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.CMD HEADER_FOOTER 1 0 *empty* |F *empty* 0 1 Created:^Spring^1997|NModified:^August^1997 |F Author;^Sidney^H.^Young^&Theresa^Julia^Zielinski|NPage^^|P
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.TXT 6 1 317 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard\qc {\fs24\b 
A Summary of Statistical Thermodynamic Calculation}{\fs24\b s}{\cf2\fs28 ©}
\par \par by}
.TXT 7 10 944 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard\qc Sidney Young\par 
\pard\qc Department of Chemistry\par University of South Alabama\par 
Mobile AL 36688-0002\par \par syoung@jaguar1.usouthal.edu}
.TXT 0 35 942 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Theresa Julia Zielinski
\par Department of Chemistry\par Niagara University\par Niagara 
University NY, 14109\par \par theresaz@localnet.com}
.TXT 4 -8 943 0 0
Cg a36.000000,36.000000,3
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard and}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\fs18 © 
Copyright Sidney H. Young and Theresa Julia Zielinksi,  1997.  All 
rights reserved.}{\cf2\fs18  You are welcome to use this document in 
your own classes but commercial use is not allowed without the 
permission of the author.}}
.TXT 7 3 947 0 0
Cg a70.250000,70.250000,1972
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\b Introduction}:  In 
this document the fundamental equations for calculating the 
thermodynamic properties of molecules and equilibrium constants for 
reactions are presented. You will see how the translation, rotational, 
vibrational and electronic states of molecules contribute to their 
thermodynamic properties.  After studying this document you will be 
able to use the Mathcad equations as templates for determining the 
thermodynamic properties of molecules and reactions of interest to you 
and for your own projects. \par \par {\b Goal}: To provide students 
with the basic statistical thermodynamic equations for computing 
thermodynamic properties of molecules and reactions.\par \par {\b 
Prerequisites:}\par 1.   Basic skills with Mathcad up to and including 
taking derivatives and using\par       evaluate from the Symbolic drop 
down menu. \par 2.   Knowledge of the existence of quantum mechanical 
energy levels; prior study \par       of the classical ideas of 
translation, rotation, and vibrational modes that \par       
contribute to heat capacity. \par 3.   Be able to distinguish between 
linear and nonlinear molecules.\par 4.   Knowledge of the relationship 
between the Gibbs free energy and the \par       equilibrium constant.
\par \par {\b Performance Object}{\b ives}:  \par 1.   Compute the 
translational, rotational, vibrational, and electronic partition 
functions. \par 2.   Compute A, U, S, G,   etc. thermodynamic 
functions from the translational, rotational, \par       vibrational, 
and electronic partition functions.\par 3.   Compute the equilibrium 
constant for a reaction from the partition functions of \par       the 
reactants and products.\par 4.   Explain the shape of heat capacity as 
a function of temperature curves. \par 5.   Identify the contribution 
from rotation, vibration, translation, and electronic \par       
degrees of freedom of the molecule or atom.\par 6.   Explain the shape 
of heat capacity curves in terms of the degree of importance of the
\par       rotation, vibration, translation, and electronic degrees of 
freedom in the atom or molecule.}
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.TXT 4 4 544 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b Part I.}{
\cf2 \par }{\cf2\b \par Thermodynamic properties of molecular nitrogen 
from statistical thermodynamics}}
.TXT 8 -1 543 0 0
Cg a67.000000,67.000000,273
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The first substance that 
we will study in this statistical mechanics exploration is molecular 
nitrogen. We will step through the process of determining the 
thermodynamic functions and heat capacity as a function of 
temperature. We will liberally use units in this exercise.}
.TXT 9 0 366 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard First define some constants:}
.EQN 4 0 948 0 0
{0:N}NAME:6.02*(10)^(23)
.TXT 0 15 367 0 0
Cg a55.000000,55.000000,17
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Avogadro's number}
.EQN 5 -15 368 0 0
{0:k}NAME:1.38*(10)^(-23)*{0:joule}NAME*({0:K}NAME)^(-1)
.EQN 0 22 369 0 0
{0:k}NAME={0}?_n_u_l_l_
.TXT 0 28 370 0 0
Cg a20.000000,20.000000,17
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Boltzman constant}
.EQN 5 -50 371 0 0
{0:h}NAME:6.626*(10)^(-34)*{0:joule}NAME*{0:sec}NAME
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{0:h}NAME={0}?_n_u_l_l_
.TXT 0 28 545 0 0
Cg a20.000000,20.000000,17
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Planck's constant}
.EQN 7 -50 376 0 0
{0:mass}NAME:(.028)/({0:N}NAME)*{0:kg}NAME
.EQN 0 22 377 0 0
{0:mass}NAME={0}?_n_u_l_l_
.TXT 0 26 546 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Mass of Nitrogen}
.EQN 6 -49 411 0 0
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.EQN 0 23 412 0 0
{0:c}NAME={0}?_n_u_l_l_
.TXT 0 25 414 0 0
Cg a23.000000,23.000000,14
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Speed of light}
.EQN 5 -47 397 0 0
{0:R}NAME:8.314*{0:joule}NAME*({0:K}NAME)^(-1)
.EQN 0 18 398 0 0
{0:R}NAME={0}?_n_u_l_l_
.TXT 0 29 399 0 0
Cg a23.000000,23.000000,16
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The gas constant}
.TXT 5 -48 396 0 0
Cg a70.000000,70.000000,69
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now set the properties 
of the gas, i.e. the temperature and pressure }
.EQN 5 1 393 0 0
{0:T}NAME:300*{0:K}NAME
.EQN 4 0 394 0 0
{0:P}NAME:1*{0:atm}NAME
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.EQN 5 -17 400 0 0
{0:V}NAME:({0:R}NAME*{0:T}NAME)/({0:P}NAME)
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{0:V}NAME={0}?_n_u_l_l_
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Compute the volume}
.TXT 8 -37 405 0 0
Cg a67.000000,67.000000,77
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b A. 
