Chapter 17, Problem 27P

(a)

To determine

The rms speed of $H_2$.

Explanation of Solution

Given:

The escape speed of gas molecules is $60\text{km}/\text{s}$.

The temperature at the Jupiter surface is $-150\text{°C}$.

Formula used:

Write the expression for the rms speed of the molecule.

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

Write the expression for the relation between Celsius and kelvin.

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

Substitute $-150\text{°C}$ in equation (2).

  $\begin{array}{l}T\left(K\right)=-150\text{°C}+273.15\\ T\left(K\right)=123.15\text{K}\end{array}$

Calculation:

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ , $123.15\text{K}$ for $T$ and $2\times {10}^{-3}\text{kg}/\text{mol}$ for $M$ in equation (1).

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.314\frac{\text{J}}{\text{mol}}\text{K}\right)\left(123.15\text{K}\right)}{2\times {10}^{-3}\text{kg}/\text{mol}}}\\ v_{\text{rms}}=1.24\text{km}/\text{s}\end{array}$

Conclusion:

The rms speed of the $H_2$ molecule is $1.24\text{km}/\text{s}$

(b)

To determine

The root mean square speed of $O_2$.

Explanation of Solution

Given:

The escape speed of gas molecules is $60\text{km}/\text{s}$.

The temperature at the Jupiter surface is $-150\text{°C}$.

Formula used:

Write the expression for the rms speed of the molecule.

  $v_{rms}=\sqrt{\frac{3RT}{M}}$

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

Calculation:

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ , $123.15\text{K}$ for $T$ and $32\times {10}^{-}{}^3\text{kg}/\text{mol}$ for $M$ in equation (1).

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(123.15\text{K}\right)}{32\times {10}^{-3}\text{kg}/\text{mol}}}\\ v_{\text{rms}}=310\text{m}/\text{s}\end{array}$

Conclusion:

The root mean square speed of oxygen molecule is $310\text{m}/\text{s}$ .

(c)

To determine

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

Explanation of Solution

Given:

The escape speed of gas molecules is $60\text{km}/\text{s}$.

The temperature at the Jupiter surface is $-150\text{°C}$.

Formula used:

Write the expression for the rms speed of the molecule.

  $v_{rms}=\sqrt{\frac{3RT}{M}}$

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

Calculation:

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ , $123.15\text{K}$ for $T$ and $44\times {10}^{-3}\text{kg}/\text{mol}$ for $M$ in equation (1).

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(123.15\text{K}\right)}{44\times {10}^{-3}\text{kg}/\text{mol}}}\\ v_{\text{rms}}=264\text{m}/\text{s}\end{array}$

Conclusion:

The root mean square speed of the $C{O}_2$ molecule is $264\text{m}/\text{s}$.

(d)

To determine

The elements found in the atmosphere of the Jupiter.

Explanation of Solution

Given:

The escape speed of gas molecules is $60\text{km}/\text{s}$.

The temperature at the Jupiter surface is $-150\text{°C}$.

Formula used:

Write the expression for the rms speed of the molecule.

  $v_{rms}=\sqrt{\frac{3RT}{M}}$

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

Calculate $20\%$ of escape velocity for Jupiter.

  $v=\frac{1}{5}{v}_{\text{escape}}$ ........ (3)

Here, $v_{\text{escape}}$ is the escape velocity of the Jupiter.

Calculation:

Substitute $60\text{km}/\text{s}$ for $v_{\text{escape}}$ in equation (3).

  $\begin{array}{l}v=\frac{1}{5}\left(60\text{km}/\text{s}\right)\\ v=12\text{km}/\text{s}\end{array}$

Conclusion:

The velocity on the Jupiter is greater than root mean square speed for the $O_{2,}$

  $C{O}_2$ , $H_2$ . Thus, these molecules will be found in Jupiter.