Gas cylinders of $\mathrm{CO}_2$ are sold in terms of weight of $\mathrm{CO}_2$. A cylinder contains $50 \mathrm{lb}$ (22.7 $\mathrm{kg}$ ) of $\mathrm{CO}_2$. How long can this cylinder be used in an experiment that requires flowing $\mathrm{CO}_2$ at $293 \mathrm{~K}(\eta=146 \mu \mathrm{P})$ through a 1.00-m-long tube (diameter $\left.=0.75 \mathrm{~mm}\right)$ with an input pressure of $1.05 \mathrm{~atm}$ and output pressure of $1.00 \mathrm{~atm}$ ? The flow is measured at the tube output.
Using Equation (17.45), the gas flow rate $\Delta V / \Delta t$ is
$$
\begin{aligned}
\frac{\Delta V}{\Delta t}= & \frac{\pi r^4}{16 \eta L P_0}\left(P_2^2-P_1^2\right) \\
= & \frac{\pi\left(0.375 \times 10^{-3} \mathrm{~m}\right)^4}{16\left(1.46 \times 10^{-5} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}\right)(1.00 \mathrm{~m})(101,325 \mathrm{~Pa})} \\
& \times\left((106,391 \mathrm{~Pa})^2-(101,325 \mathrm{~Pa})^2\right) \\
= & 2.76 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}
\end{aligned}
$$
Converting the $\mathrm{CO}_2$ contained in the cylinder to the volume occupied at $298 \mathrm{~K}$ and 1 atm pressure, we get
$$
\begin{aligned}
n_{\mathrm{CO}_2} & =22.7 \mathrm{~kg}\left(\frac{1}{0.044 \mathrm{~kg} \mathrm{~mol}^{-1}}\right)=516 \mathrm{~mol} \\
V & =\frac{n R T}{P}=1.24 \times 10^4 \mathrm{~L}\left(\frac{10^{-3} \mathrm{~m}^3}{\mathrm{~L}}\right)=12.4 \mathrm{~m}^3
\end{aligned}
$$
Given the effective volume of $\mathrm{CO}_2$ contained in the cylinder, the duration over which the cylinder can be used is
$$
\frac{12.4 \mathrm{~m}^3}{2.76 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}}=4.49 \times 10^6 \mathrm{~s}
$$
This time corresponds to roughly 52 days.