Calculating the Partition Functions}{\cf2 \par \par }1. The 
translation partition function}
.EQN 10 1 406 0 0
{0:qt}NAME:(((2*{0:\p}NAME*{0:mass}NAME*{0:k}NAME*{0:T}NAME)/(({0:h}NAME)^(2))))^(1.5)*{0:V}NAME
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{0:qt}NAME={0}?_n_u_l_l_
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.TXT 4 4 408 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard 2. The rotation 
partition function}
.EQN 6 3 32 0 0
{0:B}NAME:2.010*({0:cm}NAME)^(-1)
.EQN 1 17 549 0 0
{0:B}NAME={0}?_n_u_l_l_
.TXT 0 13 550 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Rotational Wave Numbers}
.EQN 7 -30 421 0 0
{0:Thetar}NAME:({0:B}NAME*{0:c}NAME*{0:h}NAME)/({0:k}NAME)
.TXT 4 34 551 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Thetar  is the 
rotational temperature}
.EQN 1 -17 424 0 0
{0:Thetar}NAME={0}?_n_u_l_l_
.EQN 8 -16 419 0 0
{0:qr}NAME:({0:T}NAME)/(2*{0:Thetar}NAME)
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The 
rotational partition function tells us the approximate number of 
populated states.\par \par {\cf2 On a separate worksheet calculate the 
rotational partition function for a series of diatomic gases and 
determine how the number of populated state varies as a function of 
mass and bond length. Why is bond length important here? Explain.}}
.EQN 7 -18 425 0 0
{0:qr}NAME={0}?_n_u_l_l_
.TXT 15 0 50 0 0
Cg a65.000000,65.000000,36
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard 3. The vibration 
partition function.}
.EQN 6 0 552 0 0
{0:nu}NAME:2359.61*({0:cm}NAME)^(-1)
.TXT 0 27 884 0 0
Cg a23.000000,23.000000,57
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard nu is the vibrational 
frequency of the nitrogen molecule.}
.EQN 5 -19 558 0 0
{0:nu}NAME={0}?_n_u_l_l_
.EQN 4 -9 434 0 0
{0:Thetav}NAME:({0:h}NAME*{0:nu}NAME*{0:c}NAME)/({0:k}NAME)
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Thetav  is the 
vibrational temperature.  }
.EQN 6 -16 559 0 0
{0:Thetav}NAME={0}?_n_u_l_l_
.EQN 9 -8 562 0 0
{0:qv}NAME:(1)/(1-({0:e}NAME)^((-{0:Thetav}NAME)/({0:T}NAME)))
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard qv is the 
vibrational partition function.\par \par {\cf2 What is the 
significance of having a large vibrational temperature? How does this 
correlate with the result computed here to the left that shows that 
the vibrational partition function qv=1.   } }
.EQN 8 -19 563 0 0
{0:qv}NAME={0}?_n_u_l_l_
.TXT 8 -8 565 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard 4. The electronic 
partition function}
.EQN 6 2 593 0 0
{0:qe}NAME:1
.TXT 0 17 594 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 Why is 
the electronic partition function equal to 1?}}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard We explicitly include qe 
here to retain generality of the method in this document.}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The total partition 
function is the product of the four contributing partition functions.      }
.EQN 6 -20 588 0 0
{0:q}NAME={0}?_n_u_l_l_
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Q is the canonical 
ensemble molar partition function. It is this quantity that is 
connected to the thermodynamic property known as the Helmholtz free 
energy. Q is so large that we can only calculate lnQ.}
.EQN 12 -22 602 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 Using 
Sterling's approximation show that }{\object{\*\objclass \eqn}
\rsltpict{\*\objdata 
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{0:Q}NAME÷(({0:q}NAME)^({0:n}NAME))/(({0:n}NAME)!)
}}{\cf2  leads to the equation on the left.}}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now that we have lnQ we 
can calculate the thermodynamic properties of nitrogen gas at the 
temperature and pressure  specified near the top of this document. }
.EQN 3 -22 831 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b B. 
