Degrees of Freedom of Gas Molecules: Translational, Rotational & Vibrational Modes, Constraints & JEE Problems | Class 11 Physics

Welcome to Lecture 105 of our comprehensive JEE Physics Master Series! Continuing in Unit 9: Kinetic Theory of Gases, we explore the structural cornerstone of statistical thermodynamics: the Degrees of Freedom of Gas Molecules. The concept of degrees of freedom bridges molecular geometry and classical mechanics to explain how energy distributes across microscopic modes of motion. We classify translational, rotational, and vibrational modes, formulate the constraint relation $f = 3N – k$, trace the temperature-dependent quantum freezing of rotational and vibrational states in diatomic gases, calculate effective degrees of freedom in gas mixtures, and master high-yield JEE Advanced problem archetypes.

Detailed Concept Card: Degrees of Freedom in Molecular Systems

1. Definition & Fundamental Constraint Formulation

The degrees of freedom ($f$) of a dynamical system is defined as:

1. Geometrical Definition: The minimum number of independent coordinates or variables required to specify the position, configuration, and orientation of the system completely in three-dimensional space.
2. Energetic Definition: The total number of independent quadratic (squared) terms appearing in the expression for the total energy (Hamiltonian) of the system.

For a molecule consisting of $N$ point atoms subject to $k$ independent geometric constraints (such as fixed bond lengths and fixed bond angles):

$$\mathbf{f = 3N – k}$$

2. Classification of Molecular Modes of Motion

Total degrees of freedom are partitioned into three distinct physical categories:

$$\mathbf{f = f_{\text{trans}} + f_{\text{rot}} + f_{\text{vib}}}$$

A. Translational Degrees of Freedom ($f_{\text{trans}} = 3$)

• Associated with the motion of the molecular center of mass along the three independent Cartesian directions ($x, y, z$).
• Kinetic energy expression: $$E_{\text{trans}} = \frac{1}{2} m v_x^2 + \frac{1}{2} m v_y^2 + \frac{1}{2} m v_z^2$$
• Universal Rule: Every gas molecule—whether monoatomic, diatomic, or polyatomic—possesses exactly $3$ translational degrees of freedom ($f_{\text{trans}} = 3$).

B. Rotational Degrees of Freedom ($f_{\text{rot}}$)

• Associated with the rotation of the molecule about independent principal axes passing through its center of mass.
• Monoatomic Gas ($f_{\text{rot}} = 0$): An atom is treated as a point mass with virtually all mass concentrated in the tiny nucleus ($r_{\text{nucleus}} \sim 10^{-15}\text{ m}$). Its moment of inertia about any axis is practically zero ($I \approx 0$), making rotational kinetic energy $\frac{1}{2}I\omega^2 \approx 0$.
• Linear Molecules ($f_{\text{rot}} = 2$): In diatomic ($\text{O}_2, \text{N}_2, \text{H}_2, \text{CO}$) and linear polyatomic ($\text{CO}_2, \text{C}_2\text{H}_2$) molecules, all nuclei lie along a single internuclear line (the $z$-axis).
  • Moment of inertia about the bond axis is negligible: $I_z \approx 0$.
  • Moments of inertia about the two perpendicular transverse axes are significant: $I_x = I_y \neq 0$.
  • Rotational energy: $E_{\text{rot}} = \frac{1}{2} I_x \omega_x^2 + \frac{1}{2} I_y \omega_y^2 \implies \mathbf{f_{\text{rot}} = 2}$.
• Non-Linear Polyatomic Molecules ($f_{\text{rot}} = 3$): In bent or three-dimensional geometries ($\text{H}_2\text{O}, \text{SO}_2, \text{NH}_3, \text{CH}_4$), atoms do not lie along a line. The molecule has three non-zero principal moments of inertia ($I_x, I_y, I_z \neq 0$): $$E_{\text{rot}} = \frac{1}{2} I_x \omega_x^2 + \frac{1}{2} I_y \omega_y^2 + \frac{1}{2} I_z \omega_z^2 \implies \mathbf{f_{\text{rot}} = 3}$$

