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Hurricane Wilma of 2005 is the most intense hurricane on record in the Atlantic basin, with a low-pressure reading of 882 mbar (millibars). Convert this reading into \((\mathbf{a})\) atmospheres, \((\mathbf{b})\) torr, and \((\mathbf{c})\) inches of Hg.

Short Answer

Expert verified
The pressure of Hurricane Wilma in different units is approximately: \(\mathbf{a})\) 0.8707 atmospheres, \(\mathbf{b})\) 893.42 Torr, and \(\mathbf{c})\) 26.0571 inches of Hg.

Step by step solution

01

Convert from millibars to atmospheres

To convert the pressure reading of Hurricane Wilma from millibars to atmospheres, we can use the conversion factor: 1 atm = 1013.25 mbar. This means that we need to divide the given pressure (882 mbar) by the conversion factor. \(\mathrm{atm} = \frac{882\,\mathrm{mbar}}{1013.25\,\mathrm{mbar}}\) Now, we can calculate the pressure in atmospheres.
02

Calculate the pressure in atmospheres

Now we can calculate the pressure using the formula from Step 1: \(\mathrm{atm} = \frac{882}{1013.25}\) \(\mathrm{atm} ≈ 0.8707\) So, the pressure of Hurricane Wilma in atmospheres is approximately 0.8707 atm.
03

Convert from millibars to torr

Now we will convert the 882 mbar pressure into torr using the conversion factor: 1 Torr = 0.98692 mbar. To accomplish this, we must divide the given pressure value by the conversion factor: \(\mathrm{Torr} = \frac{882\,\mathrm{mbar}}{0.98692\,\mathrm{mbar}}\) Now, we can calculate the pressure in torr.
04

Calculate the pressure in torr

Now we can calculate the pressure using the formula from Step 3: \(\mathrm{Torr} = \frac{882}{0.98692}\) \(\mathrm{Torr} ≈ 893.42\) So, the pressure of Hurricane Wilma in torr is approximately 893.42 Torr.
05

Convert from millibars to inches of Hg

Finally, we will convert the 882 mbar pressure into inches of Hg using the conversion factor: 1 in Hg = 33.8638 mbar. To do this, we must divide the given pressure value by the conversion factor: \(\mathrm{in\, Hg} = \frac{882\,\mathrm{mbar}}{33.8638\,\mathrm{mbar}}\) Now, we can calculate the pressure in inches of Hg.
06

Calculate the pressure in inches of Hg

Now we can calculate the pressure using the formula from Step 5: \(\mathrm{in\, Hg} = \frac{882}{33.8638}\) \(\mathrm{in\, Hg} ≈ 26.0571\) So, the pressure of Hurricane Wilma in inches of Hg is approximately 26.0571 in Hg. After solving all the required conversions, we found that the pressure of Hurricane Wilma was: - \(\mathbf{a})\) 0.8707 atmospheres - \(\mathbf{b})\) 893.42 Torr - \(\mathbf{c})\) 26.0571 inches of Hg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atmospheres
Atmospheric pressure is often measured in atmospheres (atm), especially when discussing meteorological events like hurricanes. The atmosphere unit is based on the average pressure at sea level on Earth, which is approximately 1013.25 millibars (mbar). This value serves as a standard reference point for other pressure units.

To convert from millibars to atmospheres, you divide the pressure value in millibars by 1013.25. In this way, you can determine how many times the given pressure fits into the standard atmospheric pressure. For instance, converting 882 mbar to atmospheres, as seen for Hurricane Wilma, results in about 0.8707 atm.

Here's the formula used:
  • Atmospheres: \( \text{atm} = \frac{\text{mbar}}{1013.25} \)
The value reveals that Wilma's pressure dropped significantly below the average sea level pressure, indicating its extreme intensity.
Torr
The torr is another common unit for measuring pressure, named after the Italian scientist Evangelista Torricelli. One torr defines the pressure exerted by a one millimeter column of mercury, correlating closely with the concept of a millimeter of mercury (mmHg).

Converting millibars to torr requires using a conversion factor where 1 torr equals approximately 0.98692 millibars. This makes the torr slightly larger than a millibar but smaller than an atmosphere.

For example, transforming 882 mbar into torr involves dividing the millibars by 0.98692, resulting in approximately 893.42 torr. This computation allows us to understand how lowering atmospheric pressure in hurricanes, like Wilma's, can be expressed accurately in different units.
  • Torr: \( \text{Torr} = \frac{\text{mbar}}{0.98692} \)
Inches of Hg
Pressure measurements sometimes use inches of mercury (in Hg), originating from mercury columns in barometers. This method involves measuring how high the atmospheric pressure can push mercury up a tube. One inch of mercury is equivalent to 33.8638 millibars.

Converting from millibars to inches of Hg then involves dividing the given pressure by this conversion factor. In our example with Hurricane Wilma, the conversion from 882 mbar to approximately 26.0571 inches of Hg provides insight into the severely low pressure characteristic of powerful hurricanes.

The conversion calculation is straightforward:
  • Inches of Hg: \( \text{in Hg} = \frac{\text{mbar}}{33.8638} \)
Understanding how the units convert is key to interpreting pressure data effectively in meteorology and related fields.

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Most popular questions from this chapter

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and Cl. The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces 83.16 \(\mathrm{mL}\) of \(\mathrm{CO}_{2}\) and 73.30 \(\mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor expressed at STP. A separate analysis shows that the sample also contains 16.44 \(\mathrm{mg}\) of Cl. (a) Determine the percentage of the composition of the substance. ( b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?

Imagine that the reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,(\mathbf{b})\) the volume increases by \(33 \%,\) (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\) e) the volume decreases by 50\(\% .[\) Sections 10.3 and 10.4\(]\)

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? \begin{equation}\begin{array}{l}{\text { (a) Because Xe is a noble gas, there is less tendency for the Xe }} \\ {\text { atoms to repel one another, so they pack more densely in }} \\ {\text { the gaseous state. }} \\ {\text { (b) Xe atoms have a higher mass than } \mathrm{N}_{2} \text { molecules. Because }} \\ {\text { both gases at STP have the same number of molecules per }} \\ {\text { unit volume, the Xe gas must be denser. }}\\\\{\text { (c) The Xe atoms are larger than } \mathrm{N}_{2} \text { molecules and thus take }} \\ {\text { up a larger fraction of the space occupied by the gas. }} \\\ {\text { (d) Because the Xe atoms are much more massive than the }} \\\ {\mathrm{N}_{2} \text { molecules, they move more slowly and thus exert }} \\\ {\text { less upward force on the gas container and make the gas }} \\ {\text { appear denser. }}\end{array}\end{equation}

Which of the following statements is false? \begin{equation}\begin{array}{l}{\text { (a) Gases are far less dense than liquids. }} \\ {\text { (b) Gases are far more compressible than liquids. }} \\\ {\text { (c) Because liquid water and liquid carbon tetrachloride do }} \\\ {\text { not mix, neither do their vapors. }} \\ {\text { (d) The volume occupied by a gas is determined by the volume }} \\ {\text { of its container. }}\end{array}\end{equation}

(a) What are the mole fractions of each component in a mixture of 15.08 \(\mathrm{g}\) of \(\mathrm{O}_{2}, 8.17 \mathrm{g}\) of \(\mathrm{N}_{2},\) and 2.64 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) ? (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)

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