Chapter 6: Q 6.28 (page 237)

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 $k T / ϵ$ ranging from 0 to 3. Have you kept enough terms in Z to give accurate results within this temperature range?

Short Answer

Expert verified

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

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

Step by step solution

Given information

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

j = 6

Explanation

Rotational partition function is:

$Z = \sum (2 j + 1) e^{- j (j + 1) ϵ / k T}$

Let,

$t = \frac{k T}{ϵ}$

Therefore,

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

The average energy is:

$\bar{E} = - \frac{1}{Z} \frac{\partial Z}{\partial β} (2)$

By chain rule:

$\frac{\partial Z}{\partial β} = \frac{\partial Z}{\partial t} \frac{\partial t}{\partial β} \frac{\partial Z}{\partial β} = \frac{\partial Z}{\partial t} {(\frac{\partial β}{\partial t})}^{- 1}$

But $β = 1 / k T, hence β = 1 / t ϵ$ hence,

$\frac{\partial Z}{\partial β} = \frac{\partial}{\partial t} (\sum (2 j + 1) e^{- j (j + 1) / t}) {(\frac{\partial}{\partial t} (\frac{1}{t ϵ}))}^{- 1} \frac{\partial Z}{\partial β} = (\sum j (j + 1) (2 j + 1) e^{- j (j + 1) / t}) (\frac{1}{t^{2}}) {(- \frac{1}{t^{2} ϵ})}^{- 1} \frac{\partial Z}{\partial β} = - \sum ϵ j (j + 1) (2 j + 1) e^{- j (j + 1) / t}$

Substitute into (2)

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

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

$C = \frac{\partial \bar{E}}{\partial T} C = \frac{k}{ϵ} \frac{\partial \bar{E}}{\partial t}$

Substitute from (3):

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