91Ó°ÊÓ

In this problem you will investigate the behavior of ordinary hydrogen, $H_2$, at low temperatures. The constant $\mathrm{Î\mu }$is$0.0076eV$. As noted in the text, only half of the terms in the rotational partition function, $equation6.3$, contribute for any given molecule. More precisely, the set of allowed $j$values is determined by the spin configuration of the two atomic nuclei. There are four independent spin configurations, classified as a single "singlet" state and three "triplet" states. The time required for a molecule to convert between the singlet and triplet configurations is ordinarily quite long, so the properties of the two types of molecules can be studied independently. The singlet molecules are known as parahydrogen while the triplet molecules are known as orthohydrogen.

(a) For parahydrogen, only the rotational states with even values of j are allowed.Use a computer (as in $Problem6.28$) to calculate the rotational partition function, average energy, and heat capacity of a parahydrogen molecule. Plot the heat capacity as a function of $kT/t$

(b) For orthohydrogen, only the rotational states with odd values of $j$are allowed. Repeat part (a) for orthohydrogen.

(c) At high temperature, where the number of accessible even-j states is essentially the same as the number of accessible odd-j states, a sample of hydrogen gas will ordinarily consist of a mixture of $1/4$parahydrogen and $3/4$orthohydrogen. A mixture with these proportions is called normal hydrogen. Suppose that normal hydrogen is cooled to low temperature without allowing the spin configurations of the molecules to change. Plot the rotational heat capacity of this mixture as a function of temperature. At what temperature does the rotational heat capacity fall to half its hightemperature value (i.e., to $k/2$per molecule)?

(d) Suppose now that some hydrogen is cooled in the presence of a catalyst that allows the nuclear spins to frequently change alignment. In this case all terms in the original partition function are allowed, but the odd-j terms should be counted three times each because of the nuclear spin degeneracy. Calculate the rotational partition function, average energy, and heat capacity of this system, and plot the heat capacity as a function of $kT/t$.

(e) A deuterium molecule, $D_2$, has nine independent nuclear spin configurations, of which six are "symmetric" and three are "antisymmetric." The rule for nomenclature is that the variety with more independent states gets called "ortho-," while the other gets called "para-." For orthodeuterium only even-j rotational states are allowed, while for paradeuterium only oddj states are allowed. Suppose, then, that a sample of $D_2$gas, consisting of a normal equilibrium mixture of $2/3$ortho and $1/3$para, is cooled without allowing the nuclear spin configurations to change. Calculate and plot the rotational heat capacity of this system as a function of temperature.*

Step by step solution

01

Part(a )Step 1:Given information

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

02

Part(a) Step 2: Simplify

The steps we will get is:

$Z=∑(2j+1)e^{-j(j+1)/t}$

The average energy is given by:

$\stackrel{¯}{E}=-\frac{1}{Z}\frac{∂Z}{∂\mathrm{β}}$

By using the chain rule :

$\frac{∂Z}{∂\mathrm{β}}=\frac{∂Z}{∂t}\frac{∂t}{∂\mathrm{β}}$

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

By substitution:

$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}$

03

Part(b) Step 1:Given information

We have been given that$C_{mix}=\frac{1}{4}{C}_{para}+\frac{3}{4}{C}_{ortha}$

04

Part(b) Step 2:simplify

The terms used to find the expression:

$\frac{kT}{\mathrm{Ï\mu }}=1.67$

$T=\frac{1.67\mathrm{Ï\mu }}{k}$

$T=\frac{1.67(0.0076\mathrm{eV})}{8.62×{10}^{-5}\mathrm{eV}/\mathrm{K}}$

$T=147\mathrm{K}$

05

Part(c) Step 1:Given information

We have been given that$Z_{eq}=Z_{\text{para}}+3Z_{\text{ortha}}$

06

Part(c) Step 2: Simplify

Parahydrogen is multiplied by factor of $1$and orthohydrogen is by $3$.

07

Part(d) Step 1 : Given information

We have been given that$Z_{eq}=Z_{\text{para}}+3Z_{\text{ortha}}$

08

Part(d) Step 2: Simlify

In this case,the function is in equilibrium$C/ka$

09

Part(e) Step 1: GIven information

We have been given that$C_D=\frac{2}{3}{C}_{para}+\frac{1}{3}{C}_{ortha}$

10

Part(e) Step 2: Simplify

In a normal deuterium and orthohydrogen we have $1/4and3/4$respectively.

the plot of the garph

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!