Law of equipartition of energy

This section sits inside the CBSE Kinetic Theory chapter, directly after the kinetic interpretation of temperature ($\tfrac12m\overline{v^2} = \tfrac32k_BT$) and directly before specific heat capacities of mono-, di-, and polyatomic gases. CBSE prescribes the law of equipartition of energy strictly as a statement to know and apply — you are not expected to derive it from the kinetic theory of gases or from statistical mechanics, only to state it correctly, count degrees of freedom, and use $U = \tfrac{f}{2}RT$. Everything you build here — the values $f = 3, 5, 6$ (and the qualitative extension to $7$ for a vibrating diatomic) — feeds straight into the specific-heat formulas $C_v = \tfrac{f}{2}R$ and $\gamma = C_p/C_v$ that follow, so getting the counting solid now saves you re-deriving it three more times.

A molecule in a gas is free to move in three-dimensional space, so specifying its translational state needs three independent numbers — its velocity components $v_x$, $v_y$, $v_z$. Each of these is a degree of freedom, and each carries a kinetic energy term that is quadratic in the coordinate: $\tfrac12mv_x^2$, $\tfrac12mv_y^2$, $\tfrac12mv_z^2$. "Quadratic" simply means the energy goes as the square of a velocity or displacement — this is the technical condition the law of equipartition actually needs.

where: $k_B = 1.38\times10^{-23}\ \text{J/K}$ is Boltzmann's constant and $T$ is the absolute temperature in kelvin. The physical reading is blunt: it does not matter whether the quadratic term is $\tfrac12mv_x^2$ (translation), $\tfrac12I\omega^2$ (rotation), or $\tfrac12kx^2$ (a vibrating bond's potential energy) — thermal agitation deposits exactly the same average energy, $\tfrac12k_BT$, into each one. A molecule with $f$ such independent quadratic terms therefore carries a total average energy per molecule of

where: $f$ is the total count of quadratic degrees of freedom (translational + rotational +, if active, vibrational). Setting $f = 3$ recovers exactly the kinetic-interpretation-of-temperature result $\overline{E} = \tfrac32k_BT$ for a point mass — equipartition does not contradict that earlier result, it explains it as one special case. Counting $f$ for the molecule types on your syllabus:

• Monatomic (He, Ar, Ne): a single atom modelled as a point mass has only translational motion — 3 independent velocity components, no meaningful rotational kinetic energy about its own centre (the moment of inertia of a point about its own axis is essentially zero). So $f = 3$.
• Diatomic, rigid rotor model (O$_2$, N$_2$, near room temperature): 3 translational degrees of freedom plus 2 rotational ones — rotation about the two axes perpendicular to the bond line. Rotation about the bond axis itself is left out of the count because the moment of inertia along that thin axis is negligibly small, so that rotational energy term contributes essentially nothing. So $f = 5$.
• Diatomic, with vibration active (only at higher temperatures — CBSE treats this qualitatively): the bond can also stretch and compress like a tiny spring, and a vibrating spring contributes two quadratic terms — kinetic energy of the oscillation and potential energy of the stretched bond — so one active vibrational mode adds $2 \times \tfrac12k_BT = k_BT$, i.e. 2 more degrees of freedom. So $f = 7$ once vibration switches on.
• Polyatomic, non-linear (e.g. H$_2$O, NH$_3$): 3 translational + 3 rotational (three independent axes of rotation, since the molecule is not a simple line) $= 6$, before counting any vibrational modes, each of which would add 2 more.

Each individual gas molecule carries average energy $\tfrac{f}{2}k_BT$; a mole contains Avogadro's number of them, $N_A = 6.022\times10^{23}$. Since Boltzmann's constant and the universal gas constant are related by $R = N_Ak_B$, scaling up from one molecule to one mole converts $k_B$ into $R$ automatically:

where: $U$ is the total internal (thermal) energy of one mole of the gas, $f$ is the degrees-of-freedom count appropriate to the molecule (3, 5, 6, or 7), $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$, and $T$ is the absolute temperature. This is the single formula that unlocks the specific heat capacities in the next sections: since $C_v = dU/dT$ at constant volume, $C_v = \tfrac{f}{2}R$ follows immediately, and $C_p = C_v + R$, $\gamma = C_p/C_v$ come along with it.

Board level - internal energy of monatomic helium at 300 K

Find the total average thermal (internal) energy of one mole of a monatomic ideal gas, such as helium, at $T = 300\ \text{K}$, using the law of equipartition of energy.

Step 1. Count degrees of freedom. Helium is monatomic, so only translational motion is possible: $f = 3$.

Step 2. Apply $U = \dfrac{f}{2}RT$ with $f = 3$, $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$, $T = 300\ \text{K}$:

Step 3. Evaluate. $\dfrac{3}{2} \times 8.314 = 12.471$, and $12.471 \times 300 = 3741.3\ \text{J}$.

Final answer: one mole of helium at 300 K carries a total internal (thermal) energy of about $\mathbf{3741\ J}$, entirely as translational kinetic energy, since it has no rotational or vibrational degrees of freedom to share it with.

