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 $N_2=28g$ )

Answer 1

Answer: 2.42

Linear momentum $p=m{c}_{rms}=m\sqrt{\frac{3k_{B}T}{m}}$
$\therefore p=\sqrt{3m{k}_{B}T}$
Now,
One mole of $N_2$ weighs $28g$,
$\therefore$ One molecule of $N_2$ weighs $\left(\frac{28}{N_A}\right)g$ $=\left(\frac{28}{6\times 10^{23}}\right)g$
$\therefore m=\left(\frac{28\times 10^{-3}}{6\times 10^{23}}\right)kg=\left(\frac{28}{6}\right)\times 10^{-26}kg$
$\therefore p=\sqrt{3\times \frac{28}{6}\times 10^{-26}\times 1.4\times 10^{-23}\times (27+273)}$
$=\sqrt{14\times 10^{-26}\times 14\times 10^{-24}\times 300}$
$=\sqrt{14\times 14\times 10^{-50}\times 300}$
$=14\times 10^{-25}\times 10\sqrt{3}$
$=14\times 1.73\times 10^{-24}$
$\therefore p=2.42\times 10^{-23}kgm{s}^{-1}$