Calculating the Thermodynamic Properties}}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard 1. Helmholtz free energy}
.EQN 4 7 485 0 0
{0:A}NAME:-{0:k}NAME*{0:T}NAME*{0:lnQ}NAME
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{0:A}NAME={0}?_n_u_l_l_
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\fs16 Notice how u is 
defined here}}
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{0:E}NAME:(3)/(2)*{0:R}NAME*{0:T}NAME+{0:R}NAME*{0:T}NAME+({0:R}NAME*{0:Thetav}NAME)/(({0:e}NAME)^({0:u}NAME)-1)
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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\dn v }and C{\dn p}  }
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard See below for the 
derivation of the of the vibration heat capacity term.}
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
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free energy.}
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.EQN 3 -18 621 0 0
{0:G}NAME:{0:H}NAME-{0:T}NAME*{0:S}NAME
.EQN 1 18 624 0 0
{0:G}NAME={0}?_n_u_l_l_
.TXT 5 -25 628 0 0
Cg a68.000000,68.000000,35
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard 6. The Helmholtz free 
energy again.}
.EQN 3 6 629 0 0
{0:A}NAME:{0:E}NAME-{0:T}NAME*{0:S}NAME
.EQN 0 14 841 0 0
{0:A}NAME={0}?_n_u_l_l_
.TXT 4 -1 844 0 0
Cg a39.625000,39.625000,90
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 
Compare with above. This is a redundancy confirmation to check that 
everything is working.}}
.TXT 5 -23 635 0 0
C x1,1,0,0
.TXT 6 5 318 0 0
Cg a60.500000,60.500000,197
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b Part 
II.  Mathematical Interlude}:\par \par Below we see the derivation of 
the vibrational contribution to the heat capacity. This shows how the 
vibration term arises from the vibration partition function. }
.EQN 14 -1 517 0 0
{0:qv}NAME÷(1)/(1-({0:e}NAME)^((-{0:Thetav}NAME)/({0:T}NAME)))
.TXT 0 20 636 0 0
Cg a44.000000,44.000000,116
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Start with the 
definition of the vibration partition function and use the ctrl = key 
sequence to write the equation.}
.EQN 12 -20 520 0 0
{0:E}NAME÷{0:R}NAME*({0:T}NAME)^(2)*{0:T}NAME"{0:ln}NAME({0:qv}NAME)
.TXT 0 20 887 0 0
Cg a44.000000,44.000000,81
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we write the 
definition of E in terms of the vibrational partition function.}
.TXT 10 5 638 0 0
Cg a37.375000,37.375000,170
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard You can obtain this 
equation by replacing qv by is equivalent defined above using cut and 
paste. Then symbolically evaluate the derivative as described 
immediately below.}
.EQN 1 -25 516 0 0
{0:E}NAME÷{0:R}NAME*({0:T}NAME)^(2)*{0:T}NAME"{0:ln}NAME((1)/(1-({0:e}NAME)^((-{0:Thetav}NAME)/({0:T}NAME))))
.TXT 15 28 640 0 0
Cg a36.000000,36.000000,215
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard This is 
obtained by placing the selection box around the derivative part of 
the equation just above and choosing Evaluate followed by using 
Evaluate Symbolically from the Symbolic menu. {\cf2 Try these steps 
for practice.}}
.EQN 2 -30 639 0 0
{0:E}NAME÷-{0:R}NAME*{0:Thetav}NAME*({0:exp}NAME((-{0:Thetav}NAME)/({0:T}NAME)))/((-1+{0:exp}NAME((-{0:Thetav}NAME)/({0:T}NAME))))
.TXT 12 1 888 0 0
Cg a65.000000,65.000000,23
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard by substitution, yields}
.TXT 7 34 647 0 0
Cg a24.375000,24.375000,54
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Use cut and past to 
generate the equation to the left.}
.EQN 1 -35 645 0 0
{0:E}NAME÷(({0:R}NAME)/((1-{0:exp}NAME(-{0:u}NAME)))*{0:u}NAME*{0:T}NAME*{0:exp}NAME(-{0:u}NAME)÷({0:R}NAME*{0:u}NAME*{0:T}NAME)/(({0:e}NAME)^({0:u}NAME)-1))
.TXT 5 -2 661 0 0
C x1,1,0,0
.TXT 3 3 655 0 0
Cg a69.000000,69.000000,70
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard We can now extend the 
derivation to obtain the heat capacity equation.}
.TXT 6 1 656 0 0
Cg a69.000000,69.000000,13
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Starting with}
.EQN 0 14 657 0 0
{0:E}NAME÷(5)/(2)*{0:R}NAME*{0:T}NAME+({0:R}NAME*{0:u}NAME*{0:T}NAME)/((({0:e}NAME)^({0:u}NAME)-1))
.TXT 0 17 659 0 0
Cg a38.000000,38.000000,5
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard where}
.EQN 0 9 660 0 0
{0:u}NAME÷({0:Thetav}NAME)/({0:T}NAME)
.TXT 8 -40 663 0 0
Cg a69.000000,69.000000,26
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Substituting for u 
yields }
.EQN 1 21 665 0 0
{0:E}NAME÷(5)/(2)*{0:R}NAME*{0:T}NAME+{0:R}NAME*({0:Thetav}NAME)/(({0:exp}NAME(({0:Thetav}NAME)/({0:T}NAME))-1))
.TXT 10 -21 666 0 0
Cg a69.000000,69.000000,52
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Next take the derivative 
with respect to T to obtain}
.TXT 1 49 845 0 0
Cg a20.000000,20.000000,34
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Do this step to verify 
the result.}
.EQN 6 -45 112 0 0
{0:Cv}NAME÷(5)/(2)*{0:R}NAME+{0:R}NAME*(({0:Thetav}NAME)^(2))/(((({0:exp}NAME(({0:Thetav}NAME)/({0:T}NAME))-1))^(2)*({0:T}NAME)^(2)))*{0:exp}NAME(({0:Thetav}NAME)/({0:T}NAME))
.EQN 1 43 117 0 0
{0:Thetav}NAME÷{0:u}NAME*{0:T}NAME
.TXT 10 -44 667 0 0
Cg a65.000000,65.000000,23
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard by substitution, yields}
.EQN 7 3 670 0 0
{0:Cv}NAME÷(5)/(2)*{0:R}NAME+{0:R}NAME*(({0:u}NAME)^(2))/((({0:exp}NAME({0:u}NAME)-1))^(2))*{0:exp}NAME({0:u}NAME)
.TXT 9 -10 671 0 0
C x1,1,0,0
.TXT 6 2 672 0 0
Cg a69.000000,69.000000,90
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b Part 
III. Obtaining the equilibrium constant for N}{\cf2\b\dn 2}{\cf2\b  
<---> 2 N as a function of temperature}}
.TXT 6 2 673 0 0
Cg a69.000000,69.000000,138
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard First obtain the 
partition function for each component of the equilibrium. In this case 
we need the partition function for both N{\dn 2} and N. }
.EQN 10 0 143 0 0
{0:qN2}NAME({0:T}NAME):(((2*{0:\p}NAME*{0:mass}NAME*{0:k}NAME*{0:T}NAME)/(({0:h}NAME)^(2))))^(1.5)*{0:V}NAME*({0:T}NAME)/(2*{0:Thetar}NAME)*(1)/(1-({0:e}NAME)^((-{0:Thetav}NAME)/({0:T}NAME)))*1
.TXT 0 43 674 0 0
Cg a19.750000,19.750000,29
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The partition function 
for N{\dn 2}}
.EQN 9 -42 157 0 0
{0:massN}NAME:(.014)/({0:N}NAME)*{0:kg}NAME
.TXT 0 22 684 0 0
Cg a46.000000,46.000000,27
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The mass of atomic nitrogen}
.TXT 6 -21 686 0 0
Cg a47.625000,47.625000,54
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now compute the atomic 
partition function for atomic N}
.EQN 7 -3 160 0 0
{0:qN}NAME({0:T}NAME):(((2*{0:\p}NAME*{0:massN}NAME*{0:k}NAME*{0:T}NAME)/(({0:h}NAME)^(2))))^(1.5)*{0:V}NAME*(4+6*({0:e}NAME)^((-27658.7*{0:K}NAME)/({0:T}NAME))+4*({0:e}NAME)^((-27671.7*{0:K}NAME)/({0:T}NAME))+6*({0:e}NAME)^((-41492.4*{0:K}NAME)/(
{0:T}NAME)))
.TXT 6 6 889 0 0
Cg a62.500000,62.500000,435
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Immediately 
above is the partition function for atomic nitrogen. It includes 
contributions from the first three excited states.\par \par Note:  
This can be coupled to determination of the term symbols for the 
states of nitrogen as a link between statistical thermodynamics and 
spectroscopy. {\cf2 Recompute the partition function without the 
highest energy level term. Decide if this energy level is essential in 
the final heat capacity calculation }}
.EQN 18 -6 161 0 0
{0:lnQN2}NAME({0:T}NAME):{0:N}NAME*{0:ln}NAME({0:qN2}NAME({0:T}NAME))-{0:N}NAME*{0:ln}NAME({0:N}NAME)+{0:N}NAME
.TXT 0 36 324 0 0
Cg a22.125000,22.125000,66
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we are computing 
the natural log of molar partition functions}
.EQN 4 -36 162 0 0
{0:lnQN}NAME({0:T}NAME):{0:N}NAME*{0:ln}NAME({0:qN}NAME({0:T}NAME))-{0:N}NAME*{0:ln}NAME({0:N}NAME)+{0:N}NAME
.EQN 7 1 165 0 0
{0:AN2}NAME({0:T}NAME):-{0:k}NAME*{0:T}NAME*{0:lnQN2}NAME({0:T}NAME)
.TXT 0 28 849 0 0
Cg a36.000000,36.000000,80
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now we calculate the 
Helmholtz free energy for each component of the equilibrium}
.EQN 5 -28 166 0 0
{0:AN}NAME({0:T}NAME):-{0:k}NAME*{0:T}NAME*{0:lnQN}NAME({0:T}NAME)
.EQN 6 1 699 0 0
{0:GN2}NAME({0:T}NAME):{0:AN2}NAME({0:T}NAME)+{0:R}NAME*{0:T}NAME
.TXT 0 28 702 0 0
Cg a36.000000,36.000000,199
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we calculate the 
Gibbs free energy. Using what you know and your text explain why the 
equation to calculate the Gibbs free energy is different from that 
used for the molecular nitrogen example. }
.EQN 6 -28 700 0 0
{0:GN}NAME({0:T}NAME):{0:AN}NAME({0:T}NAME)+{0:R}NAME*{0:T}NAME
.TXT 5 -5 745 0 0
C x1,1,0,0
.TXT 7 4 744 0 0
Cg a69.000000,69.000000,21
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now calculate delta G}
.EQN 4 0 169 0 0
{0:Delta_G}NAME({0:T}NAME):(2*{0:GN}NAME({0:T}NAME)-{0:GN2}NAME({0:T}NAME))+(9.756*1.6*(10)^(-19)*{0:N}NAME*{0:joule}NAME)
.TXT 5 8 891 0 0
Cg a52.875000,52.875000,250
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The third term in the 
expression for the change in Gibbs free energy is necessary because 
the bottoms of the potential energy functions of the products and 
reactants do not match. One is displaced from the of the other by the 
dissociation energy D{\dn o}. }
.TXT 14 16 850 0 0
Cg a42.500000,42.500000,196
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 
Calculate the equilibrium constant  and the value of the equilibrium 
constant at several temperatures. How does K vary with temperature? 
