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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 ofkT/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 E¯=ϵj(j+1)(2j+1)e-j(j+1)/t(2j+1)e-j(j+1)/tand

The average heat capacity is C=k(2j+1)e-j(j+1)/tj2(j+1)2(2j+1)e-j(j+1)/tt(2j+1)e-j(j+1)/t2-kj(j+1)(2j+1)e-j(j+1)/t2t(2j+1)e-j(j+1)/t2

Step by step solution

01

Given information

Heat capacity for values kT/ϵranging from 0 to 3.

j = 6

02

Explanation

Rotational partition function is:

Z=(2j+1)e-j(j+1)ϵ/kT

Let,

t=kTϵ

Therefore,

Z=(2j+1)e-j(j+1)/t(1)

The average energy is:

E¯=-1ZZβ(2)

By chain rule:

Zβ=ZttβZβ=Ztβt-1

But β=1/kT, henceβ=1/tϵhence,

Zβ=t(2j+1)e-j(j+1)/tt1tϵ-1Zβ=j(j+1)(2j+1)e-j(j+1)/t1t2-1t2ϵ-1Zβ=-ϵj(j+1)(2j+1)e-j(j+1)/t

Substitute into (2)

role="math" localid="1647453692824" E¯=ϵj(j+1)(2j+1)e-j(j+1)/t(2j+1)e-j(j+1)/t(3)

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

C=E¯TC=kϵE¯t

Substitute from (3):

role="math" localid="1647454207193" C=ϵktϵj(j+1)(2j+1)e-j(j+1)/t(2j+1)e-j(j+1)/tC=k(2j+1)e-j(j+1)/tj2(j+1)2(2j+1)e-j(j+1)/tt(2j+1)e-j(j+1)/t2-kj(j+1)(2j+1)e-j(j+1)/t2t(2j+1)e-j(j+1)/t2

03

Explanation

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

The graph is:

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