Chapter 21, Problem 61AP

(a)

To determine

The rms speed of the molecule for a particular diameter.

Answer to Problem 61AP

The rms speed of the molecule is $\left(4.81\times {10}^{-12}\right)d^{-3/2}$.

Explanation of Solution

Write the expression for volume of the particle.

  $V=\frac{4}{3}\pi r^3$                                                                                                 (I)

Here, $r$ is radius of the particle.

Write the expression for mass in terms of density and volume.

  $m=\rho V$                                                                                                     (II)

Here, $m$ is the mass of the particle, $\rho$ is the density of the particle, and $V$ is the volume.

Write the expression for kinetic energy of the particle.

  $\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}kT$                                                                                           (III)

Here, $v_{rms}$ is the rms speed of the particle, $k$ is the Boltzmann constant, and $T$ is the temperature.

Conclusion:

Substitute equation (I) in equation (II).

  $\begin{array}{c}m=\rho \frac{4}{3}\pi r^3\\ =\rho \frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\\ =\frac{\rho \pi d^3}{6}\end{array}$

Here, $d$ is the diameter of the particle.

Substitute above equation in the equation (III) instead of mass.

  $\begin{array}{c}\frac{1}{2}\left(\frac{\rho \pi d^3}{6}\right)v_{rms}^2=\frac{3}{2}kT\\ v_{rms}=\sqrt{\frac{18kT}{\rho \pi d^3}}\end{array}$

Substitute $1.38\times {10}^{-23}\text{J/K}$ for $k$. $293\text{K}$ for $T$, and $1000{\text{kg/m}}^3$ for $\rho$ in the above equation to find $v_{rms}$.

  $\begin{array}{c}{v}_{rms}={\left(\frac{18\left(1.38\times {10}^{-23}\text{J/K}\right)\left(293\text{K}\right)}{\left(1000{\text{kg/m}}^3\right)\left(3.14\right)}\right)}^{1/2}\left(d^{-3/2}\right)\\ =\left(4.81\times {10}^{-12}\right)\left(d^{-3/2}\right)\end{array}$

Here, $v_{rms}$ is in meters per second and $d$ is in meters.

Therefore, the rms speed of the molecule is $\left(4.81\times {10}^{-12}\right)d^{-3/2}$.

(b)

To determine

The time interval of the particle’s actual motion.

Answer to Problem 61AP

The time interval of the particle’s actual motion is $\left(2.08\times {10}^{11}\right)d^{5/2}$.

Explanation of Solution

Write the expression for the distance of the particle equal to its dimeter.

  $v=\frac{d}{\Delta t}$

Here, $v$ is the velocity of the particle, $d$ is the distance of the moving particle, and $\Delta t$ is the time interval of the particle’s actual motion.

Rearrange the above equation for $\Delta t$.

  $\Delta t=\frac{d}{v}$                                                                                             (IV)

Conclusion:

Substitute $\left(4.81\times {10}^{-12}\right)d^{-3/2}$ for $v$ in equation (IV) to find $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{d}{\left(4.81\times {10}^{-12}\right)d^{-3/2}}\\ =\left(2.08\times {10}^{11}\right)d^{5/2}\end{array}$

Here, $\Delta t$ is in second and $d$ is in meters.

Therefore, the number of moles of air in the pump is $9.97\times {10}^{-3}\text{mol}$.

(c)

To determine

The rms speed and time interval of the particle for diameter $3.00\text{μm}$.

Answer to Problem 61AP

The rms speed is $0.926\text{mm/s}$ and time interval of the particle is $3.24\text{ms}$.

Explanation of Solution

Write the expression for rms speed of the particle.

  $v_{rms}=\sqrt{\frac{18kT}{\rho \pi d^3}}$                                                                                                     (V)

Conclusion:

Substitute $1.38\times {10}^{-23}\text{J/K}$ for $k$. $293\text{K}$ for $T$, $3.00\text{μm}$ for $d$, and $1000{\text{kg/m}}^3$ for $\rho$ in the equation (V) to find $v_{rms}$.

  $\begin{array}{c}{v}_{rms}=\sqrt{\frac{18\left(1.38\times {10}^{-23}\text{J/K}\right)\left(293\text{K}\right)}{\left(1000{\text{kg/m}}^3\right)3.14{\left[\left(3.00\text{μm}\right)\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right]}^3}}\\ =0.926\times {10}^{-3}\text{m/s}\left(\frac{{10}^3\text{mm/s}}{1\text{m/s}}\right)\\ =0.926\text{mm/s}\end{array}$

Substitute $0.926\times {10}^{-3}\text{m/s}$ for $v$ and $3\times {10}^{-6}\text{m}$ for $d$ in equation (IV) to find $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{3\times {10}^{-6}\text{m}}{0.926\times {10}^{-3}\text{m/s}}\\ =3.24\times {10}^{-3}\text{s}\left(\frac{{10}^3\text{ms}}{1\text{s}}\right)\\ =3.24\text{ms}\end{array}$

Therefore, the rms speed is $0.926\text{mm/s}$ and time interval of the particle is $3.24\text{ms}$.

(d)

To determine

The rms speed and time interval of the particle for a sphere.

Answer to Problem 61AP

The rms speed is $1.32\times {10}^{-11}\text{m/s}$ and time interval of the particle for a sphere is $3.88\times {10}^{10}\text{s}$.

Explanation of Solution

Write the expression for mass of an object in terms of density and volume.

  $m=\frac{\rho \pi d^3}{6}$                                                                                                            (VI)

Rearrange the above expression for $d$.

  $\begin{array}{c}{d}^3=\frac{6m}{\rho \pi }\\ d={\left(\frac{6m}{\rho \pi }\right)}^{1/3}\end{array}$                                                                                                   (VII)

Conclusion:

Substitute $70.0\text{kg}$ for $m$ and $1000{\text{kg/m}}^3$ for $\rho$ in equation (VII) to find $d$.

  $\begin{array}{c}d={\left(\frac{6\left(70.0\text{kg}\right)}{\left(1000{\text{kg/m}}^3\right)3.14}\right)}^{1/3}\\ ={\left(0.134{\text{m}}^3\right)}^{1/3}\\ =0.511\text{m}\end{array}$

Substitute $1.38\times {10}^{-23}\text{J/K}$ for $k$. $293\text{K}$ for $T$, $\text{0}\text{.511}\text{m}$ for $d$, and $1000{\text{kg/m}}^3$ for $\rho$ in the equation (V) to find $v_{rms}$.

  $\begin{array}{c}{v}_{rms}=\sqrt{\frac{18\left(1.38\times {10}^{-23}\text{J/K}\right)\left(293\text{K}\right)}{\left(1000{\text{kg/m}}^3\right)3.14{\left(0.511\text{m}\right)}^3}}\\ =1.32\times {10}^{-11}\text{m/s}\end{array}$

Substitute $1.32\times {10}^{-11}\text{m/s}$ for $v$ and $0.511\text{m}$ for $d$ in equation (IV) to find $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{0.511\text{m}}{1.32\times {10}^{-11}\text{m/s}}\\ =3.88\times {10}^{10}\text{s}\end{array}$

Therefore, the rms speed is $1.32\times {10}^{-11}\text{m/s}$ and time interval of the particle for a sphere is $3.88\times {10}^{10}\text{s}$.