91影视

Heat capacity for values $kT/系$ranging from 0 to 3.

j = 6

Rotational partition function is:

$Z=鈭�(2j+1)e^{-j(j+1)系/kT}$

Let,

$t=\frac{kT}{系}$

Therefore,

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

The average energy is:

$\stackrel{炉}{E}=-\frac{1}{Z}\frac{鈭�Z}{鈭�尾}\left(2\right)$

By chain rule:

$$\begin{gathered} \frac{鈭�Z}{鈭�尾}=\frac{鈭�Z}{鈭�t}\frac{鈭�t}{鈭�尾} \\ \frac{鈭�Z}{鈭�尾}=\frac{鈭�Z}{鈭�t}{\left(\frac{鈭�尾}{鈭�t}\right)}^{-1} \end{gathered}$$

But $尾=1/kT\text{, hence}尾=1/t系$hence,

$$\begin{gathered} \frac{鈭�Z}{鈭�尾}=\frac{鈭�}{鈭�t}\left(鈭�(2j+1)e^{-j(j+1)/t}\right){\left(\frac{鈭�}{鈭�t}\left(\frac{1}{t系}\right)\right)}^{-1} \\ \frac{鈭�Z}{鈭�尾}=\left(鈭�j(j+1)(2j+1)e^{-j(j+1)/t}\right)\left(\frac{1}{t^2}\right){\left(-\frac{1}{t^2系}\right)}^{-1} \\ \frac{鈭�Z}{鈭�尾}=-鈭�系j(j+1)(2j+1)e^{-j(j+1)/t} \end{gathered}$$

Substitute into (2)

role="math" localid="1647453692824" $\stackrel{炉}{E}=\frac{鈭�系j(j+1)(2j+1)e^{-j(j+1)/t}}{鈭�(2j+1)e^{-j(j+1)/t}}\left(3\right)$

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

$$\begin{gathered} C=\frac{鈭�\stackrel{炉}{E}}{鈭�T} \\ C=\frac{k}{系}\frac{鈭�\stackrel{炉}{E}}{鈭�t} \end{gathered}$$

Substitute from (3):

role="math" localid="1647454207193" $$\begin{gathered} C=\frac{系}{k}\frac{鈭�}{鈭�t}\left(\frac{鈭�系j(j+1)(2j+1)e^{-j(j+1)/t}}{鈭�(2j+1)e^{-j(j+1)/t}}\right) \\ C=\frac{k鈭�(2j+1)e^{-j(j+1)/t}鈭�j^2(j+1{)}^2(2j+1)e^{-j(j+1)/t}}{{\left(t鈭�(2j+1)e^{-j(j+1)/t}\right)}^2}-\frac{k{\left(鈭�j(j+1)(2j+1)e^{-j(j+1)/t}\right)}^2}{{\left(t鈭�(2j+1)e^{-j(j+1)/t}\right)}^2} \end{gathered}$$

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

The graph is: