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\viewkind4\uc1\pard\cf1\f0\fs16 Created: Spring 1997\par
Modified: June 2006\par
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\viewkind4\uc1\pard\qr\cf1\f0\fs16 Theresa Julia Zielinski\par
Sidney H. Young\par
\cf0\f1\fs20\par
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					<c val="#000080">First define some constants:</c>
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					<c val="#000080">Planck's constant</c>
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					<c val="#000080">Mass of One Mole of Nitrogen</c>
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					<c val="#000080">Speed of light</c>
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					<c val="#000080">The gas constant</c>
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					<c val="#000080">Now set the properties of the gas, i.e. the temperature and pressure </c>
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					<c val="#000080">Compute the volume</c>
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					<b>
						<c val="#800000">A. Calculating the Partition Functions</c>
					</b>
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					<c val="#000080">1. </c>
					<b>
						<c val="#000080">The</c>
					</b>
					<b>
						<c val="#000080"> translation partition function</c>
					</b>
					<c val="#000080"> (see your text<sp count="2"/>for more information on this equation)</c>
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					<c val="#000080">The translational partition function, the equation to the left, is derived by considering a particle in a three dimensional box which is chosen for convenience because the actual shape of the box has no effect on thermodynamic properties of a gas. </c>
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					<c val="#000080">Another requirement for using the equation for qt is that the difference between adjacent energy levels be much less than kT. This means that neighboring states almost identical populations. </c>
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					<c val="#000080">qt is a measure of the number of thermally accessible states. This means that states with energy larger than kT are at most only lightly populated if at all and therefor inaccessible. In other words not all translational states are occupied even if in the derivation of the qt equation the full range from 0- </c>
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					<c val="#000080">
						<sp count="2"/>is used as limits of the integration that leads to qt. </c>
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					<c val="#000080">Note that we have here a large number of thermally accessible states at 300K. </c>
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					<b>
						<c val="#f00">Exercise 1: Describe how the number of states available to nitrogen molecules varies with temperature and volume of the container. A quality answer would be presentation of plots and explanation of the significance of the plots. </c>
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					<b>
						<c val="#000080">2. The rotation partition function</c>
					</b>
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					<c val="#000080">Rotational Wave Numbers</c>
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					<c val="#000080">Thetar<sp count="2"/>is the rotational temperature</c>
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					<c val="#000080"> is</c>
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						<c val="#000080"> the symmetry for linear molecules. It is 1 for unsymmetical molecules and 2 for symmetrical molecules. Please refer to your text for elaboration of this concept.</c>
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							<f size="9">Note</f>
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							<c val="#000080">: for more complex molecules the symmetry factor may not be easy to choose. For H</c>
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								<sub>2</sub>
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							<c val="#000080">O and NH</c>
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								<sub>3</sub>
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							<c val="#000080"> these factors are 2 and 3 respectively. For BF</c>
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								<f size="9">3</f>
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							<c val="#000080"> and CH</c>
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								<sub>4</sub>
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							<c val="#000080"> these factors are 6 and 12 respectively. One must be careful in choosing these factors. Symmetry factors will be provided for student calculations </c>
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					<c val="#000080">The rotational partition function tells us the approximate number of populated rotational states.</c>
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					<b>
						<c val="#f00">Exercise 2. 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.</c>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">The 7th energy level is the maximum population.</p>
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					<b>
						<c val="#f00">Exercise 3. Write the expression for the Boltzmann population for rotational energy levels (see Eq 18.35 in McQuarrie and Simon). Compute the Boltzmann population for the first 30 energy levels for T=300K.<sp count="2"/>Prepare a plot of the Boltzmann population as a function of rotational quantum number. For which energy level is the population a maximum? Repeat for T=1000K. </c>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">Fraction of molecules in</p>
				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit"> each rotational level at 300 K</p>
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					<b>
						<c val="#000080">3. The vibration partition function.</c>
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					<c val="#000080">nu is the vibrational frequency of the nitrogen molecule.</c>
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					<c val="#000080">Thetav<sp count="2"/>is the vibrational temperature in Kelvin.<sp count="2"/>
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					<ml:id xml:space="preserve">qv</ml:id>
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					<c val="#000080">qv is the vibrational partition function for diatomic molecules when the vibrational energy is measured from but does not include the zero-point energy. This corresponds to a shift the zero in the energy level scale and the value of E</c>