How does the equilibrium shift with increasing temperature?}}
.EQN 1 -22 706 0 0
{0:Keq}NAME({0:T}NAME):({0:e}NAME)^((-{0:Delta_G}NAME({0:T}NAME))/({0:R}NAME*{0:T}NAME))
.EQN 6 0 708 0 0
{0:Keq}NAME(300*{0:K}NAME)={0}?_n_u_l_l_
.EQN 6 0 720 0 0
{0:Keq}NAME(1000*{0:K}NAME)={0}?_n_u_l_l_
.EQN 4 0 722 0 0
{0:Keq}NAME(8000*{0:K}NAME)={0}?_n_u_l_l_
.TXT 6 15 893 0 0
Cg a45.000000,45.000000,126
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 At 
what temperature does this reaction become spontaneous? What is K at 
the point where the reaction just becomes spontaneous?}}
.TXT 12 6 851 0 0
Cg a42.000000,42.000000,485
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard To answer this question 
you want to find the temperature where the equilibrium constant equals 
1. To do this we can set up a solve block. Make an initial guess for 
this temperature based on your observations on how the equilibrium 
constant varies with temperature. Here we have chosen T to be 8000 K.  
Next set up the solve block using the Given statement. Set the 
condition for the solution and then use the Find function. If you have 
a good guess then the temperature will be found. }
.EQN 3 -17 894 0 0
{0:T}NAME:8000*{0:K}NAME
.EQN 4 -6 733 0 0
{0:Given}NAME
.EQN 4 6 896 0 0
{0:Keq}NAME({0:T}NAME)÷1
.EQN 5 0 895 0 0
{0:Tout}NAME:{0:Find}NAME({0:T}NAME)
.EQN 8 0 741 0 0
{0:Tout}NAME={0}?_n_u_l_l_
.TXT 0 21 742 0 0
Cg a42.000000,42.000000,58
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we have the 
temperature and the equilibrium constant.}
.EQN 4 -21 740 0 0
{0:Keq}NAME({0:Tout}NAME)={0}?_n_u_l_l_
.TXT 2 -10 743 0 0
C x1,1,0,0
.TXT 5 5 199 0 0
Cg a68.000000,68.000000,67
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b Part 
IV. Heat Capacity of a Linear Polyatomic, CO}{\cf2\b\dn 2}{\cf2\b , at 
15 C, 1 atm.}}
.EQN 5 0 751 0 0
{0:T}NAME:293*{0:K}NAME
.EQN 5 0 752 0 0
{0:Cvt}NAME:(3)/(2)*{0:R}NAME
.TXT 0 33 897 0 0
Cg a34.000000,34.000000,130
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 How 
many modes of vibration are present in CO}{\cf2\dn 2}{\cf2 ? Are there 
degenerate modes? What are the vibrational temperatures for these modes?}}
.EQN 3 -19 754 0 0
{0:Thetav}NAME:({4,1}ö960*{0:K}NAMEö960*{0:K}NAMEö3380*{0:K}NAMEö1997*{0:K}NAME)
.EQN 2 -14 753 0 0
{0:Cvr}NAME:{0:R}NAME
.TXT 14 31 852 0 0
Cg a28.000000,28.000000,134
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we compute the 
vibrational contribution to the heat capacity. This is an extension of 
the formula we used for nitrogen gas above.}
.EQN 1 -32 219 0 0
{0:Cvv}NAME:{0:R}NAME*((0,3,{0:i}NAME,((((({0:Thetav}NAME)[({0:i}NAME))/({0:T}NAME)))^(2)*({0:e}NAME)^(((({0:Thetav}NAME)[({0:i}NAME))/({0:T}NAME))))/(((({0:e}NAME)^(((({0:Thetav}NAME)[({0:i}NAME))/({0:T}NAME)))-1))^(2))){64})
.EQN 10 0 756 0 0
{0:Cvv}NAME={0}?_n_u_l_l_
.EQN 6 0 762 0 0
{0:Cvtot}NAME:{0:Cvt}NAME+{0:Cvr}NAME+{0:Cvv}NAME
.EQN 4 0 758 0 0
{0:Cvtot}NAME={0}?_n_u_l_l_
.TXT 0 30 900 0 0
Cg a32.750000,32.750000,40
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we compute the 
total heat capacity.}
.TXT 4 -29 899 0 0
Cg a31.000000,31.000000,10
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 Lit: 28.09}}
.TXT 5 -5 767 0 0
C x1,1,0,0
.TXT 5 3 766 0 0
Cg a68.000000,68.000000,69
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b Part 
V. Heat Capacity of a nonlinear Polyatomic, H}{\cf2\b\dn 2}{\cf2\b O}{
\cf2\b , at 15 C, 1 atm.}}
.EQN 6 1 230 0 0
{0:Cvt}NAME:(3)/(2)*{0:R}NAME
.TXT 0 22 901 0 0
Cg a40.000000,40.000000,46
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Translation contribution 
to the heat capacity.}
.EQN 7 -22 768 0 0
{0:Cvt}NAME={0}?_n_u_l_l_
.TXT 7 22 902 0 0
Cg a40.000000,40.000000,140
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Rotational contribution 
to the heat capacity. Why is the rotational contribution to the heat 
capacity for water different from that for CO{\dn 2}?}
.EQN 2 -21 231 0 0
{0:Cvr}NAME:(3)/(2)*{0:R}NAME
.EQN 7 0 770 0 0
{0:Cvr}NAME={0}?_n_u_l_l_
.TXT 8 26 903 0 0
Cg a31.625000,31.625000,100
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 How 
are the number of modes of vibration for H}{\cf2\dn 2}{\cf2 O 
different from the number of modes for CO}{\cf2\dn 2}{\cf2  and why?}}
.EQN 2 -25 232 0 0
{0:Thetav}NAME:({3,1}ö5404*{0:K}NAMEö2995*{0:K}NAMEö5254*{0:K}NAME)
.EQN 15 -1 778 0 0
{0:Cvv}NAME:{0:R}NAME*((0,2,{0:i}NAME,((((({0:Thetav}NAME)[({0:i}NAME))/({0:T}NAME)))^(2)*({0:e}NAME)^(((({0:Thetav}NAME)[({0:i}NAME))/({0:T}NAME))))/(((({0:e}NAME)^(((({0:Thetav}NAME)[({0:i}NAME))/({0:T}NAME)))-1))^(2))){64})
.TXT 0 32 780 0 0
Cg a27.875000,27.875000,71
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we are computing 
the vibrational contribution to the heat capacity}
.EQN 12 -31 777 0 0
{0:Cvv}NAME={0}?_n_u_l_l_
.EQN 8 1 238 0 0
{0:Cvtot}NAME:{0:Cvt}NAME+{0:Cvr}NAME+{0:Cvv}NAME
.TXT 0 31 785 0 0
Cg a35.000000,35.000000,40
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we compute the 
total heat capacity.}
.EQN 5 -31 783 0 0
{0:Cvtot}NAME={0}?_n_u_l_l_
.TXT 4 0 784 0 0
Cg a31.000000,31.000000,10
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 Lit: 25.28}}
.TXT 4 -7 786 0 0
C x1,1,0,0
.TXT 4 8 787 0 0
Cg a64.000000,64.000000,124
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red255\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2\b Part 
VI. The strange case of NO: The effect of a low-lying electronic \par         
     energy level on the heat capacity of NO.}}
.EQN 6 0 788 0 0
{0:Cvt}NAME:(3)/(2)*{0:R}NAME
.TXT 0 19 789 0 0
Cg a46.000000,46.000000,29
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The translation contribution.}
.TXT 5 0 791 0 0
Cg a46.000000,46.000000,28
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The rotational contribution.}
.EQN 1 -19 790 0 0
{0:Cvr}NAME:{0:R}NAME
.EQN 6 0 927 0 0
{0:Thetav}NAME:(1904.20*({0:cm}NAME)^(-1)*{0:c}NAME*{0:h}NAME)/({0:k}NAME)
.TXT 1 30 795 0 0
Cg a35.000000,35.000000,28
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard The vibrational temperature.}
.EQN 5 -30 928 0 0
{0:Thetav}NAME={0}?_n_u_l_l_
.EQN 7 -4 904 0 0
{0:Cvv}NAME({0:T}NAME):{0:R}NAME*(({0:Thetav}NAME)^(2))/(((({0:exp}NAME(({0:Thetav}NAME)/({0:T}NAME))-1))^(2)*({0:T}NAME)^(2)))*{0:exp}NAME(({0:Thetav}NAME)/({0:T}NAME))
.TXT 0 41 905 0 0
Cg a24.000000,24.000000,58
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard This is the vibrational 
contribution to the heat capacity.}
.TXT 10 -42 906 0 0
Cg a70.000000,70.000000,643
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now we will consider the 
electronic partition function for the NO molecule. This molecule has 
one unpaired electron in the highest occupied molecular orbital. The 
molecular term symbol is a doublet. Also there is a low lying excited 
state molecular orbital. This level is sufficiently low so that a 
significant population of molecules have the single electron in this 
energy level. This energy level will be singly occupied and thus have 
a doublet molecular term symbol. Since both of these levels are 
doublets they have a degeneracy of 2. Higher energy molecular orbitals 
are not populated so they do not contribute to the partition function.}
.TXT 18 22 911 0 0
Cg a41.625000,41.625000,228
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we write the 
Mathcad numerical definition of the partition function for electronic 
partition function. The first term on the right hand side of this 
equation is for the ground state. Why is this term simply the integer 
2?  }
.EQN 2 -21 909 0 0
