Chapter 21, Problem 21.61AP

(a)

To determine

The rms speed for a particle of diameter $d$.

Answer to Problem 21.61AP

The rms speed for a particle of diameter $d$ is $\left(4.82\times {10}^{-12}\right)d^{-\frac{3}{2}}\text{m}/\text{s}$.

Explanation of Solution

Given info: Density of spherical particle is $1.00\times {10}^3\text{kg}/{\text{m}}^3$, temperature is $20.0°\text{C}$ and diameter of particle is $d$.

Write the expression for the rms speed:

$v_{\text{rms}}=\sqrt{\frac{3k_{\text{B}}T}{m}}$ (1)

Here,

$v_{\text{rms}}$ is rms speed of particle.

$k_{\text{B}}$ is boltzmann’s constant.

$T$ is temperature.

$m$ is mass of gas molecule.

Write the formula for mass:

$m=\rho \times V$ (2)

Here,

$\rho$ is density of spherical particle.

$V$ is volume of spherical particle.

Write the formula for volume of spherical particle:

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

It is given that diameter of spherical particle is $d$. So, the radius of spherical particle is $\frac{d}{2}$.

Substitute $3.14$ for $\pi$ and $\frac{d}{2}\text{m}$ for $r$ in equation (3).

$\begin{array}{c}V=\frac{4}{3}\left(3.14\right){\left(\frac{d}{2}\text{m}\right)}^3\\ =0.52d^3{\text{m}}^3\end{array}$

Substitute $1.00\times {10}^3\text{kg}/{\text{m}}^3$ for $\rho$ and $0.52d^3{\text{m}}^3$ for $V$ in equation (2).

$\begin{array}{c}m=\left(1.00\times {10}^3\text{kg}/{\text{m}}^3\right)\left(0.52d^3{\text{m}}^3\right)\\ =520d^3\text{kg}\end{array}$

Since, boltzmann’s constant is $1.38\times {10}^{-23}\text{J}/\text{K}$.

Substitute $520d^3\text{kg}$ for $m$, $20.0°\text{C}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$ in equation (1).

$\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J}/\text{K}\times \frac{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{J}}\right)\left(20.0+273\text{K}\right)}{520d^3\text{kg}}}\\ =\left(4.82\times {10}^{-12}\right)d^{-\frac{3}{2}}\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the rms speed for a particle of diameter $d$ is $\left(4.82\times {10}^{-12}\right)d^{-\frac{3}{2}}\text{m}/\text{s}$.

(b)

To determine

The time interval for particle to move a distance equal to its own diameter.

Answer to Problem 21.61AP

The time interval for particle to move a distance equal to its own diameter is $\left(2.08\times {10}^{11}\right)d^{\frac{5}{2}}\text{s}$.

Explanation of Solution

Given info: Density of spherical particle is $1.00\times {10}^3\text{kg}/{\text{m}}^3$ and temperature is $20.0°\text{C}$.

Write the expression for the time interval related to rms:

$t=\frac{d^{\prime }}{v_{\text{rms}}}$ (4)

Here,

$t$ is time interval.

$d^{\prime }$ is distance.

$v_{\text{rms}}$ is rms speed.

Since particle is moving equal to its diameter. So, the distance travelled by the particles is $d$.

Substitute $d$ for $d^{\prime }$ and $\left(4.82\times {10}^{-12}\right)d^{-\frac{3}{2}}\text{m}/\text{s}$ for $v_{\text{rms}}$ as calculated in above part in equation (3).

$\begin{array}{c}t=\frac{d\text{m}}{\left(4.82\times {10}^{-12}\right)d^{-\frac{3}{2}}\text{m}/\text{s}}\\ =\left(2.08\times {10}^{11}\right)d^{\frac{5}{2}}\text{s}\end{array}$

Conclusion:

Therefore, the time interval for particle to move a distance equal to its own diameter is $\left(2.08\times {10}^{11}\right)d^{\frac{5}{2}}\text{s}$.

(c)

To determine

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

Answer to Problem 21.61AP

The rms speed for a particle of diameter $3.00\text{μm}$ is $0.926\text{mm}/\text{s}$ and the time interval for a particle of diameter $3.00\text{μm}$ is $3.24\text{ms}$.

Explanation of Solution

Given info: Density of spherical particle is $1.00\times {10}^3\text{kg}/{\text{m}}^3$, temperature is $20.0°\text{C}$ and diameter of particle is $3.00\text{μm}$.

The rms speed for a particle of diameter $d$ as calculated in above part is,

$v_{\text{rms}}= \left(4.82\times {10}^{-12}\right)d^{-\frac{3}{2}}\text{m}/\text{s}$

Substitute $3.00\text{μm}$ for $d$ in above equation.