C. Vibrational Degrees of Freedom ($f_{\text{vib}}$) & The Double-Energy Rule

• Associated with the periodic oscillation of atoms about their equilibrium bond distances.
• For an $N$-atom molecule:
  • Linear Molecule: Number of vibrational normal modes $= 3N – 5$.
  • Non-Linear Molecule: Number of vibrational normal modes $= 3N – 6$.
• Crucial Energetic Insight for JEE: In simple harmonic oscillation, each vibrational mode involves two independent quadratic energy terms: $$\epsilon_{\text{vib}} = \underbrace{\frac{1}{2} \mu \dot{x}^2}_{\text{Kinetic Energy}} + \underbrace{\frac{1}{2} k x^2}_{\text{Potential Energy}}$$ Therefore, each active vibrational mode contributes $2$ degrees of freedom to the internal energy!

3. Master Summary: Degrees of Freedom & Heat Capacities by Geometry

Gas Type	Example	$f_{\text{trans}}$	$f_{\text{rot}}$	$f_{\text{vib}}$ (Normal Modes)	Total $f$ (Room Temp)	$C_v = \frac{f}{2}R$	$C_p = C_v + R$	$\gamma = 1 + \frac{2}{f}$
Monoatomic	$\text{He}, \text{Ne}, \text{Ar}$	$3$	$0$	$0$	$\mathbf{3}$	$\frac{3}{2}R$	$\frac{5}{2}R$	$\frac{5}{3} \approx 1.67$
Diatomic (Rigid)	$\text{H}_2, \text{O}_2, \text{N}_2$	$3$	$2$	$0$ (Frozen)	$\mathbf{5}$	$\frac{5}{2}R$	$\frac{7}{2}R$	$\frac{7}{5} = 1.40$
Diatomic (with Vibration)	$\text{Cl}_2$ (High Temp)	$3$	$2$	$1$ mode ($= 2$ terms)	$\mathbf{7}$	$\frac{7}{2}R$	$\frac{9}{2}R$	$\frac{9}{7} \approx 1.29$
Linear Polyatomic (Rigid)	$\text{CO}_2, \text{C}_2\text{H}_2$	$3$	$2$	$0$ (Frozen)	$\mathbf{5}$	$\frac{5}{2}R$	$\frac{7}{2}R$	$\frac{7}{5} = 1.40$
Non-Linear Polyatomic (Rigid)	$\text{H}_2\text{O}, \text{NH}_3, \text{CH}_4$	$3$	$3$	$0$ (Frozen)	$\mathbf{6}$	$3R$	$4R$	$\frac{4}{3} \approx 1.33$

4. Quantum Freezing of Degrees of Freedom

Classical statistical mechanics assumes energy is continuous, predicting that all degrees of freedom must be equally active at all temperatures. In reality, energy levels are quantized with discrete level spacings:

• Rotational spacing: $\Delta E_{\text{rot}} \sim \frac{\hbar^2}{2I}$ (characteristic temperature $\Theta_{\text{rot}} \sim 10\text{ K} – 100\text{ K}$).
• Vibrational spacing: $\Delta E_{\text{vib}} \sim \hbar \omega$ (characteristic temperature $\Theta_{\text{vib}} \sim 1000\text{ K} – 6000\text{ K}$).

The Excitation Threshold Principle:
A degree of freedom is excited and participates in heat capacity only when thermal energy is comparable to or exceeds the quantum level spacing ($k_B T \gtrsim \Delta E$).
– Low Temperature ($T < 70\text{ K}$ for $\text{H}_2$): $k_B T < \Delta E_{\text{rot}}$. Rotations freeze out! The gas behaves as monoatomic ($f = 3, C_v = 1.5R, \gamma = 5/3$).
– Moderate / Room Temperature ($250\text{ K} – 750\text{ K}$): $\Delta E_{\text{rot}} \ll k_B T \ll \Delta E_{\text{vib}}$. Rotations are fully active, but vibrations remain frozen ($f = 5, C_v = 2.5R, \gamma = 7/5$).
– High Temperature ($T > 1000\text{ K}$): $k_B T \gtrsim \Delta E_{\text{vib}}$. Vibrational modes unlock ($f = 7, C_v = 3.5R, \gamma = 9/7$).

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Four Graded Solved Examples

Example 1 (JEE Main Level: Degrees of Freedom & Internal Energy Across Geometries)

Problem Statement: Consider three separate containers, each containing $n = 2.0\text{ moles}$ of an ideal gas at temperature $T = 300\text{ K}$:
– Container 1 contains Helium ($\text{He}$, monoatomic).
– Container 2 contains Nitrogen ($\text{N}_2$, diatomic rigid).
– Container 3 contains Water vapor ($\text{H}_2\text{O}$, non-linear triatomic rigid).
Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
1. Specify $f_{\text{trans}}, f_{\text{rot}}$, total $f$, and the adiabatic index $\gamma$ for each gas.
2. Calculate the total internal energy ($U$) of the gas in each container.

Solution:

Step 1: Degrees of Freedom & Adiabatic Index.
– Helium ($\text{He}$): Single atom ($N = 1$). Point mass with zero rotation: $$f_{\text{trans}} = 3, \quad f_{\text{rot}} = 0 \implies \mathbf{f = 3}$$ $$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \mathbf{\frac{5}{3} \approx 1.67}$$ – Nitrogen ($\text{N}_2$): Diatomic molecule ($N = 2$). Two atoms along an internuclear axis: $$f_{\text{trans}} = 3, \quad f_{\text{rot}} = 2 \implies \mathbf{f = 5}$$ $$\gamma = 1 + \frac{2}{5} = \mathbf{\frac{7}{5} = 1.40}$$ – Water vapor ($\text{H}_2\text{O}$): Non-linear bent triatomic molecule ($N = 3$): $$f_{\text{trans}} = 3, \quad f_{\text{rot}} = 3 \implies \mathbf{f = 6}$$ $$\gamma = 1 + \frac{2}{6} = \mathbf{\frac{4}{3} \approx 1.33}$$

Step 2: Total Internal Energy Calculations.
For an ideal gas, total internal energy is $U = n \left(\frac{f}{2}\right) R T$:
– For Helium ($f = 3$): $$U_{\text{He}} = 2.0 \times \left(\frac{3}{2}\right) \times (8.314\text{ J/(mol}\cdot\text{K)}) \times (300\text{ K}) = 3.0 \times 2494.2\text{ J} = \mathbf{7482.6\text{ J}}$$ – For Nitrogen ($f = 5$): $$U_{\text{N}_2} = 2.0 \times \left(\frac{5}{2}\right) \times (8.314) \times (300) = 5.0 \times 2494.2\text{ J} = \mathbf{12471.0\text{ J}}$$ – For Water vapor ($f = 6$): $$U_{\text{H}_2\text{O}} = 2.0 \times \left(\frac{6}{2}\right) \times (8.314) \times (300) = 6.0 \times 2494.2\text{ J} = \mathbf{14965.2\text{ J}}$$

Takeaway: Even though all three containers hold identical mole quantities at the same temperature, water vapor holds twice the thermal internal energy of helium because it possesses twice as many active degrees of freedom ($f = 6$ vs. $f = 3$).

Example 2 (Constraint Counting & Dimensional Analysis: $f = 3N – k$)

Problem Statement: Apply the constraint formula $f = 3N – k$ to rigorously determine the degrees of freedom for:
1. A system of two point particles connected by a rigid, massless rod of fixed length $L$.
2. A system of three point particles situated at the vertices of a rigid triangular plate whose three side lengths are fixed.
3. A system of three point particles maintained collinear with fixed adjacent distances.
4. Any general three-dimensional rigid body consisting of an arbitrary number of atoms $N \ge 3$.

Solution:

Case 1: Diatomic Molecule Model ($N = 2$).
Unconstrained coordinates: $3N = 3(2) = 6$.
The fixed internuclear bond length represents $k = 1$ independent geometric constraint: $(x_1 – x_2)^2 + (y_1 – y_2)^2 + (z_1 – z_2)^2 = L^2$.
$$f = 3N – k = 6 – 1 = \mathbf{5} \quad (3\text{ translational} + 2\text{ rotational})$$

