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Thermal Physics
Degrees of Freedom and Law of Equipartition of Energy
Degrees of freedom represent the independent ways in which a dynamical system can store energy, corresponding to the coordinates necessary to specify its configuration. The law of equipartition of energy states that in thermal equilibrium, the total energy of a system is shared equally among all active degrees of freedom, with each degree contributing 1/2 k_B T per molecule.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Calculation of degrees of freedom for monoatomic, diatomic, and triatomic gas molecules; derivation of specific heat ratios (C_p / C_v); basic applications of the law of equipartition of energy.
PGT extension
Advanced treatment of vibrational and rotational degrees of freedom at high temperatures, degrees of freedom of solids (Dulong-Petit law), and temperature-dependent behavior of specific heats of gases.
Prerequisites
Concept of Temperature
Kinetic theory of gases
Law of equipartition of energy
Definition and Translational Degrees of Freedom
Degrees of freedom (f) are defined as the independent coordinates required to specify the position and configuration of a mechanical system.
Any free gas molecule moving in three-dimensional space possesses 3 translational degrees of freedom corresponding to motion along the X, Y, and Z axes.
Translational kinetic energy is given by 1/2 m v_x^2 + 1/2 m v_y^2 + 1/2 m v_z^2, yielding 1/2 k_B T per degree of freedom.
Rotational and Vibrational Degrees of Freedom
Rotational degrees of freedom arise from rigid rotation about orthogonal axes passing through the center of mass.
Monoatomic gases have negligible moment of inertia about the atomic center, hence 0 rotational degrees of freedom.
Diatomic molecules have 2 rotational degrees of freedom at room temperature, while non-linear polyatomic molecules possess 3 rotational degrees of freedom.
Law of Equipartition of Energy
Formulated by Boltzmann, this law states that for any dynamical system in thermal equilibrium, the total energy is equally distributed among all active degrees of freedom.
Each translational and rotational degree of freedom contributes an average energy of 1/2 k_B T per molecule (or 1/2 R T per mole).
Each vibrational degree of freedom contributes two terms (kinetic and potential), resulting in an average energy of k_B T per molecule (or R T per mole).
Molar Heat Capacities and Adiabatic Exponent
Internal energy U of 1 mole of an ideal gas with f degrees of freedom is expressed as $U = (f / 2) R T$.
Molar heat capacity at constant volume C_v is defined as dU/d$T = (f / 2) R$.
Molar heat capacity at constant pressure C_p is given by C_v + $R = (f / 2 + 1) R$, leading to the ratio gamm$a = C_p / C_v = 1 + 2/f$.
Derivations
Derivation of Specific Heat Ratio (gamma) from Degrees of Freedom
Conditions: Ideal gas with f degrees of freedom in thermal equilibrium
Write the internal energy for one mole of an ideal gas: U = \frac{f}{2} R T.
Differentiate internal energy with respect to temperature to find molar heat capacity at constant volume: $C_v = \frac{dU}{dT} = \frac{f}{2} R$.
Apply Mayer's relation for ideal gases to find molar heat capacity at constant pressure: $C_p = C_v + R = \left(\frac{f}{2} + 1\right) R$.
Take the ratio of C_p to C_v to obtain the adiabatic exponent: gamm$a = \frac{C_p}{C_v} = \frac{f/2 + 1}{f/2} = 1 + \frac{2}{f}$.
Result: $\gamma = 1 + \frac{2}{f}$
Total Internal Energy of a Gas Mixture
Conditions: Non-reacting mixture of ideal gases containing n_1, n_2 moles with degrees of freedom f_1, f_2
Express the internal energy of the individual components as $U_1 = \frac{f_1}{2} n_1 R T$ and $U_2 = \frac{f_2}{2} n_2 R T$.
Sum the internal energies to find the total internal energy of the mixture: $U_{total} = U_1 + U_2 = \left(\frac{f_1 n_1 + f_2 n_2}{2}\right) R T$.
Define the effective degrees of freedom f_{mix} for the mixture using $U_{total} = \frac{f_{mix}}{2} (n_1 + n_2) R T$.
Equate and solve for f_{mix} to obtain the weighted average of degrees of freedom.
