Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 19: Statistical Physics VI: Energies</b><br>
Tue 24 Mar 2026 12:00
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== Average Kinetic Energy ==

In 3D, we derived last lecture: <div class="pn-import-equation"><nowiki>$$\langle E_k\rangle = \frac{3}{2} k_B T$$</nowiki></div>

Each component contributes equally, so: <div class="pn-import-equation"><nowiki>$$E_{kx} = E_{ky} = E_{kz} = \frac{1}{2} k_B T$$</nowiki></div>

Remember from the second module of the course we have: <div class="pn-import-equation"><nowiki>$$c = \frac{dQ}{dT}$$</nowiki></div>

And the first law of thermodynamics: <div class="pn-import-equation"><nowiki>$$dU = dQ - pdV \implies dQ = dU + pdV$$</nowiki></div>

At a constant pressure or volume: <div class="pn-import-equation"><nowiki>$$C_v = \left(\frac{\partial Q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V$$</nowiki></div>

We can define an additional value called Enthalpy (free energy): <div class="pn-import-equation"><nowiki>$$dH = dU + P dV$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$C_p = \left(\frac{\partial Q}{\partial T}\right)_P = \left(\frac{\partial H}{\partial T}\right)_V$$</nowiki></div>

For ideal gases, internal energy comes solely from kinetic energy. As we now have average kinetic energy as a function of <nowiki>$T$</nowiki>, we can now say (for per-atom and per-mole cases respectively): <div class="pn-import-equation"><nowiki>$$C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{3}{2} k_B$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{3}{2} R$$</nowiki></div>

This works perfectly for monatomic gases, where Mayer’s relation also lets us define <nowiki>$C_P$</nowiki>:

[[File:TM-Y1S2-lec19-01.png|thumb|center|800px]]

But evidently breaks for diatomic molecules… They seem to have <nowiki>$C_V = 5R/2$</nowiki>? We’ll investigate why later.

== Law of Equipartition of Energy ==

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The Law of Equipartition says that any substance in equilibrium at temperature <nowiki>$T$</nowiki> has an average energy of <nowiki>$k_B T / 2$</nowiki> per <b>degree of freedom</b>.

A <b>degree of freedom</b> is any term in the expression for the energy of the system which has a squared position or squared velocity term.

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In 1D, our expression for kinetic energy is: <div class="pn-import-equation"><nowiki>$$E_k = \frac{1}{2} mv_x^2 = \frac{1}{2}k_B T \implies C_V = \frac{1}{2}R$$</nowiki></div>

And in 3D: <div class="pn-import-equation"><nowiki>$$E_k = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2 = \frac{3}{2}k_B T \implies C_V = \frac{3}{2}R$$</nowiki></div>

=== Example ===

[[File:TM-Y1S2-lec19-02.png|thumb|center|600px]]

Consider a gas in a harmonic potential, acting as an oscillator in 1D: <div class="pn-import-equation"><nowiki>$$E(V_x, x) = \frac{1}{2}m v_x^2 + \frac{1}{2}kx^2 = k_BT$$</nowiki></div>

Or in a linear potential: <div class="pn-import-equation"><nowiki>$$E(v_x, x) = \frac{1}{2}m v_x^2 + Kx = \frac{1}{2}k_B Ts$$</nowiki></div>

=== Dulong-Petit Heat Capacity of a Solid ===

We model the inter-atomic forces between atoms as springs with spring constant <nowiki>$K$</nowiki>. Each spring has vibrational energy given by: <div class="pn-import-equation"><nowiki>$$E = \frac{1}{2}kx^2 = \frac{1}{2}mv^2 = k_B T$$</nowiki></div>

If we assume that the substance forms a perfect cubic lattice where every atom is connected to six other atoms, each spring is shared by two atoms so we say: <div class="pn-import-equation"><nowiki>$$E = 3k_B T \implies c_V = 3R \approx 24.9 \mathrm{mol^{-1} K^{-1}}$$</nowiki></div>

This ends up being a surprisingly good fit for pure elements:

[[File:TM-Y1S2-lec19-03.png|thumb|center|800px]]

== Diatomic Molecules ==

Diatomic molecules can be modelled as being connected by a single spring: <div class="pn-import-equation"><nowiki>$$E_\text{vibrational} = \frac{1}{2}kx^2 + \frac{1}{2}m v_x^2 \tag{2 DoF}$$</nowiki></div>

However, they are also free to move around the volume in which they occupy, this gives some extra translational energy: <div class="pn-import-equation"><nowiki>$$E_\text{KE} = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2 \tag{3 DoF}$$</nowiki></div>

Finally, they’re also able to rotate around the axis joining them (perpendicularly): <div class="pn-import-equation"><nowiki>$$E_\text{rot} = \frac{1}{2}I \omega^2_y + \frac{1}{2}I \omega^2_z \tag{2 DoF}$$</nowiki></div>

In theory this means that a diatomic molecule has 7 degrees of freedom. We saw earlier however that the values for <nowiki>$C_V$</nowiki> for a diatomic molecule is <nowiki>$C_V = 5R/2$</nowiki>, only implying 5 DoF? Lets look at some actual data:

[[File:TM-Y1S2-lec19-04.png|thumb|center|600px]]

Our monatomic gases behave well, with values close to <nowiki>$3R/2$</nowiki>, but our diatomic gases seem to be more varied, with Chlorine having a much higher value between <nowiki>$5R/2$</nowiki> and very roughly <nowiki>$7R/2$</nowiki>?

We can’t say that our assumption of 7DoF is wrong, because Chlorine sits nicely between them. It means that this cannot, for example, simply lack rotational energy.

The states where a molecule has vibrational energy occurs at larger energies. In order to get a reasonable probability of measuring a state with a certain energy from the Boltzmann factor:

<div class="pn-import-equation"><nowiki>$$\Pr(E_i) \propto \exp \left(- \frac{E_i}{k_B T}\right)$$</nowiki></div>

We require <nowiki>$k_BT \not\ll E_i$</nowiki>. This gives a transition, where for some temperatures there is some intermediary probability where there’s a good chance of measuring particles with and without sufficient energy for the final degrees of freedom.

This gives a plot like this:

[[File:TM-Y1S2-lec19-05.png|thumb|center|600px]]

The translational component is always present, as our gas molecules are always present. The vibrational and rotational modes only appear at certain levels where there is sufficient energy to “unlock” them.

The degrees of freedom can be considered fully excited if: <div class="pn-import-equation"><nowiki>$$k_BT \gg E_1 - E_0$$</nowiki></div>

And if: <div class="pn-import-equation"><nowiki>$$k_BT \ll E_1 - E_0$$</nowiki></div> There is insufficient energy to expect to encounter any molecules with sufficient energy for the motion associated with the degree of freedom, and it is said to be frozen out.

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Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc