Question

Show that the entropy of a two-state paramagnet, expressed as a function of temperature, is $S=Nk\left[\ln \right(2\cosh x)-x\tanh x]$, where $x=\mu B/kT$. Check that this formula has the expected behavior as $T\to 0$and $T\to \infty$.

Short answer

The entropy of the two-state paramagnet as the function of temperature is expressed as$S=Nk\left[\ln \right(2\cosh x)-x\tanh x]$,where $x=\mu B/kT$.

The entropy at temperature $T\to 0$and $T\to \infty$shows expected behavior with consistency with the third law of thermodynamics.

Step-by-step solution

Given Information

The given substance is a two-state paramagnet of which the graph of entropy as a function of temperature is to be sketched.

We have to show that:

$S=Nk\left[\ln \right(2\cosh x)-x\tanh x]$

Where,

$x=\mu B/kT$

Calculation

The magnetization for two-state paramagnet is given as:

$M=\mu N\tanh \left(\frac{\mu B}{kT}\right)$

Also, magnetization is given as:

$M=\mu (N_{\uparrow }-N_{\downarrow })$

On comparing both the above equations, we get,

$$\begin{gathered} N\tanh \left(\frac{\mu B}{kT}\right)=(N_{\uparrow }-N_{\downarrow }) \\ \tanh \left(\frac{\mu B}{kT}\right)=\frac{2N_{\uparrow }-N}{N} \\ \tanh \left(\frac{\mu B}{kT}\right)=\frac{2N_{\uparrow }}{N}-1 \\ \tanh \left(\frac{\mu B}{kT}\right)=2n-1 \\ n=\frac{1}{2}\{1+\tanh (x\left)\right\} \end{gathered}$$

Where,

$x=\frac{\mu B}{kT}$

The entropy of a two-state paramagnet is given as:

$S=k[N\ln N-N_{\uparrow }\ln N_{\uparrow }-N_{\downarrow }\ln N_{\downarrow }]$

On solving the above equation,

$$\begin{gathered} S=k\left[N\ln N-N_{\uparrow }\ln N_{\uparrow }-N_{\downarrow }\ln N_{\downarrow }\right] \\ \frac{S}{k}=N\ln N-\frac{N_{\uparrow }N}{N}\ln \frac{N_{\uparrow }N}{N}-\left(N-\frac{N_{\uparrow }N}{N}\right)\ln \left(N-\frac{N_{\uparrow }N}{N}\right) \\ \frac{S}{Nk}=\ln N-\frac{N_{\uparrow }}{N}\ln \frac{N_{\uparrow }N}{N}-\left(1-\frac{N_{\uparrow }}{N}\right)\ln N\left(1-\frac{N_{\uparrow }}{N}\right) \\ \frac{S}{Nk}=\ln N-n\ln \left(nN\right)-(1-n)\ln N(1-n) \\ \frac{S}{Nk}=\ln N-n\ln n-n\ln N-(1-n)\ln N-(1-n)\ln (1-n) \\ \frac{S}{Nk}=(1-n)\ln N-n\ln n-(1-n)\ln N-(1-n)\ln (1-n) \\ \frac{S}{Nk}=-n\ln n-(1-n)\ln (1-n) \\ \frac{S}{Nk}=-n\ln n-\ln (1-n)+n\ln (1-n) \\ \frac{S}{Nk}=n\ln \frac{(1-n)}{n}-\ln (1-n)\dots \dots \left(1\right) \end{gathered}$$

