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Law of Equipartition of Energy


The law of equipartition of energy states that each degree of freedom have an average energy $kT/2$, where $k$ is the Boltzmann constant and $T$ is the temperature of the system (in thermal equilibrium).

The degrees of freedom refer to the different ways in which the energy can be stored in a system. For example, in a gas, the energy can be stored as kinetic energy of the particles in three dimensions (x, y, and z) and as potential energy due to the intermolecular forces between the particles.

degree of freedom

The law of equipartition of energy can be used to calculate the average energy of a system and to determine the specific heat capacity of materials.

Problems from IIT JEE

Problem (IIT JEE 1997): The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are $6.21\times{10}^{-21}$ J and 484 m/s, respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)

  1. $12.42\times10^{-21}$ J, 968 m/s
  2. $8.78\times10^{-21}$ J, 684 m/s
  3. $6.21\times10^{-21}$ J, 968 m/s
  4. $12.42\times10^{-21}$ J, 684 m/s

Solution: A molecule of mass $m$ in a gas kept at temperature $T$ has average translational kinetic energy \begin{align} E=\frac{3}{2}kT \end{align} and rms speed \begin{align} v_\text{rms}=\sqrt{3kT/m}. \end{align} Thus, \begin{align} E_{600}&=2E_{300}\\ &=12.42\times{10}^{-21}\;\mathrm{J} \end{align} and \begin{align} v_\text{rms,600}&=\sqrt{2}v_\text{rms,300} \\ &=1.414\times484\\ &=684\;\mathrm{m/s}. \end{align}

Problem (IIT JEE 1997): The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules (in eV) at the same temperature is

  1. 0.0015
  2. 0.003
  3. 0.048
  4. 0.768

Solution: Average translational kinetic energy, $E=\frac{3}{2}kT$, depends only on $T$.

Problem (IIT jEE 1999): A gas mixture consists of 2 mol of oxygen and 4 mol of argon at temperature $T$. Neglecting all vibrational modes, the total internal energy of the system is

  1. $4 RT$
  2. $15 RT$
  3. $9 RT$
  4. $11 RT$

Solution: The internal energy of $n$ moles of an ideal gas at temperature $T$, with each molecule having $f$ degrees of freedom, is given by \begin{align} U=n\frac{f}{2}RT. \end{align} Oxygen is a diatomic gas with $f_\mathrm{O_2}=5$ (three translational, two rotational) and argon is a monatomic gas with $f_\mathrm{Ar}=3$ (three translational). Substitute the values in above equation to get the total internal energy of the system as \begin{align} U&=U_\mathrm{O_2}+U_\mathrm{Ar}\\ &=2\times\frac{5}{2}RT+4\times\frac{3}{2}RT\\ &=11RT.\nonumber \end{align}


  1. Kinetic Theory of Gases
  2. RMS speed of gas molecules
  3. Specific Heats Cv and Cp for Monatomic and Diatomic Gases New
  4. Mean Free Path
JEE Physics Solved Problems in Mechanics