Chapter 10, Problem 146AP

Interpretation Introduction

Interpretation:

The speed at which the target is moving and the time taken for the target to travel are to be calculated. The speed of the $\text{Bi}$ atoms is to be determined.

Concept introduction:

The circumference of a cylinder is calculated as follows:

$C=2\pi r$

Here, $r$ is the radius and $C$ is the circumference.

The time taken iscalculated as follows:

$t=\frac{\text{distance}}{\text{speed}}$

The root mean square speed for molecules is calculated as

$u_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

Here, $u_{\text{rms}}$ represents the root mean square speed of the molecule, $R$ represents the gas constant, $T$ represents the temperature, and $M$ represents its molar mass.

Answer to Problem 146AP

Solution:

(a) $61.2\text{ m}/\text{s}$

(b) $4.58\times {10}^{-4}\text{ s}$

(c) $366\text{ m}/\text{s}$

Explanation of Solution

Given information:

Temperature $T=850°\text{C}$

Distance $d=2.8\text{ cm}$

$\text{Diameter}=15\text{ cm}$

$\text{Revolutions}=130\text{ revolutions}/\sec$

a)The speed at which the target is moving

Calculate the radius as follows:

$r=\frac{\text{diameter}}{2}$

Substitute $15\text{ cm}$ for the diameter in the above equation

$\begin{array}{c}r=\frac{15}{2}\text{ cm}\\ =7.5\text{ cm}\end{array}$

The circumference of a cylinder is calculated as

$C=2\pi r$

Substitute $7.5\text{ cm}$ for $r$ in the above equation

$\begin{array}{c}C=2\pi \left(7.5\text{ cm}\right)\\ =2\times 3.14\times 7.5\text{ cm}\\ =47.1\text{ cm}\end{array}$

In 1centimetre, there are ${\text{10}}^{-2}\text{ m}$.

Convert centimetre to meter as follows:

$\begin{array}{c}47.1\text{ cm}=\left(47.1\cancel{\text{cm}}\right)\left(\frac{{\text{10}}^{-2}\text{m}}{1\cancel{\text{cm}}}\right)\\ =47.1\times {10}^{-2}\text{m}\end{array}$

The target speed can be calculated as follows:

$s=\text{circumference}\times \text{revolutions}/\text{sec}$

Substitute $47.1\times {10}^{-2}\text{m}$ for the circumference and $130$ for $\text{revolutions}/\text{sec}$ in the above equation

$\begin{array}{c}s=47.1\times {10}^{-2}\text{m}\times 130\\ =6123\times {\text{10}}^{-2}\text{ m}/\text{s}\\ =61.2\text{ m}/\text{s}\end{array}$

Hence, the speed of the target is $61.2\text{ m}/\text{s}$.

b) The time it takes for target to cover distance $2.8\text{ cm}$.

In 1centimetre, there are ${\text{10}}^{-2}\text{ m}$.

Convert centimetre to meter as follows:

$\begin{array}{c}\text{2}\text{.8 cm}=\left(\text{2}\text{.8 }\cancel{\text{cm}}\right)\left(\frac{{\text{10}}^{-2}\text{m}}{1\cancel{\text{cm}}}\right)\\ =2.8\times {10}^{-2}\text{m}\end{array}$

The time taken iscalculated as

$t=\frac{\text{distance}}{\text{speed}}$

Substitute $2.8\times {10}^{-2}\text{m}$ for distance and $61.23\text{ m}/\text{s}$ for $\text{speed}$ in the above equation

$\begin{array}{c}t=\frac{2.8\times {10}^{-2}\text{ m}}{61.23\text{ m}/\text{s}}\\ =0.0458\times {10}^{-2}\text{ s}\\ =4.58\times {10}^{-4}\text{ s}\end{array}$

Hence, the time taken by the target to travel $2.8\text{ cm}$ is $4.58\times {10}^{-4}\text{ s}$.

c) $u_{rms}\text{of Bi at 850}°\text{C}$

The distance is $\text{15 cm}$.

In 1centimetre, there are ${\text{10}}^{-2}\text{ m}$.

Convert centimetre to meter as follows:

$\begin{array}{c}\text{15 cm}=\left(\text{15 }\cancel{\text{cm}}\right)\left(\frac{{\text{10}}^{-2}\text{m}}{1\cancel{\text{cm}}}\right)\\ =15\times {10}^{-2}\text{m}\end{array}$

Calculate the speed as follows:

$\text{Speed}=\frac{\text{distance}}{t}$

Substitute $15\times {10}^{-2}\text{m}$ for distance and $4.58\times {10}^{-4}\text{ s}$ for $t$ in the above equation

$\begin{array}{c}\text{Speed}=\frac{15\times {10}^{-2}\text{ m}}{4.58\times {10}^{-4}\text{ s}}\\ =3.28\times {10}^2\text{ m}/\text{s}\\ =328\text{ m}/\text{s}\end{array}$

The temperature is $850°\text{C}$.

The conversion of temperature from degree Celsius to Kelvin is done by using the formula given below:

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(850+273.15\right)\\ =1123.15\text{ K}\end{array}$

The molar mass of bismuth is $209\times {10}^{-3}\text{kg}/\text{mol}$.

The root mean square speed for a moleculeis calculated as

$u_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

Substitute $1123.15\text{ K}$ for $T$, $209\times {10}^{-3}\text{kg}/\text{mol}$ for $M_{\text{Bi}}$, and $8.314\text{ J}/\text{K}\text{.mol}$ for $R$ in the above equation

$\begin{array}{c}{u}_{\text{rms}}=\sqrt{\frac{3\left(8.314\text{ J}/\text{K}\text{.mol}\right)\left(1123.15\text{ K}\right)}{209\times {10}^{-3}\text{kg}/\text{mol}}}\\ =\sqrt{\frac{28013.6}{209\times {10}^{-3}}}\text{ m}/\text{s}\\ =\sqrt{134\times {10}^{\text{3 }}}\text{ m}/\text{s}\\ =\sqrt{1340\times {10}^{\text{2 }}}\text{ m}/\text{s}\end{array}$

So, the rms value of speed for bismuth is

$\begin{array}{c}{u}_{\text{rms}}=36.6\times {10}^1\text{ m}/\text{s}\\ =366\text{ m}/\text{s}\end{array}$

Hence, the rmsvalue of the speed for bismuth is $366\text{ m}/\text{s}$.