$k_B = Boltzmann constant$, T = Absolute temperature Joules (J)
Ideal monoatomic gas
Energy per degree of freedom
$$E_{DOF} = \frac{1}{2} k_B T$$
$k_B = Boltzmann constant$, T = Absolute temperature Joules (J)
Thermal equilibrium
Internal Energy of 1 mole
$$U = \frac{f}{2} R T$$
f = Degrees of freedom, R = Universal gas constant, T = Temperature Joules per mole (J mol^{-1})
Ideal gas
Molar heat capacity at constant volume (C_v)
$$C_v = \frac{f}{2} R$$
f = Degrees of freedom, R = Universal gas constant J mol^{-1} K^{-1}
Ideal gas
Molar heat capacity at constant pressure (C_p)
$$C_p = \left(\frac{f}{2} + 1\right) R$$
f = Degrees of freedom, R = Universal gas constant J mol^{-1} K^{-1}
Ideal gas
Ratio of specific heats
$$\gamma = 1 + \frac{2}{f}$$
$\gamma = C_p / C_v$, f = Degrees of freedom Dimensionless
Ideal gas
Vibrational Energy Contribution
$$E_{vib} = k_B T$$
$k_B = Boltzmann constant$, T = Temperature Joules (J)
High temperature (both kinetic and potential active)
Effective degrees of freedom for gas mixture
$$f_{mix} = \frac{\sum n_i f_i}{\sum n_i}$$
$n_i = moles of i-th gas$, $f_i = degrees of freedom of i-th gas$ Dimensionless
Non-reacting gas mixture
Graphs and experiments
Temperature Dependence of Molar Heat Capacity (C_v) for Diatomic Gas
Draw: A graph of C_v / R versus absolute temperature T, showing three distinct plateaus at 1.5R (low T, translational only), 2.5R (room T, translational + rotational), and 3.5R (high T, translational + rotational + vibrational).
Interpret: Degrees of freedom are activated sequentially as thermal energy increases, overcoming rotational and vibrational energy level spacings.
Exam use: Tested in conceptual multiple-choice questions regarding C_v values of diatomic gases at cryogenic versus high temperatures.
Maxwell-Boltzmann Energy Distribution
Draw: Asymmetric bell-shaped curve plotting fraction of molecules versus molecular energy, illustrating the continuous spread of energy among molecules.
Interpret: While individual molecular energies fluctuate widely, the average energy per degree of freedom remains strictly fixed at 1/2 k_B T at a given temperature.
Exam use: Used to link statistical mechanics principles with macroscopic equipartition theorems.
Adiabatic versus Isothermal P-V Curves
Draw: P-V diagram comparing isothermal (PV = constant) and adiabatic (P$V^\gamma = constant) curves originating from the same point$, where the adiabatic curve is steeper due to gamma > 1.
Interpret: The steepness of the adiabatic curve directly reflects the value of gamma, which is determined by the degrees of freedom of the gas.
Exam use: Problem-solving involving work done during adiabatic expansions of monoatomic versus diatomic gases.
Solved numericals
TGT core
Find the degrees of freedom and the ratio of specific heats (gamma) for a diatomic gas molecule at room temperature.
Identify the molecular structure: A diatomic gas molecule at room temperature has 3 translational and 2 rotational degrees of freedom.
Calculate total degrees of freedom: $f = 3 + 2 = 5$.
Use the formula for the ratio of specific heats: gamm$a = 1 + \frac{2}{f}$.
A mixture of 2 moles of helium (monoatomic, f = 3) and 3 moles of hydrogen (diatomic, f = 5, neglecting vibration) is contained in a vessel. Calculate the equivalent degrees of freedom and the ratio gamma for the mixture.
Identify parameters for gas 1 (Helium): $n_1 = 2$, $f_1 = 3$.
Identify parameters for gas 2 (Hydrogen): $n_2 = 3$, $f_2 = 5$.
Calculate effective degrees of freedom using $f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$.
Calculate the molar heat capacity at constant volume (C_v) and constant pressure (C_p) for a non-linear triatomic gas molecule at high temperature where its vibrational mode is fully active. Take $R = 8.314 J mol^{-1} K^{-1}$.
Determine degrees of freedom for a non-linear triatomic molecule: 3 translational, 3 rotational, and let the number of vibrational modes be $s = 3N - 6 = 3(3) - 6 = 3 vibrational modes$.
Each vibrational mode contributes 2 degrees of freedom (kinetic and potential). Total degrees of freedom $f = 3 (trans) + 3 (rot) + 2 \times 3 (vib) = 12$.
Calculate C_v: $C_v = \frac{f}{2} R = \frac{12}{2} R = 6 R$.
Calculate C_p: $C_p = C_v + R = 6R + R = 7R$.
Substitute numerical value of R: $C_v = 6 \times 8.314 = 49.88 J mol^{-1} K^{-1}$ and $C_p = 7 \times 8.314 = 58.20 J mol^{-1} K^{-1}$.
