Chapter 13, Problem 67P

To determine

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules.

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are $1360{\text{ ms}}^{-1}$, $515{\text{ ms}}^{-1}$, $1920{\text{ ms}}^{-1}$ and $482{\text{ ms}}^{-1}$ respectively.

Explanation of Solution

The given temperature is $25°\text{C}$.

Find the temperature in Kelvin scale

$T(\text{K})=273.15\text{K}+T(°\text{C)}$                                (I)

Here, $T(\text{K)}$ is the temperature in Kelvin and $T(°\text{C)}$ is the temperature in degree Celsius.

Substitute $25$ for $T(°\text{C)}$ in (I) to find $T(\text{K)}$

$\begin{array}{c}T(\text{K})=273.15\text{K}+25\text{K}\\ =298.15\text{}\text{K}\end{array}$

Write the expression for rms speed of atoms and molecules

$v_{\text{rms}}=\sqrt{\frac{3kT}{m}}$                                                           (II)

Here, $v_{\text{rms}}$ is the rms speed, $k$ is the Boltzmann constant, $T$ is the temperature and $m$ is the molecular mass.

The atomic mass of helium is $4.002602\text{u}$.

Substitute $4.002602\text{u}$ for $m$, $1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $k$ and $298.15\text{}\text{K}$ for $T$ in (II) to find $v_{\text{rms}}$ of helium atom.

$\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(298.15\text{}\text{K}\right)}{4.002602\text{u}}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}\text{J}\right)\left(298.15\text{}\right)}{4.002602\text{ u}\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}{\text{kgm}}^{\text{2}}{\text{s}}^{-2}\right)\left(298.15\text{}\right)}{4.002602\times 1.66\times {10}^{-27}\text{}\text{kg}}}\\ =1363{\text{ ms}}^{-1}\\ \approx 1360{\text{ ms}}^{-1}\end{array}$

Thus, the rms speed of helium atom is $1360{\text{ ms}}^{-1}$.

The molecular mass of nitrogen is $2\left(14.00674\text{u}\right)$.

Substitute $2\left(14.00674\text{u}\right)$ for $m$, $1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $k$ and $298.15\text{}\text{K}$ for $T$ in (II) to find $v_{\text{rms}}$ of nitrogen molecule.

$\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(298.15\text{}\text{K}\right)}{2\left(14.00674\text{u}\right)}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}\text{J}\right)\left(298.15\text{}\right)}{2\left(14.00674\text{u}\right)\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}{\text{kgm}}^{\text{2}}{\text{s}}^{-2}\right)\left(298.15\text{}\right)}{2\left(14.00674\right)\times 1.66\times {10}^{-27}\text{}\text{kg}}}\\ =515{\text{ ms}}^{-1}\end{array}$

Thus, the rms speed of nitrogen molecule is $515{\text{ ms}}^{-1}$.

The molecular mass of hydrogen is $2\left(1.00794\text{u}\right)$.

Substitute $2\left(1.00794\text{u}\right)$ for $m$, $1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $k$ and $298.15\text{}\text{K}$ for $T$ in (II) to find $v_{\text{rms}}$ of hydrogen molecule.

$\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(298.15\text{}\text{K}\right)}{2\left(1.00794\text{u}\right)}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}\text{J}\right)\left(298.15\text{}\right)}{2\left(1.00794\text{u}\right)\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}{\text{kgm}}^{\text{2}}{\text{s}}^{-2}\right)\left(298.15\text{}\right)}{2\left(1.00794\right)\times 1.66\times {10}^{-27}\text{}\text{kg}}}\\ =1920{\text{ms}}^{-1}\end{array}$

Thus, the rms speed of hydrogen molecule is $1920{\text{ms}}^{-1}$.

The molecular mass of oxygen is $2\left(15.9994\text{u}\right)$.

Substitute $2\left(15.9994\text{u}\right)$ for $m$, $1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $k$ and $298.15\text{}\text{K}$ for $T$ in (II) to find $v_{\text{rms}}$ of oxygen molecule.

$\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(298.15\text{}\text{K}\right)}{2\left(15.9994\text{u}\right)}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{}\text{J}\right)\left(298.15\text{}\right)}{2\left(15.9994\text{u}\right)\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}{\text{kgm}}^{\text{2}}{\text{s}}^{-2}\right)\left(298.15\text{}\right)}{2\left(15.9994\right)\times 1.66\times {10}^{-27}\text{}\text{kg}}}\\ =482{\text{ ms}}^{-1}\end{array}$

Thus, the rms speed of oxygen molecule is $482{\text{ ms}}^{-1}$.