Welcome to Lecture 106 of our comprehensive JEE Physics Master Series! Continuing in Unit 9: Kinetic Theory of Gases, we explore one of the crowning triumphs of classical statistical mechanics: the Law of Equipartition of Energy. Building upon the geometric and physical degrees of freedom formulated in Lecture 105, this universal law governs how thermal energy partitions among microscopic quadratic modes of motion at absolute temperature $T$. We derive the fundamental $\frac{1}{2} k_B T$ energy theorem per quadratic degree of freedom, calculate total internal energy and molar heat capacities ($C_v, C_p, \gamma$) for monoatomic, diatomic, and polyatomic gases, generalize the principle to 3D crystalline solids (the Dulong-Petit Law), analyze temperature-dependent quantum deviations, and solve high-yield JEE Advanced problems.
Detailed Concept Card: The Law of Equipartition of Energy
1. Statement & Theoretical Foundation
The Law of Equipartition of Energy:
For any dynamical system in thermal equilibrium at absolute temperature $T$, the total internal energy is distributed equally among all its active quadratic degrees of freedom, and the mean energy associated with each independent quadratic term in the total energy (Hamiltonian) expression is exactly:$$\mathbf{\langle \epsilon \rangle = \frac{1}{2} k_B T \quad \text{(per molecule)}} = \mathbf{\frac{1}{2} R T \quad \text{(per mole)}}$$
where $k_B = 1.3806 \times 10^{-23}\text{ J/K}$ is Boltzmann’s constant, and $R = N_A k_B = 8.314\text{ J/(mol}\cdot\text{K)}$ is the universal gas constant.
2. Energy Partition Across Motion Categories
Any molecular coordinate contributing a squared term to total energy receives $\frac{1}{2}k_B T$ on average:
A. Translational Motion ($f_{\text{trans}} = 3$)
- Kinetic energy decomposes into three mutually orthogonal Cartesian coordinates: $$E_{\text{trans}} = \frac{1}{2} m v_x^2 + \frac{1}{2} m v_y^2 + \frac{1}{2} m v_z^2$$
- By isotropic spatial symmetry in equilibrium: $$\left\langle \frac{1}{2} m v_x^2 \right\rangle = \left\langle \frac{1}{2} m v_y^2 \right\rangle = \left\langle \frac{1}{2} m v_z^2 \right\rangle = \mathbf{\frac{1}{2} k_B T}$$
- Total Average Translational Kinetic Energy per Molecule: $$\mathbf{\langle E_{\text{trans}} \rangle = \frac{3}{2} k_B T} \quad \text{(Universal for all gas molecules)}$$
B. Rotational Motion ($f_{\text{rot}}$)
- Rotational energy about principal axes with non-zero moment of inertia ($I_i$): $$E_{\text{rot}} = \frac{1}{2} I_x \omega_x^2 + \frac{1}{2} I_y \omega_y^2 + \frac{1}{2} I_z \omega_z^2$$
- Each active rotational axis contributes: $\left\langle \frac{1}{2} I_i \omega_i^2 \right\rangle = \mathbf{\frac{1}{2} k_B T}$.
- Monoatomic Gas ($f_{\text{rot}} = 0$): $\langle E_{\text{rot}} \rangle = 0$.
- Linear Molecules ($f_{\text{rot}} = 2$): $\langle E_{\text{rot}} \rangle = 2 \times \left(\frac{1}{2} k_B T\right) = \mathbf{k_B T}$.
- Non-Linear Molecules ($f_{\text{rot}} = 3$): $\langle E_{\text{rot}} \rangle = 3 \times \left(\frac{1}{2} k_B T\right) = \mathbf{\frac{3}{2} k_B T}$.