Further,

$$\begin{gathered} \frac{(1-n)}{n}=\frac{1-\frac{1}{2}\{1+\tanh (x\left)\right\}}{\frac{1}{2}\{1+\tanh (x\left)\right\}} \\ \frac{{\displaystyle (1-n)}}{{\displaystyle n}}=\frac{\frac{1}{2}\{1-\tanh (x\left)\right\}}{\frac{1}{2}\{1+\tanh (x\left)\right\}} \\ \frac{{\displaystyle (1-n)}}{{\displaystyle n}}=\frac{\{1-\tanh (x){\}}^2}{\left\{1-{\tanh }^2\left(x\right)\right\}} \\ \frac{{\displaystyle (1-n)}}{{\displaystyle n}}=\{1-\tanh (x){\}}^2{\cosh }^{2}x \\ \frac{{\displaystyle (1-n)}}{{\displaystyle n}}={\left\{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}\right\}}^2\frac{e^x+e^{-x}}{2} \\ \frac{{\displaystyle (1-n)}}{{\displaystyle n}}=\frac{{\left(2e^{-x}\right)}^2}{2} \\ \frac{{\displaystyle (1-n)}}{{\displaystyle n}}=2e^{-2x} \\ \ln \frac{(1-n)}{n}=-2x \end{gathered}$$

Also,

$$\begin{gathered} (1-n)=1-\frac{1}{2}\{1+\tanh (x\left)\right\} \\ (1-n)=\frac{1}{2}\{1-\tanh (x\left)\right\} \\ (1-n)=\frac{1}{2}\left\{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}\right\} \\ (1-n)=\frac{1}{2}\left\{\frac{2e^{-x}}{e^x+e^{-x}}\right\} \\ (1-n)=\frac{e^{-x}}{2\left(e^x+e^{-x}/2\right)} \\ (1-n)=\frac{e^{-x}}{2\cosh x} \\ \ln (1-n)=-x-2\ln \left(\cosh x\right) \end{gathered}$$

Now, by substituting these values in equation (1), we get,

$$\begin{gathered} \frac{S}{Nk}=n\ln \frac{(1-n)}{n}-\ln (1-n) \\ \frac{S}{Nk}=\frac{1}{2}\{1+\tanh (x\left)\right\}(-2x)-\{-x-2\ln (\cosh x\left)\right\} \\ \frac{S}{Nk}=-x-x\tanh \left(x\right)+x+2\ln \left(\cosh x\right) \\ \frac{S}{Nk}=2\ln \left(\cosh x\right)-x\tanh \left(x\right) \\ \frac{S}{Nk}=2\ln \left\{\cosh \left(\frac{\mu B}{kT}\right)\right\}-\frac{\mu B}{kT}\tanh \left(\frac{\mu B}{kT}\right) \end{gathered}$$

At$T\to 0$,

$x=\frac{\mu B}{kT}\to \infty$

Hence, the entropy becomes

$$\begin{gathered} \frac{S}{Nk}=2\ln \left(\cosh \frac{\mu B}{kT}\right)-x\tanh \left(\frac{\mu B}{kT}\right) \\ \frac{S}{Nk}=2\ln \left(\cosh x\right)-x\tanh \left(x\right) \\ \frac{S}{Nk}=2\ln \left(\frac{e^x+e^{-x}}{2}\right)-x\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) \\ \frac{S}{Nk}=\ln \left(2\times \frac{e^x}{2}\right)-x\left(\frac{e^x}{e^x}\right) \\ \frac{S}{Nk}=x\ln e-x \\ \frac{S}{Nk}=x-x \\ \frac{S}{Nk}=0 \end{gathered}$$

At $T\to \infty$,

$x=\frac{\mu B}{kT}\to 0$

Hence, the entropy becomes,

$$\begin{gathered} \frac{S}{Nk}=2\ln \left(\cosh \frac{\mu B}{kT}\right)-x\tanh \left(\frac{\mu B}{kT}\right) \\ \frac{S}{Nk}=2\ln \left(\cosh x\right)-x\tanh \left(x\right) \\ \frac{S}{Nk}=2\ln \left(\cosh 0\right)-x\tanh \left(0\right) \\ \frac{S}{Nk}=\ln \left(2\cosh 0\right) \\ \frac{S}{Nk}=\ln (2\times 1) \\ \frac{S}{Nk}=0.693 \end{gathered}$$

Final answer

The entropy of the two-state paramagnet as the function of temperature is $S=Nk\left[\ln \right(2\cosh x)-x\tanh x]$with $x=\mu B/kT$.

The entropy at temperature $T\to 0$and $T\to \infty$shows expected behavior along with consistency with the third law of thermodynamics.