Competitive level - specific heats and gamma for rigid diatomic nitrogen

A rigid diatomic gas such as nitrogen (no vibration active, near room temperature) is held at $T = 500\ \text{K}$. Find (a) its total internal energy per mole, (b) $C_v$ and $C_p$, and (c) $\gamma = C_p/C_v$; then compare $U$ at $500\ \text{K}$ to the internal energy a monatomic gas would have at the same temperature.

Step 1. Count degrees of freedom for a rigid diatomic molecule: 3 translational + 2 rotational, so $f = 5$.

Step 2. Internal energy per mole. $U_{\text{diatomic}} = \dfrac{5}{2}RT = \dfrac{5}{2} \times 8.314 \times 500 = 10392.5\ \text{J}$.

Step 3. Molar specific heats follow directly from $U = \tfrac{f}{2}RT$, since $C_v = dU/dT$ at constant volume: $C_v = \dfrac{f}{2}R = \dfrac{5}{2} \times 8.314 = 20.785\ \text{J/mol}\,\cdot\,\text{K}$. Using Mayer's relation, $C_p = C_v + R = 20.785 + 8.314 = 29.099\ \text{J/mol}\,\cdot\,\text{K}$.

Step 4. Ratio of specific heats. $\gamma = \dfrac{C_p}{C_v} = \dfrac{29.099}{20.785} = 1.40$, matching the textbook value $\gamma = 7/5$ for a rigid diatomic gas.

Step 5. Compare to a monatomic gas at the same $T = 500\ \text{K}$: $U_{\text{monatomic}} = \dfrac{3}{2}RT = \dfrac{3}{2} \times 8.314 \times 500 = 6235.5\ \text{J}$. The ratio is $\dfrac{U_{\text{diatomic}}}{U_{\text{monatomic}}} = \dfrac{5/2}{3/2} = \dfrac{5}{3}$, a direct consequence of the diatomic molecule having 2 extra (rotational) degrees of freedom on top of the same 3 translational ones, each sharing the identical $\tfrac12RT$ per mole.

Final answer: $U_{\text{diatomic}} = \mathbf{10392.5\ J}$, $C_v = \mathbf{20.785\ J/mol\cdot K}$, $C_p = \mathbf{29.099\ J/mol\cdot K}$, $\gamma = \mathbf{1.40}$; the diatomic gas stores $\tfrac53$ times the internal energy of a monatomic gas at the same temperature. Timing note: this is a two-mark-turned-four-mark question in disguise — once you have $f$, every other number (U, $C_v$, $C_p$, $\gamma$) is one substitution away. Write $f=5$ first and the rest should take under ninety seconds; don't re-derive $C_p = C_v + R$ from scratch, just quote Mayer's relation.

• Assuming every gas has the same $f$ regardless of molecular structure. Monatomic is $f=3$, rigid diatomic is $f=5$, non-linear polyatomic (before vibration) is $f=6$ — mixing these up is the single most common error in this topic.
• Forgetting that a vibrational mode contributes two quadratic terms, not one — kinetic energy of oscillation plus potential energy of the stretched bond — so an active vibrational mode adds $k_BT$ (not $\tfrac12k_BT$) per mole per mode, i.e. 2 degrees of freedom, not 1.
• Counting three rotational degrees of freedom for a diatomic molecule. Only 2 are counted, because rotation about the bond axis itself has a negligibly small moment of inertia and contributes essentially no energy at ordinary temperatures.
• Trying to derive the law of equipartition of energy from kinetic theory in an exam answer. CBSE prescribes it strictly as a statement to be applied — quote it, don't attempt to prove it; only the per-molecule-to-per-mole scaling step ($U = \tfrac{f}{2}RT$) is expected to be shown as working.
• Using $f=5$ for a diatomic gas at very high temperature where vibration has switched on — in that regime $f=7$ and $U = \tfrac72RT$; the CBSE-level treatment of this switch is qualitative, but you should still be able to state which $f$ applies at which regime.

• State the law of equipartition of energy in words: each quadratic degree of freedom carries average energy $\tfrac12k_BT$ per molecule in thermal equilibrium.
• Count degrees of freedom $f$ correctly for monatomic (3), rigid diatomic (5), vibrating diatomic (7, qualitative), and non-linear polyatomic (6) molecules.
• Derive the per-mole formula $U = \tfrac{f}{2}RT$ starting from the per-molecule statement, using $R = N_Ak_B$.
• Compute total internal energy per mole for a given gas and temperature, and compare it across molecule types at the same $T$.
• Use $f$ to obtain $C_v = \tfrac{f}{2}R$, $C_p = C_v + R$, and $\gamma = C_p/C_v$ for the specific-heat sections that follow.
• Recognise, and avoid, the exam trap of attempting to derive the law itself rather than just applying it.

Ask the tutor: Want a quick drill comparing internal energy, $C_v$, $C_p$, and $\gamma$ across monatomic, diatomic, and polyatomic gases at the same temperature, or help remembering which degrees of freedom to count for a given molecule shape? Ask the AI tutor for a recap table or extra practice problems.