Question

Calculate ${\mathbf{v}}_{\mathbf{ms}}$for a helium atom and for a nitrogen molecule ( ${\mathbf{N}}_{\mathbf{2}}$; molecular mass $\mathbf{28}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}$) in the room you’re in (whose temperature is probably about $\mathbf{293}\mathbf{}\mathbf{K}$).

Short answer

The value of root mean square velocity of helium atom and nitrogen molecule is $1351.6\mathrm{m}/\mathrm{s}$and $510.81\mathrm{m}/\mathrm{s}$respectively.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The molecular mass of nitrogen is,$M=28\mathrm{g}/\mathrm{mol}$.
• The temperature of room is,$T=293\mathrm{K}$

Concept/Significance of vms

The root-mean-square speed accounts for both molecular weight and temperature, two parameters that have a direct impact on a material's kinetic energy.

Determination of the vms a helium atom and for a nitrogen molecule

The root mean square of a particle is given by,

$v_{ms}=\sqrt{\frac{3k_{B}T}{m}}$

Here,$k_B$is Boltzmann constant whose value is localid="1657876582667" $1.38\times {10}^{-23}\mathrm{J}/\mathrm{mol}·\mathrm{K}$,T is the temperature and m, is the mass of gas particle.

Substitute values in the above for Helium atom.

$$\begin{gathered} v_{\mathrm{ms},\mathrm{He}}=\sqrt{\frac{3(1.38\times {10}^{-23}\mathrm{J}/\mathrm{mol}‐\mathrm{K})(293\mathrm{K})}{6.64\times {10}^{-27}\mathrm{kg}}} \\ =\sqrt{1826837.349{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =1351.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the of Helium is 1351.6 m/s .

Substitute values in the ${\mathrm{v}}_{\mathrm{ms}}$for Nitrogen molecule.

localid="1657876645880" $$\begin{gathered} v_{\mathrm{ms},\mathrm{N}}\mathit{=}\sqrt{\frac{3(1.38\times {10}^{-23}J/mol.K)\left(293K\right)}{4.65\times {10}^{-26}kg}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\sqrt{260909.26{\mathrm{m}}^2}/{\mathrm{s}}^2 \\ =510.81\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the $v_{\mathrm{rms}}$for Nitrogen molecule is $510.81\mathrm{m}/\mathrm{s}$.