Chapter 1, Problem 1.36P

(a)

Interpretation Introduction

Interpretation:

The volume of a bacterial cell in cubic millimeters $\left({\text{mm}}^3\right)$ is to be calculated.

Concept introduction:

Volume is a physical quantity and its SI unit is cubic meter $\left({\text{m}}^{\text{3}}\right)$. The volume of an object is equal to the product of the height, length, and breadth of the object. Length, breadth, and height of an object is measured in the units of length. Therefore, volume has the unit of length to the cubic power. The other unit of the volume are liter, milliliter, cubic decimeter, microliter.

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$\left(\cancel{\text{beginning unit}}\right)\left(\frac{\text{Final unit}}{\cancel{\text{beginning unit}}}\right)=\text{Final unit}$

Answer to Problem 1.36P

The volume of a bacterial cell in cubic millimeters $\left({\text{mm}}^3\right)$ is $2.56\times {10}^{-9}{\text{ mm}}^{\text{3}}\text{/cell}$.

Explanation of Solution

The volume of a bacterial cell in cubic micrometer $\left({\text{μm}}^3\right)$ is $\text{2}{\text{.56 μm}}^3$.

The road map to calculate the volume of a bacterial cell in cubic millimeters $\left({\text{mm}}^3\right)$ is as follows:

The conversion factor to convert volume from cubic micrometer $\left({\text{μm}}^3\right)$ to cubic meter $\left({\text{m}}^3\right)$ is,

${\left(1\text{ μm}\right)}^3={\left(1\times {10}^{-6}\text{ m}\right)}^3$

The conversion factor to convert volume from cubic meter $\left({\text{m}}^3\right)$ to cubic millimeters $\left({\text{mm}}^3\right)$  is,

${\left(1\times {10}^{-3}\text{ m}\right)}^3={\left(1\text{ mm}\right)}^3$

Convert the volume from cubic micrometer $\left({\text{μm}}^3\right)$  to cubic millimeters $\left({\text{mm}}^3\right)$ as:

$\begin{array}{c}\text{Volume}\left({\text{mm}}^3\right)=\text{2}{\text{.56 μm}}^3\left(\frac{{\left(1\times {10}^{-6}\text{ m}\right)}^3}{{\left(1\text{ μm}\right)}^3}\right)\left(\frac{{\left(1\text{ mm}\right)}^3}{{\left(1\times {10}^{-3}\text{ m}\right)}^3}\right)\\ =2.56\times {10}^{-9}{\text{ mm}}^{\text{3}}\text{/cell}\end{array}$

Conclusion

The volume of a bacterial cell in cubic millimeters $\left({\text{mm}}^3\right)$ is $2.56\times {10}^{-9}{\text{ mm}}^{\text{3}}\text{/cell}$.

(b)

Interpretation Introduction

Interpretation:

The volume of ${10}^5$ bacterial cell in liter $\left(\text{L}\right)$ is to be calculated.

Concept introduction:

Volume is a physical quantity and its SI unit is cubic meter $\left({\text{m}}^{\text{3}}\right)$. The volume of an object is equal to the product of the height, length, and breadth of the object. Length, breadth, and height of an object is measured in the units of length. Therefore, volume has the unit of length to the cubic power. The other unit of the volume are liter, milliliter, cubic decimeter, microliter.

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$\left(\cancel{\text{beginning unit}}\right)\left(\frac{\text{Final unit}}{\cancel{\text{beginning unit}}}\right)=\text{Final unit}$

Answer to Problem 1.36P

The volume of ${10}^5$ bacterial cell in liter $\left(\text{L}\right)$ is ${10}^{-10}\text{ L}$.

Explanation of Solution

The volume of a bacterial cell in cubic micrometer $\left({\text{μm}}^3\right)$ is $\text{2}{\text{.56 μm}}^3$.

The road map to calculate the volume of ${10}^5$ bacterial cell in liter $\left(\text{L}\right)$ is as follows:

The conversion factor to convert volume from cubic micrometer $\left({\text{μm}}^3\right)$ to cubic meter $\left({\text{m}}^3\right)$ is,

${\left(1\text{ μm}\right)}^3={\left(1\times {10}^{-6}\text{ m}\right)}^3$

The conversion factor to convert volume from cubic meter $\left({\text{m}}^3\right)$ to cubic centimeters $\left({\text{cm}}^3\right)$  is,

${\left(1\times {10}^{-2}\text{ m}\right)}^3={\left(1\text{ cm}\right)}^3$

The conversion factor to convert volume from cubic centimeters $\left({\text{cm}}^3\right)$ to milliliters $\left(\text{mL}\right)$ is,

${\left(1\text{ cm}\right)}^3=1\text{ mL}$

The conversion factor to convert volume from milliliters $\left(\text{mL}\right)$ to liter $\left(\text{L}\right)$ is,

$1\text{ mL}={10}^{-3}\text{ L}$

Convert the volume from cubic micrometer $\left({\text{μm}}^3\right)$ to liter $\left(\text{L}\right)$ as:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=\text{2}{\text{.56 μm}}^3\left(\frac{{\left(1\times {10}^{-6}\text{ m}\right)}^3}{{\left(1\text{ μm}\right)}^3}\right)\left(\frac{{\left(1\text{ cm}\right)}^3}{{\left(1\times {10}^{-2}\text{ m}\right)}^3}\right)\left(\frac{1\text{ mL}}{{\left(1\text{ cm}\right)}^3}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =2.56\times {10}^{-15}\text{ L/cell}\end{array}$

Calculate the volume of ${10}^5$ bacterial cell in liter $\left(\text{L}\right)$ as:

 $\begin{array}{c}\text{Volume}\left(\text{L}\right)=\left({10}^5\text{ cell}\right)\left(2.56\times {10}^{-15}\text{ L/cell}\right)\\ =2.56\times {10}^{-10}\text{ L}\\ ={10}^{-10}\text{ L}\end{array}$

Conclusion

The volume of ${10}^5$ bacterial cell in liter $\left(\text{L}\right)$ is ${10}^{-10}\text{ L}$.