C. Vibrational Motion ($f_{\text{vib}}$): The Dual-Quadratic Rule
- Each normal vibrational mode behaves as a one-dimensional simple harmonic oscillator with reduced mass $\mu$ and force constant $k$: $$E_{\text{vib}} = \underbrace{\frac{1}{2} \mu \dot{x}^2}_{\text{Kinetic Energy}} + \underbrace{\frac{1}{2} k x^2}_{\text{Potential Energy}}$$
- Because a harmonic oscillator contains two independent quadratic terms, each active vibrational mode receives: $$\langle \epsilon_{\text{vib}} \rangle = \left\langle \frac{1}{2} \mu \dot{x}^2 \right\rangle + \left\langle \frac{1}{2} k x^2 \right\rangle = \frac{1}{2} k_B T + \frac{1}{2} k_B T = \mathbf{k_B T}$$
- JEE Alert: One active vibrational mode contributes $k_B T$ per molecule ($R T$ per mole), effectively counting as $2$ degrees of freedom in heat capacity expressions!
3. Total Internal Energy & Molar Heat Capacities of Ideal Gases
For $n$ moles of an ideal gas consisting of molecules with $f$ total active quadratic degrees of freedom:
$$\mathbf{U = n \left(\frac{f}{2}\right) R T = N \left(\frac{f}{2}\right) k_B T}$$
Differentiating with respect to temperature at constant volume yields the molar heat capacities:
$$\mathbf{C_v = \left(\frac{\partial U_m}{\partial T}\right)_v = \frac{f}{2} R}$$
$$\mathbf{C_p = C_v + R = \left(\frac{f}{2} + 1\right) R = \left(\frac{f + 2}{2}\right) R}$$
$$\mathbf{\gamma = \frac{C_p}{C_v} = \frac{\frac{f + 2}{2} R}{\frac{f}{2} R} = 1 + \frac{2}{f}}$$
4. Extension to Solids: The Dulong-Petit Law
In a crystalline solid containing $N_A$ atoms per mole, atoms do not undergo translational or rotational motion through space. Instead, each atom oscillates around its fixed lattice equilibrium site in three independent dimensions ($x, y, z$):
- Each atom acts as three independent 1D harmonic oscillators, possessing $3$ kinetic terms and $3$ potential terms: $$E_{\text{atom}} = \sum_{i=1}^3 \left(\frac{1}{2} m v_i^2 + \frac{1}{2} k_i x_i^2\right) \implies 6 \text{ quadratic terms}$$
- Average energy per atom: $\langle \epsilon \rangle = 6 \times \left(\frac{1}{2} k_B T\right) = 3 k_B T$.
- Total internal energy per mole ($N_A$ atoms): $$U_m = N_A (3 k_B T) = \mathbf{3 R T}$$
- Molar Heat Capacity of Solids (Dulong-Petit Law): $$\mathbf{C = \frac{dU_m}{dT} = 3 R \approx 3 \times 8.314\text{ J/(mol}\cdot\text{K)} \approx 24.94\text{ J/(mol}\cdot\text{K)}}$$
- Quantum Limitation: At low temperatures ($T \ll \Theta_D$, the Debye temperature), lattice vibrational modes freeze out into discrete phonons, causing $C_v \propto T^3$ (Debye’s $T^3$ Law). Classical equipartition holds only when $T \gtrsim \Theta_D$.
Four Graded Solved Examples
Example 1 (JEE Main Level: Microscopic & Macroscopic Thermal Energy Distribution)
Problem Statement: A rigid, thermally insulated container holds a gas mixture composed of $n_1 = 3.0\text{ moles}$ of Argon ($\text{Ar}$, monoatomic, $f_1 = 3$) and $n_2 = 2.0\text{ moles}$ of Oxygen ($\text{O}_2$, rigid diatomic, $f_2 = 5$) in thermal equilibrium at temperature $T = 300\text{ K}$. Given $k_B = 1.38 \times 10^{-23}\text{ J/K}$ and $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
1. Calculate the mean thermal energy possessed by a single Argon atom and by a single Oxygen molecule.
2. Calculate the total internal energy ($U_{\text{total}}$) of the gas mixture.
3. Find the heat energy ($\Delta Q$) required to raise the temperature of this mixture by $\Delta T = 50\text{ K}$ at constant volume.
