Specific heat capacity from kinetic theory

How does kinetic theory predict specific heats?

The law of equipartition of energy is, thus, well verified experimentally at ordinary temperatures.

NCERT Physics Part II (Class 11), Section 12.6, page 254

In plain words: the more ways a molecule has of storing energy, the more heat it takes to raise its temperature, because the heat is shared among all those ways. Count the ways, multiply by ½R per mole, and you have predicted its molar specific heat at constant volume without measuring it.

Monatomic gases

A monatomic molecule (He, Ne, Ar) has only 3 translational degrees of freedom, so its average energy is (3/2)k_BT. For a mole, U = (3/2)k_BT × N_A = (3/2)RT (NCERT Eq. 12.27). Then (NCERT Eqs. 12.28–12.31):

• C_v = dU/dT = (3/2)R;
• for an ideal gas C_p − C_v = R, so C_p = (5/2)R;
• γ = C_p/C_v = 5/3.

The relation C_p − C_v = R is the one derived from the first law in the previous chapter (Mayer’s relation); NCERT notes that it holds for any ideal gas, whether monatomic, diatomic or polyatomic (page 253).

Diatomic gases

A diatomic molecule treated as a rigid rotator, like a dumbbell, has 5 degrees of freedom: 3 translational and 2 rotational. So U = (5/2)RT per mole (NCERT Eq. 12.32), and

C_v = (5/2)R, C_p = (7/2)R, γ = 7/5 (NCERT Eqs. 12.33–12.34).

If its harmonic vibration is fully excited in the classical limit, the vibration adds k_BT per molecule: U = (7/2)RT, so C_v = (7/2)R, C_p = (9/2)R and γ = 9/7 (NCERT Eq. 12.35).

Polyatomic gases

In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (f) of vibrational modes.

NCERT Physics Part II (Class 11), Section 12.6.3, page 253

Each fully excited classical vibrational mode adds k_BT, so per molecule the energy is (3/2)k_BT + (3/2)k_BT + fk_BT, and a mole has U = (3 + f)k_BTN_A = (3 + f)RT, giving (NCERT Eq. 12.36)

C_v = (3 + f)R, C_p = (4 + f)R, γ = (4 + f)/(3 + f).

Careful: in NCERT’s formula f counts only the vibrational modes. It is not the f in C_v = (f/2)R at the top of this page, which counts every squared term. The two agree: 3 translations + 3 rotations + 2 terms per vibration make 6 + 2f terms, and (6 + 2f)/2 × R = (3 + f)R.

The 3 rotational degrees of freedom apply to non-linear molecules such as H_₂O and CH_₄. A linear molecule such as CO_₂ rotates about only two axes, like a diatomic one: rotation about the line of atoms does not count, for the reason NCERT gives for O_₂ (footnote, page 252). This point is beyond the NCERT text.

Theory and experiment

NCERT Table 12.1 (page 254) gives the predictions ignoring vibration: monatomic C_v = 12.5 J mol^⁻¹ K^⁻¹ and γ = 1.67; diatomic 20.8 and 1.40; triatomic 24.93 and 1.33 (this row counts 3 rotations, so it fits non-linear molecules). NCERT Table 12.2 gives illustrative measured values (do not treat them as universal constants at every temperature), for example: He 12.5, Ne 12.7, Ar 12.5; H_₂ 20.4, O_₂ 21.0, N_₂ 20.8; H_₂O 27.0 and CH_₄ 27.1 J mol^⁻¹ K^⁻¹.

The monatomic and diatomic gases agree well with the predictions. For some other gases, such as Cl_₂ (diatomic), C_₂H_₆ and many other polyatomic gases, the measured values are usually larger than the predictions of NCERT Table 12.1, which suggests including vibrational modes (page 254).

Specific heat capacity of solids

In the classical high-temperature harmonic model, picture a solid as N atoms, each vibrating about its mean position. A vibration in one dimension has average energy 2 × ½k_BT = k_BT, and in three dimensions 3k_BT. For a mole of atoms, U = U_₀ + 3RT, where U_₀ is a temperature-independent offset. Thus C_ᵥ = 3R. NCERT additionally neglects thermal expansion to approximate C_ₚ by the same value; small boundary work alone is not a general proof that C_ₚ = C_ᵥ. In that model,

NCERT Table 12.3 (page 254) shows approximate agreement for several listed solids near room temperature, with carbon an exception. At low temperatures quantum effects suppress heat capacity; 3R is a classical limit, not a rule for every solid. The same result appears in the Thermodynamics chapter.

