Degrees of freedom
The degrees of freedom of a molecule are the number of independent ways in which it can store energy of motion. NCERT frames it geometrically: a molecule free to move in space needs three coordinates to specify its location, so it has three independent directions of motion. If it were constrained to a plane it would need two; constrained to a line, just one. Motion of the body as a whole from one point to another is called translation, so a molecule moving freely in space has three translational degrees of freedom.
The crucial refinement is that what equipartition counts is not "directions" but quadratic terms in the energy expression. Each translational degree of freedom contributes a term containing the square of a velocity component — $\tfrac{1}{2}mv_x^2$, $\tfrac{1}{2}mv_y^2$, $\tfrac{1}{2}mv_z^2$. Rotation about an axis contributes a term $\tfrac{1}{2}I\omega^2$. Vibration is special: it contributes two quadratic terms, a kinetic energy term $\tfrac{1}{2}m(\mathrm{d}y/\mathrm{d}t)^2$ and a potential energy term $\tfrac{1}{2}ky^2$. This distinction — quadratic terms, not axes — is the entire game.
The law of equipartition
NCERT derives the seed of the law from the kinetic interpretation of temperature. The average translational energy of a molecule in thermal equilibrium is
$$\langle \varepsilon_t \rangle = \left\langle \tfrac{1}{2}mv_x^2 + \tfrac{1}{2}mv_y^2 + \tfrac{1}{2}mv_z^2 \right\rangle = \tfrac{3}{2}k_BT.$$Because the gas is isotropic — no direction is preferred — the three terms must be equal, so each one averages to $\tfrac{1}{2}k_BT$:
$$\left\langle \tfrac{1}{2}mv_x^2 \right\rangle = \left\langle \tfrac{1}{2}mv_y^2 \right\rangle = \left\langle \tfrac{1}{2}mv_z^2 \right\rangle = \tfrac{1}{2}k_BT.$$Maxwell proved that this equal sharing is not special to translation. The full statement, as NCERT gives it:
In equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to $\tfrac{1}{2}k_BT$. Each translational and rotational degree of freedom of a molecule contributes $\tfrac{1}{2}k_BT$ to the energy, while each vibrational frequency contributes $2 \times \tfrac{1}{2}k_BT = k_BT$, since a vibrational mode has both kinetic and potential energy modes.
Per mole, multiply by Avogadro's number $N_A$ and use $R = N_A k_B$: each quadratic degree of freedom contributes $\tfrac{1}{2}RT$ per mole. The proof itself is beyond the NEET syllabus; the application is the examinable part.
Monatomic gas — f = 3
A monatomic gas such as helium, neon or argon is, for kinetic purposes, a structureless point. It has no bond to rotate about and nothing to vibrate, so it has only the three translational degrees of freedom. Each contributes $\tfrac{1}{2}k_BT$, giving an average molecular energy of $\tfrac{3}{2}k_BT$ — exactly the translational result from kinetic theory, now seen as a special case of equipartition with $f = 3$.
Diatomic gas — f = 5 rising to 7
A diatomic molecule such as $\mathrm{O_2}$ or $\mathrm{N_2}$ still has its three translational degrees of freedom, but it can also rotate. NCERT (Fig. 12.6) identifies two independent axes of rotation, both perpendicular to the line joining the two atoms. Rotation about the bond axis itself has a negligible moment of inertia and is ignored for quantum mechanical reasons. So at moderate temperature the molecule is treated as a rigid rotator with
$$f = \underbrace{3}_{\text{translational}} + \underbrace{2}_{\text{rotational}} = 5, \qquad \langle \varepsilon \rangle = \tfrac{5}{2}k_BT.$$The rigid-rotator assumption is not always valid. NCERT notes that molecules like CO, even at moderate temperatures, have a vibrational mode in which the atoms oscillate along the interatomic axis like a one-dimensional oscillator. When that mode is active, the molecule gains a vibrational energy term carrying two quadratic contributions, and
$$f = \underbrace{3}_{\text{trans}} + \underbrace{2}_{\text{rot}} + \underbrace{2}_{\text{vib}} = 7, \qquad \langle \varepsilon \rangle = \tfrac{7}{2}k_BT.$$Vibration switches on only at high temperature, which is why the experimental specific heats of diatomic gases rise above the rigid-rotator prediction as temperature increases.
Polyatomic gas
A nonlinear polyatomic molecule — water vapour, ammonia, methane — has its centre of mass free to translate in three directions and can rotate about all three independent axes, because none of those axes carries a negligible moment of inertia. That gives $3 + 3 = 6$ degrees of freedom before vibration. In addition, NCERT writes the general polyatomic molecule as having a certain number $f$ of vibrational modes, each contributing $k_BT$. The internal energy per mole becomes
$$U = \left(\tfrac{3}{2}k_BT + \tfrac{3}{2}k_BT + f\,k_BT\right)N_A = (3 + f)RT,$$so $C_v = (3 + f)R$ — recovering $C_v = 3R$ for the rigid (non-vibrating) nonlinear polyatomic case where $f = 0$.
