Monoatomic (f=3, γ=5/3), Diatomic (f=5, γ=7/5). Why?

Degrees of Freedom and Specific Heat Ratio

Question

A gas molecule has $f$ degrees of freedom. Show that the ratio of specific heats $\gamma = C_P / C_V$ is given by:

\gamma = 1 + \frac{2}{f}

Hence find $\gamma$ for a monoatomic gas and a diatomic gas (at room temperature).

Solution — Step by Step

Each degree of freedom contributes $\frac{1}{2}k_BT$ of energy per molecule. For one mole of gas with $f$ degrees of freedom, the internal energy is:

U = f \cdot \frac{1}{2}RT = \frac{f}{2}RT

This is the key starting point — we’re treating every independent mode of motion as an equal energy partner.

$C_V$ is the heat required to raise temperature by 1 K at constant volume. Since no work is done at constant volume, all heat goes into increasing $U$ :

C_V = \frac{dU}{dT} = \frac{f}{2}R

We know from thermodynamics that $C_P - C_V = R$ (this comes directly from the first law + ideal gas equation). So:

C_P = C_V + R = \frac{f}{2}R + R = \frac{f+2}{2}R

\gamma = \frac{C_P}{C_V} = \frac{\frac{f+2}{2}R}{\frac{f}{2}R} = \frac{f+2}{f} = 1 + \frac{2}{f}

Monoatomic (He, Ar): only 3 translational degrees of freedom, $f = 3$

\gamma = 1 + \frac{2}{3} = \frac{5}{3} \approx 1.67

Diatomic (N₂, O₂) at room temperature: 3 translational + 2 rotational, $f = 5$

\gamma = 1 + \frac{2}{5} = \frac{7}{5} = 1.4

Final Answer: $\boxed{\gamma = 1 + \frac{2}{f}}$ , giving $\gamma = 5/3$ for monoatomic and $\gamma = 7/5$ for diatomic gases.

Why This Works

The entire derivation rests on one idea: energy distributes equally among all available degrees of freedom. A monoatomic atom like helium has nowhere to store energy except by moving — left-right, up-down, forward-backward. That’s it, $f = 3$ .

A diatomic molecule like N₂ can also rotate about two axes (not three — rotation about the bond axis has negligible moment of inertia, so that mode doesn’t count at room temperature). This gives it two extra places to absorb energy, so it takes more heat to raise its temperature by 1 K — hence lower $\gamma$ .

This is why $\gamma$ matters in real problems: it appears in adiabatic processes ( $PV^\gamma = \text{const}$ ), speed of sound ( $v = \sqrt{\gamma RT/M}$ ), and efficiency calculations. Every time you see $\gamma$ , ask yourself: what type of gas? That fixes $f$ , which fixes $\gamma$ .

Alternative Method

You can also get $C_V$ by thinking about energy stored per mode and working backwards. For a diatomic gas at high temperature, vibrational modes also activate — each vibration contributes 2 degrees of freedom (kinetic + potential). So $f = 7$ at high $T$ , giving $\gamma = 9/7 \approx 1.29$ .

This is tested in JEE as a conceptual trap: “at very high temperatures, $\gamma$ for diatomic gas is…”— the answer is $9/7$ , not $7/5$ .

For quick recall: monoatomic → $f=3$ , diatomic (room temp) → $f=5$ , diatomic (high temp) → $f=7$ . The pattern is always $\gamma = (f+2)/f$ .

Common Mistake

The most common error is using $f = 6$ for diatomic gases, thinking “3 translational + 3 rotational”. The rotation about the bond axis (the molecular axis itself) is excluded because the moment of inertia about that axis is essentially zero — quantum mechanics tells us that mode is frozen out at ordinary temperatures. Always use $f = 5$ for diatomic at room temperature in board exams and JEE Main. The JEE Main 2023 question on this specifically tested whether students knew to exclude that third rotational mode.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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