{0:qe}NAME:2+2*({0:e}NAME)^((-174.2*{0:K}NAME)/({0:T}NAME))
.TXT 9 22 914 0 0
Cg a38.875000,38.875000,85
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Here we have the 
electronic partition function written in the Mathcad symbolic form. }
.EQN 4 -22 917 0 0
{0:qe}NAME({0:T}NAME)÷2+2*({0:e}NAME)^((-174.2*{0:K}NAME)/({0:T}NAME))
.TXT 1 22 916 0 0
Cg a40.000000,40.000000,122
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Place the blue selection 
cursor next to the T in the equation and choose differentiate on 
variable from the Symbolic menu.}
.TXT 7 0 949 0 0
Cg a34.875000,34.875000,82
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\fs16 Note: you can 
take the derivative of a symbolic but not of a numerical definition.}}
.TXT 5 0 924 0 0
Cg a65.000000,65.000000,56
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard This yields 
the following equation. {\cf2 Try it for yourself.}}
.EQN 5 -22 925 0 0
{0:T}NAME"{0:qe}NAME({0:T}NAME)÷348.4*({0:K}NAME)/(({0:T}NAME)^(2))*{0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME))
.TXT 5 -4 929 0 0
C x1,1,0,0
.TXT 5 35 814 0 0
Cg a38.000000,38.000000,49
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Create the equation on 
the left by cut and paste.}
.EQN 5 -32 926 0 0
{0:Ue}NAME({0:T}NAME)÷({0:R}NAME*({0:T}NAME)^(2)*((348.4*{0:K}NAME)/(({0:T}NAME)^(2))*{0:exp}NAME((-174.2)/({0:T}NAME))))/(2+2*({0:e}NAME)^((-174.2*{0:K}NAME)/({0:T}NAME)))
.TXT 0 34 930 0 0
Cg a31.000000,31.000000,76
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 What 
is the electronic heat capacity in the limit of very high temperatures?}}
.TXT 10 -34 931 0 0
Cg a66.000000,66.000000,76
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Select the equation 
above and choose Symbolic Simplify to get the following:}
.EQN 10 4 825 0 0
{0:Ue}NAME({0:T}NAME)÷174.2*{0:R}NAME*{0:K}NAME*({0:exp}NAME((-174.2)/({0:T}NAME)))/((1.+{0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME))))
.TXT 11 -1 299 0 0
Cg a67.000000,67.000000,97
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard 
differentiation yields ({\cf2 Verify the result shown here by using 
the appropriate Mathcad procedure.})}
.EQN 11 0 300 0 0
{0:T}NAME"{0:Ue}NAME({0:T}NAME)÷30345.64*{0:R}NAME*(({0:K}NAME)^(2))/(({0:T}NAME)^(2))*({0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME)))/((1.+{0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME))))-30345.64*{0:R}NAME*({0:K}NAME)^(2)*(({0:exp}NAME(-174.2*({0:K}NAME)/(
{0:T}NAME)))^(2))/((((1.+{0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME))))^(2)*({0:T}NAME)^(2)))
.TXT 8 0 301 0 0
Cg a67.000000,67.000000,13
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard simplifies to}
.EQN 7 0 302 0 0
{0:T}NAME"{0:Ue}NAME({0:T}NAME)÷30345.64*{0:R}NAME*({0:K}NAME)^(2)*({0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME)))/((((1.+{0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME))))^(2)*({0:T}NAME)^(2)))
.TXT 9 39 932 0 0
Cg a26.000000,26.000000,58
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Cut and paste to create 
the numeric heat capacity equation}
.EQN 7 -39 303 0 0
{0:Cve}NAME({0:T}NAME):30345.64*{0:R}NAME*({0:K}NAME)^(2)*({0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME)))/((((1.+{0:exp}NAME(-174.2*({0:K}NAME)/({0:T}NAME))))^(2)*({0:T}NAME)^(2)))
.TXT 8 -6 950 0 0
C x1,1,0,0
.TXT 3 7 860 0 0
Cg a66.000000,66.000000,71
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now compute the total 
heat capacity and each of the contributing terms.}
.EQN 6 1 862 0 0
{0:Cvt}NAME={0}?_n_u_l_l_
.EQN 4 5 863 0 0
{0:Cvr}NAME={0}?_n_u_l_l_
.EQN 4 7 864 0 0
{0:Cvv}NAME(100*{0:K}NAME)={0}?_n_u_l_l_
.EQN 4 9 865 0 0
{0:Cve}NAME(100*{0:K}NAME)={0}?_n_u_l_l_
.EQN 5 -24 871 0 0
{0:Cv}NAME({0:T}NAME):{0:Cvt}NAME+{0:Cvr}NAME+{0:Cvv}NAME({0:T}NAME)+{0:Cve}NAME({0:T}NAME)
