Question

The magnitude of average linear momentum of a nitrogen molecule in a sample of nitrogen gas at $27^{\circ} C$ is _________$\times 10^{-23} kg - m / s$
(Boltzmann constant, $k _{ B }=1.4 \times 10^{-23} J / K$, Avogadro's number $N _{ A }=6 \times 10^{23}$, Molecular weight of $N _{2}=28\, g$ )

Answer 1

Linear momentum $p = mc _{ rms }= m \sqrt{\frac{3 k _{ B } T }{ m }}$
$\therefore p =\sqrt{3 mk _{ B } T }$
Now,
One mole of $N _{2}$ weighs $28 g$,
$\therefore $ One molecule of $N _{2}$ weighs $\left(\frac{28}{ N _{ A }}\right) g$ $=\left(\frac{28}{6 \times 10^{23}}\right) g$
$\therefore m =\left(\frac{28 \times 10^{-3}}{6 \times 10^{23}}\right) kg =\left(\frac{28}{6}\right) \times 10^{-26} kg$
$\therefore p =\sqrt{3 \times \frac{28}{6} \times 10^{-26} \times 1.4 \times 10^{-23} \times(27+273)}$
$=\sqrt{14 \times 10^{-26} \times 14 \times 10^{-24} \times 300}$
$=\sqrt{14 \times 14 \times 10^{-50} \times 300}$
$=14 \times 10^{-25} \times 10 \sqrt{3}$
$=14 \times 1.73 \times 10^{-24}$
$\therefore p =2.42 \times 10^{-23} kg\, ms ^{-1}$