Answer
$v_{rms} = 368~m/s$
Work Step by Step
A nitrogen atom consists of 7 protons and 7 neutrons.
A nitrogen molecule $N_2$ consists of 14 protons and 14 neutrons.
The mass of one nitrogen molecule is $28~u$
We can convert this mass to units of $kg$:
$m = 28~u\times \frac{1.66\times 10^{-27}~kg}{1~u} = 4.648\times 10^{-26}~kg$
We can find the rms speed of the nitrogen molecules:
$\overline{KE} = \frac{3}{2}~k~T$
$\frac{1}{2}m~v_{rms}^2 = \frac{3}{2}~k~T$
$v_{rms} = \sqrt{\frac{3~k~T}{m}}$
$v_{rms} = \sqrt{\frac{3~P~V}{N~m}}$
$v_{rms} = \sqrt{\frac{(3)(1.6)(1.01\times 10^5~Pa)(0.25~m)^3}{(2.0)(6.022\times 10^{23})(4.648\times 10^{-26}~kg)}}$
$v_{rms} = 368~m/s$
