Q4.43P The Clausius-Mossotti equation (... [FREE SOLUTION]

91影视

Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 4: Q4.43P (page 209)

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 $u = - p 路贰肠辞蝉胃$
(Eq. 4.6), where $胃$ 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,
$\exp (- u / kT)$
The average energy of the dipoles is therefore
$ = \frac{鈭� u e^{- (u / k t) 诲惟}}{鈭� e^{- (u / k T) 诲惟}}$
where $诲惟 = 蝉颈苍胃诲胃诲蠒$ , and the integration is over all orientations $(胃 : 0 鈫� 蟺; 蠒 : 0 鈫� 2 蟺)$ Use this to show that the polarization of a substance
containing N molecules per unit volume is
$P = Np [\coth (pE / kT) - (kT / pE)]$ (4.73)
That's the Langevin formula. Sketch as a function of $P E / K T$ .
(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掳C,and compare the experimental value in Table 4.2. (The dipole moment of wateris $6.1 脳 10^{- 30} C 路 m$ ) This is rather far off, because we have again neglected thedistinction between E and Eelse路 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掳C and 1 atm.

Short Answer

Expert verified

(a) The expression for the polarization is $P = N p (\coth (\frac{p E}{k T}) - \frac{k T}{p E})$

(b)The susceptibility is $蠂 = (\frac{N_{P}^{2}}{3 蔚_{0} k T})$ , for water in aqueous it is $蠂_{w} = 12$ and for gas it is $5.7 脳 10^{- 3}$ . The susceptibility for the liquid water is not in agreement with the expression for the gaseous state.

Step by step solution

Determine the expression for the polarization of the substance containing one molecule:

(a)

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

$\begin{array}{rcl} U & = & - \overset{炉}{P} 路 \overset{炉}{E} \\ = & P贰肠辞蝉胃 \end{array}$

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

Consider the average energy for the electric dipole is:

$\begin{array}{rcl} U & = & - P E <\cos 胃> \\ = & - P_{i n} E \end{array}$

Consider the expression for the resultant polarization as:

$\begin{array}{rcl} P & = & N P_{a v} \\ = & - N E \end{array}$

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

$\begin{array}{rcl} & = & \frac{{鈭�}_{- p E}^{p E} u e^{\frac{- u}{k T} d u}}{{鈭�}_{- p E}^{p E} e^{\frac{- u}{k T} d u}} \\ = & k t - p E (\frac{e^{\frac{p e}{k T}} + e^{\frac{- p e}{k T}}}{e^{\frac{p e}{k T}} - e^{\frac{- p e}{k T}}}) \\ = & k T - p E \coth (\frac{p E}{k T}) \end{array}$

Determine the expression for the Polarization as:

$\begin{array}{rcl} P & = & N \\ = & E \\ = & \frac{E}{E} \\ = & - \frac{E}{E} \end{array}$

Solve further as:

$P = N p (\coth (\frac{p E}{k T}) - \frac{k T}{p E})$

Therefore, the expression for the polarization is $P = N p (\coth (\frac{p E}{k T}) - \frac{k T}{p E})$

Prove the regime material is linear and determine the susceptibility expression, also determine the susceptibility of the water and compare it with the experimental value:

Consider the expression as:

$x = \frac{p E}{k T}$

Also,

$\frac{P}{N_{p}} = \coth (x) - \frac{1}{x}$

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

$x > 0 \frac{1}{x} < 0 \coth (x) > 1$

Consider for $k T > > p E$ solve as:

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

Short the terms by the Taylor series as:

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

Consider for $k T > > p E$ solve as:

$\begin{array}{rcl} \frac{P}{N_{P}} & = & (\frac{N_{P}^{2}}{3 k T}) E \\ \frac{P}{N_{P}} & = & 蔚_{0} (\frac{N_{P}^{2}}{3 蔚_{0} k T}) E \\ \frac{P}{N_{P}} & = & 蔚_{0} 蠂 E \{蠂 = (\frac{N_{P}^{2}}{3 蔚_{0} k T})\} \end{array}$

Consider the temperature in Kelvin is 293 K

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

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

Consider from the equation of the susceptibility solve as:

$\begin{array}{rcl} 蠂_{w} & = & (\frac{{(6.1 脳 10^{- 30})}^{2}}{3 (8.85 脳 10^{- 12} C^{2} / N m) (1.38 脳 10^{- 23}) {(293 K)}^{2}}) \\ = & 12 \end{array}$

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

Then, by the Charles law solve as:

$\begin{array}{rcl} V_{100} & = & \frac{V_{20} T_{100}}{T_{20}} \\ = & \frac{(22.4 脳 10^{- 3} m^{3}) (373 K)}{293 K} \\ = & 2.85 脳 10^{- 2} m^{3} \end{array}$

Solve for the number of molecules per unit volume as:

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

Solve for the susceptibility of the water as:

$\begin{array}{rcl} 蠂_{w} & = & \frac{((2.11 脳 10^{25}) {(6.1 脳 10^{- 30})}^{2})}{3 (8.85 脳 10^{- 12} c^{2} / N m) (1.38 脳 10^{- 23}) {(293 K)}^{2}} \\ = & 5.7 脳 10^{- 3} \end{array}$

Therefore, the susceptibility is $蠂 = (\frac{N_{P}^{2}}{3 蔚_{0} k T})$ , for water in aqueous it is $蠂_{w} = 12$ and for gas it is $5.7 脳 10^{- 3}$ . The susceptibility for the liquid water is not in agreement with the expression for the gaseous state.