The equipartition theorem assumes classical statistical mechanics, which breaks down at very low temperatures (cryogenic range) where quantum mechanical freezing of rotational and vibrational modes occurs.
Intermolecular forces are assumed to be negligible (ideal gas approximation), which fails at high pressures and low temperatures.
Common misconceptions
Common error: All degrees of freedom contribute equally to energy regardless of temperature.
Correct: Degrees of freedom are temperature-dependent and freeze out at lower temperatures.
Quantum mechanical energy level spacings exceed thermal energy kT at low temperatures, making certain modes inactive.
Common error: Monoatomic gases possess rotational degrees of freedom.
Correct: Monoatomic gases have zero rotational degrees of freedom.
Rotation of a point mass or sphere about its center of mass produces no change in spatial configuration or moment of inertia.
Common error: Vibrational energy contributes 1/2 k_B T per mode to internal energy.
Correct: Vibrational energy contributes k_B T per mode.
Vibration possesses both kinetic and potential energy terms, each contributing 1/2 k_B T according to the equipartition theorem.
Translational Kinetic Energy per molecule: $E_t = \frac{3}{2} k_B T$
Energy per degree of freedom: $E_{DOF} = \frac{1}{2} k_B T$
Must remember
Degrees of freedom ($f$) represent independent coordinates to specify configuration; monoatomic $f=3$, diatomic $f=5$ (rigid) or $7$ (vibrating), polyatomic non-linear $f=6$.
Law of Equipartition of Energy: Each translational and rotational degree of freedom contributes $\frac{1}{2} k_B T$ per molecule, while each vibrational mode contributes $k_B T$.
Molar heat capacities are given by $C_v = \frac{f}{2} R$ and $C_p = \left(\frac{f}{2} + 1\right) R$, with adiabatic exponent $\gamma = 1 + \frac{2}{f}$.
Vibrational degrees of freedom become active only at high temperatures due to quantum freeze-out effects at low temperatures.
For a mixture of non-reacting gases, the effective degrees of freedom is $f_{mix} = \frac{\sum n_i f_i}{\sum n_i}$.
Glossary
Degrees of Freedom ($f$)The number of independent ways in which a molecule can store energy or independent coordinates required to specify system configuration.
Equipartition of EnergyPrinciple stating that in thermal equilibrium, total energy is equally shared among active degrees of freedom with $\frac{1}{2} k_B T$ per degree per molecule.
Adiabatic Exponent ($\gamma$)The ratio of molar heat capacity at constant pressure to constant volume, $\gamma = C_p / C_v$.
Boltzmann Constant ($k_B$)Fundamental constant relating microscopic thermal energy to macroscopic temperature, $k_B = R / N_A$.
Characteristics of Gas Molecules Based on Atomicity
Parameter
Monoatomic Gas
Diatomic Gas (Room Temp)
Polyatomic Non-linear Gas
Degrees of Freedom ($f$)
$3$
$5$
$6$
Translational DOF
$3$
$3$
$3$
Rotational DOF
$0$
$2$
$3$
Molar Heat Capacity ($C_v$)
$\frac{3}{2} R$
$\frac{5}{2} R$
$3 R$
Adiabatic Exponent ($\gamma$)
$1.67$
$1.40$
$1.33$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the total number of degrees of freedom for a rigid diatomic gas molecule?
Explanation: A rigid diatomic molecule has $3$ translational and $2$ rotational degrees of freedom, totaling $5$ degrees of freedom at room temperature.
2. According to the law of equipartition of energy, what is the energy associated with each translational degree of freedom per molecule?
Explanation: Each active translational and rotational degree of freedom contributes an average thermal energy of $\frac{1}{2} k_B T$ per molecule.
3. What is the molar heat capacity at constant volume ($C_v$) for an ideal gas with $f$ degrees of freedom?
Explanation: Internal energy $U = \frac{f}{2} R T$, hence $C_v = \frac{dU}{dT} = \frac{f}{2} R$.
4. For a non-linear polyatomic gas molecule, how many rotational degrees of freedom are present?
Explanation: A non-linear polyatomic molecule can rotate independently about three mutually perpendicular axes passing through its center of mass, giving $3$ rotational degrees of freedom.
5. What is the adiabatic exponent $\gamma = \frac{C_p}{C_v}$ for a monoatomic ideal gas?
Explanation: For a monoatomic gas, $f = 3$. Thus, $\gamma = 1 + \frac{2}{3} = \frac{5}{3} \approx 1.67$.
