Chapter 17, Problem 26P

(a)

To determine

The rms speed of the $H_2$ molecules.

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0\text{km}/\text{s}$ .

Temperature is $0°\text{C}$ .

Formula used:

Write the expression for the rms speed of the molecule.

  $v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$ ........ (1)

Here, $v_{\text{rms}}$ is the root mean-square speed of the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.

Write the equation for the conversion of Celsius to kelvin.

  $T\left(°\text{K}\right)=T\left(°\text{C}\right)+273.15$ ........ (2)

  $T\left(°\text{C}\right)$ Is the temperature in Celsius and $T\left(°K\right)$ is the temperature in kelvin.

Calculation:

Substitute $0°\text{C}$ for $T\left(°\text{C}\right)$ in equation (2).

  $\begin{array}{l}T\left(°\text{K}\right)=\left(0°\text{C}\right)+273.15\\ T\left(°\text{K}\right)=273.15\text{K}\end{array}$

Substitute $273.15\text{K}$ for $T$ , $2\times {10}^{-3}\text{kg}/\text{mol}$ for $M$ and $8.31\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (1).

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.31{\text{Jmol}}^{\text{-1}}\text{K}\right)\left(273.15\text{K}\right)}{2\times {10}^{-3}\text{kg}/\text{mol}}}\\ v_{\text{rms}}=\text{1}\text{.84 Km}/\text{s}\end{array}$

Conclusion:

The rms speed of the $H_2$ molecules is $\text{1}\text{.84 Km}/\text{s}$ .

(b)

To determine

The rms speed of $O_2$ molecules.

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0\text{km}/\text{s}$ .

Temperature is $0°\text{C}$ .

Formula used:

Write the expression for the rms speed of the molecule.

  $v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

Calculation:

Substitute $273.15\text{K}$ for $T$ , $32\times {10}^{-3}\text{kg}/\text{mol}$ for $M$ and $8.31\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (1).

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.31{\text{Jmol}}^{\text{-1}}{\text{K}}^{-1}\right)\left(273.15\text{K}\right)}{32\times {10}^{-3}\text{kg}/\text{mol}}}\\ v_{\text{rms}}=461\text{m}/\text{s}\end{array}$

Conclusion:

The root mean square speed for the oxygen molecule is $461\text{m}/\text{s}$ .

(c)

To determine

The root mean square speed of the $C{O}_2$ molecule.

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0\text{km}/\text{s}$ .

Temperature is $0°\text{C}$ .

Formula used:

Write the expression for the rms speed of the molecule.

  $v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

Calculation:

Substitute $273.15\text{K}$ for $T$ , $44\times {10}^{-3}\text{kg}/\text{mol}$ for $M$ and $8.31\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (1).

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.31{\text{Jmol}}^{\text{-1}}{\text{K}}^{-1}\right)\left(273.15\text{K}\right)}{44\times {10}^{-3}\text{kg}/\text{mol}}}\\ v_{\text{rms}}=393\text{m}/\text{s}\end{array}$

Conclusion:

The root mean square speed for the $C{O}_2$ molecule is $393\text{m}/\text{s}$ .

(d)

To determine

Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.

Explanation of Solution

Given:

The escape speed of the gas molecules in atmosphere of Mars is $5.0\text{km}/\text{s}$ .

Temperature is $0°\text{C}$ .

Formula used:

Calculate $20$ % of the escape velocity for mars.

  $v=\frac{20\cdot v_{\text{escape}}}{100}$ ........ (2)

Here, $v$ is the $20$ % of the escape velocity.

Calculation:

Substitute $5.0\text{km}/\text{s}$ for $v_{\text{escape}}$ in equation (2).

  $\begin{array}{l}v=\frac{20\cdot \left(5.0\text{km}/\text{s}\right)}{100}\\ v=1\text{km}/\text{s}\end{array}$

Conclusion:

As $20$ % of the escape velocity is $1\text{km}/\text{s}$ , this is greater than $v_{\text{rms}}$ for $O_2$ and $C{O}_2$ but less than for $H_2$ . Thus $H_2$ should not be present in the atmosphere of mars.