Question

Fill in the missing algebraic steps to derive equations 3.30, 3.31, and 3.33.

Short answer

Thus the equations are derived to fill the missing steps.

Step-by-step solution

Given Information

The equation for the temperature of magnets is given as:

$\frac{1}{T}=-\frac{1}{2\mu B}\frac{\partial S}{\partial N_{\uparrow }}\dots ..\text{(1)}$

Where,

$T$= temperature

$B$= magnetic field

$\mu$= constant

$S$= entropy

$N_{\uparrow }$= number of up dipoles of the paramagnet

The entropy of the paramagnets from Stirling's approximation is given as:

$S=kN\ln N-N_{\uparrow }\ln N_{\uparrow }-\left(N-N_{\uparrow }\right)\ln \left(N-N_{\uparrow }\right)$

The expression for the number of up dipoles of the paramagnet is given as:

$N_{\uparrow }=\frac{N}{2}-\frac{U}{2\mu B}$

Where,

$N$= total number of the spins

$U$= total energy of the paramagnets

The expression for the heat capacity in the magnetic field of the paramagnet is given as:

role="math" localid="1647337393137" $C_B=\frac{\partial U}{\partial T}\dots ..\text{(2)}$

Where,

$C_B$= heat capacity in the magnetic field $B$of the paramagnet.

Calculation

By substituting the value of entropy in equation (1), we get,

$$\begin{gathered} \frac{1}{T}=-\frac{k}{2\mu B}\frac{\mathrm{\partial }}{\mathrm{\partial }{N}_{\uparrow }}[N\ln N-N_{\uparrow }\ln N_{\uparrow }-(N-N_{\uparrow })\ln (N-N_{\uparrow })] \\ \begin{array}{rl}\frac{1}{T}=-\frac{k}{2\mu B}[-\ln N_{\uparrow }& -\frac{N_{\uparrow }}{N_{\uparrow }}+\ln (N-N_{\uparrow })+\frac{N-N_{\uparrow }}{N-N_{\uparrow }}]\end{array} \\ \begin{array}{rl}\frac{1}{T}=& \frac{k}{2\mu B}\ln \frac{N_{\uparrow }}{N-N_{\uparrow }}\end{array} \end{gathered}$$

Where,

$k$= Boltzmann's constant

By substituting $\frac{N}{2}-\frac{U}{2\mu B}$for $N_{\uparrow }$in the above expression, we get,

$$\begin{gathered} \begin{array}{rl}\frac{1}{T}& =\frac{k}{2\mu B}\ln \frac{N/2-U/2\mu B}{N-N/2-U/2\mu B}\end{array} \\ \frac{1}{T}=\frac{k}{2\mu B}\ln \left(\frac{N-U/\mu B}{N+U/\mu B}\right) \end{gathered}$$

By taking exponential on both the sides in the above expression, it becomes,

$e^{1/T}=e^{k/2\mu B}\left(\frac{N-U/\mu B}{N+U/\mu B}\right)$

On simplifying the above expression for the total energy of the paramagnets, it becomes,

$$\begin{gathered} N-\frac{U}{\mu B}=e^{2\mu B/kT}\left(N+\frac{U}{\mu B}\right) \\ \frac{U}{\mu B}\left(1+e^{2\mu B/kT}\right)=N\left(1-e^{2\mu B/kT}\right) \\ U=N\mu B\left(\frac{1-e^{2\mu B/kT}}{1+e^{2\mu B/kT}}\right) \end{gathered}$$

On multiplying the numerator and denominator of the right-hand side by $e^{-\mu B/kT}$in the above expression, we get,

$$\begin{gathered} U=N\mu B\left(\frac{e^{-\mu B/kT}-e^{\mu B/kT}}{e^{-\mu B/kT}+e^{\mu B/kT}}\right) \\ U=N\mu B\left(\frac{-2\sinh (\mu B/kT)}{-2\cosh (\mu B/kT)}\right) \\ U=-N\mu B\tanh \left(\frac{\mu B}{kT}\right)\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} C_B=-N\mu B\frac{\partial }{\partial T}\tanh \left(\frac{\mu B}{kT}\right) \\ {C}_B=-N\mu B\frac{1}{{\cosh }^2(\mu B/kT)}\left(\frac{\mu B}{k}\right)\left(-T^{-2}\right) \\ {C}_B=Nk\frac{(\mu B/kT{)}^2}{{\cosh }^2(\mu B/kT)} \end{gathered}$$

Final answer

The missing steps for calculating the temperature, total energy, and heat capacity of the paramagnet using Stirling's approximation of the entropy are shown.