The Clausius-Mossotti equation (Prob. 4.41) tells you how to calculate the susceptibility of a nonpolar substance, in terms of the atomic polarizability α. The Langevin equation tells you how to calculate the susceptibility of a polar substance, in terms of the permanent molecular dipole momen

Question

The Clausius-Mossotti equation (Prob. 4.41) tells you how to calculate the susceptibility of a nonpolar substance, in terms of the atomic polarizability α. The Langevin equation tells you how to calculate the susceptibility of a polar substance, in terms of the permanent molecular dipole moment p. Here’s how it goes:

(a) The energy of a dipole in an external field E is u = -p · E = -pE cos θ (Eq. 4.6), where θ 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,

U = -p · E. (4.6)

exp(-u/kT ).

The average energy of the dipoles is therefore

[latex]=\frac{\int u e^{-(u / k T)} d \Omega}{\int e^{-(u / k T)} d \Omega}[/latex] ,

where [latex]d \Omega=\sin heta d heta d \phi[/latex], and the integration is over all orientations (θ :0 [latex] ightarrow \pi ; \phi: 0 ightarrow 2 \pi [/latex]). Use this to show that the polarization of a substancecontaining N molecules per unit volume is

[latex]P=N p[\operatorname{coth}(p E / k T)-(k T / p E)][/latex] (4.73)

That’s the Langevin formula. Sketch P/Np as a function of pE/kT .

(b) Notice that for large fields/low temperatures, virtually all the molecules are lined up, and the material is nonlinear. Ordinarily, however, kT is much greater than pE. Show that in this régime 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 water is [latex]6.1 imes 10^{-30} C \cdot m[/latex] .) This is rather far off, because we have again neglected the distinction between E and [latex]E _{ ext {else }}[/latex] . The agreement is better in low-density gases, for which the difference between E and [latex]E _{ ext {else }}[/latex] is negligible. Try it for water vapor at 100°C and 1 atm.

Material	Dielectric Constant	Material	Dielectric Constant
Vacuum	1	Benzene	2.28
Helium	1.000065	Diamond	5.7-5.9
Neon	1.00013	Salt	5.9
Hydrogen [latex]\left( H _{2} ight)[/latex]	1.000254	Silicon	11.7
Argon	1.000517	Methanol	33.0
Air (dry)	1.000536	Water	80.1
Nitrogen [latex]\left( N _{2} ight)[/latex]	1.000548	Ice (-30° C)	104
Water vapor (100° C)	1.00589	KTaNb[latex]O _{3}\left(0^{\circ} C ight)[/latex]	34,000

Table 4.2

Accepted Answer

(a) Doing the (trivial) [latex]\phi[/latex] integral, and changing the remaining integration variable from [latex] heta ext { to } u(d u=p E \sin heta d heta)[/latex] ,

[latex]\langle u angle=\frac{\int_{-p E}^{p E} u e^{-u / k T} d u}{\int_{-p E}^{p E} e^{-u / k T} d u}=\frac{\left.(k T)^{2} e^{-u / k T}[-(u / k T)-1] ight|_{-p E} ^{p E}}{-\left.k T e^{-u / k T} ight|_{-p E} ^{p E}}[/latex]

[latex]=k T\left\{\frac{\left[e^{-p E / k T}-e^{p E / k T} ight]+\left[(p E / k T) e^{-p E / k T}+(p E / k T) e^{p E / k T} ight]}{e^{-p E / k T}-e^{p E / k T}} ight\}[/latex]

[latex]=k T-p E\left[\frac{e^{p E / k T}+e^{-p E / k T}}{e^{p E / k T}-e^{-p E / k T}} ight]=k T-p E \operatorname{coth}\left(\frac{p E}{k T} ight)[/latex] .

[latex]P =N\langle p angle ; p =\langle p \cos heta angle \hat{ E }=\langle p \cdot E angle(\hat{ E } / E)=-\langle u angle(\hat{ E } / E) ; P=N p \frac{-\langle u angle}{p E}=N p\left\{\operatorname{coth}\left(\frac{p E}{k T} ight)-\frac{k T}{p E} ight\}[/latex] .

Let [latex]y \equiv P / N p, x \equiv p E / k T ext {. Then } y=\operatorname{coth} x-1 / x . ext { As } x ightarrow 0, y=\left(\frac{1}{x}+\frac{x}{3}-\frac{x^{3}}{45}+\cdots ight)-\frac{1}{x}=\frac{x}{3}-\frac{x^{3}}{45}+\cdots ightarrow 0[/latex] , so the graph starts at the origin, with an initial slope of 1/3. As [latex]x ightarrow \infty, y ightarrow \operatorname{coth}(\infty)=1[/latex] , so the graph goes asymptotically to y = 1 (see Figure).

(b) For small [latex]x, y \approx \frac{1}{3} x, ext { so } \frac{P}{N p} \approx \frac{p E}{3 k T}, ext { or } P \approx \frac{N p^{2}}{3 k T} E=\epsilon_{0} \chi_{e} E \Rightarrow P[/latex] is proportional to E, and [latex]\chi_{e}=\frac{N p^{2}}{3 \epsilon_{0} k T}[/latex] .

For water at [latex]20^{\circ}=293 K , p=6.1 imes 10^{-30} Cm ; N=\frac{ ext { molecules }}{ ext { volume }}=\frac{ ext { molecules }}{ ext { mole }} imes \frac{ ext { moles }}{ ext { gram }} imes \frac{ ext { grams }}{ ext { volume }}[/latex] .

[latex]N=\left(6.0 imes 10^{23} ight) imes\left(\frac{1}{18} ight) imes\left(10^{6} ight)=0.33 imes 10^{29} ; \chi_{e}=\frac{\left(0.33 imes 10^{29} ight)\left(6.1 imes 10^{-30} ight)^{2}}{(3)\left(8.85 imes 10^{-12} ight)\left(1.38 imes 10^{-23} ight)(293)}=12 .[/latex] Table 4.2 gives an experimental value of 79, so it’s pretty far off.

For water vapor at 100° = 373 K, treated as an ideal gas, [latex]\frac{ ext { volume }}{ ext { mole }}=\left(22.4 imes 10^{-3} ight) imes\left(\frac{373}{293} ight)=2.85 imes 10^{-2} m ^{3}[/latex] .

[latex]N=\frac{6.0 imes 10^{23}}{2.85 imes 10^{-2}}=2.11 imes 10^{25} ; \quad \chi_{e}=\frac{\left(2.11 imes 10^{25} ight)\left(6.1 imes 10^{-30} ight)^{2}}{(3)\left(8.85 imes 10^{-12} ight)\left(1.38 imes 10^{-23} ight)(373)}=5.7 imes 10^{-3}[/latex] .

Table 4.2 gives [latex]5.9 imes 10^{-3}[/latex] , so this time the agreement is quite good.

Question 4.43: The Clausius-Mossotti equation (Prob. 4.41) tells you how to...

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