Solution:
Step 1: Microscopic Mean Energy per Molecule.
According to the law of equipartition of energy, each active degree of freedom contributes $\frac{1}{2} k_B T$ per molecule:
– For Argon (monoatomic, $f = 3$ translational):
$$\langle \epsilon_{\text{Ar}} \rangle = \frac{3}{2} k_B T = 1.5 \times (1.38 \times 10^{-23}\text{ J/K}) \times (300\text{ K}) = \mathbf{6.21 \times 10^{-21}\text{ J}}$$
– For Oxygen (rigid diatomic, $f = 3\text{ trans} + 2\text{ rot} = 5$):
$$\langle \epsilon_{\text{O}_2} \rangle = \frac{5}{2} k_B T = 2.5 \times (1.38 \times 10^{-23}\text{ J/K}) \times (300\text{ K}) = \mathbf{1.035 \times 10^{-20}\text{ J}}$$
Step 2: Total Internal Energy of the Mixture.
Total internal energy is the sum of internal energies of each component:
$$U_{\text{Ar}} = n_1 \left(\frac{f_1}{2} R T\right) = 3.0 \times \left(\frac{3}{2}\right) \times (8.314) \times (300) = 4.5 \times 2494.2 = 11223.9\text{ J}$$
$$U_{\text{O}_2} = n_2 \left(\frac{f_2}{2} R T\right) = 2.0 \times \left(\frac{5}{2}\right) \times (8.314) \times (300) = 5.0 \times 2494.2 = 12471.0\text{ J}$$
$$U_{\text{total}} = U_{\text{Ar}} + U_{\text{O}_2} = 11223.9\text{ J} + 12471.0\text{ J} = \mathbf{23694.9\text{ J}}$$
Step 3: Heat Input at Constant Volume.
Since volume is fixed, work done $W = 0$, so by the First Law of Thermodynamics:
$$\Delta Q = \Delta U = (n_1 C_{v1} + n_2 C_{v2}) \Delta T$$
$$n_1 C_{v1} + n_2 C_{v2} = \left(3.0 \times \frac{3}{2} R\right) + \left(2.0 \times \frac{5}{2} R\right) = (4.5 + 5.0) R = 9.5 R$$
$$\Delta Q = 9.5 \times (8.314\text{ J/(mol}\cdot\text{K)}) \times (50\text{ K}) = \mathbf{3949.15\text{ J}}$$
Takeaway: Even though the mass of an Argon atom ($40\text{ u}$) is greater than an Oxygen molecule ($32\text{ u}$), the lighter Oxygen molecule carries $66.7\%$ more thermal energy because its rotational degrees of freedom participate equally in equipartition.
Example 2 (Mathematical Formulation: Polytropic Molar Heat Capacity & Equipartition)
Problem Statement: An ideal gas whose molecules possess $f$ active degrees of freedom undergoes a reversible polytropic process governed by $P V^x = \text{constant}$, where $x$ is the polytropic exponent ($x \neq 1$).
1. Derive an analytical expression for the molar heat capacity $C$ of the gas during this process in terms of $f, R$, and $x$.
2. Identify the specific values of $x$ for which $C = 0$ (adiabatic condition) and $C \to \infty$ (isothermal condition).
Solution:
Step 1: General First Law Derivation.
For $n$ moles of gas absorbing heat $dQ$ with temperature change $dT$:
$$dQ = dU + dW \implies n C dT = n C_v dT + P dV$$
Dividing by $n dT$:
$$C = C_v + \frac{P}{n} \frac{dV}{dT}$$
Differentiating the ideal gas equation $P V = n R T$:
$$P dV + V dP = n R dT$$
Differentiating the polytropic equation $P V^x = \text{constant}$ logarithmically:
$$\frac{dP}{P} + x \frac{dV}{V} = 0 \implies V dP = -x P dV$$
Substituting $V dP$ into the differentiated ideal gas equation:
$$P dV – x P dV = n R dT \implies P dV (1 – x) = n R dT \implies \frac{P}{n} \frac{dV}{dT} = \frac{R}{1 – x}$$
By the law of equipartition of energy, $C_v = \frac{f}{2} R$. Substituting back:
$$\mathbf{C = \frac{f}{2} R + \frac{R}{1 – x} = R \left(\frac{f}{2} + \frac{1}{1 – x}\right)}$$
Step 2: Special Thermodynamic Limits.
– Adiabatic Condition ($C = 0$):
$$\frac{f}{2} + \frac{1}{1 – x} = 0 \implies \frac{1}{x – 1} = \frac{f}{2} \implies x – 1 = \frac{2}{f} \implies \mathbf{x = 1 + \frac{2}{f} = \gamma}$$
The polytropic process with $x = \gamma$ recovers the standard adiabatic relation $P V^\gamma = \text{constant}$!
– Isothermal Condition ($C \to \infty$):
As $x \to 1$, the denominator $(1 – x) \to 0$, giving $C \to \pm \infty$. This corresponds to $P V = \text{constant}$ where finite heat is transferred with zero temperature change ($dT = 0$).
Example 3 (JEE Advanced Level: Dulong-Petit Law & Lattice Thermal Vibrations)
Problem Statement: Using the classical law of equipartition of energy, consider an elemental crystalline solid consisting of $N_A$ atoms per mole held in a rigid 3D lattice.
1. Derive the Dulong-Petit law for the molar heat capacity $C = 3R$ by modeling each atom as a 3D harmonic oscillator.
2. Calculate the specific heat capacity in $\text{J/(kg}\cdot\text{K)}$ of metallic Copper ($\text{Cu}$, molar mass $M = 63.55\text{ g/mol}$) predicted by this classical law.
3. Explain why the Dulong-Petit law fails at low temperatures and state the limiting temperature dependence according to Debye’s quantum theory.
Solution:
Step 1: Equipartition Derivation of the Dulong-Petit Law.
In a crystal lattice, each atom oscillates about its fixed equilibrium site along three mutually perpendicular directions ($x, y, z$). The total energy of an atom is:
$$E_{\text{atom}} = \underbrace{\frac{1}{2}m(v_x^2 + v_y^2 + v_z^2)}_{\text{3 Kinetic Terms}} + \underbrace{\frac{1}{2}(k_x x^2 + k_y y^2 + k_z z^2)}_{\text{3 Potential Terms}}$$
The Hamiltonian contains exactly $6$ independent quadratic terms. By the law of equipartition of energy, each quadratic term contributes an average thermal energy of $\frac{1}{2} k_B T$:
$$\langle E_{\text{atom}} \rangle = 6 \times \left(\frac{1}{2} k_B T\right) = \mathbf{3 k_B T}$$
For one mole ($N_A$ atoms), total lattice internal energy is:
$$U_m = N_A \langle E_{\text{atom}} \rangle = N_A (3 k_B T) = 3 (N_A k_B) T = \mathbf{3 R T}$$
The molar heat capacity is:
$$\mathbf{C = \frac{dU_m}{dT} = 3 R \approx 3 \times 8.314\text{ J/(mol}\cdot\text{K)} = \mathbf{24.94\text{ J/(mol}\cdot\text{K)}}}$$
Step 2: Specific Heat of Copper.
Specific heat capacity per unit mass ($c$) is:
$$c = \frac{C}{M} = \frac{24.942\text{ J/(mol}\cdot\text{K)}}{0.06355\text{ kg/mol}} \approx \mathbf{392.5\text{ J/(kg}\cdot\text{K)}}$$
Remark: This agrees closely with the experimental room-temperature value of copper ($\approx 385\text{ J/(kg}\cdot\text{K)}$).
Step 3: Quantum Breakdown at Cryogenic Temperatures.
Classical equipartition assumes energy is continuous and that lattice oscillators can be excited by arbitrarily small thermal energies. In quantum mechanics, vibrational modes are quantized into phonons of discrete energy quanta $\hbar \omega$.
– At room temperature ($T \gg \Theta_D$, the Debye temperature), $k_B T \gg \hbar \omega$, all phonon modes are populated, and $C \to 3R$.
– At low temperatures ($T \to 0\text{ K}$), $k_B T \ll \hbar \omega$, high-frequency modes freeze out. According to Debye’s quantum theory, the heat capacity drops sharply as:
$$\mathbf{C_v \propto T^3 \quad (\text{Debye } T^3 \text{ Law})}$$
Example 4 (JEE Advanced Level: Thermal Mixing & Adiabatic Compression of Coupled Gases)
Problem Statement: A thermally insulated, rigid container is divided into two compartments by a removable internal partition. Compartment 1 contains $n_1 = 1.0\text{ mole}$ of Helium ($\text{He}$, monoatomic, $f_1 = 3$) at temperature $T_1 = 400\text{ K}$. Compartment 2 contains $n_2 = 3.0\text{ moles}$ of Oxygen ($\text{O}_2$, rigid diatomic, $f_2 = 5$) at temperature $T_2 = 300\text{ K}$.
1. The partition is removed, allowing the gases to mix without chemical reaction. Determine the final equilibrium temperature $T_f$ of the mixture.
2. Find the effective adiabatic index $\gamma_{\text{mix}}$ of the mixed gas.
3. If the combined gas is now reversibly and adiabatically compressed to half of its mixed volume ($V’ = V / 2$), calculate the final temperature $T_2’$ in terms of $T_f$.
Solution:
Step 1: Equilibrium Mixing Temperature.
Because the container is rigid ($W = 0$) and thermally insulated ($Q = 0$), internal energy is conserved: $\Delta U = 0 \implies U_f = U_1 + U_2$.
$$(n_1 C_{v1} + n_2 C_{v2}) T_f = n_1 C_{v1} T_1 + n_2 C_{v2} T_2$$
Using $C_{v1} = \frac{3}{2} R$ and $C_{v2} = \frac{5}{2} R$:
$$\left(1.0 \times \frac{3}{2} R + 3.0 \times \frac{5}{2} R\right) T_f = \left(1.0 \times \frac{3}{2} R \times 400\right) + \left(3.0 \times \frac{5}{2} R \times 300\right)$$
$$(1.5 + 7.5) T_f = 600 + 2250 \implies 9.0 T_f = 2850$$
$$\mathbf{T_f = \frac{2850}{9.0} = \frac{950}{3}\text{ K} \approx \mathbf{316.67\text{ K}}}$$
Step 2: Effective Adiabatic Index ($\gamma_{\text{mix}}$).
The effective degrees of freedom of the mixture:
$$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{(1.0 \times 3) + (3.0 \times 5)}{1.0 + 3.0} = \frac{3 + 15}{4} = \frac{18}{4} = \mathbf{4.5 = \frac{9}{2}}$$
$$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{9/2} = 1 + \frac{4}{9} = \mathbf{\frac{13}{9} \approx 1.444}$$
Step 3: Adiabatic Compression Temperature.
For a reversible adiabatic process:
$$T_f V^{\gamma_{\text{mix}} – 1} = T_2′ (V’)^{\gamma_{\text{mix}} – 1}$$
$$T_2′ = T_f \left(\frac{V}{V’}\right)^{\gamma_{\text{mix}} – 1} = T_f (2)^{\frac{13}{9} – 1} = T_f (2)^{4/9}$$
Evaluating numerically:
$$2^{4/9} \approx 2^{0.4444} \approx 1.3608$$
$$T_2′ \approx 316.67\text{ K} \times 1.3608 \approx \mathbf{430.9\text{ K}}$$
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. According to the law of equipartition of energy, the average kinetic energy associated with each translational degree of freedom of an ideal gas molecule at absolute temperature $T$ is:
(A) $\frac{1}{2} k_B T$
(B) $k_B T$
(C) $\frac{3}{2} k_B T$
(D) $\frac{5}{2} k_B T$
Q2. An ideal gas molecule possesses $f$ active degrees of freedom. The ratio of its molar heat capacity at constant pressure to that at constant volume ($\gamma = C_p / C_v$) is given by:
(A) $1 + \frac{1}{f}$
(B) $1 + \frac{2}{f}$
(C) $1 + \frac{f}{2}$
(D) $\frac{f}{f + 2}$
Q3. For a rigid non-linear triatomic gas (such as water vapor, $\text{H}_2\text{O}$) at moderate temperatures where vibrational modes are frozen, the molar heat capacity at constant volume $C_v$ is:
(A) $\frac{3}{2} R$
(B) $\frac{5}{2} R$
(C) $3 R$
(D) $4 R$
Q4. At high temperatures, a diatomic gas molecule undergoes translational motion, rotational motion, and has one active vibrational normal mode. The mean total internal energy per molecule is:
(A) $\frac{5}{2} k_B T$
(B) $3 k_B T$
(C) $\frac{7}{2} k_B T$
(D) $\frac{9}{2} k_B T$
Q5. According to the classical Dulong-Petit law, treating each atom of a 3D crystalline solid as a harmonic oscillator vibrating about fixed lattice positions, the molar heat capacity of the solid at room temperature is:
(A) $\frac{3}{2} R$
(B) $3 R$
(C) $\frac{9}{2} R$
(D) $6 R$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements regarding the law of equipartition of energy and molecular motion is/are correct?
(A) Each independent quadratic term in the Hamiltonian contributes $\frac{1}{2} k_B T$ to the average molecular energy at thermal equilibrium.
(B) A single active vibrational normal mode contributes $k_B T$ to the average molecular energy because it incorporates both kinetic and potential quadratic terms.
(C) At room temperature ($300\text{ K}$), the vibrational modes of common diatomic gases ($\text{N}_2, \text{O}_2$) do not contribute noticeably to specific heat due to quantum freeze-out.
(D) The average translational kinetic energy per molecule at temperature $T$ is $\frac{3}{2} k_B T$, which is completely independent of molecular mass, shape, and chemical composition.
Q7. An ideal gas mixture consists of $n_1$ moles of Helium ($\text{He}$, monoatomic, $f_1 = 3$) and $n_2$ moles of Nitrogen ($\text{N}_2$, rigid diatomic, $f_2 = 5$) at thermal equilibrium at absolute temperature $T$. Which of the following statements is/are correct?
(A) The total internal energy of the mixture is $U = \left(\frac{3}{2} n_1 + \frac{5}{2} n_2\right) R T$.
(B) The effective molar heat capacity at constant volume is $C_{v,\text{mix}} = \frac{3 n_1 + 5 n_2}{2(n_1 + n_2)} R$.
(C) If the mixture contains equimolar amounts ($n_1 = n_2$), the adiabatic exponent of the mixture is $\gamma_{\text{mix}} = 1.50$.
(D) The average translational kinetic energy per mole is identical for Helium and Nitrogen.
Q8. Which of the following statements correctly account for the thermodynamic behavior of crystalline solids and the limits of classical equipartition?
(A) In classical theory, each atom in a crystal lattice possesses 6 quadratic energy terms (3 kinetic and 3 potential).
(B) The Dulong-Petit law predicts that the molar heat capacity of elemental solids at high temperatures approaches $3R \approx 25\text{ J/(mol}\cdot\text{K)}$.
(C) Classical equipartition fails at low temperatures because vibrational energy states are quantized into phonons with discrete energy spacing $\hbar \omega$.
