Chapter 20, Problem 10PQ

(a)

To determine

The average speed of the 12 atoms in the cylinder.

Answer to Problem 10PQ

The average speed of the 12 atoms in the cylinder is $\underset{\_}{4.30\text{km}/\text{s}}$.

Explanation of Solution

Write the expression for average speed.

$v_{\text{avg}}=\frac{{\displaystyle \sum _i^N{v}_i}}{N}$

Here, $v_i$ is the velocity and $N$ is the number of observations.

Conclusion:

Substitute $1.50\text{km}/\text{s}$ for $v_1$, $5.00\text{km}/\text{s}$ for $v_2$, $8.00\text{km}/\text{s}$ for $v_3$, $1.25\text{km}/\text{s}$ for $v_4$, $2.95\text{km}/\text{s}$ for $v_5$, $4.40\text{km}/\text{s}$ for $v_6$, $5.60\text{km}/\text{s}$ for $v_7$, $3.50\text{km}/\text{s}$ for $v_8$, $7.80\text{km}/\text{s}$ for $v_9$, $2.70\text{km}/\text{s}$ for $v_{10}$, $2.90\text{km}/\text{s}$ for $v_{11}$, $6.00\text{km}/\text{s}$ for $v_{12}$ and $12$ for $N$ in the above equation.

    $\begin{array}{c}{v}_{\text{avg}}=\frac{1}{12}\left(1.50\text{km}/\text{s}+5.00\text{km}/\text{s}+8.00\text{km}/\text{s}+1.25\text{km}/\text{s}+2.95\text{km}/\text{s}+4.40\text{km}/\text{s}+5.60\text{km}/\text{s}+3.50\text{km}/\text{s}+7.80\text{km}/\text{s}+2.70\text{km}/\text{s}+2.90\text{km}/\text{s}+6.00\text{km}/\text{s}\right)\\ =4.30\text{km}/\text{s}\end{array}$

Therefore, the average speed of the 12 atoms in the cylinder is $\underset{\_}{4.30\text{km}/\text{s}}$.

(b)

To determine

The rms speed of the 12 atoms in the cylinder.

Answer to Problem 10PQ

The rms speed of the 12 atoms in the cylinder is $\underset{\_}{4.80\text{km}/\text{s}}$.

Explanation of Solution

Write the expression for rms speed.

$v_{\text{rms}}=\sqrt{\frac{{\displaystyle \sum _i^N{v}_i^2}}{N}}$

Here, $v_i$ is the velocity and $N$ is the number of observations.

Conclusion:

Substitute $1.50\text{km}/\text{s}$ for $v_1$, $5.00\text{km}/\text{s}$ for $v_2$, $8.00\text{km}/\text{s}$ for $v_3$, $1.25\text{km}/\text{s}$ for $v_4$, $2.95\text{km}/\text{s}$ for $v_5$, $4.40\text{km}/\text{s}$ for $v_6$, $5.60\text{km}/\text{s}$ for $v_7$, $3.50\text{km}/\text{s}$ for $v_8$, $7.80\text{km}/\text{s}$ for $v_9$, $2.70\text{km}/\text{s}$ for $v_{10}$, $2.90\text{km}/\text{s}$ for $v_{11}$, $6.00\text{km}/\text{s}$ for $v_{12}$ and $12$ for $N$ in the above equation.

    $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{1}{12}\left({\left(1.50\text{km}/\text{s}\right)}^2+{\left(5.00\text{km}/\text{s}\right)}^2+{\left(8.00\text{km}/\text{s}\right)}^2+{\left(1.25\text{km}/\text{s}\right)}^2+{\left(2.95\text{km}/\text{s}\right)}^2+{\left(4.40\text{km}/\text{s}\right)}^2+{\left(5.60\text{km}/\text{s}\right)}^2+{\left(3.50\text{km}/\text{s}\right)}^2+{\left(7.80\text{km}/\text{s}\right)}^2+{\left(2.70\text{km}/\text{s}\right)}^2+{\left(2.90\text{km}/\text{s}\right)}^2+{\left(6.00\text{km}/\text{s}\right)}^2\right)}\\ =4.80\text{km}/\text{s}\end{array}$

Therefore, the rms speed of the 12 atoms in the cylinder is $\underset{\_}{4.80\text{km}/\text{s}}$.