The whole payoff of equipartition is computing molar heat capacities — see specific heat from kinetic theory for $C_v$, $C_p$ and $\gamma$ across all gas types.
Degrees of freedom across gas types
The one table to commit to memory. Every entry below follows from the single rule $\tfrac{1}{2}k_BT$ per quadratic term, with vibration counted twice.
| Gas type | Translational | Rotational | Vibrational | Total f | Energy / molecule |
|---|---|---|---|---|---|
| Monatomic (He, Ne, Ar) | 3 | 0 | 0 | 3 | $\tfrac{3}{2}k_BT$ |
| Diatomic, rigid — moderate T (O2, N2) | 3 | 2 | 0 | 5 | $\tfrac{5}{2}k_BT$ |
| Diatomic, vibrating — high T (CO at high T) | 3 | 2 | 2 | 7 | $\tfrac{7}{2}k_BT$ |
| Polyatomic, nonlinear, rigid (H2O, NH3) | 3 | 3 | 0 | 6 | $3k_BT$ |
| Polyatomic with vibration | 3 | 3 | $2f_{\text{vib}}$ | $6 + 2f_{\text{vib}}$ | $(3 + f_{\text{vib}})k_BT$ |
Read the table as a single rule applied repeatedly: count one quadratic term for each translational direction, one for each genuine rotational axis, and two for each active vibrational mode. Multiply the total $f$ by $\tfrac{1}{2}k_BT$ for the energy per molecule, or by $\tfrac{1}{2}RT$ for the energy per mole.
Internal energy U = (f/2)nRT
Because an ideal gas has no intermolecular potential energy, its internal energy is purely the sum of these molecular energy modes. For $n$ moles of a gas whose molecules each have $f$ active quadratic degrees of freedom,
$$\boxed{\,U = \frac{f}{2}\,n R T\,}.$$Two consequences NEET tests directly. First, $U$ depends only on temperature (and on $f$), never on pressure or volume — a fact that underlies the result that the internal energy of an ideal gas is a function of $T$ alone. Second, for a mixture of non-reacting ideal gases the total internal energy is simply the sum over the components, each evaluated with its own $f$. That additivity is exactly what the 2017 PYQ below exploits.
A vessel holds 3 mol of a monatomic gas and 2 mol of a rigid diatomic gas at the same temperature $T$. Find the total internal energy of the mixture.
Monatomic part. $f = 3$, so $U_1 = \tfrac{3}{2}n_1 RT = \tfrac{3}{2}(3)RT = \tfrac{9}{2}RT$.
Diatomic part. $f = 5$, so $U_2 = \tfrac{5}{2}n_2 RT = \tfrac{5}{2}(2)RT = 5RT$.
Total. $U = U_1 + U_2 = \tfrac{9}{2}RT + 5RT = \tfrac{19}{2}RT = 9.5\,RT$. Each gas keeps its own $f$; never apply a single $f$ to the whole mixture.
Counting traps
Score every degree-of-freedom count against this checklist before you compute energy or specific heat.
- Vibration counts twice. One vibrational mode = 2 quadratic terms = $k_BT$, never $\tfrac{1}{2}k_BT$.
- Diatomic rotation has 2 axes, not 3. The bond-axis rotation is dropped, so rigid diatomic $f = 5$.
- Vibration is temperature-gated. At room temperature most diatomics are rigid ($f = 5$); vibration only switches on at high $T$, raising $f$ toward 7.
- Mixtures keep separate $f$ values. Sum $\tfrac{f_i}{2}n_i RT$ over each component; do not average the atomicities.
- Translational KE is universal. $\tfrac{3}{2}k_BT$ per molecule for any ideal gas; only total $U$ depends on $f$.
- Per molecule vs per mole. $\tfrac{1}{2}k_BT$ per molecule equals $\tfrac{1}{2}RT$ per mole — check which the question wants.
Equipartition in one breath
- Each quadratic term in a molecule's energy carries $\tfrac{1}{2}k_BT$ per molecule, or $\tfrac{1}{2}RT$ per mole, in thermal equilibrium.
- Translational and rotational modes give one quadratic term each; a vibrational mode gives two (kinetic + potential), so it counts as $k_BT$.
- Monatomic $f = 3$; rigid diatomic $f = 5$; vibrating diatomic $f = 7$; nonlinear polyatomic $f = 6$ (plus $2f_\text{vib}$ if vibrating).
- Internal energy of $n$ moles: $U = \tfrac{f}{2}nRT$, a function of temperature alone for fixed $f$.
- Average translational KE is $\tfrac{3}{2}k_BT$ for every ideal gas; only the total energy grows with $f$.
- In a mixture, add $\tfrac{f_i}{2}n_i RT$ over each component separately.