Question 16.3: Employing basic transport theory, determine the following tr...
Employing basic transport theory, determine the following transport properties for gaseous argon: (a) its dynamic viscosity at 300 K, (b) its thermal conductivity at 300 K, and (c) its self-diffusion coefficient at 300 K and 1 atm.
(a) From Eq. (15.22), the mean particle speed for argon at 300 K is
\overline{V} = \left(\frac{8kT}{\pi m}\right)^{{1}/{2}} = \left[\frac{8(1.3807\times10^{-23}kg.m^{2}/s^{2}.K)(300K)}{\pi (39.948)(1.6605\times10^{-27}kg)}\right]^{{1}/{2}} = 398.8 m/s.
Hence, from Eq. (16.31) and previously calculated parameters taken from Examples 16.1 and 16.2, we find that the dynamic viscosity becomes
\eta = \frac{1}{3}nm\ell\overline{V} (16.31)
= \frac{1}{3}(2.45\times10^{25}m^{-3})(6.63\times10^{-26}kg)(7.85\times10^{-8}m)(398.8m/s)
= 1.70 × 10^{-5} kg/m · s,
which is equivalent to 1.70 × 10^{-5} N · s/m².
(b) From Eqs. (16.31) and (16.32), the thermal conductivity for argon at 300 K is
\lambda = \frac{1}{3}\left(\frac{nc_{\nu}}{N_{A}}\right)\ell\overline{V}; (16.32)
\lambda = \left(\frac{c_{\nu}}{M}\right)\eta = \frac{3}{2}\left(\frac{R}{M}\right)\eta
= \frac{3}{2}\left(\frac{8.3145\times10^{3}J/K·kmol}{39.948_\ kg/kmol}\right)(1.70 × 10^{-5} kg/m · s)
= 5.31 × 10^{-3} W/m · K.
(c) From Eq. (16.34), the self-diffusion coefficient for argon at 300 K and 1 atm becomes
D = \frac{1}{3}\ell\overline{V} = \frac{1}{3}(7.85 × 10^{-8}m)(398.8 m/s) = 1.04 × 10^{-5} m²/s.