The key idea is the equipartition of energy: each quadratic degree of freedom contributes $\frac{1}{2}R$ to the molar specific heat at constant volume ($C_v$). For a monatomic gas, only 3 translational degrees of freedom exist, giving $C_v = \frac{3}{2}R \approx 2.98\ \text{cal mol}^{-1}\ \text{K}^{-1}$ (close to the observed $2.92$).

Diatomic gases like those in the table have additional rotational degrees of freedom. At room temperature, a diatomic molecule has 3 translational and 2 rotational degrees of freedom (vibrational modes are frozen out), so $C_v = \frac{5}{2}R \approx 4.97\ \text{cal mol}^{-1}\ \text{K}^{-1}$. The observed values for hydrogen, nitrogen, oxygen, nitric oxide, and carbon monoxide all cluster near this figure, confirming the equipartition prediction. …

The equipartition of energy is the key idea here. It says that each quadratic term in a molecule’s energy contributes $\frac{1}{2}k_BT$ per molecule (or $\frac{1}{2}RT$ per mole) to the internal energy. A monatomic gas has only three translational degrees of freedom — motion along x, y, and z — so its molar internal energy is $U = \frac{3}{2}RT$, giving $C_v = \frac{3}{2}R \approx 2.98\ \text{cal mol}^{-1}\ \text{K}^{-1}$. The observed value of $2.92$ is close, confirming this.

Now look at the gases in the table: hydrogen, nitrogen, oxygen, nitric oxide, carbon monoxide, and chlorine. All are diatomic molecules (two atoms). A diatomic molecule can do more than just translate. It can also rotate about two perpendicular axes (like a dumbbell spinning), adding two rotational degrees of freedom. Each contributes $\frac{1}{2}RT$ to the molar internal energy. So for a diatomic gas with translation and rotation active:

Then:

With $R \approx 1.987\ \text{cal mol}^{-1}\ \text{K}^{-1}$, this gives:

That matches the values for nitrogen, oxygen, nitric oxide, and carbon monoxide almost exactly. Hydrogen’s $4.87$ is slightly lower — a subtle quantum effect: at room temperature, hydrogen’s rotational levels are not fully populated because its moment of inertia is very small, so the equipartition prediction isn’t fully realised.

1. Why the difference from monatomic gases?

   Monatomic gases have only 3 translational degrees of freedom. Diatomic gases have 3 translational + 2 rotational = 5 active degrees of freedom at room temperature. Each degree contributes $\frac{1}{2}R$ to $C_v$, so diatomic $C_v$ is $\frac{5}{2}R \approx 4.97$, while monatomic $C_v$ is $\frac{3}{2}R \approx 2.98$. The table confirms this: most diatomic gases cluster around $4.97$, far above $2.92$.

2. What about chlorine’s larger value ($6.17$)? …

Shortcut — read degrees of freedom directly off the ratio $C_v/R$. Equipartition gives $C_v = \frac{f}{2}R$, so $f = 2C_v/R$. Using $R \approx 1.987\ \text{cal mol}^{-1}\text{K}^{-1}$:

• Monatomic: $C_v/R \approx 2.92/1.987 \approx 1.5 \Rightarrow f=3$ (translation only).
• H$_2$, N$_2$, O$_2$, NO, CO: $C_v/R \approx 4.97/1.987 \approx 2.5 \Rightarrow f=5$ (3 translational + 2 rotational) — the near-identical ratio across five chemically different gases is itself strong evidence that rotation, not mass or bond strength, is what's being added. …