Answer
343.35 m/s
Work Step by Step
Here we use equations 14.6 $\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT$ and the ideal gas law $PV=nRT$ to find the rms speed.
$\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT=>v_{rms}=\sqrt {\frac{3kT}{m}}$
$v_{rms}=\sqrt {\frac{3k(\frac{PV}{nR})}{m}}=\sqrt {\frac{3kPV}{nmR}}-(1)$
The value of m for $SO_{2}$ can be calculated as follows.
$m=\{32.07\space u+2(15.9994\space u)\}(\frac{1.6605\times10^{-27}kg}{1\space u})=1.064\times10^{-25}kg$
Let's plug known values into equation (1),
$v_{rms}=\sqrt {\frac{3(1.38\times10^{-23}J/K)(2.12\times10^{4}Pa)(50\space m^{3})}{(421\space mol)(1.064\times10^{-25}kg)(8.31\space J/mol\space K)}}=343.35\space m/s$
