Chapter 10, Problem 3P

(a)

To determine

The distance from A of the impact points of molecules.

Answer to Problem 3P

The distance covered by molecules travelling at $\langle v\rangle$ is $1.58\text{cm}$ ; distance covered by molecules travelling at $v_{\text{rms}}$ is $1.45\text{cm}$ and distance covered by molecules travelling at $v_{\text{mp}}$ is $1.58\text{cm}$ .

Explanation of Solution

The molecule covers d distance in t time with the speed of v.

Write the expression for angular displacement.

  $\theta =\omega t$        (I)

Here, $\theta$ is the angular displacement, $\omega$ is the angular velocity and $t$ is the required time.

Substitute $d/v$ for t and $2\pi N/60$ for in equation (I).

  $\theta =\left(\frac{2\pi N}{60}\right)\frac{d}{v}$

Here, d is the diameter of cylinder and v is the velocity of molecule.

Write the expression for distance covered.

  $s=\frac{d}{2}\theta$        (II)

Here, s is the distance covered.

Substitute $\frac{2\pi Nd}{60v}$ for $\theta$ in equation (II).

  $s=\frac{\pi d^{2}N}{60v}$        (III)

Write the expression for average velocity.

  $\langle v\rangle =\sqrt{\frac{8k_{\text{B}}T}{\pi m}}$        (IV)

Here, m is the mass of gas molecule, $k_{\text{B}}$ is Boltzmann’s constant, T is the absolute temperature and $\langle v\rangle$ is the average velocity.

Write the expression for root mean square velocity.

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

Here, $v_{\text{rms}}$ is the root mean square velocity.

Write the expression for most probable velocity.

  $v_{\text{mp}}=\sqrt{\frac{2k_{\text{B}}T}{m}}$        (VI)

Here, $v_{\text{mp}}$ is the most probable velocity.

Conclusion:

Substitute $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$, $850\text{K}$ for T and $6.94\times {10}^{-25}\text{ kg}$ for $m$ in equation (IV).

  $\begin{array}{c}\langle v\rangle =\sqrt{\frac{8\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(850\text{K}\right)}{\pi \left(6.94\times {10}^{-25}\text{kg}\right)}}\\ =207\text{m}/\text{s}\end{array}$

Substitute $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$, $850\text{K}$ for T and $6.94\times {10}^{-25}\text{ kg}$ for $m$ in equation (V).

  $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(850\text{K}\right)}{\left(6.94\times {10}^{-25}\text{kg}\right)}}\\ =225\text{m}/\text{s}\end{array}$

Substitute $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k_{\text{B}}$, $850\text{K}$ for T and $6.94\times {10}^{-25}\text{ kg}$ for $m$ in equation (V).

  $\begin{array}{c}{v}_{\text{mp}}=\sqrt{\frac{2\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(850\text{K}\right)}{\left(6.94\times {10}^{-25}\text{kg}\right)}}\\ =184\text{m}/\text{s}\end{array}$

Substitute 6250 for N, $0.10\text{m}$ for d and $207\text{m}/\text{s}$ for v in equation (III)

  $\begin{array}{c}{s}_{\langle v\rangle }=\frac{\pi {\left(0.10\text{m}\right)}^2\left(6250\right)}{\left(60\text{s}\right)\left(207\text{m}/\text{s}\right)}\\ =0.0158\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =1.58\text{cm}\end{array}$

Substitute 6250 for N, $0.10\text{m}$ for d and $225\text{m}/\text{s}$ for v in equation (III)

  $\begin{array}{c}{s}_{\text{rms}}=\frac{\pi {\left(0.10\text{m}\right)}^2\left(6250\right)}{\left(60\text{s}\right)\left(225\text{m}/\text{s}\right)}\\ =0.0145\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =1.45\text{cm}\end{array}$

Substitute 6250 for N, $0.10\text{m}$ for d and $184\text{m}/\text{s}$ for v in equation (III)

  $\begin{array}{c}{s}_{\text{mp}}=\frac{\pi {\left(0.10\text{m}\right)}^2\left(6250\right)}{\left(60\text{s}\right)\left(184\text{m}/\text{s}\right)}\\ =0.0178\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =1.78\text{cm}\end{array}$

Thus, the distance covered by molecules travelling at $\langle v\rangle$ is $1.58\text{cm}$ ; distance covered by molecules travelling at $v_{\text{rms}}$ is $1.45\text{cm}$ and distance covered by molecules travelling at $v_{\text{mp}}$ is $1.58\text{cm}$ .

(b)

To determine

The reason why ${\text{Bi}}_2$ was used instead of ${\text{O}}_2$ or ${\text{N}}_2$ .

Answer to Problem 3P

The molecules of ${\text{Bi}}_2$ is massive compared to ${\text{O}}_2$ or ${\text{N}}_2$ .

Explanation of Solution

${\text{Bi}}_2$ is metal, heavier in mass; ${\text{O}}_2$ and ${\text{N}}_2$ are non-metals, lighter in weight. The cylinder is rotating very rapidly. The molecules striking the cylinder should have greater momentum; it means the molecules should have large mass.

The momentum of ${\text{Bi}}_2$ is greater than the momentum of ${\text{O}}_2$ and ${\text{N}}_2$ ; so it is used in place of ${\text{O}}_2$ or ${\text{N}}_2$ .

Conclusion:

Thus, the molecules of ${\text{Bi}}_2$ is massive compared to ${\text{O}}_2$ or ${\text{N}}_2$ .