Question 18.2: Determine the collision frequency and mean free path for neo...
Determine the collision frequency and mean free path for neon at 273 K and 0.113 MPa. The molecular mass of neon is 20.183 kg/kgmole.
For neon,
m = \frac{M}{N_o} = \frac{20.183 \text{kg/kgmole}}{6.022 × 10^{26} \text{molecules/kgmole}} = 3.35 × 10^{−26} \text{kg/molecule}
Then, the root mean square velocity of the neon molecules is given by Eq. (18.13) as
V_\text{rms} =\sqrt{3RT} = \sqrt{\frac{3kT}{m}} (18.13)
V_\text{rms} = (\frac{3kT}{m})^{1/2} = [\frac{3[1.38 × 10^{−23} \text{J/(molecule.K)}] (273 \text{K})}{3.35 × 10^{−26} \text{kg/molecule}}]^{1/2} = 581 \text{m/s}
From Table 18.1, we find that the radius of the neon molecule is 1.30 × 10^{−10} \text{m},
| Table 18.1 Typical Values of the Effective Molecular Radius | |
| Molecule | Effective Radius, r (m) |
| \text{He} | 1.37 × 10^{−10} |
| \text{Ne} | 1.30 × 10^{−10} |
| \text{Ar} | 1.82 × 10^{−10} |
| \text{H}_2 | 0.74 × 10^{−10} |
| \text{N}_2 | 1.10 × 10^{−10} |
| \text{O}_2 | 1.21 × 10^{−10} |
| \text{Br}_2 | 2.28 × 10^{−10} |
| \text{Cl}_2 | 1.99 × 10^{−10} |
| \text{F}_2 | 1.41 × 10^{−10} |
| \text{I}_2 | 2.67 × 10^{−10} |
| \text{HBr} | 1.41 × 10^{−10} |
| \text{HCl} | 1.27 × 10^{−10} |
| \text{HF} | 0.92 × 10^{−10} |
| \text{HI} | 1.60 × 10^{−10} |
| \text{H}_2\text{O} | 1.50 × 10^{−10} |
| \text{CO} | 1.13 × 10^{−10} |
| \text{NO} | 1.15 × 10^{−10} |
| \text{CO}_2 | 2.30 × 10^{−10} |
| \text{NH}_3 | 2.22 × 10^{−10} |
| \text{CH}_4 | 2.07 × 10^{−10} |
| Source: Material drawn from the JANAF Thermochemical Tables, first ed., 1961, Thermal Research Laboratory, Dow Chemical Corporation, Midland, I. Reprinted by permission. | |
so the collision cross-section is
σ = 4πr^2 = 4π(1.30 × 10^{−10} \text{m})^2 = 2.12 × 10^{−19} \text{m}^2
and the collision frequency is
\mathcal{F} = σV_\text{rms} \frac{N}{\sout{V}} (\frac{8}{3π})^{1/2}
where
\frac{N}{\sout{V}} = \frac{p}{kT} = \frac{0.113 × 10^6 \text{N/m}^2}{[1.380 × 10^{−23} \text{N.m/(molecule.K)}] (273 \text{K})} = 3.00 × 10^{25} \text{molecules/m}^3
then
\mathcal{F} = σV_\text{avg} \frac{N}{V} (\frac{8}{3π})^{1/2} = (3.00 × 10^{25} \text{molecules/m}^3) (\frac{8}{3π})^{1/2} (2.12 × 10^{−19} \text{m}^2) (581 \text{m/s})
= 3.40 × 10^9 \text{collisions/s}
so that the molecular mean free path is given by Eq. (18.15) as
λ = \frac{1}{(N/\sout{V})σ} = \frac{1}{(3.00 × 10^{25} \text{molecules/m}^3) (2.12 × 10^{−19} \text{m}^2)} = 1.57 × 10^{−7} m