Specific heat capacity of diatomic gases

This section sits directly inside the CBSE Class 11 Kinetic Theory chapter, right after degrees of freedom and the (statement-only) law of equipartition of energy, and right after the monatomic specific-heat result. The prescribed board scope asks you to apply the same equipartition reasoning to a diatomic molecule modelled as a rigid rotor, and to know — qualitatively only — that vibration can raise $f$ at high temperature; the syllabus does not ask for a derivation of equipartition itself, nor for the Maxwell–Boltzmann speed distribution, nor for a numeric vibrational correction unless a question states one. Everything below stays inside those limits.

Picture a diatomic molecule as two small spheres (the atoms) joined by a stiff, weightless rod (the bond) — a rigid dumbbell, free to move as a whole. Set up axes through its centre of mass: one along the bond, and two perpendicular to it. The molecule can translate along any of the three ordinary spatial directions, contributing the same 3 translational degrees of freedom every gas has, monatomic or not. But unlike a point particle, this dumbbell can also tumble: it can rotate about either of the two axes perpendicular to the bond, and each such rotation carries its own kinetic energy, $\tfrac12 I\omega^2$, which is a quadratic term just like $\tfrac12 mv^2$ and so earns its own $\tfrac12 k_BT$ under equipartition. What about rotation about the bond axis itself — spinning the dumbbell like a drill bit along its own length? Classically that motion exists, but the two atoms are treated as point masses sitting essentially on that axis, so the moment of inertia about it is vanishingly small compared with the other two axes. A negligible moment of inertia means a negligible contribution to kinetic energy for any ordinary angular speed, so this third rotational term is dropped from the count. That is the one deliberate exclusion that separates a diatomic molecule's $f=5$ from what a naive "three axes, three rotations" guess would give.

where: $f_{\text{translation}}$ counts motion of the centre of mass along the three ordinary directions of space, $f_{\text{rotation}}$ counts tumbling about the two axes perpendicular to the bond (rotation about the bond itself excluded), and $f_{\text{total}}$ is the degrees of freedom used in the equipartition sum below. Physically: every one of these five is an independent way the molecule can store kinetic energy, and the law of equipartition treats each one exactly alike, handing it $\tfrac12 k_BT$ of average energy regardless of what kind of motion it represents.

Board level - heating nitrogen gas at constant pressure

Find the heat required to raise the temperature of 3 mol of nitrogen gas (diatomic, treated as a rigid rotor) from 280 K to 320 K at constant pressure.

Step 1. Identify the process and the right specific heat. Heating at constant pressure means the heat supplied equals $\Delta Q = nC_p\Delta T$, using $C_p$ rather than $C_v$ because the gas is also doing work as it expands against constant pressure, and that extra work must be paid for out of the same heat input.

Step 2. Use the diatomic value derived above, $C_p = \tfrac72R = 29.10\ \text{J/mol}\,\cdot\,\text{K}$, since nitrogen is diatomic and well described as a rigid rotor at these ordinary temperatures (no vibration active).

Step 3. Substitute the numbers: $n=3\ \text{mol}$, $\Delta T = 320-280 = 40\ \text{K}$.

The heat required is 3492 J, or about 3.49 kJ, for this constant-pressure heating of nitrogen.

Competitive level - adiabatic compression of a diatomic gas

An ideal diatomic gas (rigid rotor, $\gamma = 7/5$) occupies volume $V$ at pressure $P$. It undergoes an adiabatic process in which its volume is halved. Find the ratio of the final pressure to the initial pressure, and compare with what an isothermal compression to the same volume would give.

Step 1. Write the adiabatic condition. For a quasi-static adiabatic process on an ideal gas, $PV^\gamma = \text{constant}$, so between the initial and final states, $P_1V_1^\gamma = P_2V_2^\gamma$, which rearranges to $\dfrac{P_2}{P_1} = \left(\dfrac{V_1}{V_2}\right)^\gamma$.

Step 2. Substitute the given compression. Halving the volume means $V_2 = V_1/2$, so $V_1/V_2 = 2$, giving $\dfrac{P_2}{P_1} = 2^{\gamma} = 2^{7/5}$.

Step 3. Evaluate $2^{7/5}$ using logarithms. Since $\log_{10}2 = 0.301$, $1.4 \times 0.301 = 0.4214$, so $2^{1.4} = 10^{0.4214} \approx 2.64$.

Step 4. Compare with isothermal compression to the same final volume. For an isothermal process, $P_1V_1 = P_2V_2$ ($\gamma$ effectively replaced by 1), so halving the volume would only double the pressure, $P_2/P_1 = 2$. The adiabatic case gives a final pressure about 2.64 times the initial pressure — noticeably more than the isothermal factor of 2, because in the adiabatic process no heat escapes, so the compression work itself raises the temperature and pushes the pressure up further than an isothermal squeeze would. Timing note: in an exam, don't re-derive $PV^\gamma=\text{constant}$ from scratch — quote it directly, plug in the given volume ratio, and evaluate the power using logarithms (or recognise standard values like $2^{1.4}\approx 2.64$ from practice) in well under two minutes; this whole problem should take about 90 seconds once $\gamma=7/5$ is recalled.

• Miscounting rotational axes. Students often assume a diatomic molecule has 3 rotational degrees of freedom, one about each axis, forgetting that rotation about the bond axis has negligible moment of inertia and is excluded — the correct count is $f=5$, not $f=6$.
• Reusing the monatomic $\gamma$. Writing $\gamma = 5/3$ out of habit for any ideal gas, instead of switching to the diatomic value $\gamma = 7/5 = 1.4$ once the problem says nitrogen, oxygen, hydrogen, or any other diatomic gas.
• Applying $f=5$ at temperatures where vibration matters. The rigid-rotor result $\gamma = 7/5$ assumes vibration is frozen out, which is true near room temperature for most diatomic gases but not at very high temperature; CBSE expects only a qualitative awareness of this, never a numeric switch mid-problem unless the question explicitly asks for it.
• Confusing the isothermal and adiabatic gas laws. Using $PV = \text{constant}$ (isothermal) when the process described is adiabatic, or vice versa — the exponent $\gamma$ appears only in the adiabatic relation, and it is exactly the diatomic $\gamma=7/5$ that must be used there.
• Forgetting Mayer's relation links $C_p$ and $C_v$. Some students try to derive $C_p$ from equipartition directly instead of using $C_p = C_v + R$, which is both faster and less error-prone once $C_v$ is known.

• Count the degrees of freedom of a rigid diatomic molecule ($f=5$) and explain clearly why rotation about the bond axis is excluded.
• Derive $C_v = \tfrac52R$, $C_p = \tfrac72R$, and $\gamma = 7/5$ for a diatomic gas starting from the law of equipartition of energy.
• State, without deriving, how vibrational modes at high temperature would raise $f$ and lower $\gamma$ further.
• Apply $\Delta Q = nC_v\Delta T$ or $\Delta Q = nC_p\Delta T$ correctly, choosing the right specific heat for the process described.
• Apply $PV^\gamma = \text{constant}$ for an adiabatic process on a diatomic gas, and distinguish it from the isothermal relation $PV=\text{constant}$.
• Compare the diatomic result against the monatomic case and explain, in words, why more degrees of freedom means a smaller $\gamma$.

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