91Ó°ÊÓ

The equation for the temperature of magnets is given as:

$\frac{1}{T}=-\frac{1}{2\mathrm{μ}B}\frac{∂S}{∂N_{↑}}…..\text{(1)}$

Where,

$T$= temperature

$B$= magnetic field

$\mathrm{μ}$= constant

$S$= entropy

$N_{↑}$= number of up dipoles of the paramagnet

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

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

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

$N_{↑}=\frac{N}{2}-\frac{U}{2\mathrm{μ}B}$

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" $C_B=\frac{∂U}{∂T}…..\text{(2)}$

Where,

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

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

$$\begin{gathered} \frac{1}{T}=−\frac{k}{2\mathrm{μ}B}\frac{\mathrm{∂}}{\mathrm{∂}{N}_{↑}}[N\ln N−N_{↑}\ln N_{↑}−(N−N_{↑})\ln (N−N_{↑})] \\ \begin{array}{rl}\frac{1}{T}=−\frac{k}{2\mathrm{μ}B}[−\ln N_{↑}& −\frac{N_{↑}}{N_{↑}}+\ln (N−N_{↑})+\frac{N−N_{↑}}{N−N_{↑}}]\end{array} \\ \begin{array}{rl}\frac{1}{T}=& \frac{k}{2\mathrm{μ}B}\ln \frac{N_{↑}}{N−N_{↑}}\end{array} \end{gathered}$$

Where,

$k$= Boltzmann's constant

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

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

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

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

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

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

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

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

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

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

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