91Ó°ÊÓ

Heat capacity for values $kT/\mathrm{Ï\mu }$ranging from 0 to 3.

j = 6

Rotational partition function is:

$Z=∑(2j+1)e^{-j(j+1)\mathrm{Ï\mu }/kT}$

Let,

$t=\frac{kT}{\mathrm{Ï\mu }}$

Therefore,

$Z=∑(2j+1)e^{-j(j+1)/t}\left(1\right)$

The average energy is:

$\stackrel{¯}{E}=-\frac{1}{Z}\frac{∂Z}{∂\mathrm{β}}\left(2\right)$

By chain rule:

$$\begin{gathered} \frac{∂Z}{∂\mathrm{β}}=\frac{∂Z}{∂t}\frac{∂t}{∂\mathrm{β}} \\ \frac{∂Z}{∂\mathrm{β}}=\frac{∂Z}{∂t}{\left(\frac{∂\mathrm{β}}{∂t}\right)}^{-1} \end{gathered}$$

But $\mathrm{β}=1/kT\text{, hence}\mathrm{β}=1/t\mathrm{Ï\mu }$hence,

$$\begin{gathered} \frac{∂Z}{∂\mathrm{β}}=\frac{∂}{∂t}\left(∑(2j+1)e^{-j(j+1)/t}\right){\left(\frac{∂}{∂t}\left(\frac{1}{t\mathrm{Ï\mu }}\right)\right)}^{-1} \\ \frac{∂Z}{∂\mathrm{β}}=\left(∑j(j+1)(2j+1)e^{-j(j+1)/t}\right)\left(\frac{1}{t^2}\right){\left(-\frac{1}{t^2\mathrm{Ï\mu }}\right)}^{-1} \\ \frac{∂Z}{∂\mathrm{β}}=-∑\mathrm{Ï\mu }j(j+1)(2j+1)e^{-j(j+1)/t} \end{gathered}$$

Substitute into (2)

role="math" localid="1647453692824" $\stackrel{¯}{E}=\frac{∑\mathrm{Ï\mu }j(j+1)(2j+1)e^{-j(j+1)/t}}{∑(2j+1)e^{-j(j+1)/t}}\left(3\right)$

The partial derivative of total energy with respect to temperature equals the heat capacity, which is:

$$\begin{gathered} C=\frac{∂\stackrel{¯}{E}}{∂T} \\ C=\frac{k}{\mathrm{Ï\mu }}\frac{∂\stackrel{¯}{E}}{∂t} \end{gathered}$$

Substitute from (3):

role="math" localid="1647454207193" $$\begin{gathered} C=\frac{\mathrm{Ï\mu }}{k}\frac{∂}{∂t}\left(\frac{∑\mathrm{Ï\mu }j(j+1)(2j+1)e^{-j(j+1)/t}}{∑(2j+1)e^{-j(j+1)/t}}\right) \\ C=\frac{k∑(2j+1)e^{-j(j+1)/t}∑j^2(j+1{)}^2(2j+1)e^{-j(j+1)/t}}{{\left(t∑(2j+1)e^{-j(j+1)/t}\right)}^2}-\frac{k{\left(∑j(j+1)(2j+1)e^{-j(j+1)/t}\right)}^2}{{\left(t∑(2j+1)e^{-j(j+1)/t}\right)}^2} \end{gathered}$$

Using python to plot a function between t and $C/k$. The code is given below:

The graph is: