{2}kT$$

	For the non-linear polyatomic molecules $H_2O$ and $CH_4$, the number of degrees of freedom is $i = 6$. Therefore, the average energy of a single molecule is:		For the non-linear polyatomic molecules $H_2O$ and $CH_4$, the number of degrees of freedom is $i = 6$. Therefore, the average energy of a single molecule is:
	$$\bar{\varepsilon}_{H_2O} = \bar{\varepsilon}_{CH_4} = \frac{6}{2}kT = 3kT$$		$$\bar{\varepsilon}_{H_2O} = \bar{\varepsilon}_{CH_4} = \frac{6}{2}kT = 3kT$$

	@@ -57,7 +56,3 @@Solution
	#### Answer		#### Answer
	$$\bar{\varepsilon}_{H_2} = \bar{\varepsilon}_{N_2} = \frac{5}{2}kT$$		$$\bar{\varepsilon}_{H_2} = \bar{\varepsilon}_{N_2} = \frac{5}{2}kT$$
	$$\bar{\varepsilon}_{H_2O} = \bar{\varepsilon}_{CH_4} = 3kT$$		$$\bar{\varepsilon}_{H_2O} = \bar{\varepsilon}_{CH_4} = 3kT$$

	#### Answer

	[Insert a concise answer or boxed result]