Case 2: Rigid Triangular Molecule ($N = 3$, Non-Linear).
Unconstrained coordinates: $3N = 3(3) = 9$.
Fixing the lengths of the three sides ($L_{12}, L_{23}, L_{31}$) imposes $k = 3$ independent constraints.
$$f = 3N – k = 9 – 3 = \mathbf{6} \quad (3\text{ translational} + 3\text{ rotational})$$

Case 3: Rigid Collinear Triatomic Molecule ($N = 3$, Linear like $\text{CO}_2$).
Unconstrained coordinates: $3N = 9$.
Geometric constraints: 2 bond lengths ($r_{12}, r_{23}$) plus 2 independent angle constraints to maintain collinearity (e.g., $\vec{r}_{12} \times \vec{r}_{23} = 0$, which specifies two independent component equations), giving $k = 4$ independent constraints.
$$f = 3N – k = 9 – 4 = \mathbf{5} \quad (3\text{ translational} + 2\text{ rotational})$$

Case 4: General 3D Rigid Body ($N \ge 3$).
The position of particle 1 requires 3 coordinates. Particle 2 requires 2 coordinates (constrained by fixed distance to 1). Particle 3 requires 1 coordinate (constrained by fixed distances to 1 and 2). Once three non-collinear particles are fixed, the positions of all other $N – 3$ particles are completely determined relative to them. Thus:
$$\mathbf{f = 3 + 2 + 1 = 6} \quad (3\text{ translational} + 3\text{ rotational})$$ Every three-dimensional rigid body possesses exactly $6$ degrees of freedom.

Example 3 (JEE Advanced Level: Temperature-Dependent Plateaus of Diatomic Hydrogen)

Problem Statement: One mole of Hydrogen gas ($\text{H}_2$) is examined across three distinct temperature regimes: $T_1 = 50\text{ K}$, $T_2 = 300\text{ K}$, and $T_3 = 2500\text{ K}$. Given that the rotational characteristic temperature of $\text{H}_2$ is $\Theta_{\text{rot}} \approx 85\text{ K}$ and the vibrational characteristic temperature is $\Theta_{\text{vib}} \approx 6000\text{ K}$:
1. State the active degrees of freedom $f$ and adiabatic index $\gamma$ in each of the three regimes.
2. Calculate the molar heat capacity at constant volume ($C_v$) in each regime in terms of $R$.
3. Explain why the rotational modes freeze out below $70\text{ K}$ using quantum energy spacing.

Solution:

Step 1 & 2: Active Modes, $C_v$, and $\gamma$ Across Regimes.
– Regime I ($T_1 = 50\text{ K}$): Cryogenic Regime.
Since $T_1 < \Theta_{\text{rot}} = 85\text{ K}$, thermal energy $k_B T$ is insufficient to excite the first rotational state. Rotations and vibrations are both quantum-mechanically frozen. Only translational motion survives: $$\mathbf{f = 3}, \quad C_v = \frac{3}{2}R, \quad \gamma = 1 + \frac{2}{3} = \mathbf{\frac{5}{3} \approx 1.67}$$ - Regime II ($T_2 = 300\text{ K}$): Room Temperature Regime.
Here $\Theta_{\text{rot}} \ll T_2 \ll \Theta_{\text{vib}}$. Rotations are fully excited ($f_{\text{rot}} = 2$), while vibrations remain frozen ($f_{\text{vib}} = 0$): $$\mathbf{f = 3 + 2 = 5}, \quad C_v = \frac{5}{2}R, \quad \gamma = 1 + \frac{2}{5} = \mathbf{\frac{7}{5} = 1.40}$$ – Regime III ($T_3 = 2500\text{ K}$): High Temperature Regime.
At $2500\text{ K}$, thermal energy is comparable to vibrational spacing ($k_B T \sim \hbar\omega$). The vibrational mode unlocks, contributing $2$ quadratic terms (kinetic + potential): $$\mathbf{f = 3 + 2 + 2 = 7}, \quad C_v = \frac{7}{2}R, \quad \gamma = 1 + \frac{2}{7} = \mathbf{\frac{9}{7} \approx 1.29}$$

