Why Helium and Oxygen Need Different Heat to Warm Up

Give the same heat to one mole of helium and one mole of oxygen, and the helium gets hotter. Oxygen molecules can spin, so part of the energy goes into rotation instead of raising temperature.

Counting the independent ways a molecule can store energy explains specific heats, the ratio $\gamma$, and more.

This part covers degrees of freedom, the law of equipartition, predicting $C_p$, $C_v$ and $\gamma$, and mean free path. Take $R=8.314$ J/mol K.

**Degrees of freedom are the independent ways a molecule can move and store energy; a monatomic molecule has $3$ translational, a rigid diatomic molecule has $3$ translational plus $2$ rotational, and a rigid non-linear polyatomic molecule has $3$ translational plus $3$ rotational.

- Monatomic** (He, Ne, Ar): $f=3$
- Rigid diatomic (O${}_2$, N${}_2$) at ordinary temperatures: $f=5$ — rotation about the bond axis stores negligible energy
- Diatomic with vibration at high temperatures: $f=7$ — the vibration adds kinetic and potential energy terms
- Rigid non-linear polyatomic (such as H${}_2$O, CH${}_4$): $f=6$

Worked example — counting with constraints. Two atoms moving freely have $2\times 3=6$ coordinates. A fixed bond length removes one, leaving $6-1=5$ — matching the rigid diatomic count.

An everyday example. A cricket ball bowled with spin both moves forward and rotates, storing energy in translation and rotation at once.

The substance. A vibrational mode counts twice in energy terms, because it has both kinetic and potential energy.

**The law of equipartition states that in thermal equilibrium, each degree of freedom that stores energy as a squared term gets an average energy of $\frac{1}{2}{k}_{B}T$ per molecule, so one mole of an ideal gas with $f$ such terms has internal energy $U=\frac{f}{2}RT$.

- Monatomic**: $U=\frac{3}{2}RT$ per mole
- Rigid diatomic: $U=\frac{5}{2}RT$ per mole
- Vibrating diatomic: $U=\frac{7}{2}RT$ per mole

Worked example 1 — oxygen. $2$ mol of oxygen at $300$ K:

$$U=2\times \frac{5}{2}\times 8.314\times 300\approx 1.25\times 10^4\text{ J}$$

Worked example 2 — helium. $2$ mol of helium at $300$ K:

$$U=2\times \frac{3}{2}\times 8.314\times 300\approx 7.48\times 10^3\text{ J}$$

Worked example 3 — per molecule. One oxygen molecule at $300$ K has, on average,

$$\frac{5}{2}{k}_{B}T=2.5\times 1.38\times 10^{-23}\times 300\approx 1.04\times 10^{-20}\text{ J}$$

An everyday example. Warming the air in a room raises not only the speed of its nitrogen and oxygen molecules but also their spinning.

The substance. Equipartition is a classical result — at low temperatures, rotational and vibrational modes stop taking their share.

**Since $U=\frac{f}{2}RT$ per mole, the molar specific heats are $C_v=\frac{f}{2}R$ and $C_p=C_v+R$, giving $\gamma =\frac{C_p}{C_v}=1+\frac{2}{f}$.

Predictions:

- Monatomic** ($f=3$): $C_v=\frac{3}{2}R\approx 12.5$, $C_p\approx 20.8$ J/mol K, $\gamma =\frac{5}{3}\approx 1.67$
- Rigid diatomic ($f=5$): $C_v\approx 20.8$, $C_p\approx 29.1$ J/mol K, $\gamma =1.4$
- Vibrating diatomic ($f=7$): $C_v\approx 29.1$, $C_p\approx 37.4$ J/mol K, $\gamma \approx 1.29$
- Rigid non-linear polyatomic ($f=6$): $C_v=3R\approx 24.9$, $C_p=4R\approx 33.3$ J/mol K, $\gamma \approx 1.33$

Worked example — a mixture. $1$ mol of helium is mixed with $1$ mol of oxygen.

$$C_v=\frac{1\times \frac{3}{2}R+1\times \frac{5}{2}R}{2}=2R{C}_p=3R\gamma =\frac{3}{2}=1.5$$

An everyday example. Sound travels faster in helium than in air, partly because helium's larger $\gamma$ makes it springier.

The substance. **A measured value of $\gamma$ reveals the shape of a gas's molecules** — $1.67$ means single atoms, $1.4$ means two-atom molecules.

**Mean free path is the average distance a molecule travels between collisions, $\lambda =\frac{1}{\sqrt{2}\pi d^{2}n}$, so it is shorter for bigger molecules and for denser gases.**

Here $d$ is the molecular diameter and $n$ the number of molecules per unit volume. The number density links to Avogadro's number through $n=\frac{N_A}{V_m}$, where $V_m$ is the molar volume.

**Worked example 1 — number density at $0$ °C and $1$ atm.** With $V_m=22.4$ L:

$$n=\frac{6.022\times 10^{23}}{0.0224}\approx 2.69\times 10^{25}{\text{ m}}^{-3}$$

Worked example 2 — mean free path in air. With $d=3\times 10^{-10}$ m:

$$\lambda =\frac{1}{\sqrt{2}\times \pi \times (3\times 10^{-10}{)}^2\times 2.69\times 10^{25}}\approx 9.3\times 10^{-8}\text{ m}$$

At about $500$ m/s, a molecule collides roughly $\frac{500}{9.3\times 10^{-8}}\approx 5.4\times 10^9$ times a second.

An everyday example. Perfume sprayed in one corner takes a while to reach the far side of a room, since each molecule's path is broken by constant collisions.

The substance. Halving the pressure doubles the mean free path, which is why molecules in a high vacuum can cross a whole chamber without colliding.