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In the United States, barometric pressures are generally reported in inches of mercury (in. Hg). On a beautiful summer day in Chicago the barometric pressure is 30.45 in. Hg. (a) Convert this pressure to torr. (b) Convert this pressure to atm. (c) A meteorologist explains the nice weather by referring to a "high- pressure area." In light of your answer to parts (a) and (b), explain why this term makes sense.

Short Answer

Expert verified
To convert the given barometric pressure (30.45 in. Hg) to torr and atm, we follow these steps: 1. Convert inches of mercury to millimeters of mercury: Pressure in mm Hg = 30.45 in. Hg × 25.4 = 773.43 mm Hg. 2. Convert millimeters of mercury to torr: Pressure in torr = 773.43 mm Hg (since 1 mm Hg = 1 torr). 3. Convert torr to atmospheres: Pressure in atm = 773.43 torr ÷ 760 = 1.018 atm. Meteorologists refer to nice weather as a "high-pressure area" because high-pressure areas are associated with clear, calm weather due to sinking air suppressing cloud formation and precipitation. Higher atmospheric pressure correlates with better weather conditions.

Step by step solution

01

Convert inches of mercury to millimeters of mercury

To convert the given pressure in inches to millimeters, we'll multiply it by the conversion factor 25.4 mm Hg/in. Hg: Pressure in mm Hg = Pressure in in. Hg × 25.4
02

Convert millimeters of mercury to torr

1 mm Hg is equivalent to 1 torr, so the pressure in torr will be the same as in mm Hg. Pressure in torr = Pressure in mm Hg #b. Convert pressure to atm#
03

Convert torr to atmospheres

To convert pressure in torr to atmospheres, we'll divide the pressure in torr by the conversion factor 760 torr/atm: Pressure in atm = Pressure in torr ÷ 760 #c. Explain the term "high-pressure area"#
04

Relate barometric pressure to meteorology

High-pressure areas are associated with clear, calm weather because they are characterized by sinking air that suppresses cloud formation and precipitation. Low-pressure areas, on the other hand, are characterized by rising air, which facilitates the formation of clouds and precipitation. Thus, a meteorologist would refer to nice weather as being associated with a "high-pressure area" because higher atmospheric pressure correlates with better weather conditions.

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Key Concepts

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

Barometric Pressure
Barometric pressure is a way of measuring atmospheric pressure, usually using a mercury barometer. A barometer consists of an open-ended glass tube standing in a mercury-filled container. When the barometer equilibrates, the height of mercury in the tube represents the atmospheric pressure against the Earth's surface.
In the United States, this measurement is often recorded in inches of mercury (in. Hg). Barometric pressure indicates how much weight the atmosphere holds above us. On a beautiful day in Chicago, you might see the barometric pressure reported as 30.45 in. Hg.
Understanding barometric pressure is essential for predicting weather patterns and is crucial for meteorologists when forecasting clear or stormy weather. By converting these measurements into more standard units like torr or atmospheres, we can understand international weather reports more clearly.
Torr Conversion
Converting pressure into torr involves understanding the relationship between different units of pressure measurements. One inch of mercury is equivalent to 25.4 millimeters of mercury (mm Hg), and since the unit of torr is identical to mm Hg, this conversion becomes quite straightforward.
If you have a barometric pressure reading in inches, like 30.45 in. Hg, converting this to torr involves multiplying by the conversion factor of 25.4:
  • Pressure in mm Hg = 30.45 × 25.4
Once you calculate the pressure in mm Hg, it directly converts to torr since 1 mm Hg equals 1 torr. This ease of conversion helps when pressure readings need to be communicated in scientific contexts, as torr is a preferred unit in many scientific applications.
Atmospheres
An atmosphere (atm) is a unit of pressure based on the average atmospheric pressure at sea level. It is a convenient and intuitive unit commonly used in various scientific disciplines. One atmosphere is equivalent to 760 torr.
To convert a pressure reading from torr to atmosphere, you divide the pressure in torr by 760 because:
  • 1 atm = 760 torr
For example, a pressure converted to 772.5 torr would be divided by 760 to convert it into atmospheres:
  • Pressure in atm = 772.5 ÷ 760
Understanding how to convert between these units allows scientists and those in meteorology to discuss atmospheric conditions in a universally understood way, aiding collaboration and study across cultures and regions.
Weather and Pressure
Weather conditions are heavily influenced by atmospheric pressure regions. High-pressure areas, like the one describing the nice weather in Chicago, often lead to clear and stable weather conditions because the descending air suppresses the formation of clouds and precipitation.
Conversely, low-pressure areas are associated with rising air, which can lead to the development of clouds and possibly rain or storms. This is why weather forecasts often correlate high pressure with nice weather and low pressure with unsettled conditions.
  • High pressure: Clear and calm weather
  • Low pressure: Cloudy with potential precipitation
By understanding the relation between these pressure systems and weather, meteorologists can predict and explain weather patterns more effectively, providing valuable insights for daily planning and safety.

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Most popular questions from this chapter

You have an evacuated container of fixed volume and known mass and introduce a known mass of a gas sample. Measuring the pressure at constant temperature over time, you are surprised to see it slowly dropping. You measure the mass of the gas-filled container and find that the mass is what it should be-gas plus container-and the mass does not change over time, so you do not have a leak. Suggest an explanation for your observations.

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if 1.50 mol has a pressure of 1.25 atm at a temperature of 6C;(b) the absolute temperature of the gas at which 3.33×103 mol occupies 478 mL at 750 torr; (c) the pressure, in atmospheres, if 0.00245 mol occupies 413 mL at 138C;(d) the quantity of gas, in moles, if 126.5 L at 54C has a pressure of 11.25kPa.

(a) A liquid and a gas are moved to larger containers. How does their behavior differ once they are in the larger containers? Explain the difference in molecular terms. (b) Although liquid water and carbon tetrachloride, CCl4(l), do not mix, their vapors form a homogeneous mixture. Explain. (c) Gas densities are generally reported in grams per liter, whereas liquid densities are reported in grams per milliliter. Explain the molecular basis for this difference.

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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 0C. (Assume for simplicity that the ideal-gas equation still holds.)

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