(D) Diamond exhibits an unusually low heat capacity at room temperature because its very strong carbon-carbon bonds lead to an exceptionally high Debye temperature ($\Theta_D \sim 2200\text{ K}$).
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A rigid container holds $n = 2.0\text{ moles}$ of a non-linear polyatomic ideal gas whose molecules have $3$ translational and $3$ rotational degrees of freedom (all vibrational modes frozen). When $\Delta Q = 498.84\text{ J}$ of heat is supplied to the gas at constant volume, determine the resulting temperature rise ($\Delta T$) of the gas in Kelvin. (Take $R = 8.314\text{ J/(mol}\cdot\text{K)}$).
Q10. A gas mixture contains $1.0\text{ mole}$ of a monoatomic gas ($f_1 = 3$) and $1.0\text{ mole}$ of a diatomic gas with $1$ active vibrational mode ($f_2 = 3\text{ trans} + 2\text{ rot} + 2\text{ vib} = 7$) at high temperature. The effective adiabatic index of the mixture is expressed in simplest fractional form as $\gamma_{\text{mix}} = \frac{p}{q}$, where $p$ and $q$ are co-prime positive integers ($\gcd(p, q) = 1$). Find the value of $(p + q)$.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) $\frac{1}{2} k_B T$ | Energy per Translational Degree of Freedom |
| Q2 | (B) $1 + \frac{2}{f}$ | Adiabatic Exponent from Equipartition |
| Q3 | (C) $3 R$ | Molar Heat Capacity of Rigid Non-Linear Polyatomic Gas |
| Q4 | (C) $\frac{7}{2} k_B T$ | Energy of Diatomic Molecule with Active Vibration |
| Q5 | (B) $3 R$ | Dulong-Petit Law for 3D Lattice Vibrations |
| Q6 | (A), (B), (C), (D) | Equipartition Principles & Molecular Dynamics |
| Q7 | (A), (B), (C), (D) | Gas Mixture Equipartition & Heat Capacity |
| Q8 | (A), (B), (C), (D) | Dulong-Petit Law & Quantum Breakdown in Solids |
| Q9 | 10 | Isochoric Heating of Polyatomic Gas ($C_v = 3R$) |
| Q10 | 12 | Mixture Exponent $\gamma = 7/5 \implies p + q = 12$ |
Complete Step-by-Step Solutions
Solution 1:
By the law of equipartition of energy, each independent coordinate that appears as a quadratic term in the energy expression receives an average energy of $\frac{1}{2} k_B T$. Since translation along any single Cartesian direction involves the term $\frac{1}{2} m v_i^2$, the energy per translational degree of freedom is strictly $\frac{1}{2} k_B T$.
Correct Option: (A)
Solution 2:
For $f$ active degrees of freedom, the molar heat capacities are:
$$C_v = \frac{f}{2} R, \quad C_p = C_v + R = \left(\frac{f}{2} + 1\right) R$$
Taking the ratio:
$$\gamma = \frac{C_p}{C_v} = \frac{\frac{f}{2} R + R}{\frac{f}{2} R} = 1 + \frac{2}{f}$$
Correct Option: (B)
Solution 3:
A rigid non-linear triatomic molecule has $3$ translational and $3$ rotational degrees of freedom, giving $f = 3 + 3 = 6$. Therefore:
$$C_v = \frac{f}{2} R = \frac{6}{2} R = \mathbf{3 R}$$
Correct Option: (C)
Solution 4:
For a diatomic molecule with active vibration:
– Translational degrees of freedom: $3 \times \left(\frac{1}{2} k_B T\right) = \frac{3}{2} k_B T$.
– Rotational degrees of freedom: $2 \times \left(\frac{1}{2} k_B T\right) = k_B T$.
– Vibrational normal mode ($1$ mode $= 2$ quadratic terms): $2 \times \left(\frac{1}{2} k_B T\right) = k_B T$.
