Answer
745.4 K
Work Step by Step
Here we use equation 14.6 $\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT$ to find the temperature of the carbon dioxide.
$\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT=>T=\frac{mv_{rms}^{2}}{3k}$
Let's plug known values into this equation.
$T=\frac{44\space u(\frac{1.66\times10^{-27}kg}{1\space u})(650\space m/s)^{2}}{3(1.38\times10^{-23}J/K)}=745.4\space k$
