Answer
(a) decrease
(b) $920m/s$
Work Step by Step
(a) We know that if the number of molecules of gas is increased, then then the temperature is decreased. We also know that the rms speed is directly proportional to the square root of the temperature, that is $v_{rms}\propto \sqrt{T}$. Thus, the rms speed of the molecules will be decreased due to decreased temperature.
(b) We can find the final rms speed as follows:
The initial rms speed is given as $v_{rms,1}=\sqrt{\frac{3RT_1}{M}}$
$v_{rms, 2}=\sqrt{\frac{3RT_2}{M}}$
$v_{rms,2}=\sqrt{\frac{3R(\frac{T_1}{2})}{M}}$
$v_{rms,2}=\frac{1}{\sqrt2}\sqrt{\frac{3RT_1}{M}}$
$v_{rms,2}=\frac{1}{\sqrt2}v_{rms,1}$
We plug in the known values to obtain:
$v_{rms,2}=\frac{1}{\sqrt{2}}(1300m/s)$
$v_{rms,2}=920m/s$