Step 3: Quantum Mechanism of Freezing.
Rotational energy levels are quantized: $$E_J = \frac{\hbar^2}{2I} J(J + 1), \quad J = 0, 1, 2, \dots$$ The energy required to excite the first rotational state ($J = 0 \to J = 1$) is: $$\Delta E_{\text{rot}} = E_1 – E_0 = \frac{\hbar^2}{I} = k_B \Theta_{\text{rot}}$$ Because the moment of inertia $I$ of $\text{H}_2$ is exceptionally tiny (due to the minuscule mass of hydrogen atoms), its energy gap $\Delta E_{\text{rot}}$ is unusually large. When $T < 70\text{ K}$, thermal collision energy $k_B T < \Delta E_{\text{rot}}$, so collisions cannot supply the quantum of energy needed to spin the molecule. Molecules remain locked in the ground state $J = 0$, completely freezing the rotational degree of freedom!

Example 4 (JEE Advanced Level: Effective Degrees of Freedom of a Gas Mixture)

Problem Statement: A thermal enclosure contains a non-reacting mixture of $n_1 = 1.0\text{ mole}$ of Helium ($\text{He}$, monoatomic, $f_1 = 3$) and $n_2 = 2.0\text{ moles}$ of Nitrogen ($\text{N}_2$, rigid diatomic, $f_2 = 5$) at temperature $T$. Determine:
1. The effective degrees of freedom $f_{\text{mix}}$ of the mixture.
2. The effective molar heat capacities $C_{v, \text{mix}}$ and $C_{p, \text{mix}}$ in terms of $R$.
3. The effective adiabatic index $\gamma_{\text{mix}}$ of the mixture expressed in simplest fractional form.

Solution:

Step 1: Effective Degrees of Freedom ($f_{\text{mix}}$).
The total internal energy of the mixture is the sum of the internal energies of the individual gases: $$U_{\text{mix}} = U_1 + U_2 = n_1 \left(\frac{f_1}{2} R T\right) + n_2 \left(\frac{f_2}{2} R T\right)$$ Equating this to $(n_1 + n_2)\left(\frac{f_{\text{mix}}}{2} R T\right)$: $$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{(1.0 \times 3) + (2.0 \times 5)}{1.0 + 2.0} = \frac{3 + 10}{3} = \mathbf{\frac{13}{3} \approx 4.333}$$

Step 2: Effective Molar Heat Capacities.
Molar heat capacity at constant volume: $$C_{v, \text{mix}} = \frac{f_{\text{mix}}}{2} R = \frac{13/3}{2} R = \mathbf{\frac{13}{6} R \approx 2.167 \, R}$$ Molar heat capacity at constant pressure: $$C_{p, \text{mix}} = C_{v, \text{mix}} + R = \frac{13}{6} R + R = \mathbf{\frac{19}{6} R \approx 3.167 \, R}$$

Step 3: Effective Adiabatic Index ($\gamma_{\text{mix}}$).
$$\gamma_{\text{mix}} = \frac{C_{p, \text{mix}}}{C_{v, \text{mix}}} = \frac{19/6 \, R}{13/6 \, R} = \mathbf{\frac{19}{13} \approx 1.462}$$ Cross-check using degrees of freedom formula: $$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{13/3} = 1 + \frac{6}{13} = \frac{19}{13} \quad (\text{Exact verification!})$$

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Practice Worksheet: 10 JEE Problems

Questions 1 to 5: JEE Main Level (Single Correct)

Q1. The number of translational degrees of freedom possessed by a non-linear triatomic gas molecule (such as $\text{H}_2\text{O}$) at room temperature is:

(A) $1$
(B) $2$
(C) $3$
(D) $6$

Q2. The number of rotational degrees of freedom for Carbon Dioxide ($\text{CO}_2$), which is a linear triatomic molecule, at room temperature is:

(A) $0$
(B) $1$
(C) $2$
(D) $3$

Q3. The relationship between the adiabatic index $\gamma = C_p / C_v$ and the total active degrees of freedom $f$ of an ideal gas molecule is given by:

(A) $f = \frac{2}{\gamma – 1}$
(B) $f = \frac{\gamma – 1}{2}$
(C) $f = \frac{2}{\gamma + 1}$
(D) $f = \frac{\gamma + 1}{2}$

Q4. An ideal gas has an adiabatic index $\gamma = 1.33 = 4/3$. The active degrees of freedom of the gas molecules is:

(A) $3$
(B) $5$
(C) $6$
(D) $7$

Q5. At an extremely low temperature of $40\text{ K}$, the active degrees of freedom of Hydrogen gas ($\text{H}_2$) is observed to be:

(A) $3$
(B) $5$
(C) $6$
(D) $7$

Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)

Q6. Which of the following statements regarding the vibrational modes of gas molecules is/are correct?

