Chapter 17, Problem 25P

(a)

To determine

The temperature at which the root mean square speeds for $O_2$ is equal to $15\%$ of the escape speed for earth.

Explanation of Solution

Given:

The escape speed of the surface of the planet is $\sqrt{2gR}$ .

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 value o the molecule, $R$ is the gas constant, $T$ is the temperature and $M$ is the molecular mass of the molecule.

Equate $15\%$ of $v_e$ to $v_{\text{rms}}$ .

  $0.15\sqrt{2g{R}_E}=\sqrt{\frac{3RT}{M}}$

Here, $R_E$ is the radius of earth and $g$ is acceleration due to gravity.

Solve the above equation for $T$ .

  $T=\frac{0.045g{R}_{E}M}{3R}$ ......... (2)

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $6.37\times {10}^6{R}_E$ , $32\times {10}^{-3}\text{kg}/\text{mol}$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ in equation (2).

  $\begin{array}{l}T=\frac{0.045\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(6.37\times {10}^6\text{m}\right)\left(32\times {10}^{-3}\text{kg}/\text{mol}\right)}{3\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ T=3.61\times {10}^3\text{K}\end{array}$

Conclusion:

The temperature is $3.61\times {10}^3\text{K}$ .

(b)

To determine

The temperature at which the root mean square speeds for $H_2$ is equal to $15\%$ of the escape speed for earth.

Explanation of Solution

Given:

The escape speed of the surface of the planet is $\sqrt{2gR}$ .

Formula used:

Write the expression for the rms speed of the molecule.

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

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

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $6.37\times {10}^6$ for $R_E$ and $2\times {10}^{-3}\text{kg}/\text{mol}$ for $M$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (2).

  $\begin{array}{l}T=\frac{0.045\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(6.37\times {10}^6\right)\left(2\times {10}^{-3}\text{kg}/\text{mol}\right)}{3\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ T=225\text{K}\end{array}$

Conclusion:

The temperature is $225\text{K}$ .

(c)

To determine

The reason for the low hydrogen in the earth atmosphere.

Explanation of Solution

Given:

The escape speed of the surface of the planet is $\sqrt{2gR}$ .

Introduction:

All the giant planets like Jupiter, Saturn, Uranus and Neptune contains helium and hydrogen in the atmosphere of the giant planets.

Earth contains low amount of hydrogen in the atmosphere because hydrogen gas is very light as compared to other elements and has high escape speed at the high temperature and thus hydrogen gas escapes from the earth’s atmosphere.

Conclusion:

The hydrogen gas molecules attain greater escape speed than other elements and therefore earth’s atmosphere contains low amount of hydrogen.

(d)

To determine

Explanation of Solution

Given:

The escape speed of the surface of the planet is $\sqrt{2gR}$ .

Gravity on moon is one-sixth the gravity on moon.

Radius of earth is $1738\text{km}$ .

Formula used:

Substitute $g_{\text{moon}}$ for $g$ and $R_{\text{moon}}$ for $R_E$ in equation (2).

  $T=\frac{0.045g_{\text{moon}}{R}_{\text{moon}}M}{3R}$

Substitute $\frac{1}{6}{g}_{\text{earth}}$ for $g_{\text{moon}}$ in the above equation.

  $T=\frac{0.0025\left(\frac{1}{6}{g}_E\right)R_{\text{moon}}M}{R}$ ........ (3)

Here, $g_E$ is the gravity of earth.

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g_E$ , $1.738\times {10}^6\text{m}$ for $R_{\text{moon}}$ , $32\times {10}^{-3}\text{g}/\text{mol}$ for $M$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (3).

  $\begin{array}{l}T=\frac{0.0025\left(\frac{1}{6}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1.738\times {10}^6\text{m}\right)\right)32\times {10}^{-3}\text{g}/\text{mol}}{8.314\text{J}/\text{mol}\cdot \text{K}}\\ T=164\text{K}\end{array}$

Thus, the temperature of $O_2$ is $164\text{K}$ for which the root mean square speed is equal to $15\%$ of the escape speed of earth.

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g_E$ , $1.738\times {10}^6\text{m}$ for $R_{\text{moon}}$ , $2\times {10}^{-3}\text{g}/\text{mol}$ for $M$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (3).

  $\begin{array}{l}T=\frac{0.0025\left(\frac{1}{6}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\right)\left(1.738\times {10}^6\text{m}\right)\left(2\times {10}^{-3}\text{g}/\text{mol}\right)}{8.314\text{J}/\text{mol}\cdot \text{K}}\\ T=10.3\text{K}\end{array}$

Conclusion:

The escape speed is low on moon as the acceleration due to gravity is less there. Thus the largerpercentage of molecules moves with escape speed.