Degrees of freedom and specific heats

Heat the same amount of helium and nitrogen gas by the same temperature rise, and nitrogen soaks up noticeably more heat. Both are ideal gases obeying the same kinetic theory, so the difference cannot come from the gas law — it comes from how many independent ways each kind of molecule can carry energy. A point-like helium atom can only move in three directions. A dumbbell-shaped nitrogen molecule can also tumble end over end about two axes. Each such independent "way of storing energy" is called a degree of freedom, and the law of equipartition of energy says every one of them, on average, carries exactly the same share of the gas's thermal energy. Counting degrees of freedom for monatomic, diatomic and polyatomic gases, and turning that count into $C_v$ , $C_p$ and $\gamma=C_p/C_v$ , is the whole content of this section — and it explains, quantitatively, why different gases heat up differently.

The core insight

A gas molecule's degrees of freedom $f$ (3 for monatomic, 5 for diatomic, 6 for non-linear polyatomic) fix its specific heats through equipartition: $C_v=\tfrac{f}{2}R$ , $C_p=\tfrac{f+2}{2}R$ , and $\gamma=\tfrac{f+2}{f}$ .

The core idea

Degrees of freedom count the independent coordinates needed to fully describe a molecule's motion — equivalently, the independent ways it can hold kinetic energy.

Monatomic (He, Ne, Ar): a point mass, free only to translate along $x$ , $y$ , $z$ . $f=3$ , all translational.
Diatomic ( $\text{O}_2$ , $\text{N}_2$ ): 3 translational plus rotation about the two axes perpendicular to the bond (rotation about the bond axis itself is dropped — the moment of inertia there is negligible). $f=3+2=5$ . Vibration is left out at the temperatures this syllabus covers.
Polyatomic, non-linear ( $\text{H}_2\text{O}$ , $\text{NH}_3$ ): 3 translational plus 3 rotational, about three mutually perpendicular axes. $f=6$ .

The law of equipartition of energy (stated, not derived, per syllabus) says that in thermal equilibrium every degree of freedom carries the same average energy, $\tfrac12k_BT$ per molecule, i.e. $\tfrac12RT$ per mole. Summing over all $f$ degrees of freedom of one mole of gas gives the total internal energy

U=\frac{f}{2}RT,

where: $U$ is molar internal energy, $f$ the degrees of freedom per molecule, $R$ the gas constant, $T$ the absolute temperature. This matches the earlier kinetic-theory result $\tfrac32RT$ for translation alone when $f=3$ . Because internal energy of an ideal gas depends only on $T$ , heating at constant volume (no work done) puts every joule straight into $U$ , so $C_v=dU/dT=\tfrac{f}{2}R$ . Combined with $C_p-C_v=R$ from the first law, $C_p=\tfrac{f+2}{2}R$ and $\gamma=C_p/C_v=\tfrac{f+2}{f}$ .

Gas type	$f$	$C_v$	$C_p$	$\gamma$
Monatomic	3	$\tfrac32R$	$\tfrac52R$	$5/3\approx1.67$
Diatomic	5	$\tfrac52R$	$\tfrac72R$	$7/5=1.4$
Polyatomic (non-linear)	6	$3R$	$4R$	$4/3\approx1.33$

These predicted $\gamma$ values agree closely with experiment (helium $\approx1.66$ , nitrogen $\approx1.40$ ), which is the real justification for trusting the equipartition picture: it is not merely stated, it is checked against measured gas behaviour. Notice the pattern across the table: as $f$ climbs from 3 to 5 to 6, both $C_v$ and $C_p$ climb steadily (more degrees of freedom means more places to park absorbed heat), while $\gamma$ falls steadily toward 1 (a larger denominator shrinks the fraction $(f+2)/f$ less dramatically than it grows). This is why $\gamma$ , though it looks like an afterthought, is often the quantity measured directly in the lab (from the speed of sound in a gas, for instance) and then used to identify whether an unknown gas is monatomic, diatomic or polyatomic.

Figure

Bar chart comparing degrees of freedom and gamma for monatomic diatomic and polyatomic gases

Derivation of Cv and Cp from equipartition

Derivation AMolar specific heats from degrees of freedom

The goal is to turn the equipartition statement, that each of a molecule's $f$ degrees of freedom carries average energy $\tfrac12RT$ per mole, into the two measurable heat capacities $C_v$ and $C_p$ and their ratio $\gamma$ , using only the definition of heat capacity and the first-law relation $C_p-C_v=R$ that already holds for any ideal gas.

Step 1. Total internal energy. Summing $\tfrac12RT$ over all $f$ degrees of freedom of one mole gives $U=\dfrac{f}{2}RT$ . For an ideal gas $U$ depends on $T$ alone, not on volume or pressure.

Step 2. Heat capacity at constant volume. At constant volume no work is done ( $\Delta V=0$ ), so all supplied heat raises internal energy: $C_v=\left(\dfrac{dU}{dT}\right)_V=\dfrac{f}{2}R$ .

Step 3. Heat capacity at constant pressure. At constant pressure some heat goes into expansion work, so more heat is needed for the same $\Delta T$ ; the exact excess is fixed by the first law as $C_p-C_v=R$ . Hence $C_p=C_v+R=\dfrac{f}{2}R+R=\dfrac{f+2}{2}R$ .

Step 4. Ratio of specific heats. Dividing, $\gamma=\dfrac{C_p}{C_v}=\dfrac{f+2}{f}$ . Larger $f$ makes $\gamma$ smaller and closer to 1, since a larger fraction of absorbed heat is "used up" spreading over more degrees of freedom rather than raising translational (thermometer-measured) energy alone.

Worked example

Find $C_v$ , $C_p$ and $\gamma$ for oxygen ( $\text{O}_2$ , diatomic), and use $C_v$ to find the heat needed to raise 2 moles of oxygen by $50\ \text{K}$ at constant volume. Take $R=8.314\ \text{J/mol}\,\cdot\,\text{K}$ .

Step 1. Oxygen is diatomic, so $f=5$ .

Step 2. $C_v=\dfrac{f}{2}R=\dfrac52(8.314)=20.785\ \text{J/mol}\,\cdot\,\text{K}$ ; $C_p=C_v+R=20.785+8.314=29.099\ \text{J/mol}\,\cdot\,\text{K}$ ; $\gamma=\dfrac{29.099}{20.785}=1.40$ .

Step 3. Heat at constant volume, $Q=nC_v\Delta T=(2)(20.785)(50)=2078.5\ \text{J}$ .

About $2.08\ \text{kJ}$ of heat is needed — noticeably more than the $\tfrac32R\Delta T$ per mole a monatomic gas like helium would need for the same $\Delta T$ , because oxygen has two extra (rotational) degrees of freedom absorbing energy alongside translation.

Competitive stretch

A monatomic gas and a diatomic gas, one mole each, are both heated at constant volume through the same $\Delta T=100\ \text{K}$ . Compare the heat each absorbs.

Step 1. Monatomic ( $f=3$ ): $Q_{\text{mono}}=C_v\Delta T=\tfrac32R\Delta T=\tfrac32(8.314)(100)=1247.1\ \text{J}$ .

Step 2. Diatomic ( $f=5$ ): $Q_{\text{dia}}=C_v\Delta T=\tfrac52R\Delta T=\tfrac52(8.314)(100)=2078.5\ \text{J}$ .

Step 3. Ratio: $Q_{\text{dia}}/Q_{\text{mono}}=5/3\approx1.67$ . The diatomic gas needs $5/3$ times as much heat for the same temperature rise, since equipartition spreads the absorbed energy over its two extra rotational degrees of freedom, which the monatomic gas does not have.

Common traps

Counting rotation about a diatomic molecule's own bond axis as a degree of freedom — excluded, since the moment of inertia about that axis is negligible (essentially a point mass sitting on the axis).
Forgetting the "+2" when going from $C_v$ to $C_p$ : $C_p=C_v+R$ comes from the first law of thermodynamics, not from equipartition itself, so it must be added separately.
Treating $\gamma$ as a universal constant — it depends on gas type through $f$ ; identify monatomic, diatomic or polyatomic before quoting a value.
Trying to derive the equipartition law from scratch — the syllabus asks you to state and apply it, not prove it from more fundamental statistics.

Ask the tutor: unsure why the bond-axis rotation of a diatomic molecule is dropped from its degrees of freedom, or want $\gamma$ worked out for another gas — ask the AI tutor.

Why does equipartition give exactly half of RT per degree of freedom, and not some other fraction?

The syllabus asks you to state and use the law of equipartition of energy, not derive it — that derivation needs statistical mechanics beyond this course. You can, however, check it against something you already trust: purely translational kinetic energy per mole is $\tfrac32RT$ , matching three translational degrees of freedom each contributing $\tfrac12RT$ . That agreement between the kinetic-theory result you already derived and the equipartition prediction is exactly the justification you are expected to cite in an exam, without attempting the underlying statistical proof.

Does vibration ever count as a degree of freedom for diatomic gases like oxygen or nitrogen?

At the temperatures this syllabus deals with, no — vibrational modes need much more energy to activate than translation or rotation, so they stay effectively frozen out and are left out of $f$ . At very high temperatures a diatomic molecule can start vibrating and $C_v$ rises above $\tfrac52R$ , but that regime and the extra degrees of freedom it brings are beyond what you need here. Stick to $f=5$ for diatomic gases across the temperature ranges this course examines, and treat any deviation as an advanced footnote rather than something to build into your calculations.

Why does a gas with more degrees of freedom need more heat for the same temperature rise?

Temperature is a measure of translational kinetic energy only. When you heat a gas at constant volume, equipartition spreads the absorbed energy equally over all $f$ degrees of freedom, not just the three translational ones. A diatomic gas has two extra rotational degrees of freedom soaking up energy alongside translation, so for the same rise in translational energy (same $\Delta T$ , since that is what a thermometer reads) it must absorb more total heat than a monatomic gas — in the ratio $f_{\text{diatomic}}/f_{\text{mono}}=5/3$ , exactly the factor confirmed in the competitive-stretch calculation above.