Question

Use a computer to sum the rotational partition function (equation 6.30) algebraically, keeping terms through j = 6. Then calculate the average energy and the heat capacity. Plot the heat capacity for values of$kT/ϵ$ ranging from 0 to 3. Have you kept enough terms in Z to give accurate results within this temperature range?

Short answer

The average energy is $\overline{E}=\frac{\sum ϵj(j+1)(2j+1)e^{-j(j+1)/t}}{\sum (2j+1)e^{-j(j+1)/t}}$and

The average heat capacity is $C=\frac{k\sum (2j+1)e^{-j(j+1)/t}\sum j^2(j+1{)}^2(2j+1)e^{-j(j+1)/t}}{{\left(t\sum (2j+1)e^{-j(j+1)/t}\right)}^2}-\frac{k{\left(\sum j(j+1)(2j+1)e^{-j(j+1)/t}\right)}^2}{{\left(t\sum (2j+1)e^{-j(j+1)/t}\right)}^2}$

Step-by-step solution

Given information

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

j = 6

Explanation

Rotational partition function is:

$Z=\sum (2j+1)e^{-j(j+1)ϵ/kT}$

Let,

$t=\frac{kT}{ϵ}$

Therefore,

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

The average energy is:

$\overline{E}=-\frac{1}{Z}\frac{\partial Z}{\partial \beta }\left(2\right)$

By chain rule:

$$\begin{gathered} \frac{\partial Z}{\partial \beta }=\frac{\partial Z}{\partial t}\frac{\partial t}{\partial \beta } \\ \frac{\partial Z}{\partial \beta }=\frac{\partial Z}{\partial t}{\left(\frac{\partial \beta }{\partial t}\right)}^{-1} \end{gathered}$$

But $\beta =1/kT\text{, hence}\beta =1/tϵ$hence,

$$\begin{gathered} \frac{\partial Z}{\partial \beta }=\frac{\partial }{\partial t}\left(\sum (2j+1)e^{-j(j+1)/t}\right){\left(\frac{\partial }{\partial t}\left(\frac{1}{tϵ}\right)\right)}^{-1} \\ \frac{\partial Z}{\partial \beta }=\left(\sum j(j+1)(2j+1)e^{-j(j+1)/t}\right)\left(\frac{1}{t^2}\right){\left(-\frac{1}{t^2ϵ}\right)}^{-1} \\ \frac{\partial Z}{\partial \beta }=-\sum ϵj(j+1)(2j+1)e^{-j(j+1)/t} \end{gathered}$$

Substitute into (2)

role="math" localid="1647453692824" $\overline{E}=\frac{\sum ϵj(j+1)(2j+1)e^{-j(j+1)/t}}{\sum (2j+1)e^{-j(j+1)/t}}\left(3\right)$

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

$$\begin{gathered} C=\frac{\partial \overline{E}}{\partial T} \\ C=\frac{k}{ϵ}\frac{\partial \overline{E}}{\partial t} \end{gathered}$$

Substitute from (3):

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

Explanation

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

The graph is: