Specific heat capacity of polyatomic gases

Breathe out on a cold morning and watch the fog of your breath curl into the air. That cloud is mostly water vapour, $\text{H}_2\text{O}$ — three atoms bent into a shape like a boomerang, tumbling and jiggling in ways a simple two-atom molecule never could. Carbon dioxide from a fizzy drink, ammonia in a cleaning spray, methane from a gas stove — these are all polyatomic molecules, built from three or more atoms, and they are the last and busiest member of the ideal-gas family this chapter studies. A monatomic atom like helium can only fly in straight lines. A diatomic molecule like nitrogen can fly and tumble end over end. A polyatomic molecule can fly, tumble about every axis you can imagine, and its atoms can stretch and flex against each other like tiny springs. Each of these motions is a place energy can hide, and the chapter's big promise — the law of equipartition of energy — says every one of these "places" (called degrees of freedom) grabs, on average, the same share $\tfrac12 k_BT$ of thermal energy per molecule. Count the places correctly and the molar specific heats $C_v$ and $C_p$ , and their ratio $\gamma = C_p/C_v$ , fall straight out of arithmetic — no new physics, only careful bookkeeping. This article does that bookkeeping for polyatomic gases: a non-linear molecule such as water, ammonia or methane picks up three translational and three rotational degrees of freedom, giving $C_v = 3R$ and $C_p = 4R$ even before its atoms start vibrating. Switch on vibration and the numbers climb further, and $\gamma$ slides ever closer to $1$ — the signature of a molecule with almost nowhere left to not store energy. By the end you will be able to derive these results from scratch and use them confidently in numericals, exactly as CBSE expects.

Key takeaways

A non-linear polyatomic molecule (water, ammonia, methane) has $3$ translational plus $3$ rotational degrees of freedom, giving a base count $f = 6$ before vibration is switched on.
By the law of equipartition of energy, each degree of freedom contributes $\tfrac12 R$ per mole to the internal energy, so $U = \tfrac{f}{2}RT = 3RT$ for the vibration-free case.
This gives $C_v = 3R \approx 24.9\ \text{J/mol}\,\cdot\,\text{K}$ , and, using Mayer's relation $C_p = C_v + R$ , $C_p = 4R \approx 33.3\ \text{J/mol}\,\cdot\,\text{K}$ .
The ratio $\gamma = C_p/C_v = 4/3 \approx 1.33$ for a non-linear polyatomic gas with no active vibration — the smallest $\gamma$ of the three gas types studied so far.
Each active vibrational mode adds $2$ to $f$ (one for kinetic energy, one for potential energy of the "spring"), so with $f'$ vibrational modes active, $C_v = \left(3+f'\right)R$ and $\gamma$ falls further, approaching $1$ for very complex molecules.
A linear polyatomic molecule such as $\text{CO}_2$ is the odd one out: rotating about its own axis stores negligible energy, so it behaves rotationally like a diatomic gas ( $f=5$ base), not like $f=6$ — CBSE numericals flag this explicitly.
Key idea: a non-linear polyatomic gas has six equipartition-active degrees of freedom (three translational, three rotational), giving Cv = 3R, Cp = 4R and gamma = 4/3 before vibration, with every active vibrational mode adding 2 to f and pushing gamma further down toward 1.

The core insight

A non-linear polyatomic molecule has $f=6$ equipartition degrees of freedom (translation $3$ + rotation $3$ ), so $C_v=3R$ , $C_p=4R$ , $\gamma = 4/3$ , and every active vibrational mode adds $2$ more to $f$ , always pushing $\gamma$ closer to $1$ .

Where it fits in the syllabus

This article closes the CBSE Class 11 sequence on molar specific heats, applying the same equipartition method already used for monatomic and diatomic gases to molecules with three or more atoms. It sits squarely inside the prescribed 2025-26 Kinetic Theory scope: degrees of freedom, the law of equipartition of energy (statement only, no derivation from first principles of statistical mechanics), and its application to specific heats. What the syllabus wants from you here is the base non-linear result $f=6 \Rightarrow C_v = 3R,\ C_p = 4R,\ \gamma = 4/3$ , and a qualitative sense that vibrational modes, when active, add further degrees of freedom and lower $\gamma$ — not a derivation of why particular vibrational modes switch on at particular temperatures, which belongs to quantum theory well beyond Class 11.

The physical picture

Every molecule, however complicated, first shares the three translational degrees of freedom that any point mass has — motion along $x$ , $y$ and $z$ . What distinguishes a polyatomic molecule is what happens on top of that. A non-linear polyatomic molecule — one whose atoms do not all sit on a single straight line, like the bent water molecule or the pyramidal ammonia molecule — has three genuinely independent moments of inertia, one for rotation about each of three mutually perpendicular axes through its centre of mass. Unlike a diatomic molecule (where spinning about the bond axis itself stores negligible energy, since the atoms' mass lies almost entirely on that axis), a non-linear polyatomic molecule can rotate about all three axes and store real kinetic energy each time. That gives $3$ rotational degrees of freedom, on top of the $3$ translational ones, for a base total

f = 3\ (\text{translation}) + 3\ (\text{rotation}) = 6.

where: $f$ is the number of independent quadratic energy terms (degrees of freedom) available to store thermal energy in a single molecule, counting only translation and rotation for the moment — vibration is added separately below. By the law of equipartition of energy, each of these $f$ "slots" carries, on average, the same share $\tfrac12 k_BT$ of energy per molecule (or $\tfrac12 RT$ per mole). Total internal energy of one mole of an ideal non-linear polyatomic gas (vibration switched off) is therefore

U = \frac{f}{2}RT = \frac{6}{2}RT = 3RT.

where: $U$ is the internal energy of one mole of gas, $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ is the universal gas constant, and $T$ is the absolute temperature in kelvin. The physical reading is simple: a hotter polyatomic gas is one whose molecules are, on average, translating faster and tumbling faster about all three axes — twice as many rotational "storage bins" as a diatomic molecule, so more energy is needed to raise its temperature by the same amount.

Figure

A bar chart comparing degrees of freedom for monatomic diatomic and non linear polyatomic molecules, each bar split into a translational block of three and a rotational block, showing three for monatomic, five for diatomic, and six for non linear polyatomic

From here, the specific heats follow purely from calculus and Mayer's relation, which is derived exactly the same way as for monatomic and diatomic gases and is not repeated here — it is quoted as an already-established chapter result. What is new for polyatomic molecules is the possibility of vibration: the bonds between atoms can stretch and bend like tiny springs, and each independent way a molecule can vibrate is a vibrational mode. A vibrating bond stores energy in two forms at once — kinetic energy of the moving atoms and potential energy of the stretched "spring" — so each active vibrational mode contributes two quadratic terms, not one, and therefore adds $2$ to $f$ , not $1$ . Whether a given vibrational mode is thermally "active" at a given temperature is a quantum-mechanical question outside this chapter's scope; CBSE only asks you to apply the $+2$ rule once told a mode is active.

Derivation AMolar specific heats of a non-linear polyatomic gas from equipartition

The goal of this derivation is to obtain $C_v$ , $C_p$ and $\gamma$ for one mole of an ideal, non-linear polyatomic gas (three atoms or more, not all on one line, vibration switched off for now) starting only from the degrees-of-freedom count $f=6$ and the equipartition value $\tfrac12 RT$ per mole per degree of freedom, then generalise the result to include $f'$ active vibrational modes. Every step is algebra — differentiation of a linear function of $T$ , and the already-established Mayer relation — with no calculus beyond a first derivative and no appeal to microscopic collision dynamics.

Step 1. Count degrees of freedom for a non-linear polyatomic molecule. As shown in the physical picture above, translation contributes $3$ and rotation about three independent axes contributes another $3$ , giving a base $f = 6$ , with vibration assumed inactive for this first pass.

Step 2. Write the internal energy of one mole. By equipartition, each of the $f$ quadratic energy terms per molecule carries mean energy $\tfrac12 k_BT$ ; multiplying by Avogadro's number $N_A$ gives $\tfrac12 RT$ per mole per degree of freedom (since $N_Ak_B = R$ ). Summing over all $f=6$ degrees of freedom, $U = \dfrac{f}{2}RT = \dfrac{6}{2}RT = 3RT$ .

Step 3. Differentiate to get $C_v$ . At constant volume, no work is done by the gas, so all supplied heat raises internal energy: $C_v = \left(\dfrac{dU}{dT}\right)_V = \dfrac{d}{dT}(3RT) = 3R$ . Numerically, $C_v = 3 \times 8.314 = 24.94\ \text{J/mol}\,\cdot\,\text{K}$ .

Step 4. Apply Mayer's relation to get $C_p$ . For one mole of an ideal gas, $C_p - C_v = R$ (already established earlier in the chapter from $PV=RT$ ), so $C_p = C_v + R = 3R + R = 4R = 4\times8.314 = 33.26\ \text{J/mol}\,\cdot\,\text{K}$ .

Step 5. Form the ratio $\gamma$ . $\gamma = \dfrac{C_p}{C_v} = \dfrac{4R}{3R} = \dfrac{4}{3} \approx 1.33$ . This is the smallest $\gamma$ among the three gas families covered in the chapter, consistent with the ordering $f_{\text{mono}}=3 < f_{\text{dia}}=5 < f_{\text{poly}}=6$ : more degrees of freedom spread the same heat input over more storage bins, so temperature (and hence pressure, at fixed volume) rises more sluggishly, and $C_p/C_v$ sits closer to $1$ .

Step 6. Generalise to include active vibration. If $f'$ vibrational modes are thermally active, each adds $2$ to $f$ (kinetic $+$ potential per mode), so the total is $f_{\text{total}} = 6 + 2f'$ , and the same three steps give $C_v = \left(3+f'\right)R$ , $C_p = C_v + R = \left(4+f'\right)R$ , and $\gamma = \dfrac{4+f'}{3+f'}$ . As $f'$ grows, $\gamma \to 1$ from above — the general chapter trend that more active degrees of freedom always push $\gamma$ down, never up. Result: for a non-linear polyatomic gas with no active vibration, $C_v = 3R \approx 24.9\ \text{J/mol}\,\cdot\,\text{K}$ , $C_p = 4R \approx 33.3\ \text{J/mol}\,\cdot\,\text{K}$ , $\gamma = 4/3 \approx 1.33$ ; with $f'$ vibrational modes active, $C_v = (3+f')R$ , $C_p=(4+f')R$ , $\gamma = (4+f')/(3+f')$ .

Worked examples

Board level — heat required to warm a non-linear polyatomic gas at constant volume

Find the heat needed to raise the temperature of $2\ \text{mol}$ of a non-linear polyatomic gas (treat it as ammonia-like, with vibration inactive) from $300\ \text{K}$ to $330\ \text{K}$ at constant volume.

Step 1. Identify the correct molar specific heat. For a non-linear polyatomic gas with vibration inactive, $f=6$ , so $C_v = 3R = 3\times8.314 = 24.94\ \text{J/mol}\,\cdot\,\text{K}$ .

Step 2. Apply $\Delta Q = nC_v\Delta T$ at constant volume, since no work is done and all heat goes into raising internal energy. Here $n=2\ \text{mol}$ and $\Delta T = 330-300=30\ \text{K}$ .

Step 3. Substitute and compute. $\Delta Q = 2\times24.94\times30 = 1496.4\ \text{J}$ .

Answer: about $1496\ \text{J}$ of heat is needed to raise $2\ \text{mol}$ of this non-linear polyatomic gas by $30\ \text{K}$ at constant volume.

Competitive level — comparing gamma with one active vibrational mode, and the corresponding constant-pressure heat

A polyatomic gas has $C_v = 4R$ (i.e. one vibrational mode is active in addition to the base $f=6$ , giving $f_{\text{total}}=8$ ). Find $C_p$ and $\gamma$ for this gas, compare $\gamma$ to the vibration-free polyatomic value, and find the heat needed to raise $2\ \text{mol}$ of this gas from $300\ \text{K}$ to $330\ \text{K}$ at constant pressure.

Step 1. Apply Mayer's relation to get $C_p$ . $C_p = C_v + R = 4R + R = 5R = 5\times8.314 = 41.57\ \text{J/mol}\,\cdot\,\text{K}$ .

Step 2. Form $\gamma$ . $\gamma = \dfrac{C_p}{C_v} = \dfrac{5R}{4R} = \dfrac{5}{4} = 1.25$ .

Step 3. Compare to the vibration-free case. The vibration-free non-linear polyatomic gas has $\gamma = 4/3 \approx 1.33$ ; this vibrating gas has the lower value $\gamma = 1.25$ , because the extra vibrational mode adds $2$ more degrees of freedom (one kinetic, one potential) over which the same heat input must spread, exactly the general trend that more active degrees of freedom push $\gamma$ toward $1$ .

Step 4. Compute the constant-pressure heat. At constant pressure, $\Delta Q = nC_p\Delta T = 2\times41.57\times30$ .

Step 5. Evaluate. $\Delta Q = 2\times41.57\times30 = 2494.2\ \text{J}$ .

**Answer: $C_p = 5R \approx 41.57\ \text{J/mol}\,\cdot\,\text{K}$ , $\gamma = 5/4 = 1.25$ (lower than the vibration-free $4/3 \approx 1.33$ ), and raising $2\ \text{mol}$ of this vibrating polyatomic gas by $30\ \text{K}$ at constant pressure needs $2494.2\ \text{J}$ — noticeably more than the $1496.4\ \text{J}$ needed at constant volume in the previous example, both because $C_p > C_v$ and because this gas has one extra active mode. Timing note: this is a two-formula chain — Mayer's relation for $C_p$ , then either $\gamma = C_p/C_v$ or $\Delta Q = nC_p\Delta T$ — solvable in well under two minutes once $C_v$ is given. The only trap worth double-checking is which molar heat the problem wants: constant-volume heating uses $C_v$ , constant-pressure heating uses $C_p$ , and swapping the two is the single commonest slip under time pressure.*

