91影视

Consider the formula for the susceptibility as follows:

$\mathbf{蠂}\mathbf{=}\mathbf{N}\frac{\mathbf{伪}}{{\mathbf{蔚}}_{\mathbf{0}}}$

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

Consider the formula for the ideal gas as follows:

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

Refer to the table 4.1 in the textbook.

Solve for the susceptibility for hydrogen as:

$\begin{array}{rcl}{蠂}_{e1}& =& \frac{N鈭�1}{{蔚}_0}\\ & =& 2.7脳{10}^{-25}脳\frac{\left(4蟺{蔚}_0脳{10}^{-30}\right)尾}{{蔚}_0}\\ & =& 2.7脳{10}^{-25}脳\frac{\left(4蟺{蔚}_0脳{10}^{-30}\right)0.67}{{蔚}_0}\\ & =& 2.7脳{10}^{-4}\\ & & \end{array}$

Consider the dielectric constant for the hydrogen is:

${蔚}_{r1}=1.00025$

Since,

$\begin{array}{rcl}{蠂}_{e1}& =& {蔚}_r-1\\ & =& 1.00025-1\\ & =& 2.5脳{10}^{-4}\\ & & \end{array}$

Consider for helium ${伪}_2=\left(4蟺{蔚}_0脳{10}^{-30}\right)0.205$. Solve for the susceptibility for helium as:

$\begin{array}{rcl}{蠂}_{e2}& =& \frac{N{伪}_2}{{蔚}_0}\\ & =& 3.4脳{10}^{-4}\left(0.205\right)\\ & =& 6.97脳{10}^{-5}\\ & & \end{array}$

Consider the dielectric constant for the hydrogen is:

${蔚}_{r2}=1.000065$

Since,

$\begin{array}{rcl}{蠂}_{e2}& =& {蔚}_{r2}-1\\ & =& 1.000065-1\\ & =& 6.5脳{10}^{-5}\\ & & \end{array}$

Consider the value of $尾$for argon is 1.64.

Solve for the susceptibility as:

$\begin{array}{rcl}{蠂}_{e3}& =& 3.4脳{10}^{-4}尾\\ & =& 3.4脳{10}^{-4}\left(1.64\right)\\ & =& 5.6脳{10}^{-4}\\ & & \end{array}$

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

Solve for the susceptibility as:

$\begin{array}{rcl}{蠂}_{e3}& =& {蔚}_{r3}-1\\ & =& 1.00052-1\\ & =& 5.6脳{10}^{-4}\\ & & \end{array}$

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.