Total average energy:
$$\langle E \rangle = \frac{3}{2} k_B T + k_B T + k_B T = \mathbf{\frac{7}{2} k_B T}$$
Correct Option: (C)
Solution 5:
In the Dulong-Petit model, each of the $N_A$ atoms in a mole of crystal oscillates in 3 dimensions, possessing $3$ kinetic and $3$ potential energy terms ($6$ quadratic terms total):
$$\langle E_{\text{atom}} \rangle = 6 \times \left(\frac{1}{2} k_B T\right) = 3 k_B T \implies U_m = 3 N_A k_B T = 3 R T$$
$$C = \frac{dU_m}{dT} = \mathbf{3 R}$$
Correct Option: (B)
Solution 6:
(A) Correct: The fundamental theorem of equipartition assigns $\frac{1}{2} k_B T$ to every independent quadratic term in the Hamiltonian.
(B) Correct: Harmonic vibration has both kinetic energy ($\frac{1}{2}\mu \dot{x}^2$) and potential energy ($\frac{1}{2}kx^2$), contributing $2 \times \frac{1}{2}k_B T = k_B T$.
(C) Correct: Vibrational energy spacing $\hbar \omega \sim 3000\text{ K} \gg 300\text{ K}$, so thermal collisions cannot excite vibrations at room temperature.
(D) Correct: $\langle E_{\text{trans}} \rangle = \frac{3}{2} k_B T$ depends strictly on temperature $T$, not on mass or molecular identity.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: Internal energy is extensive: $U = n_1 \left(\frac{3}{2} R T\right) + n_2 \left(\frac{5}{2} R T\right)$.
(B) Correct: $C_{v,\text{mix}} = \frac{U}{(n_1 + n_2) T} = \frac{3 n_1 + 5 n_2}{2(n_1 + n_2)} R$.
(C) Correct: For $n_1 = n_2 = n$:
$$f_{\text{mix}} = \frac{n(3) + n(5)}{2n} = \frac{8}{2} = 4 \implies \gamma_{\text{mix}} = 1 + \frac{2}{4} = \mathbf{1.50}$$
(D) Correct: Both gases have $3$ translational degrees of freedom, so each possesses $\frac{3}{2} R T$ translational kinetic energy per mole.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: 3 spatial coordinates each give kinetic ($\frac{1}{2}m v_i^2$) and potential ($\frac{1}{2}k x_i^2$) terms, totaling 6 quadratic terms.
(B) Correct: $C = 6 \times \frac{1}{2} R = 3 R \approx 24.94\text{ J/(mol}\cdot\text{K)}$.
(C) Correct: When $k_B T < \hbar \omega$, lattice vibrational modes cannot be excited classically, freezing out heat capacity.
(D) Correct: Diamond’s strong C-C covalent bonds result in very high vibrational frequencies, making $\Theta_D \approx 2200\text{ K} \gg 300\text{ K}$, so its heat capacity at room temperature is far below $3R$.
Correct Options: (A), (B), (C), (D)
Solution 9:
For a rigid non-linear polyatomic gas, $f = 3\text{ trans} + 3\text{ rot} = 6$.
$$C_v = \frac{f}{2} R = \frac{6}{2} R = 3 R$$
The heat supplied at constant volume is:
$$\Delta Q = n C_v \Delta T = n (3 R) \Delta T$$
$$498.84 = (2.0) \times (3 \times 8.314) \times \Delta T$$
$$498.84 = 49.884 \times \Delta T \implies \Delta T = \frac{498.84}{49.884} = \mathbf{10\text{ K}}$$
Answer: 10
Solution 10:
The effective degrees of freedom of the equimolar mixture:
$$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{(1.0 \times 3) + (1.0 \times 7)}{1.0 + 1.0} = \frac{10}{2} = 5$$
The effective adiabatic exponent:
$$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{5} = \frac{7}{5}$$
Here $p = 7$ and $q = 5$, which are co-prime positive integers ($\gcd(7, 5) = 1$).
$$p + q = 7 + 5 = \mathbf{12}$$
Answer: 12