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Applications of diatomic and polyatomic ideal gases 1
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2 Example 1. Calculate the C v for H 2, HD, and D 2 at 150K, 250 K, 350 K, and 3500 K. From Table 4.1.1 read: Molecule150K250K350 K3500K H2H2 63202.5 2.5000053.267441 HD 55002.5 2.5000373.317335 D2D2 44902.52.5000052.5004423.373445 Results of C v /Nk: Since C v is dominated by the translation (1.5R) and rotation (R) contributions, at low temperatures C v is constant (2.5R=20.8 J/K). At high temperatures (3500 K), the vibrational contribution starts to take effect and we see the differences in C v between the isotopes. equation 4.2.4
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3 Example 2. Calculate the C v (in J/mol K) for N2 and Cl2 from 300K to 2700 K. How well can these results be fit with a polynomial in T? For N 2, using data from table 4.1.1, calculate U and Cv need D o and
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4 Both U and C v can be fit using a polynomial in T For N 2, using data from table 4.1.1, calculate U and Cv
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5 For Cl 2, using data from table 4.1.1, calculate U and Cv The fit for C v is less accurate. This is because the vibrational temperature ( = 814 K) is quite low for Cl 2, i.e., vibrational motions affect C v at temperature above 814 K. Since the vibrational contribution to C v cannot be described using a truncated polynomial, we see larger errors in this case.
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6 Example 3. We have used the harmonic oscillator approximation n=0,1,2,…. To include anharmonicity: n=0,1,2,…. where is a constant. Evaluate the vibrational partition function for this case.
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8 Example 4. We have used the rigid rotor approximation for the rotational energy modes: j =0.1,2,… Due to centrifugal forces, the molecules stretch slightly with increasing rotational motion. This can be corrected using an expansion about the rigid rotator model:
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9 The rotational partition function is: Then, if the rotational energy is given by At high temperatures, the summation can be replaced by an integral
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10 Substituting,
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11 Example 5. Determine C v and C p of an ideal diatomic gas for the limit T 0. A corollary from the 3 rd law of Thermodynamics is that the heat capacity of a material in the perfect crystalline state should be zero at absolute zero. Is the result of this example consistent with the 3 rd law? x 2 exp (-x) This does not agree with the 3rd law of thermodynamics. However, the reason is easy to understand in that the description here is of a diatomic gas, not a perfect crystal. Later we will study the description of crystalline solids where it is shown that the heat capacity of a perfect crystalline solid is zero at 0 K.
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12 Example 6. One assumption that is sometimes made is that for isotopic species the bond length is the same in each of the species, as is the vibrational force constant, k f. Are the rotational and vibrational temperatures of H 2, HD, and D 2 in Table 4.1.1 consistent with these assumptions? The bond separation distance can be obtained from the rotational temperature or The bond force constant can be determined from the vibrational temperature
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13 H2H2 D2D2 HD Q r /K87.543.865.8 d(A)0.7450.744 Q v /K632044905500 k f (N/m) 568.4573.8574.0 Therefore, the rotation and vibration temperatures are consistent with the equal bond length and force constant assumptions for isotopes
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