Common traps

Assuming every polyatomic molecule has $f=6$ regardless of shape. This base count is only for non-linear polyatomic molecules. A linear polyatomic molecule, such as $\text{CO}_2$ , behaves rotationally like a diatomic molecule: rotation about the linear molecular axis stores negligible energy (just as with a diatomic bond axis), so it has only $2$ rotational degrees of freedom and a base $f=5$ , not $6$ . CBSE numericals almost always specify "non-linear" explicitly for the $f=6$ formula to apply.
Forgetting that each vibrational mode contributes $2$ to $f$ , not $1$ . Counting only the kinetic part and forgetting the potential-energy part of the vibrating "spring" halves the vibrational contribution to $C_v$ and gives a wrong $\gamma$ .
Believing $\gamma$ increases as molecules get more complex. It is the reverse: more atoms mean more degrees of freedom, and more degrees of freedom always push $\gamma$ down, toward $1$ — never up.
Reaching for the monatomic or diatomic formulas out of habit. Always re-derive $C_v$ , $C_p$ , $\gamma$ from the correct $f$ stated (or implied) for the molecule type in the problem, rather than recalling a memorised number that belonged to a different gas.
Mixing up $C_v$ and $C_p$ in the heat formula. Constant-volume heating uses $\Delta Q = nC_v\Delta T$ ; constant-pressure heating uses $\Delta Q = nC_p\Delta T$ — the two answers differ by the factor $C_p/C_v = \gamma$ , and picking the wrong one is a very common mark loss.

You should be able to…

Count degrees of freedom for a non-linear polyatomic molecule, arriving at $f=6$ before vibration.
Distinguish linear from non-linear polyatomic molecules and state why linear ones (like $\text{CO}_2$ ) use $f=5$ , not $f=6$ .
Derive $C_v = 3R$ , $C_p = 4R$ , $\gamma = 4/3$ for a non-linear polyatomic gas with vibration inactive, starting from equipartition.
Extend the formulas to include $f'$ active vibrational modes: $C_v=(3+f')R$ , $C_p=(4+f')R$ , $\gamma=(4+f')/(3+f')$ .
Rank monatomic, diatomic and non-linear polyatomic gases by $\gamma$ and explain the trend using degrees of freedom.
Solve heat-capacity numericals for polyatomic gases at constant volume and constant pressure, choosing $C_v$ or $C_p$ correctly.

Ask the tutor: Want a worked comparison table of monatomic, diatomic and polyatomic specific heats side by side, or extra practice numericals mixing linear and non-linear polyatomic gases?

Why does a non-linear molecule get three rotational degrees of freedom but a diatomic molecule only gets two?

It comes down to moment of inertia. A diatomic molecule's mass sits almost entirely along the line joining the two atoms, so spinning it about that line takes negligible energy — that axis is effectively "free," contributing nothing measurable, leaving only the two axes perpendicular to the bond as real rotational degrees of freedom. A non-linear molecule like water or ammonia has no such special axis: its atoms are spread out in three dimensions, so it has three genuinely different, non-negligible moments of inertia, one for each perpendicular axis through its centre of mass. All three store real rotational kinetic energy, so all three count.

Is carbon dioxide treated as f=5 or f=6 in CBSE numericals?

Carbon dioxide is a linear polyatomic molecule — oxygen, carbon and oxygen all sit on one straight line — so it behaves rotationally like a diatomic gas: $f=5$ (three translational, two rotational), not the $f=6$ used for non-linear molecules like water or ammonia. If a CBSE question gives you carbon dioxide without extra qualification, use $f=5$ unless the problem explicitly tells you to treat it as having three active rotational axes (which would be unusual and non-standard).

Do we actually need to know which vibrational modes are active at room temperature?

No — CBSE Class 11 does not ask you to predict whether a given vibrational mode is thermally active at a given temperature (that depends on quantum energy-level spacing, a much later topic). What you are expected to do is apply the $+2$ rule correctly once told how many vibrational modes are active, computing the resulting $C_v$ , $C_p$ and $\gamma$ . In practice, most CBSE numericals either keep vibration switched off entirely, or state explicitly how many modes are active, exactly as in the two worked examples above.

Why does gamma matter so much if we mostly just compute Cv and Cp anyway?

Because $\gamma = C_p/C_v$ shows up again and again in later contexts — most immediately in describing how a gas behaves during compression or expansion (adiabatic processes are governed by $PV^\gamma = \text{constant}$ , a result quoted at higher levels, though the derivation itself is beyond Class 11's Kinetic Theory chapter). Within this chapter, $\gamma$ is also the fastest way to sanity-check a specific-heat answer: since $\gamma$ must always lie between $1$ and about $1.67$ , and must decrease as a molecule gets more complex (monatomic $5/3$ greater than diatomic $7/5$ greater than polyatomic $4/3$ or lower), a computed $\gamma$ outside that pattern is an immediate signal of an arithmetic slip.