Question 18.10: Carbon dioxide is a linear triatomic molecule that has the f...
Carbon dioxide is a linear triatomic molecule that has the following characteristic temperatures
Θ_r = 0.562 \text{K}
Θ_{v1} = 1932 \text{K}
Θ_{v2} = Θ_{v3} = 960. \text{K}
Θ_{v4} = 3380 \text{K}
Determine the specific internal energy, specific enthalpy, and specific entropy of \text{CO}_2 at a temperature of 1000. \text{K} and a pressure of 1.00 \text{atm}.
The mass of the \text{CO}_2 molecule is
m = M/\text{N}_o = 44.01/6.023 × 10^{26} = 7.31 × 10^{−26} \text{kg/molecule}
and the gas constant for \text{CO}_2 is
R = ℜ/M = 8.3143/44.01 = 0.1889 \text{KJ/(kg.K)}
Equation (18.56) gives the vibrational specific internal energy at absolute zero temperature as
(u_o)_\text{vib} = \sum\limits_{i=1}^{3b−5} RΘ_{vi}/2 (18.56)
(u_o)_\text{vib} = (0.1889) (1932 + 960. + 960. + 3380)/2 = 683 \text{kJ/kg}
and Eq. (18.55a) gives the vibrational component of the specific internal energy as
u_\text{vib} = (u_o)_\text{vib} + R \sum\limits_{i=1}^{3b−5} \frac{Θ_{vi}}{[\text{exp} (Θ_{vi}/T) -1]} (18.55a)
u_\text{vib} = 683 + (0.1889) \left\{(1932) [\text{exp} (1.932) −1]^{-1} + 2 (960) [\text{exp} (0.960.) −1]^{-1} + (3380) [\text{exp} (3.380) −1]^{-1}\right\} = 992 \text{kJ/kg}
The translational and rotational components are given by Eqs. (18.49a) and (18.49b) as
u_\text{trans} = \frac{3}{2} RT = \frac{3}{2} (0.1889) (1000.) = 283.4 \text{kJ/kg}
u_\text{rot} = RT = (0.1889) (1000.) = 188.9 \text{kJ/kg}
Then,
u = u_\text{trans} + u_\text{rot} + u_\text{vib} = 283.4 + 188.9 + 992 = 1465 \text{kJ/k}
The specific enthalpy is now given simply by
h = u + RT = 1465 + (0.1889) (1000.) = 1654 \text{kJ/kg}
The translational and rotational specific entropy values are calculated from Eqs. (18.54a) and (18.54b). First, we calculate
s_\text{trans} = R \left\{\text{ln} [(2πm/ħ^2)^{3/2} (kT)^{5/2} /p] + \frac{5}{2}\right\} (18.54a)
s_\text{rot} = R \left\{\text{ln} [T/ (σ Θ_r)] + 1\right\} (18.54b)
(2πm/ħ^2)^{3/2} (kT)^{5/2} /p = [ 2π(7.31 × 10^{–26}) / (6.626 × 10^{–34})^{2}]^{3/2}
× [(1.38 × 10^{–23}) (1000)]^{5/2} /101,325
= 2.36 × 10^8 \text{per molecule}
and then Eq. (18.54a) gives
s_\text{trans} =(0.1889) [\text{ln} (2.36 × 10^8) + \frac{5}{2}] = 4.11 \text{kJ/(kg.K)}
and Eq. (18.54b) with σ = 2 from Table 18.9 gives
| Table 18.9 Rotational Symmetry Number for Some Simple Materials | |
| Material | σ |
| Any diatomic molecule with two different atoms (e.g., \text{HCl, HI}, or \text{NO}) | 1 |
| Any diatomic molecule with two identical atoms (e.g., \text{H}_2, \text{O}_2, or \text{N}_2) | 2 |
| Any triatomic molecule with two different atoms forming an isosceles triangle (such as \text{H}_2\text{O}) or any linear triatomic | 2 |
| molecule (e.g., \text{CO}_2 or \text{NO}_2,) | |
| Any quatratomic molecule with two different atoms forming an equilateral triangular pyramid (e.g., \text{NH}_3) | 3 |
| Any molecule forming a plane rectangle (e.g., \text{C}_2\text{H}_4) | 4 |
| Any pentatomic molecule with two different atoms forming a regular tetrahedron with the carbon atom at the center of | 12 |
| mass (e.g., \text{CCl}_4 or \text{CH}_4) | |
s_\text{rot} =(0.1889) \left\{[\text{ln} [1000 / (2) (0.562) ]+ 1\right\} = 1.47 \text{kJ/(kg.K)}
Equation (18.55e) is then used to find the vibrational component of the specific entropy as
s_\text{vib} = R \sum\limits_{i=1}^{3b−5}\left\{\text{ln} [1− \text{exp} (−Θ_{vi}/T)]^{-1} + (Θ_{vi}/T) / [\text{exp} (Θ_{vi}/T) -1]\right\} (18.55e)
s_\text{vib} = (0.1889) \left\{\text{ln} [ 1− \text{exp} (−1.932)]^{-1} + (1.932) [\text{exp} (1.932) -1]^{-1}\right\}
+ \text{ln} [ 1− \text{exp} (−0.960)]^{-1} + (0.960) [\text{exp} (0.960) -1]^{-1}
+ \text{ln} [ 1− \text{exp} (−0.960)]^{-1} + (0.960) [\text{exp} (0.960) -1]^{-1}
+ \text{ln} [ 1− \text{exp} (− 3.380)]^{-1} + (3.380) [\text{exp} (3.380) -1]^{-1}
= 0.527 \text{kJ/(kg.K)}
Then, the specific entropy is
s = s_\text{trans} + s_\text{rot} + s_\text{vib} = 4.11 + 1.47 + 0.527 = 6.11 \text{kJ/(kg.K)}