(A) Each active vibrational normal mode contributes two quadratic energy terms (one kinetic and one potential) to the internal energy.
(B) A linear molecule with $N$ atoms possesses $3N – 5$ vibrational normal modes.
(C) A non-linear molecule with $N$ atoms possesses $3N – 6$ vibrational normal modes.
(D) At room temperature ($300\text{ K}$), vibrational modes are frozen out in common diatomic gases ($\text{N}_2, \text{O}_2$) because vibrational quantum level spacing is much larger than $k_B T$.

Q7. Which of the following statements correctly compare ideal gases of different molecular structures in thermal equilibrium?

(A) As molecular atomicity increases, the adiabatic index $\gamma$ monotonically decreases ($\gamma_{\text{mono}} > \gamma_{\text{dia}} > \gamma_{\text{poly}}$).
(B) For a fixed quantity of $n\text{ moles}$ at temperature $T$, internal energy is highest for the gas possessing the greatest degrees of freedom.
(C) The molar heat capacity at constant volume $C_v$ is directly proportional to the active degrees of freedom ($C_v = \frac{f}{2}R$).
(D) Monoatomic gases have zero rotational degrees of freedom because the nuclear moment of inertia is negligibly small.

Q8. A gas mixture contains $n_1$ moles of gas 1 with degrees of freedom $f_1$ and $n_2$ moles of gas 2 with degrees of freedom $f_2$. Which of the following relations is/are correct?

(A) The effective degrees of freedom of the mixture is $f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$.
(B) The effective molar heat capacity is $C_{v, \text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$.
(C) The effective adiabatic index satisfies $\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}}$.
(D) The effective adiabatic index satisfies $\gamma_{\text{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$.

Questions 9 to 10: Numerical / Integer Answer Type

Q9. A rigid non-linear triatomic gas molecule consists of $3$ atoms with $3$ fixed interatomic bond lengths at room temperature where all vibrational modes are frozen. Determine its total number of active degrees of freedom $f$.

Q10. A gas mixture contains $n_1 = 2.0\text{ moles}$ of a monoatomic gas ($f_1 = 3$) and $n_2 = 1.0\text{ mole}$ of a rigid diatomic gas ($f_2 = 5$) at room temperature. The adiabatic index of the mixture is expressed in simplest fractional form as $\gamma_{\text{mix}} = \frac{a}{b}$, where $a$ and $b$ are co-prime positive integers. Find the value of $(a + b)$.

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Detailed Solutions & Answer Key

Answer Key

Question	Correct Option / Answer	Topic / Concept
Q1	(C) $3$	Universal Translational Degrees of Freedom
Q2	(C) $2$	Rotational Modes of Linear Molecules
Q3	(A) $f = \frac{2}{\gamma – 1}$	Relation Between $\gamma$ and $f$
Q4	(C) $6$	Degrees of Freedom for $\gamma = 4/3$
Q5	(A) $3$	Quantum Rotational Freeze-Out in $\text{H}_2$
Q6	(A), (B), (C), (D)	Vibrational Degrees of Freedom & Quantization
Q7	(A), (B), (C), (D)	Comparative Atomicity & Thermodynamic Relations
Q8	(A), (B), (C), (D)	Gas Mixture Equivalence Formulations
Q9	6	Non-Linear Triatomic Molecule Modes ($3 + 3 = 6$)
Q10	28	Mixture $\gamma = 17/11 \implies a + b = 28$

Complete Step-by-Step Solutions

Solution 1:
The motion of the center of mass of any physical entity in 3-dimensional space is completely described by three independent coordinates ($x, y, z$). Thus, all gas molecules without exception possess exactly $3$ translational degrees of freedom.
Correct Option: (C)

