Question

Check the Clausius-Mossotti relation (Eq. 4.72) for the gases listed in Table 4.1. (Dielectric constants are given in Table 4.2.) (The densities here are so small that Eqs. 4.70 and 4.72 are indistinguishable. For experimental data that confirm the Clausius-Mossotti correction term see, for instance, the first edition of Purcell's Electricity and Magnetism, Problem 9.28.)

Short answer

The susceptibilities of gases in 4.1 and susceptibilities of gases in 4.2 are good agreement.

Step-by-step solution

Step 1:Determine the formulas:

Consider the formula for the susceptibility as follows:

$\mathbf{\chi }\mathbf{=}\mathbf{N}\frac{\mathbf{\alpha }}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

Here, N is the number of atoms per unit volume and $\alpha$is the atomic polarization.

Consider the formula for the ideal gas as follows:

$$\begin{gathered} N=\frac{Avogadronumber}{volumeatSTP} \\ N=\frac{6.02\times {10}^{23}}{22.4\text{litres}} \\ N=2.7\times {10}^{25} \end{gathered}$$

Determine the susceptibility for various atom

Refer to the table 4.1 in the textbook.

Solve for the susceptibility for hydrogen as:

$\begin{array}{rcl}{\chi }_{e1}& =& \frac{N\propto 1}{{\epsilon }_0}\\ & =& 2.7\times {10}^{-25}\times \frac{\left(4\pi {\epsilon }_0\times {10}^{-30}\right)\beta }{{\epsilon }_0}\\ & =& 2.7\times {10}^{-25}\times \frac{\left(4\pi {\epsilon }_0\times {10}^{-30}\right)0.67}{{\epsilon }_0}\\ & =& 2.7\times {10}^{-4}\\ & & \end{array}$

Consider the dielectric constant for the hydrogen is:

${\epsilon }_{r1}=1.00025$

Since,

$\begin{array}{rcl}{\chi }_{e1}& =& {\epsilon }_r-1\\ & =& 1.00025-1\\ & =& 2.5\times {10}^{-4}\\ & & \end{array}$

Consider for helium ${\alpha }_2=\left(4\pi {\epsilon }_0\times {10}^{-30}\right)0.205$. Solve for the susceptibility for helium as:

$\begin{array}{rcl}{\chi }_{e2}& =& \frac{N{\alpha }_2}{{\epsilon }_0}\\ & =& 3.4\times {10}^{-4}\left(0.205\right)\\ & =& 6.97\times {10}^{-5}\\ & & \end{array}$

Consider the dielectric constant for the hydrogen is:

${\epsilon }_{r2}=1.000065$

Since,

$\begin{array}{rcl}{\chi }_{e2}& =& {\epsilon }_{r2}-1\\ & =& 1.000065-1\\ & =& 6.5\times {10}^{-5}\\ & & \end{array}$

Consider the value of $\beta$for argon is 1.64.

Solve for the susceptibility as:

$\begin{array}{rcl}{\chi }_{e3}& =& 3.4\times {10}^{-4}\beta \\ & =& 3.4\times {10}^{-4}\left(1.64\right)\\ & =& 5.6\times {10}^{-4}\\ & & \end{array}$

Consider the value of the dielectric constant for argon is 1.00052.

Solve for the susceptibility as:

$\begin{array}{rcl}{\chi }_{e3}& =& {\epsilon }_{r3}-1\\ & =& 1.00052-1\\ & =& 5.6\times {10}^{-4}\\ & & \end{array}$

Note that the susceptibilities of the gases in 4.1 and 4.2 are in good agreement.

Therefore, the Clausius-Mossotti equation for the gases listed in the table 4.1 are determined.