					<c val="#000080">
						<sub>0</sub>
					</c>
					<c val="#000080"> and results in equations that are easier to use with no loss of generality Noggle, page 243.</c>
					<c val="#000080">. </c>
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					<c val="#000080">You might see the qvib written as on the left here. This expression arises when the bottom of the potential by including zero point energies in the derivation of the vibrational partition function. Again the text you use should be examined for further study.<sp count="3"/>
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					<ml:id xml:space="preserve">qvib</ml:id>
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									<ml:id xml:space="preserve">Thetav</ml:id>
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									<ml:real>2</ml:real>
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										<ml:id xml:space="preserve">Thetav</ml:id>
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					<c val="#000080">We will use the qv expression in the balance of the document.</c>
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					<b>
						<c val="#f00">Exercise 4. 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.<sp count="4"/>
						</c>
					</b>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">
					<b>
						<c val="#f00">Exercise 5. Compute the first five vibrational energy level populations for N</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00"> at 300K, 500K and 1000K. Compare these and then choose a temperature where you expect the population of the second vibrational energy level to contribute noticeably to the vibrational partition function. What generalization can you draw from this set of calculations?</c>
					</b>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">The 13th energy level is the maximum population.</p>
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					<b>
						<c val="#000080">4. The electronic partition function</c>
					</b>
				</p>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">Array below is for 3:</p>
				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">Fraction of molecules in</p>
				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit"> each rotational level at 1000K</p>
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					<c val="#000080">The general expression for the electronic partition function is:</c>
				</p>
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									<ml:real>...</ml:real>
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								<sp count="2"/>where g0, g1, and g2 are the degeneracies of electronic levels 0, 1, and 2. The </c>
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							<c val="#000080">Q </c>
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							<c val="#000080"> values are the electronic energy levels in Kelvin (i.e., the energies divided by kB). If the higher electronic states are far above the ground state then only the first term contributes to the partition function.<sp count="2"/>The nitrogen molecule the partition function is:</c>
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					<b>
						<c val="#f00">Exercise 6. Why is the electronic partition function equal to 1? Verify this result by using the first four electronic energy level values and computing the contribution of each to the partition function at 300K. Repeat the calculations at 2000K and determine if any higher states need to included in the partition function at this temperature. For N</c>
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					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
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						<c val="#f00">, the energy levels (in cm</c>
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					<b>
						<c val="#f00">
							<sup>-1</sup>
						</c>
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					<b>
						<c val="#f00">) are 0, 50203.6, 59619.3, 59808, and 66272.4.</c>
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					<b>
						<c val="#000080">5. The total partition function.</c>
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					<c val="#000080">The total partition function is the product of the four contributing partition functions.<sp count="6"/>
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					<c val="#000080">We explicitly include qe here to retain generality of the method in this document.</c>
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					<c val="#000080">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. Q=Q(N,V,T). For indistinguishable independent molecules </c>
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					<c val="#000080">(such as a mole of nitrogen</c>
					<c val="#000080"> molecules) Q is defined as </c>
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							<c val="#000080">. Consult your text for discussions about the structure of canonical ensembles. Note how the canonical partition function is related to the molecular partition function, q.</c>
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					<c val="#000080">Now that we have lnQ we can calculate the thermodynamic properties of nitrogen gas at the temperature and pressure<sp count="2"/>specified near the top of this document. We will come back to this in a few weeks. </c>
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					<b>
						<c val="#800000">Part II. Heat Capacity of a Linear Polyatomic, CO</c>
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						<c val="#800000">
							<sub>2</sub>
						</c>
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						<c val="#800000">, at 15 C, 1 atm.</c>
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				<p style="Normal" margin-left="0" margin-right="0" text-indent="0" text-align="left" list-style-type="none" tabs="inherit">At 10,000 K about 14.6% of the vibrations are in the second state. Thus we can say that the energy difference between the ground state and the first excited state is a large amount.<sp count="2"/>To excite Nitrogen we need a large temperature due to the triple bond and the linear configuration.<sp count="2"/>Other linear strong bonded molecules will have similarly large energy gaps to excite them.</p>
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						<c val="#f00">Exercise 7. How many vibrational modes are present for CO</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00">? Identify the degenerate modes. Locate the vibrational frequencies (hint: remember the heat of combustion lab?) and confirm the Thetav values. </c>
