The Clausius-Mossotti equation (Prob. 4.41) tells you how to calculatethe susceptibility of a nonpolar substance, in terms of the atomic polariz-ability. The Langevin equation tells you how to calculate the susceptibility of apolar substance, in terms of the permanent molecular dipole moment p. Here's howit goes:

(a) The energy of a dipole in an external field E is$\mathbf{u}\mathbf{=}\mathbf{-}\mathbf{p}·\mathbf{Ecos\theta }$

(Eq. 4.6), where$\mathbf{\theta }$ is the usual polar angle, if we orient the z axis along E.

Statistical mechanics says that for a material in equilibrium at absolute temperature

T, the probability of a given molecule having energy u is proportional to

the Boltzmann factor,

$\mathbf{exp}\left(-u/\mathrm{kT}\right)$

The average energy of the dipoles is therefore

$<u>\mathbf{=}\frac{\mathbf{\int }\mathbf{u}{\mathbf{e}}^{\mathbf{-}\left(u/kt\right)\mathbf{d\Omega }}}{\mathbf{\int }{\mathbf{e}}^{\mathbf{-}\left(u/kT\right)\mathbf{d\Omega }}}$

where $\mathbf{d\Omega }\mathbf{=}\mathbf{sin\theta d\theta d}\varphi$, and the integration is over all orientations $\left(\mathbf{\theta }\mathbf{:}\mathbf{0}\to \mathbf{\pi };\varphi :\mathbf{0}\to \mathbf{2}\mathbf{\pi }\right)$Use this to show that the polarization of a substance

containing N molecules per unit volume is

$\mathbf{P}\mathbf{=}\mathbf{Np}\left[\coth \left(\mathrm{pE}/\mathrm{kT}\right)-\left(\mathrm{kT}/\mathrm{pE}\right)\right]$ (4.73)

That's the Langevin formula. Sketch as a function of$\mathit{P}\mathit{E}\mathbf{/}\mathit{K}\mathit{T}$ .

(b) Notice that for large fields/low temperatures, virtually all the molecules arelined up, and the material is nonlinear. Ordinarily, however, kT is much greaterthan p E. Show that in this regime the material is linear, and calculate its susceptibility,in terms of N, p, T, and k. Compute the susceptibility of water at 20°C,and compare the experimental value in Table 4.2. (The dipole moment of wateris $\mathbf{6}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{30}}\mathbf{C}·\mathbf{m}$) This is rather far off, because we have again neglected thedistinction between E and Eelse· The agreement is better in low-density gases,for which the difference between E and Eelse is negligible. Try it for water vapor

at 100°C and 1 atm.

(a)

Consider the formula for the energy of the dipole in the external field as:

$\begin{array}{rcl}\mathbf{U}& \mathbf{=}& \mathbf{-}\overline{\mathbf{P}}·\overline{\mathbf{E}}\\ & \mathbf{=}& \mathbf{PEcos\theta }\\ & & \end{array}$

Here, P is the polarization energy and E is the electric intensity.

Consider the average energy for the electric dipole is:

Consider the expression for the resultant polarization as:

Write the average energy function from the Boltzmann distribution function as:

Determine the expression for the Polarization as:

Solve further as:

$P=Np\left(\coth \left(\frac{pE}{kT}\right)-\frac{kT}{pE}\right)$

Therefore, the expression for the polarization is$P=Np\left(\coth \left(\frac{pE}{kT}\right)-\frac{kT}{pE}\right)$

Consider the expression as:

$x=\frac{pE}{kT}$

Also,

$\frac{P}{N_p}=\coth \left(x\right)-\frac{1}{x}$

Consider for E to be very large and Tto be small solve as:

$$\begin{gathered} x>0 \\ \frac{1}{x}<0 \\ \coth \left(x\right)>1 \end{gathered}$$

Consider for $kT>>pE$solve as:

$\begin{array}{rcl}\coth \left(x\right)& =& \frac{e^x+e^{-x}}{e^x-e^{-x}}\\ & =& \frac{2\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\dots \right)}{2\left(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots \right)}\\ & & \end{array}$

Short the terms by the Taylor series as:

$\begin{array}{rcl}\coth \left(x\right)& =& \left(\frac{1}{x}\right)+\left(\frac{x}{3}\right)-\left(\frac{x^2}{45}\right)+\text{Higher}\text{terms}\\ & =& \frac{1}{x}+\frac{x}{3}\\ & & \backslash \end{array}$

Consider for $kT>>pE$solve as:

$\begin{array}{rcl}\frac{P}{N_P}& =& \left(\frac{N_P^2}{3kT}\right)E\\ \frac{P}{N_P}& =& {\epsilon }_0\left(\frac{N_P^2}{3{\epsilon }_{0}kT}\right)E\\ \frac{P}{N_P}& =& {\epsilon }_0\chi E\left\{\chi =\left(\frac{N_P^2}{3{\epsilon }_{0}kT}\right)\right\}\\ & & \end{array}$

Consider the temperature in Kelvin is 293 K

Consider the total number of molecules per unit volume is obtained as:

$\begin{array}{rcl}N& =& \left(6.0\times {10}^{23}\right)\left(\frac{1}{18}\right)\left({10}^4\right)\\ & =& 0.33\times {10}^{29}\\ & & \end{array}$

Consider from the equation of the susceptibility solve as:

$\begin{array}{rcl}{\chi }_w& =& \left(\frac{{\left(6.1\times {10}^{-30}\right)}^2}{3\left(8.85\times {10}^{-12}{C}^2/Nm\right)\left(1.38\times {10}^{-23}\right){\left(293\text{K}\right)}^2}\right)\\ & =& 12\end{array}$

Consider for the water vapor at 100 degree Celsius the temperature is 393 K.

Then, by the Charles law solve as:

$\begin{array}{rcl}{V}_{100}& =& \frac{V_{20}{T}_{100}}{T_{20}}\\ & =& \frac{\left(22.4\times {10}^{-3}{\text{m}}^3\right)\left(373\text{K}\right)}{293\text{K}}\\ & =& 2.85\times {10}^{-2}{\text{m}}^3\\ & & \end{array}$

Solve for the number of molecules per unit volume as:

$\begin{array}{rcl}N& =& \frac{6.023\times {10}^{23}\text{mol}}{2.85\times {10}^{-2}{\text{m}}^3}\\ & =& 2.11\times {10}^{25}\\ & & \end{array}$

Solve for the susceptibility of the water as:

$\begin{array}{rcl}{\chi }_w& =& \frac{\left(\left(2.11\times {10}^{25}\right){\left(6.1\times {10}^{-30}\right)}^2\right)}{3\left(8.85\times {10}^{-12}{c}^2/Nm\right)\left(1.38\times {10}^{-23}\right){\left(293\text{K}\right)}^2}\\ & =& 5.7\times {10}^{-3}\end{array}$

Therefore, the susceptibility is $\chi =\left(\frac{N_P^2}{3{\epsilon }_{0}kT}\right)$, for water in aqueous it is ${\chi }_w=12$and for gas it is $5.7\times {10}^{-3}$. The susceptibility for the liquid water is not in agreement with the expression for the gaseous state.