$\begin{array}{c}{v}_{\text{rms}}=\left(4.82\times {10}^{-12}\right){\left(3.00\text{μm}\times \frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}^{-\frac{3}{2}}\text{m}/\text{s}\\ =9.27\times {10}^{-4}\text{m}/\text{s}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ \cong 0.926\text{mm}/\text{s}\end{array}$

The time interval for a particle of diameter $d$ as calculated in above part is,

$t=\left(2.08\times {10}^{11}\right)d^{\frac{5}{2}}\text{s}$

Substitute $3.00\text{μm}$ for $d$ in above equation.

$\begin{array}{c}t=\left(2.08\times {10}^{11}\right){\left(3.00\text{μm}\times \frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}^{\frac{5}{2}}{\text{m}}^{-1}\text{s}\\ \text{=0}\text{.00324}\text{s}\times \frac{{10}^3\text{ms}}{1\text{s}}\\ \text{=3}\text{.24}\text{ms}\end{array}$

Conclusion:

Therefore, the rms speed for a particle of diameter $3.00\text{μm}$ is $0.926\text{mm}/\text{s}$ and the time interval for a particle of diameter $3.00\text{μm}$ is $3.24\text{ms}$.

(d)

To determine

The rms speed and the time interval for a sphere of $70.0\text{kg}$.

Answer to Problem 21.61AP

The rms speed for a sphere of $70.0\text{kg}$ is $1.32\times {10}^{-11}\text{m}/\text{s}$ and the time interval for a sphere of $70.0\text{kg}$ is $3.88\times {10}^{10}\text{s}$.

Explanation of Solution

Given info: Density of spherical particle is $1.00\times {10}^3\text{kg}/{\text{m}}^3$, temperature is $20.0°\text{C}$ and mass of the sphere is $70.0\text{kg}$.

Write the expression for the rms speed:

$v_{\text{rms}}=\sqrt{\frac{3k_{\text{B}}T}{m}}$ (5)

Here,

$v_{\text{rms}}$ is rms speed of particle.

$k_{\text{B}}$ is boltzmann’s constant.

$T$ is temperature.

$m$ is mass of sphere.

Substitute $70.0\text{kg}$ for $m$, $20.0°\text{C}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$ in equation (5).

$\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J}/\text{K}\times \frac{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}{1\text{J}}\right)\left(20.0+273\text{K}\right)}{70.0\text{kg}}}\\ =1.32\times {10}^{-11}\text{m}/\text{s}\end{array}$

Thus, the rms speed of the partical is $1.32\times {10}^{-11}\text{m}/\text{s}$.

Write the formaula for volume in relation to mass and density.

$V=\frac{m}{D}$

Here,

$D$ is density.

Substitute $70.0\text{kg}$ for $m$ and $1.00\times {10}^3\text{kg}/{\text{m}}^3$ in above equation.

$\begin{array}{c}V=\frac{70.0\text{kg}}{1.00\times {10}^3\text{kg}/{\text{m}}^3}\\ =0.07{\text{m}}^3\end{array}$

Write the formula for the volume of sphere:

$V=\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$

Here,

$V$ is volume of sphere.

$d$ is diameter of sphere.

Rearrange above equation for $d$.

$\begin{array}{c}{\left(\frac{d}{2}\right)}^3=\frac{3V}{4\pi }\\ d=2{\left(\frac{3V}{4\pi }\right)}^{\frac{1}{3}}\end{array}$ (6)

Substitute $0.07{\text{m}}^3$ for $V$ and $3.14$ for $\pi$ in equation (6).

$\begin{array}{c}d=2{\left(\frac{3\times 0.07{\text{m}}^3}{4\times 3.14}\right)}^{\frac{1}{3}}\\ =0.51\text{m}\end{array}$

Write the expression for the time interval related to rms:

$t=\frac{d^{\prime }}{v_{\text{rms}}}$

Substitute $d$ for $d^{\prime }$ in above equation.

$t=\frac{d}{v_{\text{rms}}}$

Substitute $0.51\text{m}$ for $d$ and $1.32\times {10}^{-11}\text{m}/\text{s}$ for $v_{\text{rms}}$ in above equation.

$\begin{array}{c}t=\frac{0.51\text{m}}{1.32\times {10}^{-11}\text{m}/\text{s}}\\ =3.86\times {10}^{10}\text{s}\end{array}$

Conclusion:

Therefore, the rms speed for a sphere of $70.0\text{kg}$ is $1.32\times {10}^{-11}\text{m}/\text{s}$ and the time interval for a sphere of $70.0\text{kg}$ is $3.88\times {10}^{10}\text{s}$.