Average dipole energy of HCl molecule

HCl is a polar molecule with a permanent dipole moment $p_0$. In a field $E$, a dipole at angle $\theta$ to the field has potential energy

$$U = -p_0E\cos\theta$$

Thermal motion randomizes the dipoles, so the number of dipoles in an energy state follows the Boltzmann distribution, $\propto e^{-U/kT} = e^{p_0E\cos\theta/kT}$.

Average of $\cos\theta$

The number of dipoles in the solid angle between $\theta$ and $\theta + d\theta$ is proportional to $2\pi\sin\theta\,d\theta\, e^{x\cos\theta}$, where $x = p_0E/kT$.

$$\langle\cos\theta\rangle = \frac{\int_0^{\pi}\cos\theta\, e^{x\cos\theta}\sin\theta\,d\theta}{\int_0^{\pi} e^{x\cos\theta}\sin\theta\,d\theta}$$

Put $u = \cos\theta$:

$$\langle\cos\theta\rangle = \frac{\int_{-1}^{1}u\,e^{xu}du}{\int_{-1}^{1}e^{xu}du} = \coth x - \frac{1}{x} = L(x)$$

$L(x)$ is the Langevin function.

Small-field approximation

At normal fields $p_0E \ll kT$ ($x \ll 1$), so $L(x) \approx x/3$:

$$\langle\cos\theta\rangle = \frac{p_0E}{3kT}$$

Average dipole energy

$$\begin{aligned} \langle U\rangle &= -p_0E\langle\cos\theta\rangle \\ &= -p_0E\cdot\frac{p_0E}{3kT} \\ &= -\frac{p_0^2E^2}{3kT} \end{aligned}$$

Related results

• Average dipole moment along the field: $p_{av} = p_0\langle\cos\theta\rangle = \dfrac{p_0^2E}{3kT}$
• Orientational polarizability: $\alpha_d = \dfrac{p_0^2}{3kT}$

The average energy is negative (alignment lowers energy), grows as $E^2$, and falls as $1/T$ because heat opposes alignment.