/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Hurricane Wilma of 2005 is the m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 (a) atmospheres, \((\mathbf{b})\) torr, and \((\mathbf{c})\) inches of \(\mathrm{Hg}\).

Short Answer

Expert verified
Conversion results: (a) 0.870 atm, (b) 661.705 torr, (c) 26.036 inches of Hg.

Step by step solution

01

Convert millibars to atmospheres

First, remember that 1 atmosphere (atm) is equivalent to 1013.25 millibars (mbar). To convert 882 mbar to atmospheres, use the conversion formula:\[\text{atmospheres} = \frac{\text{millibars}}{1013.25}\]Substitute 882 mbar for millibars in the formula:\[\text{atmospheres} = \frac{882}{1013.25} \approx 0.870 \, \text{atm}\]
02

Convert millibars to torr

The torr and millibar can be converted directly using the knowledge that 1 mbar = 0.750062 torr. Use this conversion to find torr:\[\text{torr} = 882 \, \text{mbar} \times 0.750062 \, \text{torr/mbar}\]Calculate the conversion:\[\text{torr} \approx 661.705 \, \text{torr}\]
03

Convert millibars to inches of Hg

To convert the pressure from millibars to inches of Hg, use the conversion factor where 1 mbar is approximately equal to 0.02953 inches of Hg. Use this to calculate:\[\text{inches of Hg} = 882 \, \text{mbar} \times 0.02953 \, \text{inches of Hg/mbar}\]Calculate the conversion:\[\text{inches of Hg} \approx 26.036 \, \text{inches of Hg}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Converting Millibars to Atmospheres
To convert millibars to atmospheres, it’s essential to know the relationship between these two units of pressure. One atmosphere (atm) is defined as 1013.25 millibars (mbar). The atmospheric pressure at sea level is typically around 1 atm, which provides a baseline for comparison. When converting from millibars to atmospheres, you need to divide the number of millibars by 1013.25. This calculation indicates how many atmospheres the given pressure amount corresponds to. This is particularly useful in meteorology, where atmospheric conditions are often expressed in millibars owing to their precision and convenience. In practice, using the formula:\[\text{atmospheres} = \frac{\text{millibars}}{1013.25}\]you substitute 882 millibars to get approximately 0.870 atmospheres. This means that the pressure during Hurricane Wilma was slightly less than the average atmospheric pressure at sea level.
Converting Millibars to Torr
The torr is a unit of pressure based on an absolute scale and is frequently used in measuring partial vacuums or low-pressure systems. The connection between millibars and torr can be quickly made by remembering that 1 millibar equals approximately 0.750062 torr. To perform the conversion, multiply the number of millibars by this conversion factor:- 1 millibar = 0.750062 torrUsing the hurricane data: \[\text{torr} = 882 \, \text{mbar} \times 0.750062 \, \text{torr/mbar}\]results in approximately 661.705 torr. This calculation highlights a different perspective of the hurricane's pressure, useful in various scientific and engineering applications due to the precision torr provides.
Converting Millibars to Inches of Mercury
Inches of mercury (inHg) is a measure of pressure commonly used in meteorology for weather forecasts and in aviation settings. It denotes the height of a column of mercury that pressure can support. Each inch of mercury equates to approximately 33.86389 millibars, or conversely, 1 millibar is about 0.02953 inches of mercury. To convert millibars to inches of mercury, the procedure involves multiplying the number of millibars by 0.02953:- 1 millibar = 0.02953 inches of HgFor Hurricane Wilma's pressure, this calculation is:\[\text{inches of Hg} = 882 \, \text{mbar} \times 0.02953 \, \text{inches of Hg/mbar}\]which comes out to approximately 26.036 inches of Hg. This gives a clear picture of the storm's intensity, as lower values of inHg reflect a lower pressure associated with more severe weather conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which one or more of the following statements are true? (a) \(\mathrm{O}_{2}\) will effuse faster than \(\mathrm{Cl}_{2}\). (b) Effusion and diffusion are different names for the same process. (c) Perfume molecules travel to your nose by the process of effusion. (d) The higher the density of a gas, the shorter the mean free path.

It turns out that the van der Waals constant \(b\) equals four times the total volume actually occupied by the molecules of a mole of gas. Using this figure, calculate the fraction of the volume in a container actually occupied by Ar atoms (a) at STP, (b) at 20.27 MPa pressure and \(0^{\circ} \mathrm{C}\). (Assume for simplicity that the ideal-gas equation still holds.)

(a) Calculate the density of sulfur hexafluoride gas at 94.26 \(\mathrm{kPa}\) and \(21^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and \(99.06 \mathrm{kPa}\).

You have a sample of gas at \(0^{\circ} \mathrm{C}\). You wish to increase the rms speed by a factor of 3 . To what temperature should the gas be heated?

Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$ \mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g) $$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate \(145 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is \(110 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.