Solution 2:
In any linear molecule (such as $\text{CO}_2$, $\text{O}=\text{C}=\text{O}$), the three atoms lie along a single straight line. Rotation about the molecular bond axis has a negligible moment of inertia ($I \approx 0$). Significant rotation occurs only about the two mutually perpendicular axes orthogonal to the molecular axis. Hence, a linear molecule has exactly $2$ rotational degrees of freedom.
Correct Option: (C)

Solution 3:
From $\gamma = \frac{C_p}{C_v} = \frac{\frac{f}{2}R + R}{\frac{f}{2}R} = 1 + \frac{2}{f}$: $$\gamma – 1 = \frac{2}{f} \implies \mathbf{f = \frac{2}{\gamma – 1}}$$ Correct Option: (A)

Solution 4:
Using $f = \frac{2}{\gamma – 1}$ with $\gamma = 4/3$: $$f = \frac{2}{4/3 – 1} = \frac{2}{1/3} = \mathbf{6}$$ This corresponds to a rigid non-linear polyatomic molecule ($3$ translational $+ 3$ rotational).
Correct Option: (C)

Solution 5:
At $T = 40\text{ K}$, the thermal kinetic energy $k_B T$ is strictly smaller than the first rotational quantum excitation energy ($\Delta E_{\text{rot}} \approx 85 k_B$). As a result, rotational motion freezes out, and the hydrogen molecules can only translate. Thus, $f = f_{\text{trans}} = \mathbf{3}$.
Correct Option: (A)

Solution 6:
(A) Correct: A harmonic oscillator possesses kinetic energy ($\frac{1}{2}\mu v^2$) and elastic potential energy ($\frac{1}{2}kx^2$), contributing $2$ quadratic terms.
(B) Correct: Out of $3N$ total coordinates, $3$ are translational and $2$ are rotational, leaving $3N – 5$ vibrational modes.
(C) Correct: Out of $3N$ coordinates, $3$ are translational and $3$ are rotational, leaving $3N – 6$ vibrational modes.
(D) Correct: For $\text{N}_2$, $\Theta_{\text{vib}} \sim 3300\text{ K} \gg 300\text{ K}$, so vibrations remain locked in the ground state at room temperature.
Correct Options: (A), (B), (C), (D)

Solution 7:
(A) Correct: $\gamma = 1 + 2/f$. As $f$ increases ($3 \to 5 \to 6$), $\gamma$ decreases ($1.67 \to 1.40 \to 1.33$).
(B) Correct: $U = \frac{f}{2}nRT$. Higher $f$ directly scales higher internal energy.
(C) Correct: $C_v = \frac{dU}{dT} = \frac{f}{2}R$.
(D) Correct: Nuclear radius $\sim 10^{-15}\text{ m}$ gives $I \approx 0$, making rotational energy negligible.
Correct Options: (A), (B), (C), (D)

Solution 8:
(A) Correct: By conservation of internal energy: $f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$.
(B) Correct: Molar heat capacity is a mole-weighted average.
(C) Correct: Directly follows from $\gamma_{\text{mix}} = 1 + 2/f_{\text{mix}}$.
(D) Correct: Ratio of total $C_p$ to total $C_v$ gives the adiabatic index.
Correct Options: (A), (B), (C), (D)

Solution 9:
For a rigid non-linear triatomic molecule (such as $\text{H}_2\text{O}$): $$f = f_{\text{trans}} + f_{\text{rot}} = 3 + 3 = \mathbf{6}$$ Alternatively, using $f = 3N – k$ with $N = 3$ atoms and $k = 3$ rigid bond constraints: $$f = 3(3) – 3 = 9 – 3 = \mathbf{6}$$ Answer: 6

Solution 10:
Effective degrees of freedom of the mixture: $$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{(2.0 \times 3) + (1.0 \times 5)}{2.0 + 1.0} = \frac{6 + 5}{3} = \frac{11}{3}$$ Adiabatic index of the mixture: $$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{11/3} = 1 + \frac{6}{11} = \frac{17}{11}$$ Here $a = 17$ and $b = 11$, which are co-prime positive integers ($\gcd(17, 11) = 1$).
$$a + b = 17 + 11 = \mathbf{28}$$ Answer: 28