Given below are observations on molar specific heats at room temperature of some common gases. | Gas | Molar specific heat $(C_v)$ $(\text{cal mol}^{-1}\ \text{K}^{-1})$ | | --- | --- | | Hydrogen | $4.87$ | | Nitrogen | $4.97$ | | Oxygen | $5.02$ | | Nitric oxide | $4.99$ | | Carbon monoxide | $5.01$ | | Chlorine | $6.17$ | The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is $2.92\ \text{cal/mol K}$. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Given below are observations on molar specific hea – Class 11 Physics

Exercises · 10.15

Given below are observations on molar specific heats at room temperature of some common gases.

Gas	Molar specific heat $(C_v)$ $(\text{cal mol}^{-1}\ \text{K}^{-1})$
Hydrogen	$4.87$
Nitrogen	$4.97$
Oxygen	$5.02$
Nitric oxide	$4.99$
Carbon monoxide	$5.01$
Chlorine	$6.17$

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is $2.92\ \text{cal/mol K}$ . Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

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Concept understanding — Equipartition Of Energy

Equipartition of Energy: From Intuition to the Precise Statement

Imagine you have a box of gas — a bunch of tiny molecules zipping around, spinning, and vibrating. Each molecule has some energy. But here's the question: How is that total energy shared among all the different ways a molecule can move?

That's exactly what the equipartition of energy answers.

The Core Intuition

Think of a room full of people dancing. Some are jumping, some are spinning, some are waving their arms. If the music is steady and the room is crowded, eventually everyone will be moving with roughly the same average energy per "type" of motion. No one type of motion hogs all the energy.

In a gas at thermal equilibrium, nature does the same thing. Every independent way a molecule can store energy — every "degree of freedom" — gets the same average amount of energy.

Note

A degree of freedom is an independent way a molecule can move or store energy. For a single atom, that's just three directions of translation (x, y, z). For a diatomic molecule, you also get rotations and vibrations.

The Precise Statement

\text{Average energy per degree of freedom} = \frac{1}{2} k_B T

Where:

$k_B$ is Boltzmann's constant ( $1.38 \times 10^{-23} \ \text{J/K}$ )
$T$ is the absolute temperature in Kelvin

So if a molecule has $f$ degrees of freedom, its total average energy is:

\langle E \rangle = \frac{f}{2} k_B T

Why "Half $k_B T$ "?

This comes from a deeper statistical mechanics result. For any quadratic term in the energy expression (like $\frac{1}{2}mv_x^2$ for translation, or $\frac{1}{2}I\omega^2$ for rotation), the average energy contributed by that term is exactly $\frac{1}{2}k_B T$ .

Tip

| Molecule type | Degrees of freedom ( $f$ ) | Average energy |

|:---|:---:|:---:|

| Monatomic (He, Ar) | 3 (all translational) | $\frac{3}{2}k_B T$ |

| Diatomic (N₂, O₂) at room temp | 5 (3 trans + 2 rot) | $\frac{5}{2}k_B T$ |

| Diatomic at high temp | 7 (3 trans + 2 rot + 2 vib) | $\frac{7}{2}k_B T$ |

A Concrete Example

Take a helium atom in a box at 300 K. It can only move in three directions. Each direction contributes $\frac{1}{2}k_B T$ of energy.

So the average kinetic energy of one helium atom is:

\langle E \rangle = 3 \times \frac{1}{2} k_B T = \frac{3}{2} k_B T

For a mole of helium (Avogadro's number $N_A$ of atoms), the total internal energy becomes:

U = N_A \cdot \frac{3}{2} k_B T = \frac{3}{2} RT

where $R = N_A k_B$ is the universal gas constant. …

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The equipartition of energy explains why diatomic gases have higher molar specific heats than monatomic gases because they have additional rotational degrees of freedom. Chlorine’s larger value suggests its vibrational modes are also partially excited at room temperature.

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:

U = \frac{3}{2}RT + \frac{2}{2}RT = \frac{5}{2}RT

Then:

C_v = \frac{dU}{dT} = \frac{5}{2}R

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

C_v = \frac{5}{2} \times 1.987 \approx 4.97\ \text{cal mol}^{-1}\ \text{K}^{-1}

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.

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 $2.92$ .
What about chlorine’s larger value ( $6.17$ )? …

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