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Chapter 10: Problem 22

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 in different units is approximately: (a) 0.87 atmospheres, (b) 660.6 torr, and (c) 0.260 inches of Hg.

Step by step solution

01

(a) Conversion to atmospheres (atm)

To convert the pressure from mbar to atm, we'll use the conversion factor for atmospheres: 1 atm = 1013.25 mbar First, we will divide the pressure in mbar (882 mbar) by the conversion factor: \(atm = \frac{882 \ mbar}{1013.25 \ mbar}\) Now, calculate the value of atm: \(atm = 0.870 \ atm\approx 0.87 \ atm\)
02

(b) Conversion to torr

To convert the pressure from mbar to torr, we'll use the conversion factors for atmospheres and torr: 1 atm = 1013.25 mbar, 1 torr = \(\frac{1}{760}\) atm First, convert the pressure from mbar to atm as we did in part (a): \(atm = \frac{882 \ mbar}{1013.25 \ mbar} \approx 0.870 \ atm\) Now, we'll convert atm to torr using the conversion factor: \(torr = 0.87 \ atm \times 760 \ \frac{torr}{atm} \) Now, calculate the value of torr: \(torr \approx 660.6 \ torr\)
03

(c) Conversion to inches of Hg

To convert the pressure from mbar to inches of Hg, we'll use the conversion factor for inches of Hg: 1 inch of Hg = 3386.39 mbar First, divide the pressure in mbar (882 mbar) by the conversion factor: \(inches \ of \ Hg = \frac{882 \ mbar}{3386.39 \ mbar}\) Now, calculate the value of inches of Hg: \(inches \ of \ Hg \approx 0.260 \ inches \ of \ Hg\) So, the pressure of Hurricane Wilma is approximately: - (a) 0.87 atmospheres, - (b) 660.6 torr, - (c) 0.260 inches of Hg.

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is the force exerted on a surface by the weight of the air above it. At sea level, this pressure is around 1013.25 millibars (mbar) or 1 atmosphere (atm).
It is a fundamental concept in meteorology and affects everyday weather conditions.

When discussing pressure, it's crucial to understand the average atmospheric pressure values that serve as baselines for calculations and conversions. For instance, in the context of hurricanes like Wilma, readings are significantly lower, which indicates intense storm systems.
  • Standard atmospheric pressure at sea level: 1013.25 mbar
  • Conversion to other units like atm or Pascals helps in understanding and comparing different pressure scenarios.
Understanding how pressure is converted between different units will enhance comprehension of how atmospheric conditions are measured and analyzed.
Torr
The torr is a unit of pressure, traditionally used in measuring partial vacuums. One torr is defined as 1/760 of an atmosphere.
It鈥檚 named after Evangelista Torricelli, an Italian physicist and mathematician who is credited with the invention of the barometer.

The pressure of gases, like air, can be measured using this unit which helps in applications ranging from industrial vacuum systems to scientific research.
When converting atmospheric pressure readings into torr, it鈥檚 critical to utilize proper conversion factors to ensure accuracy.
  • 1 atm = 1013.25 mbar = 760 torr
  • It鈥檚 beneficial, especially for chemistry and physics students, to learn to convert between different units as it aids in solving various practical and theoretical problems.
The concept of converting atmospheric pressure into torr provides insight into how pressures are standardized across different scientific fields.
Inches of Mercury
Inches of mercury (in Hg) is another unit used to measure pressure, particularly in weather forecasting and aviation. The unit originates from the pressure required to support a column of mercury one inch high in a barometer.
It is a traditional unit of measure and provides an easy-to-understand way to express atmospheric pressure.

It's still commonly used in weather reports across the United States and in some elements of aviation, thus its familiarity benefits various sectors.
To convert pressure into inches of mercury, conversion factors are necessary to provide equivalent values in different units.
  • Standard sea level pressure is 29.92 inches of mercury.
  • 1 inch of Hg = 3386.39 mbar, which gives an idea of the pressure in more relatable terms for observant areas.
Knowing how to convert pressure readings to inches of mercury can help in understanding forecasts and communicating with professionals in meteorology and aviation.

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Most popular questions from this chapter

The physical fitness of athletes is measured by \({ }^{~} V_{\mathrm{O}_{2}}\) max, " which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of \(45 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min, but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of \(88.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min. (a) Calculate the volume of oxygen, in mL, consumed in \(1 \mathrm{hr}\) by an average man who weighs \(185 \mathrm{lbs}\) and has a \(V_{\mathrm{O}_{2}}\) max reading of \(47.5 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min. (b) If this man lost \(20 \mathrm{lb}\), exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to \(65.0 \mathrm{~mL} \mathrm{O} / \mathrm{kg}\) body mass/min, how many \(\mathrm{mL}\) of oxygen would he consume in \(1 \mathrm{hr}\) ?

Hydrogen has two naturally occurring isotopes, \({ }^{1} \mathrm{H}\) and \({ }^{2} \mathrm{H}\). Chlorine also has two naturally occurring isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Thus, hydrogen chloride gas consists of four distinct types of molecules: \({ }^{1} \mathrm{H}^{35} \mathrm{Cl},{ }^{1} \mathrm{H}^{37} \mathrm{Cl},{ }^{2} \mathrm{H}^{35} \mathrm{Cl}\), and \({ }^{2} \mathrm{H}^{37} \mathrm{Cl}\). Place these four molecules in order of increasing rate of effusion.

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in a blimp?

Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose \(120.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}(\mathrm{~g})\) is stored in a \(1100.0\) - L metal cylinder at \(280^{\circ} \mathrm{C}\). (a) Calculate the pressure of the gas, assuming idealgas behavior. (b) By using the data in Table \(10.3\), calculate the pressure of the gas according to the van der Waals equation. (c) Under the conditions of this problem, which correction dominates, the one for finite volume of gas molecules or the one for attractive interactions?

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 200 atm pressure and \(0{ }^{\circ} \mathrm{C}\). (Assume for simplicity that the ideal-gas equation still holds.)
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