Answer
327.4 m/s
Work Step by Step
Here we use equation 14.5 $PV=\frac{2}{3}N(\frac{1}{2}mv_{rms}^{2})$ to find the final translational rms speed.
Since the pressure and volume of the gas are kept constant, while the number of molecules is doubled, we can write,
$P_{1}V_{1}=P_{2}V_{2}$
$\frac{2}{3}N_{1}(\frac{1}{2}mv_{rms,1}^{2})=\frac{2}{3}N_{2}(\frac{1}{2}mv_{rms,2}^{2})$
$N_{1}v_{rms,1}^{2}=N_{2}v_{rms,2}^{2}=>v_{rms,2}=v_{rms,1}\sqrt {\frac{N_{1}}{N_{2}}}$
Let's plug known values into this equation.
$v_{rms,2}=(463\space m/s)\sqrt {\frac{N_{1}}{2N_{1}}}=327.4\space m/s$
Final translation speed = 327.4 m/s
