Question

Determine the root mean square velocity of nitrogen at a

temperature of 300 K.

Answer 1

Gives

T=300k

Nitrogen

$M_N=0.028kg$

R=8.314$Jmol^-1k^-1$

$$V_{RMS}=\sqrt\frac{3RT}{M_N}=\sqrt\frac{3\times8.314\times300}{0.028}=517m/sec$$