Chapter 13, Problem 119P

(a)

To determine

The distance travelled by the oxygen molecule between the collisions with other molecules.

Answer to Problem 119P

The distance travelled by the oxygen molecule between the collisions with other molecules is $\underset{\_}{1\times {10}^{-7}\text{m}}$.

Explanation of Solution

Given that the diameter of the molecule is $0.3\text{nm}$, the temperature is $20\text{°C}$ ($293.15\text{K}$), and the time is $1.0\text{s}$.

Write the expression for the mean free path between consecutive collisions.

$\Lambda =\frac{1}{\sqrt{2}\pi d^{2}N/V}$ (I)

Here, $\Lambda$ is the mean free path, $d$ is the diameter of the molecule, $N$ is the number of molecules, and $V$ is the volume.

Write the ideal gas law equation.

$PV=NkT$ (II)

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

Solve equation (II) for $N/V$.

$\frac{N}{V}=\frac{P}{kT}$ (III)

Use equation (III) in (I).

$\Lambda =\frac{kT}{\sqrt{2}\pi d^{2}P}$ (IV)

The pressure of still air is about $1.013\times {10}^5\text{Pa}$ or $1\text{atm}$.

Conclusion:

Substitute $1.38\times {10}^{-23}\text{J/K}$ for $k$, $293.15\text{K}$ for $T$, $0.3\text{nm}$ for $d$, and $1.013\times {10}^5\text{Pa}$ for $P$ in equation (IV) to find $\Lambda$.

$\begin{array}{c}\Lambda =\frac{\left(1.38\times {10}^{-23}\text{J/K}\right)\left(293.15\text{K}\right)}{\sqrt{2}\pi {\left(0.3\text{nm}\right)}^2\left(1.013\times {10}^5\text{Pa}\right)}\\ =\frac{\left(1.38\times {10}^{-23}\text{J/K}\right)\left(293.15\text{K}\right)}{\sqrt{2}\pi {\left(0.3\text{nm}\times \frac{1\text{m}}{1\times {10}^{-9}\text{nm}}\right)}^2\left(1.013\times {10}^5\text{Pa}\right)}\\ =1\times {10}^{-7}\text{m}\end{array}$

Therefore, the distance travelled by the oxygen molecule between the collisions with other molecules is $\underset{\_}{1\times {10}^{-7}\text{m}}$.

(b)

To determine

The number of collisions in $1.0\text{s}$.

Answer to Problem 119P

The number of collisions in $1.0\text{s}$ is $\underset{\_}{5\times {10}^9}$.

Explanation of Solution

Given that the diameter of the molecule is $0.3\text{nm}$, the temperature is $20\text{°C}$ ($293.15\text{K}$), and the time is $1.0\text{s}$. It is obtained that distance travelled by the oxygen molecule between the collisions with other molecules is $1\times {10}^{-7}\text{m}$.

Write the expression for the rms speed of the molecules.

$v_{\text{rms}}=\sqrt{\frac{3kT}{m}}$ (V)

Here, $v_{\text{rms}}$ is the rms speed, and $m$ is the mass of the molecule (the mass of oxygen molecule is $15.9994\text{u}$)

In one second, it can be imagined that the molecules moving with the rms speed and leaving a trail $x$ consisting of jagged segments of length equal to the mean free path $\Lambda$.

Write the expression for the distance travelled by the molecule in $1.0\text{s}$.

$x=v_{\text{rms}}\times \left(1.0\text{s}\right)$ (VI)

Write the expression for the number of collisions $N$ in one second.

$N=\frac{x}{\Lambda }$ (VII)

Conclusion:

Substitute $1.38\times {10}^{-23}\text{J/K}$ for $k$, $293.15\text{K}$ for $T$, and $\left(2\times 15.9994\text{u}\right)$ for $m$ in equation (V) to find $v_{\text{rms}}$.

$\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(293.15\text{K}\right)}{2\times 15.9994\text{u}}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(293.15\text{K}\right)}{2\times 15.9994\text{u}\times \frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}}}\\ =478\text{m/s}\end{array}$

Substitute $478\text{m/s}$ for $v_{\text{rms}}$ in equation (VI) to find $x$.

$\begin{array}{c}x=\left(478\text{m/s}\right)\left(1.0\text{s}\right)\\ =478\text{m}\end{array}$

Substitute $478\text{m}$ for $x$, and $1\times {10}^{-7}\text{m}$ for $\Lambda$ in equation (VII) to find $N$.

$\begin{array}{c}N=\frac{478\text{m}}{1\times {10}^{-7}\text{m}}\\ =478\times {10}^7\text{m}\\ \approx 5\times {10}^9\text{m}\end{array}$

Therefore, the number of collisions in $1.0\text{s}$ is $\underset{\_}{5\times {10}^9}$.

(c)

To determine

The total distance travelled by the molecule in $1.0\text{s}$.

Answer to Problem 119P

The total distance travelled by the molecule in $1.0\text{s}$ is $\underset{\_}{500\text{m}}$.

Explanation of Solution

From part (b), it is obtained that the distance travelled by the molecule in $1.0\text{s}$ is,

$\begin{array}{c}x=478\text{m}\\ \approx 500\text{m}\end{array}$

Conclusion:

Therefore, the total distance travelled by the molecule in $1.0\text{s}$ is $\underset{\_}{500\text{m}}$.

(d)

To determine

The displacement of the molecule in $1.0\text{s}$.

Answer to Problem 119P

The displacement of the molecule in $1.0\text{s}$ is $\underset{\_}{1\text{cm}}$.

Explanation of Solution

It is obtained that the distance travelled by the oxygen molecule between the collisions with other molecules is $1\times {10}^{-7}\text{m}$, and the number of collisions in $1.0\text{s}$ is $5\times {10}^9$.

Write the expression for the component of displacement of a diffusing molecule.

$x_{\text{rms}}=\sqrt{2Dt}$ (VIII)

Here, $D$ is the mass diffusivity, and $t$ is the time.

There will be components $y_{\text{rms}}$ and $z_{\text{rms}}$ with same magnitude as $x_{\text{rms}}$ since the molecule is travelling in three dimensions.

Write the expression for the displacement of the molecule in $1.0\text{s}$.

$\text{displacement}=\sqrt{x_{\text{rms}}^2+y_{\text{rms}}^2+z_{\text{rms}}^2}$ (IX)

Conclusion:

Substitute $1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$ for $D$, and $1.0\text{s}$ for $t$ in equation (VIII) to find $x_{\text{rms}}$.

$\begin{array}{c}{x}_{\text{rms}}=\sqrt{2\left(1.8\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}\right)\left(1.0\text{s}\right)}\\ =6\times {10}^{-3}\text{m}\end{array}$

Substitute $6\times {10}^{-3}\text{m}$ for $x_{\text{rms}}$, $y_{\text{rms}}$, and $z_{\text{rms}}$ in equation (IX) to find the displacement.

$\begin{array}{c}\text{displacement}=\sqrt{{\left(6\times {10}^{-3}\text{m}\right)}^2+{\left(6\times {10}^{-3}\text{m}\right)}^2+{\left(6\times {10}^{-3}\text{m}\right)}^2}\\ =0.010\text{m}\\ =0.010\text{m}\times \frac{100\text{cm}}{1\text{m}}\\ =1.0\text{cm}\end{array}$

Therefore, the displacement of the molecule in $1.0\text{s}$ is $\underset{\_}{1\text{cm}}$.