6. Why do vibrational degrees of freedom not contribute to the specific heat of diatomic gases at room temperature?
Explanation: At room temperature, thermal energy $k_B T$ is insufficient to excite the vibrational quantum states, causing vibrational modes to freeze out.
7. What is the contribution of each vibrational mode to the internal energy of a gas molecule at high temperatures?
Explanation: A vibrational mode has both kinetic and potential energy terms, each contributing $\frac{1}{2} k_B T$, resulting in a total contribution of $k_B T$ per vibrational mode.
8. What is the internal energy of $2$ moles of an ideal diatomic gas (rigid) at temperature $T$?
Explanation: For $n = 2$ moles of a diatomic gas ($f = 5$), internal energy $U = n \frac{f}{2} R T = 2 \times \frac{5}{2} R T = 5 R T$.
9. Which of the following gases has the highest value of $\gamma = \frac{C_p}{C_v}$?
Explanation: Adiabatic exponent $\gamma = 1 + \frac{2}{f}$ is inversely proportional to $f$. Monoatomic gases have the lowest $f$ ($f=3$), hence the highest $\gamma$ ($1.67$).
10. What is the molar heat capacity at constant pressure ($C_p$) for an ideal gas with $f$ degrees of freedom?
Explanation: Using Mayer's relation $C_p = C_v + R = \frac{f}{2} R + R = \left(\frac{f}{2} + 1\right) R$.
11. A gas mixture contains $n_1$ moles of gas $1$ (degrees of freedom $f_1$) and $n_2$ moles of gas $2$ (degrees of freedom $f_2$). What is the effective degree of freedom of the mixture?
Explanation: The effective degree of freedom for a non-reacting gas mixture is given by the mole-fraction weighted average: $f_{mix} = \frac{\sum n_i f_i}{\sum n_i}$.
12. How many degrees of freedom does a linear triatomic molecule (like $CO_2$) possess at room temperature?
Explanation: A linear triatomic molecule has $3$ translational, $2$ rotational, and $4$ vibrational degrees of freedom (or $f = 5$ treating it as rigid, but with vibrational contribution included or standard textbook count $f=7$ for linear triatomic at room/moderate temp, or $3N-5 = 4$ vibrational modes + $3$ trans + $2$ rot = $7$).
13. What is the translational kinetic energy of $1$ mole of any ideal gas at absolute temperature $T$?
Explanation: Translational kinetic energy per molecule is $\frac{3}{2} k_B T$. Multiplying by Avogadro number $N_A$ gives $\frac{3}{2} R T$ per mole.
14. Which of the following statements about monoatomic gases is correct?
Explanation: Monoatomic gases have $3$ degrees of freedom (all translational) and zero rotational or vibrational DOF at normal temperatures, so their internal energy is purely translational kinetic energy.
15. What happens to the molar heat capacity $C_v$ of a diatomic gas as the temperature is raised from room temperature to very high temperatures ($> 5000\text{ K}$)?
Explanation: At very high temperatures, vibrational modes become active, increasing $f$ from $5$ to $7$, which increases $C_v$ from $\frac{5}{2} R$ to $\frac{7}{2} R$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. If the degrees of freedom of a gas molecule is $f$, then the ratio of specific heats $\gamma = \frac{C_p}{C_v}$ is given by:
2. The total internal energy of $1$ mole of a monoatomic gas at temperature $T$ is:
Explanation: For a monoatomic gas, $f = 3$. Internal energy $U = \frac{f}{2} R T = \frac{3}{2} R T$.
3. A diatomic gas has $5$ degrees of freedom at room temperature. When heated to high temperature, vibrational modes get activated and degrees of freedom become $7$. What is the percentage change in $C_v$?
4. According to the law of equipartition of energy, the energy per molecule per degree of freedom is:
Explanation: Each degree of freedom contributes $\frac{1}{2} k_B T$ per molecule to the internal energy.
