Chapter 21, Problem 21.44AP

(a)

To determine

The average speedof a molecules.

Answer to Problem 21.44AP

Answer:The average speed of a molecules is $3.90\text{km}/\text{s}$.

Explanation of Solution

Speed of molecule 1 is $3.00\text{km}/\text{s}$, Speed of molecule 2 is $4.00\text{km}/\text{s}$, Speed of molecule 3 is $5.80\text{km}/\text{s}$, Speed of molecule 4 is $2.50\text{km}/\text{s}$, Speed of molecule 5 is $3.60\text{km}/\text{s}$, Speed of molecule 6 is $1.90\text{km}/\text{s}$, Speed of molecule 7 is $3.80\text{km}/\text{s}$, Speed of molecule 8 is $6.60\text{km}/\text{s}$.

Write the expression for summation of the speeds of the molecules.

    ${\displaystyle \sum _{i=1}^N{v}_i}=\left(v_1\right)+\left(v_2\right)+\left(v_3\right)+\left(v_4\right)+\left(v_5\right)+\left(v_6\right)+\left(v_7\right)+\left(v_8\right)$                       (1)

Here,

$v_1$ is the Speed of molecule 1.

$v_2$ is the Speed of molecule 2.

$v_3$ is the Speed of molecule 3.

$v_4$ is the Speed of molecule 4.

$v_5$ is the Speed of molecule 5.

$v_6$ is the Speed of molecule 6.

$v_7$ is the Speed of molecule 7.

$v_8$ is the Speed of molecule 8.

Substitute $3.00\text{km}/\text{s}$ for $v_1$, $4.00\text{km}/\text{s}$ for $v_2$, $5.80\text{km}/\text{s}$ for $v_3$, $2.50\text{km}/\text{s}$ for $v_4$, $3.60\text{km}/\text{s}$ for $v_5$, $1.90\text{km}/\text{s}$ for $v_6$, $3.80\text{km}/\text{s}$ for $v_7$, $6.60\text{km}/\text{s}$ for $v_8$ in equation (1) to find ${\displaystyle \sum _{i=1}^N{v}_i}$.

    $\begin{array}{c}{\displaystyle \sum _{i=1}^N{v}_i}=\left\{\begin{array}{l}\left(3.00\text{km}/\text{s}\right)+\left(4.00\text{km}/\text{s}\right)+\left(5.80\text{km}/\text{s}\right)+\left(2.50\text{km}/\text{s}\right)\\ +\left(3.60\text{km}/\text{s}\right)+\left(1.90\text{km}/\text{s}\right)+\left(3.80\text{km}/\text{s}\right)+\left(6.60\text{km}/\text{s}\right)\end{array}\right\}\\ =31.20\text{km}/\text{s}\end{array}$

Thus, the value of summation of the speeds of the molecules is $31.20\text{km}/\text{s}$.

Write the formula to calculate the average speed of a molecules.

    $\overline{v}=\frac{{\displaystyle \sum _{i=1}^N{v}_i}}{N}$ (2)

Here,

$\overline{v}$ is the average speed of a molecules.

${\displaystyle \sum _{i=1}^N{v}_i}$ is the summation of the speeds of the molecules.

$N$ is the number of speeds of molecule.

Substitute $31.20\text{km}/\text{s}$ for ${\displaystyle \sum _{i=1}^N{v}_i}$, $8$ for $N$ in equation (2) to find $\overline{v}$.

    $\begin{array}{c}\overline{v}=\frac{31.20\text{km}/\text{s}}{8}\\ =3.90\text{km}/\text{s}\end{array}$

Conclusion:

Therefore, the average speed of a molecules is $3.90\text{km}/\text{s}$.

(b)

To determine

The root mean square speed of a molecules.

Answer to Problem 21.44AP

Answer:The root mean square speed of a molecules is $4.18\text{km}/\text{s}$.

Explanation of Solution

 Speed of molecule 1 is $3.00\text{km}/\text{s}$, Speed of molecule 2 is $4.00\text{km}/\text{s}$, Speed of molecule 3 is $5.80\text{km}/\text{s}$, Speed of molecule 4 is $2.50\text{km}/\text{s}$, Speed of molecule 5 is $3.60\text{km}/\text{s}$, Speed of molecule 6 is $1.90\text{km}/\text{s}$, Speed of molecule 7 is $3.80\text{km}/\text{s}$, Speed of molecule 8 is $6.60\text{km}/\text{s}$.

Write the expression for summation of the square of speeds of the molecules.

    ${\displaystyle \sum _{i=1}^N{v}_i^2}={\left(v_1\right)}^2+{\left(v_2\right)}^2+{\left(v_3\right)}^2+{\left(v_4\right)}^2+{\left(v_5\right)}^2+{\left(v_6\right)}^2+{\left(v_7\right)}^2+{\left(v_8\right)}^2$ (3)

Substitute $3.00\text{km}/\text{s}$ for $v_1$, $4.00\text{km}/\text{s}$ for $v_2$, $5.80\text{km}/\text{s}$ for $v_3$, $2.50\text{km}/\text{s}$ for $v_4$, $3.60\text{km}/\text{s}$ for $v_5$, $1.90\text{km}/\text{s}$ for $v_6$, $3.80\text{km}/\text{s}$ for $v_7$, $6.60\text{km}/\text{s}$ for $v_8$ in equation (3) to find ${\displaystyle \sum _{i=1}^N{v}_i^2}$.

    $\begin{array}{c}{\displaystyle \sum _{i=1}^N{v}_i^2}=\left\{\begin{array}{l}{\left(3.00\text{km}/\text{s}\right)}^2+{\left(4.00\text{km}/\text{s}\right)}^2+{\left(5.80\text{km}/\text{s}\right)}^2+{\left(2.50\text{km}/\text{s}\right)}^2\\ +{\left(3.60\text{km}/\text{s}\right)}^2+{\left(1.90\text{km}/\text{s}\right)}^2+{\left(3.80\text{km}/\text{s}\right)}^2+{\left(6.60\text{km}/\text{s}\right)}^2\end{array}\right\}\\ =139.46\text{km}/\text{s}\end{array}$

Thus, the value of summation of the square of speeds of the molecules is $139.46\text{km}/\text{s}$.

Write the formula to calculate the root mean square speed of a molecules.

    $v_{\text{rms}}=\sqrt{\frac{{\displaystyle \sum _{i=1}^N{v}_i^2}}{N}}$                           (4)

Here,

$v_{\text{rms}}$ is the root mean square speed of a molecules.

${\displaystyle \sum _{i=1}^N{v}_i^2}$ is the summation of the square of speeds of the molecules.

$N$ is the number of speeds of molecule.

Substitute $139.46\text{km}/\text{s}$ for ${\displaystyle \sum _{i=1}^N{v}_i^2}$, $8$ for $N$ in equation (4) to find $v_{\text{rms}}$,

    $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{139.46\text{km}/\text{s}}{8}}\\ =4.1752\text{km}/\text{s}\\ \approx 4.18\text{km}/\text{s}\end{array}$

Conclusion:

Therefore, the root mean square speed of a molecules is $4.18\text{km}/\text{s}$.