Answer
(a) Helium
(b) $1370m/s$
(c) $1.86\mu m$
Work Step by Step
(a) We can identify the required gas as follows:
$n=\frac{pV}{RT}$
We plug in the known values to obtain:
$n=\frac{50\times 10^3}{8.31\times 300}$
This simplifies to:
$n=20.06mole$
Now mass of one mole of gas is given as $\frac{0.0802}{20.06}=0.0039Kg$. This shows that the required gas is Helium.
(b) The required rms speed can be determined as follows:
$v_{rms}=\sqrt{\frac{3RT}{M}}$
We plug in the known values to obtain:
$v_{rms}=\sqrt{\frac{3\times 8.31\times 300}{4\times 10^{-3}}}$
$\implies v_{rms}=1370m/s$
(c) The mean free path can be calculated as
$\lambda=\frac{k_B T}{\sqrt{2}\pi d^2 p}$
We plug in the known values to obtain:
$\lambda=\frac{1.38\times 10^{-23}\times 300}{\sqrt{2}\pi (1\times 10^{-10})^2\times 50\times 10^3}=1.86\mu m$
