# Energy and degrees of freedom

Keep around...$$k=\frac{R}{N_A}$$

### Heat capacity of an ideal gas

The kinetic theory implies, with very few assumptions, $$\frac{3}{2}kT = \frac{1}{2}m\overline{v^2}.$$

Assuming that all of the internal energy of our ideal gas consists of its kinetic energy, we can immediately write down an equation for $U$: $$U = N\frac{1}{2}m\overline{v^2}=N\(\frac{3}{2}kT\) =N\(\frac{3}{2}\frac{R}{N_A}T\) =\frac{3}{2}nRT.$$

For a reversible process with our ideal gas at constant volume: $$\begin{align}c_v & =\( \frac{\partial u}{\partial T} \)_v =\( \frac{\partial (U/n)}{\partial T} \)_v =\( \frac{\partial (3RT/2)}{\partial T} \)_v \\ & = \frac{3}{2}R.\end{align}$$

Using Mayer's equation for $c_P$: $$c_P = c_v+R=\(\frac{3}{2}+1\)R=\frac{5}{2}R.$$

The ratio $\gamma$ is: $$\gamma= \frac{c_P}{c_v} = \frac{5}{3}=1.67.$$

This matches up *amazingly well* for monatomic gases:

But works not as well for *diatomic* gases. As the number
of atoms / molecule increases, $\gamma$ gets even further from 1.67.

On the other hand, if you cool diatomic $H_2$ gas down, you find that its heat capacity $\to \frac 32R$ below about 50 K and thus $\gamma\to 1.67$ just like a monatomic gas!

## Degrees of freedom

The question with the Gibbs phase rule was...

How many numbers do we need to uniquely specify the [thermodynamic] state of a system?.

Now, with atoms we can ask a parallel question...

How many numbers do we need to uniquely specify the "state" of a single particle (atom)?

The answer is 6:

- 3 numbers, $x$, $y$, and $z$ to specify the position of the atom, and
- 3 numbers, $v_x=(dx/dt)$, $v_y$, and $v_z$ to specify the speed of the atoms.
- That is,
**6 degrees of freedom**.

[Why don't we worry about the accelerations??]

To the extent that we have an ideal gas, there are only accelerations during collisions, and "most of the time" the atoms are moving ballistically between collisions...

For the simple billiard balls of an ideal monatomic gas, **the energy depends only on the 3 speed numbers**, not on the 3 position numbers.

$\Rightarrow$ There are 3 **"energy-related degrees of freedom"**.

If you substitute the expression $v^2=v_x^2+v_y^2+v_z^2$ into the integrals we used to find average values of things, you can readily convince yourself that $$\frac{1}{2}m\overline{v^2}= \frac{1}{2}m\(\overline{v_x^2} +\overline{v_y^2}+\overline{v_z^2}\).$$

Since the molecules are equally likely to be moving in any direction, the magnitudes are equal $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$. This means that the energy associated with each component of velocity is the same: 1/3 for each component of the overall kinetic energy... $$\frac{1}{2}m\overline{v_x^2}=\frac{1}{3}\frac{3}{2}kT=\frac{1}{2}kT.$$

## Equipartition

The **equipartition theorem** generalizes what we *know* happens with monatomic gases to those with more than one atom per molecule. It claims **there is an energy $\frac{1}{2}kT$** associated with each **energy degree of freedom $f$**. The internal energy per particle is related to the **number of energy degrees of freedom $f$ per particle** as:

$$U=Nf\frac{1}{2}kT =\frac{f}{2}nRT.$$

The heat capacity dependence on $f$ is... $$\begin{align}c_v & =\( \frac{\partial u}{\partial T} \)_v =\( \frac{\partial (fRT/2)}{\partial T} \)_v \\ & = \frac{f}{2}R.\end{align}$$

And $c_P$ is: $$c_P = c_v+R=\( \frac{f}{2}+1 \)R=\( \frac{f+2}{2} \) R.$$

The ratio $\gamma$ is: $$\gamma= \frac{c_P}{c_v} = \frac{f+2}{f}.$$

### Diatomic gases

Counting the energy-related degrees of freedom of a two-atom gas:

**linear speed**:
$v_x$, $v_y$, and $v_z$ specify the speed of the center of mass of the two atoms.

$\Rightarrow 3$.

**spinning**: If we use spherical polar coordinates $r$, $\theta$, and $\phi$ to track the position of one of the two atoms with respect to the other, then the time rates of change $\dot{\phi}$ and $\dot{\theta}$ keep track of the rotation of one atom about the other.

$\Rightarrow 2$.

**harmonic oscillator (spring)**: The third spherical coordinate is $r$. Certainly $\dot{r}$ is related to the relative motion of the two atoms. But now, for the first time, there is also energy (elastic / potential energy) related to the relative separation $r$ of the two atoms.

$\Rightarrow 2$.

We have $f=3+2+2=7$, so we'd expect... $$\gamma=\frac{9}{7} = 1.29.$$

The actual value (at room temperature) for the diatomic gases is closer to $1.4=7/5$ which would imply $f=5$ if equipartition is true.

But there are hints at several thousand degrees of the heat capacity increasing again...

The full explanation of the temperature dependence emerges from a quantum mechanical treatment of spinning and oscillating systems. Here's the cartoon version:

- The energy levels for translational motion in a large container are very closely spaced. They approximate a continuum.
- Rotational energy levels are further apart, and vibrational energy levels are even further apart.
- At sufficiently low temperatures there is not enough thermal energy around to excite any of the rotational, let alone vibrational degrees of freedom.
- See: UCDavis - ChemWiki.

Explaining the temperature dependence of the heat capacities of gases is one of the **successes of quantum mechanics** that Paul Dirac cites to convince folks of the necessity for and usefulness of the theory.

*Are there yet more degrees of freedom that are associated with the "state" of the atom? What are their energies like compared to rotational or vibrational degrees of freedom?*

I'm thinking of the energy levels associated with electrons in 1s, 2p, 3d, etc orbitals

Finally, in the limit of lots of atoms per molecule, with lots of internal degrees, $f\gg 1$. If equipartition is true, $\gamma=(f-2)/f\approx 1$, when $f$ is large. So, we'd expect values for gases in the range: $$1.67 \gt \gamma \gt 1.0.$$

[Point out that heat capacities are not (yet) temperature dependent]