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

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

<u>=∫ue−(u/kT)dΩ∫e−(u/kT)dΩ<u>=\frac{\int u e^{-(u / k T)} d \Omega}{\int e^{-(u / k T)} d \Omega} ,

where dΩ=sin⁡θdθdϕd \Omega=\sin \theta d \theta d \phi, and the integration is over all orientations (θ :0 →π;ϕ:0→2π\rightarrow \pi ; \phi: 0 \rightarrow 2 \pi ). Use this to show that the polarization of a substancecontaining N molecules per unit volume is

P=Np[coth⁡(pE/kT)−(kT/pE)]P=N p[\operatorname{coth}(p E / k T)-(k T / p E)]                         (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 6.1×10−30C⋅m6.1 \times 10^{-30} C \cdot m .) This is rather far off, because we have again neglected the distinction between E and Eelse E _{\text {else }} . The agreement is better in low-density gases, for which the difference between E and Eelse E _{\text {else }} 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 (H2)\left( H _{2}\right) 1.000254 Silicon 11.7
Argon 1.000517 Methanol 33.0
Air (dry) 1.000536 Water 80.1
Nitrogen (N2)\left( N _{2}\right) 1.000548 Ice (-30° C) 104
Water vapor (100° C) 1.00589 KTaNbO3(0∘C)O _{3}\left(0^{\circ} C \right) 34,000

Table 4.2

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(a) Doing the (trivial) ϕ\phi integral, and changing the remaining integration variable from θ to u(du=pEsin⁡θdθ)\theta \text { to } u(d u=p E \sin \theta d \theta) ,

⟨u⟩=∫−pEpEue−u/kTdu∫−pEpEe−u/kTdu=(kT)2e−u/kT[−(u/kT)−1]∣−pEpE−kTe−u/kT∣−pEpE\langle u\rangle=\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]\right|_{-p E} ^{p E}}{-\left.k T e^{-u / k T}\right|_{-p E} ^{p E}}

 

=kT{[e−pE/kT−epE/kT]+[(pE/kT)e−pE/kT+(pE/kT)epE/kT]e−pE/kT−epE/kT}=k T\left\{\frac{\left[e^{-p E / k T}-e^{p E / k T}\right]+\left[(p E / k T) e^{-p E / k T}+(p E / k T) e^{p E / k T}\right]}{e^{-p E / k T}-e^{p E / k T}}\right\}

 

=kT−pE[epE/kT+e−pE/kTepE/kT−e−pE/kT]=kT−pEcoth⁡(pEkT)=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}}\right]=k T-p E \operatorname{coth}\left(\frac{p E}{k T}\right) .

P=N⟨p⟩;p=⟨pcos⁡θ⟩E^=⟨p⋅E⟩(E^/E)=−⟨u⟩(E^/E);P=Np−⟨u⟩pE=Np{coth⁡(pEkT)−kTpE}P =N\langle p \rangle ; p =\langle p \cos \theta\rangle \hat{ E }=\langle p \cdot E \rangle(\hat{ E } / E)=-\langle u\rangle(\hat{ E } / E) ; P=N p \frac{-\langle u\rangle}{p E}=N p\left\{\operatorname{coth}\left(\frac{p E}{k T}\right)-\frac{k T}{p E}\right\} .

Let  y≡P/Np,x≡pE/kT. Then y=coth⁡x−1/x. As x→0,y=(1x+x3−x345+⋯ )−1x=x3−x345+⋯→0y \equiv P / N p, x \equiv p E / k T \text {. Then } y=\operatorname{coth} x-1 / x . \text { As } x \rightarrow 0, y=\left(\frac{1}{x}+\frac{x}{3}-\frac{x^{3}}{45}+\cdots\right)-\frac{1}{x}=\frac{x}{3}-\frac{x^{3}}{45}+\cdots \rightarrow 0 , so the graph starts at the origin, with an initial slope of 1/3. As x→∞,y→coth⁡(∞)=1x \rightarrow \infty, y \rightarrow \operatorname{coth}(\infty)=1 , so the graph goes asymptotically to y = 1 (see Figure).

(b) For small  x,y≈13x, so PNp≈pE3kT, or P≈Np23kTE=ϵ0χeE⇒Px, y \approx \frac{1}{3} x, \text { so } \frac{P}{N p} \approx \frac{p E}{3 k T}, \text { or } P \approx \frac{N p^{2}}{3 k T} E=\epsilon_{0} \chi_{e} E \Rightarrow P is proportional to E, and χe=Np23ϵ0kT\chi_{e}=\frac{N p^{2}}{3 \epsilon_{0} k T} .

For water at  20∘=293K,p=6.1×10−30Cm;N= molecules  volume = molecules  mole × moles  gram × grams  volume 20^{\circ}=293 K , p=6.1 \times 10^{-30} Cm ; N=\frac{\text { molecules }}{\text { volume }}=\frac{\text { molecules }}{\text { mole }} \times \frac{\text { moles }}{\text { gram }} \times \frac{\text { grams }}{\text { volume }} .

N=(6.0×1023)×(118)×(106)=0.33×1029;χe=(0.33×1029)(6.1×10−30)2(3)(8.85×10−12)(1.38×10−23)(293)=12.N=\left(6.0 \times 10^{23}\right) \times\left(\frac{1}{18}\right) \times\left(10^{6}\right)=0.33 \times 10^{29} ; \chi_{e}=\frac{\left(0.33 \times 10^{29}\right)\left(6.1 \times 10^{-30}\right)^{2}}{(3)\left(8.85 \times 10^{-12}\right)\left(1.38 \times 10^{-23}\right)(293)}=12 . 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,  volume  mole =(22.4×10−3)×(373293)=2.85×10−2m3\frac{\text { volume }}{\text { mole }}=\left(22.4 \times 10^{-3}\right) \times\left(\frac{373}{293}\right)=2.85 \times 10^{-2} m ^{3} .

N=6.0×10232.85×10−2=2.11×1025;χe=(2.11×1025)(6.1×10−30)2(3)(8.85×10−12)(1.38×10−23)(373)=5.7×10−3N=\frac{6.0 \times 10^{23}}{2.85 \times 10^{-2}}=2.11 \times 10^{25} ; \quad \chi_{e}=\frac{\left(2.11 \times 10^{25}\right)\left(6.1 \times 10^{-30}\right)^{2}}{(3)\left(8.85 \times 10^{-12}\right)\left(1.38 \times 10^{-23}\right)(373)}=5.7 \times 10^{-3} .

Table 4.2 gives  5.9×10−35.9 \times 10^{-3} , so this time the agreement is quite good.

4.43

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