					</b>
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					<c val="#000080">Next we compute the vibrational contribution to the heat capacity, Cvv. This is an extension of the formula we used for nitrogen gas above.</c>
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					<c val="#000080">Here we compute the total heat capacity.</c>
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					<c val="#f00">Lit: 28.09</c>
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					<b>
						<c val="#800000">Part III. Heat Capacity of a Nonlinear Polyatomic, H</c>
					</b>
					<b>
						<c val="#800000">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#800000">O, at 15 C, 1 atm.</c>
					</b>
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					<b>
						<c val="#f00">Exercise 8. Why is the rotational contribution to the heat capacity for water different from that for CO</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00">?</c>
					</b>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">The electronic partition function is equal to 1 because the first level is equal to the degeneracy (here g = 1). The other levels do not play a role because the energy gaps between the levels are so large that the probability of filling them is extremely low. </p>
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					<b>
						<c val="#f00">Exercise 9. How are the number of modes of vibration for H</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00">O different from the number of modes for CO</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00"> and why?</c>
					</b>
				</p>
			</text>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="0" text-align="inherit" list-style-type="none" tabs="inherit">
					<b>
						<c val="#f00">Exercise 10. Repeat the calculations illustrated for CO</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00"> above for H</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00">O. Some of the steps are outlined below. You may use copy and paste commands where appropriate and useful. Be sure to include both the definitions and the evaluations of the parameters as done above.</c>
					</b>
				</p>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">7. </p>
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						<ml:real>288.15</ml:real>
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							<ml:real>3</ml:real>
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					<c val="#000080">Translation contribution to the heat capacity.</c>
				</p>
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							<ml:real>3380</ml:real>
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							<ml:mult/>
							<ml:real>960</ml:real>
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						</ml:apply>
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							<ml:mult/>
							<ml:real>960</ml:real>
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					<ml:id xml:space="preserve">Cvr</ml:id>
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					<c val="#000080">Rotational contribution to the heat capacity. </c>
				</p>
			</text>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">CO2 has four vibrational modes. The third and fourth mode listed above are degenerate (bending modes). </p>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">These are the vibrational temperatures for the modes of vibration of water at 15 C. </p>
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							<ml:real>667</ml:real>
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							<ml:real>1333.08</ml:real>
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								<ml:real>-1</ml:real>
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							<ml:real>2349.0</ml:real>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="0" text-align="inherit" list-style-type="none" tabs="inherit">
					<c val="#000080">Next compute the vibrational contribution to the heat capacity. Note the use of the three vibrational mode vibrational temperatures.</c>
				</p>
			</text>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">8. The rotational heat capacity depends on the symmetry of the molecule. CO<sub>2</sub> is linear and has only two rotational degrees of freedom. Since it is linear, it has there is only one Thetar value (the two moments of inertia are equal) and σ =2. The qr for CO<sub>2</sub> is</p>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">H<sub>2</sub>O is a nonlinar polyatomic and is an asymmetric top. This means water has three different moments of inertia and three different Thetar values. Now qr is</p>
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					<c val="#000080">Now compute the total heat capacity.</c>
				</p>
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										<ml:apply>
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										<ml:id xml:space="preserve">Thetar</ml:id>
										<ml:id xml:space="preserve">C</ml:id>
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					<c val="#f00">Lit: 25.28</c>
				</p>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="0" text-align="inherit" list-style-type="none" tabs="inherit">
					<b>
						<c val="#f00">Exercise 11. Briefly discuss the accuracy of the statistical mechanical calculations for CO</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00"> and H</c>
					</b>
					<b>
						<c val="#f00">
							<sub>2</sub>
						</c>
					</b>
					<b>
						<c val="#f00">O. </c>
					</b>
				</p>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">9. CO<sub>2</sub> is a linear molecule so it has 2 rotational degrees of freedom. This leave 3n-5 for vibrational degrees of freedom. For CO<sub>2</sub> (three atoms), there are 3*3-5 = 4 vibrational modes. For H<sub>2</sub>O, a nonlinear molecule, there are 3 rotational degrees of freedom. This leaves 3n-6 for vibrational degrees of freedom. Water also has three atoms so there are 3*3-6 = 3 vibrational modes for water. </p>
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				<p style="Normal" margin-left="inherit" margin-right="inherit" text-indent="inherit" text-align="inherit" list-style-type="inherit" tabs="inherit">11. The statistical mechanical calculations are very accurate. For water, the difference between the calculation (25.169 J/mol*K) from the experimental (25.28 J/mol*K) is 0.44%. For CO2, the difference between the calucation and experimental (28.279 and 28.09 J/mol*K respectively) is 0.67%. </p>
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