Chapter 20, Problem 27PQ

(a)

To determine

The most probable speed, average speed and rms speed for the oxygen gas.

Answer to Problem 27PQ

The most probable speed of oxygen gas is $390\text{m}/\text{s}$ , average speed of the oxygen gas is $440\text{m}/\text{s}$ and rms speed for the oxygen gas is $480\text{m}/\text{s}$ .

Explanation of Solution

Write the expression for the most probable speed of gas molecules.

  $v_{\text{mp}}=\sqrt{\frac{2k_{B}T}{m}}$                                                                                                         (I)

Here, $v_{\text{mp}}$ is the most probable speed of the gas molecules, $k_B$ is the Boltzmann constant, $T$ is the temperature and $m$ is the mass of the molecule.

Write the expression for the average speed of gas molecules.

  $v_{\text{av}}=\sqrt{\frac{8k_{B}T}{\pi m}}$                                                                                                          (II)

Here, $v_{\text{av}}$ is the average speed of the gas molecules.

Write the expression for the rms speed of gas molecules.

  $v_{\text{rms}}=\sqrt{\frac{3k_{B}T}{m}}$                                                                                                        (III)

Here, $v_{\text{av}}$ is the average speed of the gas molecules.

Conclusion:

The mass of oxygen molecule is $32\text{u}$ , temperature of the gas is $22°\text{C}$ and Boltzmann constant is $1.38\times {10}^{-23}\text{J}/\text{K}$ .

Convert mass $32\text{u}$ into kilogram.

  $\begin{array}{c}m=32\text{u}\times \frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}\\ =5.31\times {10}^{-26}\text{kg}\end{array}$

Convert $22°\text{C}$ into Kelvin scale.

  $\begin{array}{c}22°\text{C}=\left(22\text{+273}\right)\text{K}\\ =\text{295}\text{K}\end{array}$

Substitute $5.31\times {10}^{-26}\text{kg}$ for $m$ . $\text{295}\text{K}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_B$ in equation (I) to get $v_{\text{mp}}$ .

  $\begin{array}{c}{v}_{\text{mp}}=\sqrt{\frac{2\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\text{295}\text{K}\right)}{\left(5.31\times {10}^{-26}\text{kg}\right)}}\\ =390\text{m}/\text{s}\end{array}$

Substitute $5.31\times {10}^{-26}\text{kg}$ for $m$ . $\text{295}\text{K}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_B$ in equation (II) to get $v_{\text{av}}$ .

  $\begin{array}{c}{v}_{\text{av}}=\sqrt{\frac{8\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\text{295}\text{K}\right)}{\pi \left(5.31\times {10}^{-26}\text{kg}\right)}}\\ =440\text{m}/\text{s}\end{array}$

Substitute $5.31\times {10}^{-26}\text{kg}$ for $m$ . $\text{295}\text{K}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_B$ in equation (III) to get $v_{\text{rms}}$ .

  $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\text{295}\text{K}\right)}{\left(5.31\times {10}^{-26}\text{kg}\right)}}\\ =480\text{m}/\text{s}\end{array}$

Therefore, the most probable speed of oxygen gas is $390\text{m}/\text{s}$ , average speed of the oxygen gas is $440\text{m}/\text{s}$ and rms speed for the oxygen gas is $480\text{m}/\text{s}$ .

(b)

To determine

The most probable speed, average speed and rms speed for the nitrogen gas.

Answer to Problem 27PQ

The most probable speed of nitrogen gas is $420\text{m}/\text{s}$ , average speed of the nitrogen gas is $470\text{m}/\text{s}$ and rms speed for the nitrogen gas is $510\text{m}/\text{s}$ .

Explanation of Solution

Rewrite equations from part(a).

Write the expression for the most probable speed of gas molecules.

  $v_{\text{mp}}=\sqrt{\frac{2k_{B}T}{m}}$                                                                                                         (I)

Here, $v_{\text{mp}}$ is the most probable speed of the gas molecules, $k_B$ is the Boltzmann constant, $T$ is the temperature and $m$ is the mass of the molecule.

Write the expression for the most average speed of gas molecules.

  $v_{\text{av}}=\sqrt{\frac{8k_{B}T}{\pi m}}$                                                                                                          (II)

Here, $v_{\text{av}}$ is the average speed of the gas molecules.

Write the expression for the rms speed of gas molecules.

  $v_{\text{rms}}=\sqrt{\frac{3k_{B}T}{m}}$                                                                                                        (III)

Here, $v_{rms}$ is the rms speed of the gas molecules.

Conclusion:

The mass of oxygen molecule is $28\text{u}$ .

Convert mass $28\text{u}$ into kilogram.

  $\begin{array}{c}m=28\text{u}\times \frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}\\ =4.65\times {10}^{-26}\text{kg}\end{array}$

Substitute $4.65\times {10}^{-26}\text{kg}$ for $m$ . $\text{295}\text{K}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_B$ in equation (I) to get $v_{\text{mp}}$ .

  $\begin{array}{c}{v}_{\text{mp}}=\sqrt{\frac{2\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\text{295}\text{K}\right)}{\left(4.65\times {10}^{-26}\text{kg}\right)}}\\ =420\text{m}/\text{s}\end{array}$

Substitute $4.65\times {10}^{-26}\text{kg}$ for $m$ . $\text{295}\text{K}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_B$ in equation (II) to get $v_{\text{av}}$ .

  $\begin{array}{c}{v}_{\text{av}}=\sqrt{\frac{8\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\text{295}\text{K}\right)}{\pi \left(4.65\times {10}^{-26}\text{kg}\right)}}\\ =470\text{m}/\text{s}\end{array}$

Substitute $4.65\times {10}^{-26}\text{kg}$ for $m$ . $\text{295}\text{K}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_B$ in equation (III) to get $v_{\text{rms}}$ .

  $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\text{295}\text{K}\right)}{\left(4.65\times {10}^{-26}\text{kg}\right)}}\\ =510\text{m}/\text{s}\end{array}$

Therefore, the most probable speed of nitrogen gas is $420\text{m}/\text{s}$ , average speed of the nitrogen gas is $470\text{m}/\text{s}$ and rms speed for the nitrogen gas is $510\text{m}/\text{s}$ .