The viscosities of carbon dioxide at 1 atm and 0, 490, and 850°C are 139, 330, and 436 mP, respectively. Calculate the apparent rigid sphere diameter of carbon dioxide at each of these temperatures.
Solution:
We apply the equation
\begin{equation} \eta =\frac{5}{16\pi}\frac{(MRT)^{1/2}}{N_Ad^2} \end{equation}
\begin{equation} d ^2=\frac{5}{16\pi}\frac{(MRT)^{1/2}}{N_A\eta} \end{equation}
We convert $\eta$ to the units of the international system
\begin{equation} 139\;mp\frac{1p}{1000\;mp}\frac{1Ns/m^2}{10p}=139x10^{-4}\;NS/m^2 \end{equation}
M= 0.044 kg/mole; R=8.314 J/molK T=273K; $N_A=6.023x10^{23}\;mol^{-1}$.
Substituting into the equation gives a diameter for carbon dioxide of $4.59x10^{-10}\;m$