Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
The pressure of Hurricane Wilma (882 mbar) can be converted into: (a) approximately 0.8705 atmospheres, (b) approximately 661 torr, and (c) approximately 26.04 inches of \(\mathrm{Hg}\).

Step by step solution

01

(a) Conversion to atmospheres

To convert the pressure reading from mbar to atmospheres, we can use the following proportion: \(1 \ \ \mathrm{atm} \ = \ 1013.25 \ \ \mathrm{mbar}\) Thus, to convert the pressure of 882 mbar to atmospheres, we have the equation: \[\frac{1 \ \mathrm{atm}}{1013.25 \ \mathrm{mbar}} \ = \frac{x}{882 \ \mathrm{mbar}}\] Solve for x: \[ x = \frac{882 \ \mathrm{mbar}}{1013.25 \ \mathrm{mbar}} \times 1 \ \mathrm{atm} \] \[ x \approx 0.8705 \ \mathrm{atm} \]
02

(b) Conversion to torr

To convert the pressure reading from mbar to torr, we can use the following proportion: \(1 \ \ \mathrm{atm} \ = \ 760 \ \ \mathrm{torr}\) Thus, to convert the pressure of 882 mbar to torr, we can first convert it to atm as in (a) and then multiply by the conversion factor for torr: \[ 882 \ \mathrm{mbar} \times \frac {1 \ \mathrm{atm}}{1013.25 \ \mathrm{mbar}} \times \frac {760 \ \mathrm{torr}}{1 \ \mathrm{atm}} \ = x \ \mathrm{torr} \] \[ x \approx 661 \ \mathrm{torr} \]
03

(c) Conversion to inches of \(\mathrm{Hg}\)

To convert the pressure reading from mbar to inches of \(\mathrm{Hg}\), we can use the following proportion: \(1 \ \ \mathrm{atm} \ = \ 29.92 \ \ \mathrm{in\ of\ Hg}\) Thus, to convert the pressure of 882 mbar to inches of \(\mathrm{Hg}\), we can first convert it to atm as in (a) and then multiply by the conversion factor for inches of \(\mathrm{Hg}\): \[ 882 \ \mathrm{mbar} \times \frac {1 \ \mathrm{atm}}{1013.25 \ \mathrm{mbar}} \times \frac {29.92 \ \mathrm{in\ of\ Hg}}{1 \ \mathrm{atm}} \ = x \ \mathrm{in\ of\ Hg} \] \[ x \approx 26.04 \ \mathrm{in\ of\ Hg} \] So, the pressure of Hurricane Wilma is approximately: (a) 0.8705 atmospheres, (b) 661 torr, and (c) 26.04 inches of \(\mathrm{Hg}\).

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 Vaia!

Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of air in the atmosphere on a surface. This pressure is crucial for understanding weather patterns and is often measured in terms of atmospheric units. A standard atmosphere (atm) is defined as 1013.25 millibars (mbar), or equivalently, 760 torr.

This measurement is crucial for comparing atmospheric conditions across different geographical locations. During Hurricane Wilma, the recorded atmospheric pressure was 882 mbar, lower than the standard atmosphere, reflecting the intense nature of the hurricane.

To convert from mbar to atmospheres, we use the conversion factor: \(1\ \text{atm} = 1013.25\ \text{mbar}\). This gives us a simplified way to express pressures in terms that are more universally understood in scientific and meteorological contexts.
Torr
Torr is another unit of pressure named after Evangelista Torricelli, an Italian physicist. It is equivalent to the pressure exerted by a 1 mm column of mercury and is often used in scientific applications like vacuum pressures.

The equation involves converting millibars to atmospheres first, and then using the relationship \(1\ \text{atm} = 760\ \text{torr}\), we can easily convert to torr.
  • Begin with mbar to atm conversion.
  • Use atm to torr conversion factor.
This process yields a pressure of approximately 661 torr for the hurricane, highlighting how much lower it was compared to regular atmospheric pressure.
Inches of Mercury
Inches of mercury, abbreviated as in of Hg, is a unit of measurement that originated from the mercury barometer, one of the oldest tools for measuring air pressure. It is especially known in meteorological reports in the United States.

The conversion from millibars to inches of mercury begins by converting millibars to atmospheres, similar to the previous steps, and then using the conversion factor \(1\ \text{atm} = 29.92\ \text{in hinspace of hinspace Hg}\).

This provides a way to relate atmospheric pressure to a tangible concept that is easy to grasp. For the hurricane example, converting 882 mbar results in a reading of approximately 26.04 inches of Hg.

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

An aerosol spray can with a volume of \(250 \mathrm{~mL}\) contains \(2.30 \mathrm{~g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(23{ }^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the propane occupy at STP? (c) The can's label says that exposure to temperatures above \(130^{\circ} \mathrm{F}\) may cause the can to burst. What is the pressure in the can at this temperature?

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23{ }^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal 150.0 atm? (d) What would be the pressure of the gas, in \(\mathrm{kPa},\) if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

Consider the combustion reaction between \(25.0 \mathrm{~mL}\) of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL})\) and \(12.5 \mathrm{~L}\) of oxygen \(\mathrm{gas}\) measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

Complete the following table for an ideal gas: $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{v} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & ? \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & ? \mathrm{~mol} & 27^{\circ} \mathrm{C} \\ 650 \text { torr } & ? \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ ? \mathrm{~atm} & 585 \mathrm{~mL} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

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

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.

Sign-up for free