.EQN 5 0 869 0 0
{0:Cv}NAME(40*{0:K}NAME)={0}?_n_u_l_l_
.EQN 4 0 870 0 0
{0:Cvne}NAME({0:T}NAME):{0:Cvt}NAME+{0:Cvr}NAME+{0:Cvv}NAME({0:T}NAME)
.TXT 0 27 877 0 0
Cg a40.000000,40.000000,42
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Heat capacity without 
the electronic term.}
.EQN 4 -27 872 0 0
{0:Cvnv}NAME({0:T}NAME):{0:Cvt}NAME+{0:Cvr}NAME+{0:Cve}NAME({0:T}NAME)
.TXT 0 27 876 0 0
Cg a40.000000,40.000000,43
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Heat capacity without 
the vibrational term.}
.TXT 9 -32 880 0 0
C x1,1,0,0
.TXT 8 3 879 0 0
Cg a70.000000,70.000000,79
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard Now we systematically 
vary the temperature and examine the results graphically.}
.TXT 5 27 934 0 0
Cg a36.250000,36.250000,215
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard  At very low 
temperatures the electronic contribution to the heat capacity 
dominates and saturates as shown by the peak in the graph. The heat 
capacity then declines to a minimum  until the vibration term kicks 
in. }
.EQN 1 -23 868 0 0
{0:i}NAME:20,40;1000
.EQN 11 -3 881 0 0
&&(_n_u_l_l_&_n_u_l_l_)&{0:Cv}NAME({0:i}NAME*{0:K}NAME),{0:Cvne}NAME({0:i}NAME*{0:K}NAME),{0:Cvnv}NAME({0:i}NAME*{0:K}NAME),{0:Cve}NAME({0:i}NAME*{0:K}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&{0:i}NAME
0 0 1 1 1 0 0 1 1 
0 0 1 1 1 0 0 1 1 
0 1 0 0 1 1 NO-TRACE-STRING
0 2 1 0 1 1 NO-TRACE-STRING
0 3 2 0 1 1 NO-TRACE-STRING
0 4 3 0 1 1 NO-TRACE-STRING
0 1 4 0 1 1 NO-TRACE-STRING
0 2 5 0 1 1 NO-TRACE-STRING
0 3 6 0 1 1 NO-TRACE-STRING
0 4 0 0 1 1 NO-TRACE-STRING
0 1 1 0 1 1 NO-TRACE-STRING
0 2 2 0 1 1 NO-TRACE-STRING
0 3 3 0 1 1 NO-TRACE-STRING
0 4 4 0 1 1 NO-TRACE-STRING
0 1 5 0 1 1 NO-TRACE-STRING
0 2 6 0 1 1 NO-TRACE-STRING
0 3 0 0 1 1 NO-TRACE-STRING
0 4 1 0 1 1 NO-TRACE-STRING
0 1 1 21 18 10 0 3 
.TXT 4 38 882 0 0
Cg a24.000000,24.000000,165
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\ul Figure legend}.\par 
\par Cvne is the heat capacity with no electronic term\par \par Cvnv 
is without the vibration term.\par \par Cve is electronic term alone.
\par \par Cv is the total heat capacity.}
.TXT 25 -38 883 0 0
Cg a56.000000,56.000000,225
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red128\green0\blue0;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 By 
inspection of the graph describe where the electronic contribution is 
most important and how this contribution changes with temperature. Why 
is this happening? When does the vibrational contribution become 
important? Why? }}
.TXT 10 -1 953 0 0
Cg a74.000000,74.000000,507
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\b Mastery Exercise:}
\par \par The energy levels for an electron constrained to move on a 
ring with a diameter of 0.4 nm\par  \par are given by {\object{
\*\objclass \eqn}\rsltpict{\*\objdata 
.EQN 1268 13 952 0 0
{0:E.n}NAME÷(({0:hbar}NAME)^(2)*({0:n}NAME)^(2))/(2*{0:m}NAME*({0:d}NAME)^(2))
}} where n is the quantum number identifying the energy level, m is 
the mass of the electron and d is the diameter of the ring. Write the 
expression for the partition function for the electron on the ring. At 
room temperature would the electronic partition function for this 
system contribute significantly to the heat capacity. Explain by 
providing graphical or arithmetic evidence. }
.TXT 23 -3 954 0 0
Cg b78.000000,78.000000,306
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard \par {\b Acknowlegment}: 
The authors acknowledges the National Science Foundation for support 
of the 1997 NSF-UFE Workshop on "Numerical Methods in the 
Undergraduate Chemistry Curriculum Using the Mathcad Software" and the 
organizers (Jeff Madura, Andrzej Wierzbicki and Sidney Young, 
University of South Alabama).\par }