Question

The molecular weight of hydrogen molecules is $M=2.408\times {10}^{-3}\text{kg/mol}.$ Calculate the root mean square speed of hydrogen molecules $\left(H_2\right)\text{at}300\text{K}.$

• A. $3\text{km/s}$
• B. $2\text{km/s}$
• C. $1.77\text{km/s}$
• D. $0.8\text{km/s}$

Answer 1

The correct option is D $0.8\text{km/s}$

Given that,
Molecular mass of $H_2$ molecule, $M=2.408\times {10}^{-3}\text{kg/mol}$
Let the mass of $H_2$ molecule is $m$ and Avogadro number is $N_o$.
The mass of one $H_2$ molecule can be calculated from the molecular weight as
$m=\frac{M}{N_o}=\frac{2.408\times {10}^{-3}\text{kg/mol}}{6.02\times {10}^{23}{\text{mol}}^{-1}}=4\times {10}^{-27}\text{kg}$
Then, using formula
$\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}kT$
$\therefore v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3\times 1.4\times {10}^{-23}\times 300}{4\times {10}^{-27}}}=1.77\text{km/s}$

Why this question? Tip: Molecules of ideal gases posses only translational motion. So they possess only translational kinetic energy which is equal $\frac{1}{2}m{v}_{rms}^2$