Chapter 6, Problem 6.38QP

(a)

Interpretation Introduction

Interpretation: Hurricane Irene registered an atmospheric pressure of $982\text{mbar}$ in August $2011$. Hurricane Katrina registered an atmospheric pressure of $86.2\text{kPa}$ in September $2005$. The difference in pressure between these two hurricanes in $\text{mmHg}$, $\text{atm}$, and millibar is to be calculated.

Concept introduction: Instrument that is used to measures atmospheric pressure is known as Torricelli’s barometer. It was invented by scientist called Evangelista Torricelli. He was the Italian mathematician and physicist.

Atmospheric pressure varies from place to place. Therefore, the standard atmospheric pressure measurement unit is $1\text{atm}$. In barometer this standard atmospheric pressure supports a column of mercury which is $760\text{mm}$ high. The ratio of force per unit area of an object is defined as pressure. The derived SI unit of pressure is Pascal.

To determine: The difference between two pressure of hurricanes in $\text{mm}\text{Hg}$.

Answer to Problem 6.38QP

Solution

The difference between two pressure of hurricanes in millimeter of mercury is $\underset{\_}{90mmHg}$.

Explanation of Solution

Explanation

Given

The given value of pressure in Hurricane Irene is $982\text{mbar}$.

The given value of pressure in Hurricane Katrina is $86.2\text{kPa}$.

Conversion of $1\text{mbar}$ into $\text{atm}$ is done as,

$1\text{mbar}=\frac{1}{1013.25}\text{atm}$

Conversion of $1\text{atm}$ into $\text{mm}$ of mercury $\left(\text{mm}\text{Hg}\right)$ is done as,

$1\text{atm}=\text{760}\text{mm}\text{of}\text{mercury}$

Therefore, conversion of $982\text{mbar}$ into $\text{mm}$ of mercury is,

$\begin{array}{c}982\text{mbar}=\frac{760\times 982}{1013.25}\text{mm}\text{Hg}\\ =\text{737}\text{mm}\text{Hg}\end{array}$

Conversion of $1\text{kPa}$ into $\text{atm}$ is done as,

$1\text{kPa}=\frac{1}{101.325}\text{atm}$

Conversion of $1\text{atm}$ into $\text{mm}$ of mercury $\left(\text{mm}\text{Hg}\right)$ is done as,

$1\text{atm}=\text{760}\text{mm}\text{of}\text{mercury}$

Therefore, conversion of $86.2\text{kPa}$ into $\text{mm}$ of mercury is,

$\begin{array}{c}86.2\text{kPa}=\frac{760\times 86.2}{101.325}\text{mm}\text{Hg}\\ =\text{647}\text{mm}\text{Hg}\end{array}$

The difference between two pressures of hurricanes is,

$\begin{array}{c}\text{Difference}=737\text{mm}\text{Hg}-647\text{mm}\text{Hg}\\ =\text{90}\text{mm}\text{Hg}\end{array}$

Therefore, the difference between two pressures of hurricanes is $\underset{\_}{90mmHg}$.

(b)

Interpretation Introduction

To determine: The difference between two pressure of hurricanes in $\text{atm}$.

Answer to Problem 6.38QP

Solution

The difference between two pressure of hurricanes in $\text{atm}$ is $\underset{\_}{0.118atm}$.

Explanation of Solution

Explanation

Given

The given value of pressure in Hurricane Irene is $982\text{mbar}$.

The given value of pressure in Hurricane Katrina is $86.2\text{kPa}$.

Conversion of $1\text{mbar}$ into $\text{atm}$ is done as,

$1\text{mbar}=\frac{1}{1013.25}\text{atm}$

Therefore, conversion of $982\text{mbar}$ into $\text{atm}$ is,

$\begin{array}{c}982\text{mbar}=\frac{1\times 982}{1013.25}\text{atm}\\ =\text{0}\text{.969}\text{atm}\end{array}$

Conversion of $1\text{kPa}$ into $\text{atm}$ is done as,

$1\text{kPa}=\frac{1}{101.325}\text{atm}$

Therefore, conversion of $86.2\text{kPa}$ into $\text{atm}$ is,

$\begin{array}{c}86.2\text{kPa}=\frac{1\times 86.2}{101.325}\text{atm}\\ =\text{0}\text{.851}\text{atm}\end{array}$

The difference between two pressures of hurricanes is,

$\begin{array}{c}\text{Difference}=0.969\text{atm}-0.851\text{atm}\\ =\text{0}\text{.118}\text{atm}\end{array}$

Therefore, the difference between two pressures of hurricanes is $\underset{\_}{0.118atm}$.

(c)

Interpretation Introduction

To determine: The difference between two pressure of hurricanes in $\text{mbar}$.

Answer to Problem 6.38QP

Solution

The difference between two pressures of hurricanes $\text{mbar}$ is $\underset{\_}{120mbar}$.

Explanation of Solution

Explanation

Given

The given value of pressure in Hurricane Irene is $982\text{mbar}$.

The given value of pressure in Hurricane Katrina is $86.2\text{kPa}$.

Conversion of $1\text{kPa}$ into $\text{atm}$ is done as,

$1\text{kPa}=\frac{1}{101.325}\text{atm}$

Conversion of $1\text{atm}$ into $\text{mbar}$ is done as,

$1\text{atm}=1013.25\text{mbar}$

Therefore, conversion of $86.2\text{kPa}$ into $\text{mbar}$ is,

$\begin{array}{c}86.2\text{kPa}=\frac{1013.25\times 86.2}{101.325}\text{mbar}\\ =\text{862}\text{mbar}\end{array}$

The difference between two pressures of hurricanes is,

$\begin{array}{c}\text{Difference}=982\text{mbar}-862\text{mbar}\\ =\text{120}\text{mbar}\end{array}$

Therefore, the difference between two pressures of hurricanes is $\underset{\_}{120mbar}$.

Conclusion

The difference between two pressure of hurricanes in millimeter of mercury is $\underset{\_}{90mmHg}$.

The difference in pressure between given two hurricanes is $\underset{\_}{0.118atm}$.

The difference between two pressures of hurricanes $\text{mbar}$ is $\underset{\_}{120mbar}$