Chapter 13, Problem 85P

(a)

To determine

The average distance between Ping-Pong balls at any instant.

Answer to Problem 85P

Average distance of separation is $52\text{cm}$.

Explanation of Solution

Diameter of ball is $3.75\text{cm}$ and diameter of $N_2$ molecule is $0.30\text{nm}$.

It is assumed that the $N_2$ molecules are located at the centre of a sphere.

Write the equation to find the average distance of separation between Ping-Pong balls at any instant.

$d_{av}=a{d}_{n2}$ (I)

Here, $d_{av}$ is the average distance between Ping –Pong balls (same as diameter of spheres), $a$ is the scale factor, and $d_{n2}$ is the separation between $N_2$ molecules.

Write the equation for $a$.

$a=\frac{d_{ping}}{d_{n2}}$ (I)

Here, $d_{ping}$ is the diameter of Ping-Pong balls.

Write the equation for each ping pong ball.

$\frac{V}{N}=\frac{1}{6}\pi d_{av}^3$

Here, $V$ is the total volume occupied by spheres and $N$ is the Avogadro number.

Rewrite the above relation in terms of $d_{av}$.

$d_{av}={\left(\frac{6V}{\pi N}\right)}^{1/3}$ (II)

Rewrite equation (I) by substituting equations (II) and (III).

$d_{av}=\left(\frac{d_{ping}}{d_{n2}}\right){\left(\frac{6V}{\pi N}\right)}^{1/3}$

Conclusion:

Substitute $3.75\text{cm}$ for $d_{ping}$, $0.30\text{nm}$ for $d_{n2}$, $0.0224{\text{m}}^{\text{3}}$ for $V$, and $6.022\times {10}^{23}$ for $N$ in the above equation to find $d_{av}$.

$\begin{array}{c}{d}_{av}=\left(\frac{3.75\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{0.30\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right){\left(\frac{6\left(0.0224{\text{m}}^{\text{3}}\right)}{\pi \left(6.022\times {10}^{23}\right)}\right)}^{1/3}\\ =\left(12.5\times {10}^7\right)\left(0.0416\times {10}^{-7}\text{m}\right)\\ =0.52\text{m}\left(\frac{{\text{10}}^{\text{2}}\text{cm}}{\text{1}\text{m}}\right)\\ =52\text{cm}\end{array}$

Therefore, the average distance of separation is $52\text{cm}$.

(b)

To determine

The average distance travelled by Ping-Pong balls before collision with one another.

Answer to Problem 85P

Average distance travelled is $12\text{m}$.

Explanation of Solution

Diameter of ball is $3.75\text{cm}$ and diameter of $N_2$ molecule is $0.30\text{nm}$.

It is assumed that the $N_2$ molecules are located at the centre of a sphere.

Write the equation for mean free path.

$\Lambda =\frac{1}{\sqrt{2}\pi d_{n2}^2\left(N/V\right)}$

Here,$\Lambda$ is the mean free path.

Write the relation between $N/V$ and pressure.

$\frac{N}{V}=\frac{P}{K_{B}T}$

Here, $K_B$ is the Boltzmann constant, $T$ is the temperature, and $P$ is the pressure.

Rewrite the equation for $\Lambda$ by substituting the above relation.

$\begin{array}{c}\Lambda =\frac{1}{\sqrt{2}\pi d_{n2}^2\left(\frac{P}{K_{B}T}\right)}\\ =\frac{K_{B}T}{\sqrt{2}\pi d_{n2}^{2}P}\end{array}$

Write the average distance travelled by Ping-Pong balls before collision with one another.

$d_{col}=a\Lambda$

Here, $d_{col}$ is the average distance travelled by Ping-Pong balls before collision with one another.

Rewrite the above relation by substituting $\frac{P}{\sqrt{2}\pi d_{n2}^2{K}_{B}T}$ for $\Lambda$.

$d_{col}=a\frac{K_{B}T}{\sqrt{2}\pi d_{n2}^{2}P}$

Rewrite the above relation by substituting equation (I).

$d_{col}=\left(\frac{d_{ping}}{d_{n2}}\right)\frac{K_{B}T}{\sqrt{2}\pi d_{n2}^{2}P}$

Conclusion:

Substitute $3.75\text{cm}$ for $d_{ping}$, $0.30\text{nm}$ for $d_{n2}$, $1.38\times {10}^{-23}\text{J}/\text{K}$ for $K_B$, $\text{273}\text{.15}\text{K+0}\text{.0}\text{K}$ for $T$, $3.14$ for $\pi$, $0.30\text{nm}$ for $d_{n2}$, and $1.00\text{atm}$ for $P$ in the above equation to find $d_{col}$.

$\begin{array}{c}{d}_{col}=\left(\frac{3.75\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{0.30\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)\frac{\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\text{273}\text{.15}\text{K+0}\text{.0}\text{K}\right)}{\sqrt{2}\left(3.14\right)\left(0.30\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)\left(1.00\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}/\text{atm}}{1.00\text{atm}}\right)\right)}\\ =\frac{14.13\times {10}^{-23}\text{J}\cdot \text{m}}{1.18\times {110}^{-23}\text{J}}\\ =12\text{m}\end{array}$

Therefore, the average distance travelled is $12\text{m}$.