5. Two moles of ideal monoatomic gas are mixed with three moles of ideal diatomic gas ($f=5$). Assuming no vibrational contribution, the effective molar heat capacity $C_v$ of the mixture is:
Explanation: Total internal energy $U_{mix} = n_1 U_1 + n_2 U_2 = 2 \left(\frac{3}{2} R T\right) + 3 \left(\frac{5}{2} R T\right) = 3 R T + \frac{15}{2} R T = \frac{21}{2} R T$. Thus $C_{v,mix} = \frac{U_{mix}}{(n_1+n_2)T} = \frac{\frac{21}{2} R}{5} = \frac{21}{10} R$ per mole? Wait, total $C_{v,total} = n_1 C_{v1} + n_2 C_{v2} = 2(\frac{3}{2}R) + 3(\frac{5}{2}R) = 3R + 7.5R = 10.5 R = \frac{21}{2} R$. Molar heat capacity per mole of mixture is $C_{v,mix} = \frac{\sum n_i C_{vi}}{\sum n_i} = \frac{21/2 R}{5} = \frac{21}{10} R$. Wait, let's re-verify options. If total heat capacity is $\frac{21}{2} R$, then for 5 moles it is $\frac{21}{10} R$. Let's check option B: $\frac{21}{5} R$ would be total energy divided by moles if multiplied by 2. Wait, $U = n C_v T$. Let's use $f_{mix} = \frac{2(3) + 3(5)}{2+3} = \frac{6+15}{5} = \frac{21}{5}$. Then $C_{v,mix} = \frac{f_{mix}}{2} R = \frac{21}{10} R$. Wait, let's check standard formula: $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2(\frac{3}{2}R) + 3(\frac{5}{2}R)}{5} = \frac{3R + 7.5R}{5} = \frac{10.5R}{5} = 2.1 R = \frac{21}{10} R$. Let's adjust question or option to match $\frac{21}{10} R$ or total heat capacity. If the question asks for total heat capacity of the mixture, it is $\frac{21}{2} R$. Let's modify option B to $\frac{21}{10} R$.
6. What is the ratio of root mean square velocity $v_{rms}$ to the average velocity $v_{avg}$ of gas molecules at temperature $T$?
Attempt these questions under a ten-minute limit using a separate timer.
1. If a gas has $f$ degrees of freedom and internal energy $U$, the relation between $U$ and work done $W$ in an adiabatic process ($PV^\gamma = \text{constant}$) for $n$ moles is given by $-dU = dW$. What is $\Delta U$ for an adiabatic expansion?
Explanation: In an adiabatic expansion, gas cools down ($dT < 0$), so internal energy $\Delta U = n \frac{f}{2} R \Delta T$ is negative.
2. What is the number of degrees of freedom for a rigid triatomic linear molecule?
Explanation: A linear triatomic rigid molecule has $3$ translational and $2$ rotational degrees of freedom, totaling $5$ degrees of freedom (similar to a diatomic molecule).
3. For an ideal gas, the molar heat capacity at constant pressure $C_p$ is $\frac{7}{2} R$. What is the atomicity and type of gas assuming rigid molecules?
Explanation: $C_p = \frac{7}{2} R = \left(\frac{f}{2} + 1\right) R \implies \frac{f}{2} = \frac{5}{2} \implies f = 5$. A rigid gas with $f=5$ is diatomic.
4. Which of the following molecules possesses $6$ degrees of freedom at room temperature?
Explanation: A non-linear polyatomic molecule like water ($H_2O$) has $3$ translational and $3$ rotational degrees of freedom, totaling $6$ degrees of freedom at room temperature.
5. What is the ratio of molar heat capacities $\gamma$ for a gas mixture consisting of $2$ moles of argon and $3$ moles of helium?
Explanation: Both argon and helium are monoatomic gases with $\gamma = 1.67$. Any mixture of monoatomic gases is also monoatomic with $\gamma = \frac{5}{3} \approx 1.67$.
6. According to the equipartition theorem, what is the total kinetic energy of $1$ mole of a rigid diatomic gas?
Explanation: For a rigid diatomic gas, total internal energy is purely kinetic (translational + rotational), amounting to $U = \frac{5}{2} R T$ per mole.
7. In a non-linear polyatomic molecule with $N$ atoms, what is the total number of degrees of freedom?
Explanation: A system of $N$ free atoms has $3N$ coordinates (degrees of freedom). When constrained into a molecule, it still has $3N$ total degrees of freedom distributed among translation, rotation, and vibration.
8. If the adiabatic exponent of a gas is $\frac{4}{3}$, the gas is:
Explanation: $\gamma = 1 + \frac{2}{f} = \frac{4}{3} \implies \frac{2}{f} = \frac{1}{3} \implies f = 6$. A gas with $f=6$ is a non-linear polyatomic molecule.
9. What is the contribution of vibrational kinetic energy alone to the internal energy per molecule for a single vibrational mode?
Explanation: A vibrational mode consists of both kinetic and potential energy quadratic terms. Each term contributes $\frac{1}{2} k_B T$ according to equipartition.
10. Why does Dulong-Petit law (stipulating molar heat capacity of solids as $3 R$) hold true at high temperatures?
Explanation: In a 3D solid lattice, each atom has $3$ vibrational degrees of freedom, each contributing $k_B T$ (both kinetic and potential), yielding $3 R$ per mole at high temperatures.