Question 18.9: To test the diatomic Maxwell-Boltzmann gas equations just de...
To test the diatomic Maxwell-Boltzmann gas equations just developed, compute the value of c_v/R for nitrous oxide (\text{NO}) at 20.0°\text{C} and compare it with the measured result given in Table 18.3.
| Table 18.3 Measured Values of the Specific Heats of Various Gases at 20.0ºC | |||
| Gas | c_v/R | c_p/R | k= c_p/c_v |
| Monatomic | |||
| \text{He} | 1.5 | 2.5 | 1.67 |
| \text{Ne} | 1.5 | 2.5 | 1.67 |
| \text{Ar} | 1.51 | 2.52 | 1.67 |
| \text{Kr} | 1 | 1.68 | 1.68 |
| \text{Xe} | 1.52 | 2.51 | 1.65 |
| Diatomic | |||
| \text{CO} | 2.51 | 3.51 | 1.4 |
| \text{NO} | 2.51 | 3.51 | 1.4 |
| \text{H}_2 | 2.44 | 3.42 | 1.4 |
| \text{O}_2 | 2.53 | 3.53 | 1.4 |
| \text{N}_2 | 2.5 | 3.5 | 1.4 |
| Triatomic | |||
| \text{CO}_2 | 3.48 | 4.48 | 1.29 |
| \text{SO}_2 | 3.97 | 4.97 | 1.25 |
| \text{H}_2\text{O} | 3.05 | 4.05 | 1.33 |
For a diatomic Maxwell-Boltzmann gas, the constant volume specific heat is given by Eq. (18.52d) as
c_v = \frac{5}{2} R + \frac{R(Θ_v/T)^2 \text{exp} (Θ_v /T)}{[\text{exp} (Θ_v /T) -1]^2}
then,
\frac{c_v}{R} = \frac{5}{2} + \frac{(Θ_v/T)^2 \text{exp} (Θ_v /T)}{[\text{exp} (Θ_v /T) -1]^2}
From Table 18.8, we find that Θ_v = 2740 \text{K} for \text{NO}, so
| Table 18.8 Characteristic Vibrational and Rotational emperatures of Some Common Diatomic Materials | ||
| Material | Θ_v(\text{K}) | Θ_r (\text{K}) |
| \text{H}_2 | 6140 | 85.5 |
| \text{HF} | 5954 | 30.3 |
| \text{OH} | 5360 | 27.5 |
| \text{HCl} | 4300 | 15.3 |
| \text{CH} | 4100 | 20.7 |
| \text{N}_2 | 3340 | 2.86 |
| \text{HBr} | 3700 | 12.1 |
| \text{HI} | 3200 | 9.0 |
| \text{Co} | 3120 | 2.77 |
| \text{NO} | 2740 | 2.47 |
| \text{O}_2 | 2260 | 2.09 |
| \text{Cl}_2 | 810 | 0.35 |
| \text{Br}_2 | 470 | 0.12 |
| \text{I}_2 | 309 | 0.05 |
| \text{Na}_2 | 230 | 0.22 |
| \text{K}_2 | 140 | 0.08 |
| Source: From Lee, John F., Sears, Francis W., Turcotte, Donald L. Statistical Thermodynamics, © 1963. Addison- Wesley Publishing Co., Inc, Reading, MA. Adapted from Table 10-1 on page 204. Reprinted with permission. | ||
(\frac{c_v}{R})_\text{NO} = \frac{5}{2} + \frac{(\frac{2740}{20.0 + 273.15})^2 \text{exp} (\frac{2740}{20.0 + 273.15})^2}{ [\text{exp} (\frac{2740}{20.0 + 273.15}) -1]} = 2.51
as in Table 18.3. Note that, since R_{\text{NO}} = Y/M_{\text{NO}} = 8.3143/30.01 = 0.2771 \text{kJ/kg.K}, then (c_v)_{\text{NO}} = 0.2771 × 2.51 = 0.695 \text{kJ/kg.K} at 20.0°\text{C}.