1. Applications of diatomic and polyatomic ideal gases 1

2. 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

3. 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

4. 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

5. 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.

6. 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.

7. 7

8. 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:

9. 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

10. 10 Substituting,

11. 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.

12. 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

13. 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