Solved examples

Take R = 8.31 J mol^⁻¹ K^⁻¹.

Common mistakes to avoid

Quick revision summary

• Compact notation: C_v = fR/2, C_p = C_v + R, γ = C_p/C_v, with f = 3 for monatomic and f = 5 for rigid diatomic gases in the stated classical regime. For a classical harmonic solid, C_v = 3R per mole of atoms.
• In the classical ideal-gas model with fixed f active quadratic terms per molecule: U = (f/2)RT per mole, C_v = (f/2)R, C_p = C_v + R.
• Monatomic: C_v = (3/2)R, C_p = (5/2)R, γ = 5/3 ≈ 1.67.
• Rigid diatomic: C_v = (5/2)R, C_p = (7/2)R, γ = 7/5 = 1.40; with vibration, (7/2)R, (9/2)R, 9/7.
• Polyatomic (non-linear, f vibrational modes): C_v = (3 + f)R, C_p = (4 + f)R.
• γ = 1 + 2/f for f energy modes: 5/3, 7/5 and 9/7 for f = 3, 5 and 7.
• Solids: C_ᵥ = 3R per mole of atoms in the classical harmonic limit, often approximately C_ₚ when expansion corrections are small.

Questions students ask

How do you find the degrees of freedom of a gas from γ?

An effective quadratic-term count for an ideal gas follows from γ as f = 2/(γ − 1), found by rearranging γ = C_p/C_v = 1 + 2/f, where f counts every energy mode (squared term) of a molecule. For example, a gas with γ = 4/3 has f = 2/(1/3) = 6, consistent with a rigid non-linear molecule with 3 translations and 3 rotations, but not uniquely identifying that molecular structure. Mixtures and partly excited modes can give the same heat-capacity ratio. A value of f that is not a whole number points to partly excited vibrations or to a mixture of gases: carbon dioxide at room temperature has γ ≈ 1.29, so f ≈ 2/0.29 ≈ 6.9.

What is the specific heat capacity of helium gas?

Kinetic theory gives helium, a monatomic gas, a specific heat capacity of about 3.12 kJ kg^⁻¹ K^⁻¹ at constant volume and 5.20 kJ kg^⁻¹ K^⁻¹ at constant pressure, close to the measured values. These come from dividing the molar values (3/2)R and (5/2)R by the molar mass, 0.004 kg mol^⁻¹: 1.5 × 8.314/0.004 ≈ 3118 J kg^⁻¹ K^⁻¹ and 2.5 × 8.314/0.004 ≈ 5196 J kg^⁻¹ K^⁻¹. At constant pressure that is more than water’s 4186 J kg^⁻¹ K^⁻¹, not because each helium atom needs much energy but because a kilogram of helium holds 250 mol of atoms.

Can the specific heat of a gas be zero, negative or infinite?

A process heat capacity C = ΔQ/(μΔT) can be zero or negative, or undefined when ΔT = 0. It is distinct from the material heat capacities C_ᵥ and C_ₚ. In an adiabatic process no heat flows while T changes, so C = 0; in an isothermal step with nonzero heat transfer the ratio is undefined (often described as formally infinite), not an infinite material C_ᵥ; and if a gas does more work than the heat it takes in, it cools while gaining heat, so C is negative. For a fixed amount of ideal gas following a quasi-static path PV^ⁿ = constant with constant C_ᵥ and only pressure–volume work (beyond NCERT), C = C_v + R/(1 − n), which is C_p for n = 0, zero for n = γ and negative for 1 < n < γ.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why is C_v for oxygen larger than for helium?

   Show answerHide answer

   An oxygen molecule can also rotate about two axes, so it has 5 degrees of freedom against helium’s 3. More modes share the energy, so more heat is needed per kelvin: (5/2)R against (3/2)R.

2. Q2. Find C_p for argon from kinetic theory.

   Show answerHide answer

   Argon is monatomic, so C_p = (5/2)R = 2.5 × 8.31, about 20.8 J mol^⁻¹ K^⁻¹, matching NCERT Table 12.2.

3. Q3. What value of γ does kinetic theory predict for a rigid diatomic gas, and how could γ tell you whether a gas is monatomic or diatomic?

   Show answerHide answer

   γ = 7/5 = 1.40 for a rigid diatomic gas, against 5/3 ≈ 1.67 for a monatomic one. If these are the only two candidate ideal-gas models under the stated conditions, γ distinguishes them. In general, mixtures or thermally changing internal modes prevent unique molecular identification.